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using System;
#if !NETSTANDARD1_6 && !NETSTANDARD2_0
using System.Data.Linq.Mapping;
#else
using System.Data;
#endif
using LinqToDB;
using LinqToDB.Common;
using NUnit.Framework;
namespace Tests.Linq
{
using Model;
#if !NETSTANDARD1_6 && !NETSTANDARD2_0
[System.Data.Linq.Mapping.Table(Name = "Person")]
#else
[System.ComponentModel.DataAnnotations.Schema.Table("Person")]
#endif
public class L2SPersons
{
private int _personID;
#if !NETSTANDARD1_6 && !NETSTANDARD2_0
[System.Data.Linq.Mapping.Column(
Storage = "_personID",
Name = "PersonID",
DbType = "integer(32,0)",
IsPrimaryKey = true,
IsDbGenerated = true,
AutoSync = AutoSync.Never,
CanBeNull = false)]
#else
[System.ComponentModel.DataAnnotations.Schema.Column("PersonID",
TypeName = "integer(32,0)")]
#endif
public int PersonID
{
get { return _personID; }
set { _personID = value; }
}
#if !NETSTANDARD1_6 && !NETSTANDARD2_0
[System.Data.Linq.Mapping.Column]
#else
[System.ComponentModel.DataAnnotations.Schema.Column]
#endif
public string FirstName { get; set; }
#if !NETSTANDARD1_6 && !NETSTANDARD2_0
[System.Data.Linq.Mapping.Column]
#else
[System.ComponentModel.DataAnnotations.Schema.Column]
#endif
public string LastName;
#if !NETSTANDARD1_6 && !NETSTANDARD2_0
[System.Data.Linq.Mapping.Column]
#else
[System.ComponentModel.DataAnnotations.Schema.Column]
#endif
public string MiddleName;
#if !NETSTANDARD1_6 && !NETSTANDARD2_0
[System.Data.Linq.Mapping.Column]
#else
[System.ComponentModel.DataAnnotations.Schema.Column]
#endif
public string Gender;
}
[TestFixture]
public class L2SAttributeTests : TestBase
{
[Test, IncludeDataContextSource(false, ProviderName.SQLiteClassic, ProviderName.SQLiteMS)]
public void IsDbGeneratedTest(string context)
{
using (var db = GetDataContext(context))
{
db.BeginTransaction();
var id = db.InsertWithIdentity(new L2SPersons
{
FirstName = "Test",
LastName = "Test",
Gender = "M"
});
db.GetTable<L2SPersons>().Delete(p => p.PersonID == ConvertTo<int>.From(id));
}
}
}
}
|
code
|
मुख्य बात यह हमेशा याद रखना है कि शब्द "कमाएँ" शब्द "काम" से आता है, "कारोबारी सत्रों में विदेशी मुद्रा और द्विआधारी विकल्प फ्रीबी" नहीं। आप इंटरनेट पर कमा सकते हैं, लेकिन आपको इसके लिए कड़ी मेहनत करने की एक लक्ष्य और इच्छा होना चाहिए।
कारोबारी सत्रों में विदेशी मुद्रा और द्विआधारी विकल्प - २४ऑप्शन समीक्षा और एक नई धोखाधड़ी
अक्सर घर के काम काज में महिलाएं इतनी व्यस्त हो जाती हैं कि उनके पास पति से बात करने का समय नहीं होता लेकिन जब पति सारा दिन ऑफिस में काम करके घर आए तो कुछ समय निकाल कर उनके साथ जरूर बैठें और उनके मन की बात जरूर सूनें। विश्वविद्यालय के कर्मचारी आपको रूसी संघ के आंतरिक मामलों के मंत्रालय के प्रवासन के लिए कार्यालय के एक निशान से आगमन के नोटिस का अलग-अलग हिस्सा देंगे। यह माइग्रेशन अकाउंटिंग के पंजीकरण के बारे में दस्तावेज है।वह कारोबारी सत्रों में विदेशी मुद्रा और द्विआधारी विकल्प देश में उस अवधि के लिए रहने का अधिकार देता है। शुरू में यह ३ महीने है, फिर १ वर्ष।
इसके बावजूद, सेवा में कई फायदे हैं आप उन्हें कम से कम बहुक्रियाशीलता के साथ सूचीबद्ध करना शुरू कर सकते हैं यह साइट इनपुट के लिए और धन निकालने के लिए दोनों काम करती है। इसकी सहायता से, आप अपने ऑनलाइन पैसे वितरण प्रणाली को इस तरह से व्यवस्थित कर सकते हैं कि समय के साथ सभी बिलों का भुगतान, ऋण चुकाना या भुगतान प्राप्त करने और आपके लिए सुविधाजनक मुद्रा में परिवर्तित कर सकें। इस साइट के लिए कौन सा उपयोगी हो सकता है, कारोबारी सत्रों में विदेशी मुद्रा और द्विआधारी विकल्प इसके बारे में अधिक जानकारी, हम थोड़ी अधिक जानकारी देंगे। शायद आपके पास अब भी ग्मेल पर एक मेलबॉक्स बनाने से संबंधित समस्याएं हैं या आप उनसे इस लेख में टिप्पणी में नीचे पूछ सकते हैं, साथ ही मेरे साथ फ़ॉर्म का उपयोग कर सकते हैं
मल के साथ क्षमता को ठंड में रखा जाना चाहिए, लेकिन जमे हुए नहीं। प्रयोगशाला सप्ताहांत पर काम नहीं करती है। यू की परिभाषा एस स्वास्थ्य और मानव सेवा विभाग (एचएचएस) ' स्वास्थ्य और मानव सेवा विभाग (एचएचएस) एक कैबिनेट स्तरीय सरकारी विभाग है जो स्वास्थ्य और मानव सेवा प्रदान करता है, और सामाजिक सेवाओं, चिकित्सा और सार्वजनिक स्वास्थ्य में अनुसंधान को बढ़ावा देता है । यह ११ एजेंसियों के माध्यम से यह प्राप्त करता है कि १०० से अधिक कार्यक्रमों का प्रबंधन। एजेंसियों में रोग नियंत्रण और रोकथाम (सीडीसी), खाद्य एवं औषधि प्रशासन (एफडीए) और बच्चों और परिवारों के प्रशासन (एसीएफ) के लिए केंद्र शामिल हैं। डाउन डाउन 'यू।
यांत्रिक गियरबॉक्स ऑपरेशन का सिद्धांत यह है कि टोक़ चरणों में बदल जाता है। कार इंटीरियर में हैंडल स्विच करने के बाद इंटरैक्टिंग गियर की एक जोड़ी गति में आती है। साइट पर हर दिन से नई प्रोफाइल हैंपरियोजना "माई ओपिनियन" के सहयोगी। उपयोगकर्ताओं से प्रतिक्रिया अक्सर सर्वेक्षण में भाग लेने के लिए दुर्लभ प्रस्तावों को इंगित करती है। यह क्यों हो रहा है?
बाइनरी विकल्प ट्रेडिंग की सफलता दर - कारोबारी सत्रों में विदेशी मुद्रा और द्विआधारी विकल्प
मगर, तनेजा कहते हैं कि पेट्रोल और डीजल के भाव में बढ़ोतरी इस बात पर निर्भर करेगी कि क्रूड में तेजी के दौर में भारतीय तेल कंपनियों ने कितना तेल खरीदा क्योंकि महंगाई तभी जारी रहेगी जब कंपनियों ने महंगा तेल खरीदा होगा। होटल व्यवसाय का आचरण धन का एक आकर्षक निवेश है। होटल व्यवसाय की अर्थव्यवस्था ऐसी है कि होटल उद्योग में मौजूदा पूंजी को गुणा करने के लिए महत्वपूर्ण अवसर हैं। चुनौतियों होटल उद्योग और विशेष रूप से होटल व्यवसाय के बावजूद, लाभप्रदता होटल व्यवसाय अपनी क्षमता उच्च बनी (कारोबारी सत्रों में विदेशी मुद्रा और द्विआधारी विकल्प मौसम, सेवा की गुणवत्ता, आरक्षण प्रणाली की अपूर्णता, वहाँ अत्यधिक आरक्षण की संभावना है),।
"अवसर अकसर आते हैं। जब सोने की बारिश होती है, तो बाल्टी डालें, न कि अंगूठे। हर निवेशक खरीद-और-पकड़ की रणनीति में विश्वास करता है विवाद का एकमात्र विषय यह है कि होल्डिंग अवधि कितनी देर तक चलनी चाहिए। हर कारोबारी सत्रों में विदेशी मुद्रा और द्विआधारी विकल्प किशोरी के लिए जो एक कम मात्रा वाली इक्विटी खरीदता है और ८ दशकों तक इसे बरकरार रखता है, रास्ते में लाभांश एकत्रित कर रहा है, वहाँ दर्जनों सट्टेबाज़ हैं जो एक सप्ताह से भी कम समय में अपनी स्थिति से बाहर निकलना चाहते हैं। जो न केवल स्टॉक की आवश्यकता है, बल्कि किसी लेन-देन की लागत को ऑफसेट करने के लिए पर्याप्त रूप से सराहना भी करता है। स्विंग ट्रेडिंग उस निवेशक के लिए है, जो वाकई सप्ताहांत का इंतजार नहीं कर सकता।
सभी कि आप के लिए आवश्यक है शुद्ध सर्फ करने के लिए इस विज्ञापनदाता से पैसे प्राप्त होगा। चरण ५। स्थापना को पूरा करने के बाद, त्वचा पर एक वायुरोधी झिल्ली संलग्न है। उसके कपड़े क्षैतिज रूप से फैले हुए हैं और ओएसबी को स्टेपल द्वारा स्टेपल के साथ तय किए गए हैं। जोड़ों में फिल्म चिपक जाती है और चिपकने वाला टेप के साथ चिपकाया जाता है। सामग्री को बहुत कसकर नहीं खींचा जाना चाहिए, लेकिन वहां कोई कमी नहीं होनी चाहिए।
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hindi
|
\begin{document}
\title{{\bf An FPT Algorithm Beating 2-Approximation for $k$-Cut}}
\author{ Anupam Gupta\thanks{Supported in part by NSF awards
CCF-1536002, CCF-1540541, and CCF-1617790. This work was done in
part when visiting the Simons Institute for the Theory of
Computing. } \and Euiwoong Lee\thanks{Supported by NSF award CCF-1115525, Samsung scholarship, and Simons award for graduate students in TCS.}
\and Jason Li\thanks{{\tt [email protected]} }}
\date{Computer Science Department \\ Carnegie Mellon University \\ Pittsburgh, PA 15213.}
\thispagestyle{empty}
\maketitle
\begin{abstract}
In the $k$\textsc{-Cut}\xspace problem, we are given an edge-weighted graph $G$ and an
integer $k$, and have to remove a set of edges with minimum total
weight so that $G$ has at least $k$ connected components. Prior work
on this problem gives, for all $h \in [2,k]$, a $(2-h/k)$-approximation
algorithm for $k$-cut that runs in time $n^{O(h)}$. Hence to get a
$(2 - \varepsilon)$-approximation algorithm for some absolute constant
$\varepsilon$, the best runtime using prior techniques is
$n^{O(k\varepsilon)}$. Moreover, it was recently shown that getting a
$(2 - \varepsilon)$-approximation for general $k$ is NP-hard, assuming the
Small Set Expansion Hypothesis.
If we use the size of the cut as the parameter, an FPT algorithm to
find the exact $k$\textsc{-Cut}\xspace is known, but solving the $k$\textsc{-Cut}\xspace problem exactly
is $W[1]$-hard if we parameterize only by the natural parameter of
$k$. An immediate question is: \varepsilonmph{can we approximate $k$\textsc{-Cut}\xspace better
in FPT-time, using $k$ as the parameter?}
We answer this question positively. We show that for some absolute
constant $\varepsilon > 0$, there exists a $(2 - \varepsilon)$-approximation
algorithm that runs in time $2^{O(k^6)} \cdot \widetilde{O} (n^4) $. This is the first FPT
algorithm that is parameterized only by $k$ and strictly improves the
$2$-approximation.
\varepsilonnd{abstract}
\setcounter{page}{1}
\section{Introduction}
\langlebel{sec:introduction}
We consider the $k$\textsc{-Cut}\xspace problem: given an edge-weighted graph $G =
(V,E,w)$ and an integer $k$, delete a minimum-weight set of edges so
that $G$ has at least $k$ connected components. This problem is a
natural generalization of the global min-cut problem, where the goal is
to break the graph into $k=2$ pieces.
Somewhat surprisingly, the problem has poly-time algorithms for any
constant $k$: the current best result gives an $\tilde{O}(n^{2k})$-time
deterministic algorithm~\cite{Thorup08}. On the approximation algorithms
front, several $2$-approximation algorithms are
known~\cite{SV95, NR01, RS02}. Even a trade-off result is known: for any
$h \in [1,k]$, we can essentially get a $(2-\frac{h}{k})$-approximation in
$n^{O(h)}$ time~\cite{XCY11}. Note that to get $(2-\varepsilon)$ for some
absolute constant $\varepsilon > 0$, this algorithm takes time $n^{O(\varepsilon k)}$,
which may be undesirable for large $k$. On the other hand, achieving a
$(2-\varepsilon)$-approximation is NP-hard for general $k$, assuming the
Small Set Expansion Hypothesis (SSEH)~\cite{Manurangsi17}.
What about a better \varepsilonmph{fine-grained result} when $k$ is small?
Ideally we would like a runtime of $f(k) \mathrm{poly}(n)$ so it scales better
as $k$ grows --- i.e., an FPT algorithm with parameter $k$. Sadly, the
problem is $W[1]$-hard with this parameterization~\cite{DEFPR03}. (As an
aside, we know how to compute the optimal $k$\textsc{-Cut}\xspace in time $f(|\varepsilonnsuremath{\mathsf{Opt}}\xspace|) \cdot
n^{2}$~\cite{KT11, Chitnis}, where $|\varepsilonnsuremath{\mathsf{Opt}}\xspace|$ denotes the cardinality of
the optimal $k$\textsc{-Cut}\xspace.)
The natural question suggests itself: can we give a better approximation
algorithm that is FPT in the parameter $k$?
Concretely, the question we consider in this paper is: \varepsilonmph{If we
parameterize $k$\textsc{-Cut}\xspace by $k$, can we get a $(2-\varepsilon)$-approximation for
some absolute constant $\varepsilon > 0$ in FPT time---i.e., in time
$f(k) \mathrm{poly}(n)$?} (The hard instances which show $(2-\varepsilon)$-hardness
assuming SSEH~\cite{Manurangsi17} have $k = \Omega(n)$, so such an FPT
result is not ruled out.) We answer the question positively.
\begin{theorem}[Main Theorem]
\langlebel{thm:kcut-main}
There is an absolute constant $\varepsilon > 0$ and an a
$(2-\varepsilon)$-approximation algorithm for the $k$\textsc{-Cut}\xspace problem on general
weighted graphs that runs in time $2^{O(k^6)} \cdot \tilde{O}(n^4)$.
\varepsilonnd{theorem}
Our current $\varepsilon$ satisfies $\varepsilon \geq 0.0003$ (see the calculations in
\S\ref{sec:conclusion}). We hope that our result will serve as a
proof-of-concept that we can do better than the factor of~2 in FPT$(k)$
time, and eventually lead to a deeper understanding of the trade-offs
between approximation ratios and fixed-parameter tractability for the
$k$\textsc{-Cut}\xspace problem. Indeed, our result combines ideas from approximation
algorithms and FPT, and shows that considering both settings
simultaneously can help bypass lower bounds in each individual setting,
namely the $W[1]$-hardness of an exact FPT algorithm and the
SSE-hardness of a polynomial-time $(2-\varepsilon)$-approximation.
To prove the theorem, we introduce two variants of $k$\textsc{-Cut}\xspace.
\text{Laminar}kcut{k} is a special case of $k$\textsc{-Cut}\xspace where both the graph and the
optimal solution are promised to have special properties, and \textsc{Partial VC}\xspacelong
(\textsc{Partial VC}\xspace) is a variant of $k$\textsc{-Cut}\xspace where $k - 1$ components are required to be
singletons, which served as a hard instance for both the exact
$W[1]$-hardness and the $(2 - \varepsilon)$-approximation SSE-hardness. Our
algorithm consists of three main steps where each step is modular,
depends on the previous one: an FPT-AS for \textsc{Partial VC}\xspace, an algorithm for
\text{Laminar}kcut{k}, and a reduction from $k$\textsc{-Cut}\xspace to \text{Laminar}kcut{k}. In the
following section, we give more intuition for our three steps.
\subsection{Our Techniques}
\langlebel{sec:techniques}
For this section, fix an optimal $k$-cut
${\cal S}^* = \{ S^*_1, \dots, S^*_k\}$, such that
$w(\partial{S^*_1}) \leq \dots \leq w(\partial{S^*_k})$. Let the
optimal cut value be
$\varepsilonnsuremath{\mathsf{Opt}}\xspace := w(E(S^*_1, \dots, S^*_k)) = \sum_{i=1}^k w(\partial{S^*_i}) /
2$; here $E(A_1,\cdots, A_k)$ denotes the edges that go between
different sets in this partition. The $(2 - 2/k)$-approximation iterative
greedy algorithm by Saran and Vazirani~\cite{SV95} repeatedly computes
the minimum cut in each connected component and takes the cheapest one
to increase the number of connected components by $1$. Its
generalization by Xiao et al.~\cite{XCY11} takes the minimum $h$-cut
instead of the minimum $2$-cut to achieve a $(2 - h/k)$-approximation in
time $n^{O(h)}$.
\subsubsection{Step I: \textsc{Partial VC}\xspacelong}
\langlebel{sec:overview-pvc}
The starting point for our algorithm is the $W[1]$-hardness result of
Downey et al.~\cite{DEFPR03}: the reduction from $k$-clique results in a
$k$\textsc{-Cut}\xspace instance where the optimal solution consists of $k-1$ singletons
separated from the rest of the graph. Can we approximate such instances
well? Formally, the \textsc{Partial VC}\xspace problem asks: given a edge-weighted graph, find
a set of $k-1$ vertices such that the total weight of edges hitting
these vertices is as small as possible? Extending the result of
Marx~\cite{Marx07} for the maximization version, our first conceptual
step is an FPT-AS for this problem, i.e., an algorithm that given a
$\delta >0$, runs in time $f(k,\delta)\cdot \mathrm{poly}(n)$ and gives a
$(1+\delta)$-approximation to this problem.
\subsubsection{Step II: Laminar $k$-cut}
\langlebel{sec:overview-lam}
The instances which inspire our second idea are closely related to the
hard instances above. One instance on which the greedy algorithm of
Saran and Vazirani gives a approximation no better than $2$ for large
$k$ is this: take two cliques, one with $k$ vertices and unit edge
weights, the other with $k^2$ vertices and edge weights $1/(k+1)$, so
that the weighted degree of all vertices is the same. (Pick one vertex
from each clique and identify them to get a connected graph.) The
optimal solution is to delete all edges of the small clique, at cost
$\binom{k}{2}$. But if the greedy algorithm breaks ties poorly, it will
cut out $k-1$ vertices one-by-one from the larger clique, thereby
getting a cut cost of $\approx k^2$, which is twice as large. Again we
could use \textsc{Partial VC}\xspace to approximate this instance well. But if we replace each
vertex of the above instance itself by a clique of high weight edges,
then picking out single vertices obviously does not work. Moreover, one
can construct recursive and ``robust'' versions of such instances where
we need to search for the ``right'' (near-)$k$-clique to break
up. Indeed, these instances suggest the use of dynamic programming (DP),
but what structure should we use DP on?
One feature of such ``hard'' instances is that the optimal $k$\textsc{-Cut}\xspace
${\cal S}^* = \{S_1^*, \ldots, S_k^*\}$ is composed of near-min-cuts in the
graph. Moreover, no two of these near-min-cuts cross each other. We now
define the \text{Laminar}kcut{k} problem: find a $k$\textsc{-Cut}\xspace on an instance where
none of the $(1+\varepsilon)$-min-cuts of the graph cross each other, and where
each of the cut values $w(\partial{S_i^*})$ for $i = 1,\ldots, k-1$ are
at most $(1+\varepsilon)$ times the min-cut. Because of this laminarity (i.e.,
non-crossing nature) of the near-min-cuts, we can represent the
near-min-cuts of the graph using a tree $\mathcal T$, where the nodes of $G$
sit on nodes of the tree, and edges of $\mathcal T$ represent the near-min-cuts
of $G$. Rooting the tree appropriately, the problem reduces to
``marking'' $k-1$ incomparable tree nodes and take the near-min-cuts
given by their parent edges, so that the fewest edges in $G$ are
cut. Since all the cuts represented by $\mathcal T$ are near-min-cuts and
almost of the same size, it suffices to mark $k-1$ incomparable nodes to
maximize the number of edges in $G$ both of whose endpoints lie below a
marked node. We call such edges \varepsilonmph{saved} edges. \agnote{Any
figures?} In order to get a $(2-\varepsilon)$-approximation for \text{Laminar}kcut{k},
it suffices to save $\approx \varepsilon k \mathsf{Mincut}$ weight of edges.
Note that if $\mathcal T$ is a star with $n$ leaves and each vertex in $G$ maps
to a distinct leaf, this is precisely the \textsc{Partial VC}\xspace problem, so we do not
hope to find the optimal solution (using dynamic programming, say).
Moreover, extending the FPT-AS for \textsc{Partial VC}\xspace to this more general setting
does not seem directly possible, so we take a different approach. We
call a node an \varepsilonmph{anchor} if has some $s$ children which when marked
would save $\approx \varepsilon s \mathsf{Mincut}$ weight. We take the following
``win-win'' approach: if there were $\Omega(k)$ anchors that were
incomparable, we could choose a suitable subset of $k$ of their children
to save $\approx \varepsilon k \mathsf{Mincut}$ weight. And if there were not, then all
these anchors must lie within a subtree of $\mathcal T$ with at most $k$
leaves. We can then break this subtree into $2k$ paths and guess which
paths contain anchors which are parents of the optimal solution. For
each such guess we show how to use \textsc{Partial VC}\xspace to solve the problem and save a
large weight of edges. Finally how to identify these anchors? Indeed,
since all the mincuts are almost the same, finding an anchor again
involves solving the \textsc{Partial VC}\xspace problem!
\subsubsection{Step III: Reducing $k$\textsc{-Cut}\xspace to \text{Laminar}kcut{k}}
\langlebel{sec:overview-redn}
\begin{wrapfigure}{L}{0.38\textwidth}
\centering
\includegraphics[width=0.35\textwidth]{conform}
\caption{\langlebel{fig:conform} The blue set on the right,
formed by $S_5^* \cup S_7^* \cup S_{11}^*$, conforms to the
algorithm's partition ${\cal S}$ on the left.}
\varepsilonnd{wrapfigure}
We now reduce the general $k$\textsc{-Cut}\xspace problem to \text{Laminar}kcut{k}. This
reduction is again based on observations about the
graph structure in cases where the iterative greedy algorithms do not
get a $(2 - \varepsilon$)-approximation. Suppose ${\cal S} = \{ S_1, \dots,
S_{k'} \}$ be the connected components of $G$ at some point of an
iterative algorithm ($k' \leq k$). For a subset $\varepsilonmptyset \neq U
\subsetneq V$, we say that $U$ {\varepsilonm conforms} to partition ${\cal S}$ if
there exists a subset $J \subsetneq [k']$ of parts such that $U =
\cup_{j \in J} S_j$. One simple but crucial observation is the
following: if there exists a subset $\varepsilonmptyset \neq I \subsetneq [k]$ of
indices such that $\cup_{i\in I} S^*_i$ conforms to ${\cal S}$
(i.e., $\cup_{i\in I} S^*_i = \cup_{j \in J} S_j$), we can
``guess'' $J$ to partition $V$ into the two parts $\cup_{i \in I} S^*_i$
and $\cup_{i \notin I} S^*_i$. Since the edges between these two parts
belong to the optimal cut and each of them is strictly smaller than $V$,
we can recursively work on each part without any loss.
Moreover, the number of choices for $J$ is at most
$2^{k'}$ and each guess produces one more connected component, so the
total running time can be bounded by $f(k)$ times the running time of
the rest of the algorithm, for some function $f(\cdot)$. Therefore, we
can focus on the case where none of $\cup_{i \in I} S^*_i$ conforms to
the algorithm's partition ${\cal S}$ at any point during the algorithm's
execution.
\begin{wrapfigure}{R}{0.38\textwidth}
\centering
\includegraphics[width=0.35\textwidth]{histogram}
\caption{\langlebel{fig:histo} The blue curve shows cut sizes
for algorithm's cuts, red curve shows $w(\partial{S^*_i})$
values. The blue area (and in fact all the area below
$w(\partial{S_1^*})$ and above the algorithm's curve) makes the first
inequality loose. The grey area (and in fact all the area above
$w(\partial{S_1^*})$ and below OPT's curve) makes the second
inequality loose.}
\varepsilonnd{wrapfigure}
Now consider the iterative min-cut algorithm of Saran and Vazirani, and
let $c_i$ be the cost of the min cut in the $i^{th}$ iteration ($1 \leq
i \leq k - 1$). By our above assumption about non-conformity, none of
$\cup_{i \in I} S^*_i$, and in particular the subset $S^*_1$, conform to
the current components. This implies that deleting the remaining edges
in $\partial S^*_1$ is a valid cut that increases the number of
connected components by at least $1$, so $c_i \leq
w(\partial{S^*_1})$. Then we have the following chain of inequalities:
\[
\sum_{i=1}^{k-1} c_i \leq k \cdot w(\partial{S^*_1}) \leq \sum_{i=1}^k w(\partial{S^*_i}) = 2\varepsilonnsuremath{\mathsf{Opt}}\xspace.
\]
If the iterative min-cut algorithm could not get a $(2 -
\varepsilon)$-approximation, the two inequalities above must be essentially
tight. Hence almost all our costs $c_i$ must be close to
$w(\partial{S^*_1})$ and almost all $w(\partial{S^*_i})$ must be close
to $w(\partial{S^*_1})$.
Slightly more formally, let $\mathfrak{a} \in [k]$ be the smallest integer such
that
$c_{\mathfrak{a}} \gtrsim w(\partial{S^*_1})$
---so that the first $\mathfrak{a} - 1$ cuts are ones where
we pay ``much'' less than $\partial{S^*_1}$ and make the first
inequality loose. And let $\mathfrak{b} \in [k]$ be the smallest number such that
$w(\partial{S^*_{\mathfrak{b}}}) \gtrsim w(\partial{S^*_1})$
--- so that the last
$k - \mathfrak{b}$ cuts in OPT are much larger than $\partial{S^*_1}$ and make
the second inequality loose. Then if the iterative min-cut algorithm is
no better than a $2$-approximation, we can imagine that $\mathfrak{a} = o(k)$ and
$\mathfrak{b} \geq k - o(k)$.
For simplicity, let us assume that $\mathfrak{a} = 1$ and $\mathfrak{b} = k$ here.
Indeed, instead of just considering min-cuts, suppose we also consider
min-4-cuts, and take the one with better edges cut per number of new
components. The arguments of the previous paragraph still hold, so
$\mathfrak{a} = 1$ implies that
the best min-cuts and best min-4-way cuts (divided by 3) are roughly
at least $w(\partial{S^*_1})$ in the original $G$.
Since the min-cut is also at most $w(\partial{S^*_1})$,
the weight of the min-cut is
roughly $w(\partial{S^*_1})$ and none of the near-min-cuts
cross (else we would get a good 4-way cut). I.e., the
near-min-cuts in the graph form a laminar family.
Together with the fact that $\partial{S^*_1}, \dots, \partial{S^*_{k - 1}}$ are near-min-cuts (we assumed $\mathfrak{b} = k$),
this is precisely an instance of \text{Laminar}kcut{k}, which completes
the proof!
\iffalse
\alert{I would stop here and leave the rest of the details to the main
body, maybe just move on to explaining laminar-cut.}
\varepsilonlnote{I am fine with this except that the another promise of Laminar ensure that all but one optimal components are near min cuts. }
Let $S^*_{\geq \mathfrak{b}} := \cup_{i \geq \mathfrak{b}} S^*_{i}$ so that in the
$\mathfrak{b}$-cut $\{ S^*_1, \dots, S^*_{\mathfrak{b} - 1}, S^*_{\geq \mathfrak{b}} \}$, all but
one component satisfy that the weight of their boundary is close to
$w(\partial{S^*_1})$. Another application of the conformity argument
ensures that among the current components $S_1, \dots, S_{\mathfrak{a}}$,
\varepsilonlnote{this is another cute idea... should we not explain it here? If
we do, we definitely need a figure explaining what happens when two
$S_j$'s cross one $S^*_i$. } \agnote{I would say we postpone this idea
for now, we might lose people.} there is exactly one component (say
$S_1$) intersecting every $S^*_1, \dots, S^*_{\mathfrak{b} - 1}, S^*_{\geq \mathfrak{b}}$
and all the others are strictly contained in one $S^*_i$ (or $S^*_{\geq
\mathfrak{b}}$). In particular, note that $S_1$ intersects both $S^*_1$ and $V
\setminus S^*_1$, so $\mathsf{Mincut}(G[S_1]) \leq w(\partial{S^*_1})$.
Then we can see that $G[S_1]$ satisfies the promises of
$\text{Laminar}cut{\mathfrak{b}}{\varepsilonilon'}$: there exists a $\mathfrak{b}$-cut where all but
one component are close to the min cut and no two near-min cuts cross.
We guess $S_1$, run our algorithm for $\text{Laminar}cut{\mathfrak{b}}{\varepsilonilon'}$ for
$G[S_1]$ to get $\mathfrak{b} - 1$ more components. It results in $\mathfrak{a} + \mathfrak{b} -
1$ components, and if it is still less than $k$, we finally perform the
iterative greedy algorithm (here the min $2$-cut suffices) to get $k -
\mathfrak{a} - \mathfrak{b} + 1$ more components. The conformity still ensures that we pay
at most $w(\partial{S^*_1})$ in each iteration.
The total cost is the sum of (1) the cost to get the first $\mathfrak{a}$ components (2) the cost of $\text{Laminar}cut{\mathfrak{b}}{\varepsilonilon_1}$, and (3) the cost to get the final $k - \mathfrak{a} - \mathfrak{b} + 1$ components.
Since $\mathfrak{a}$ and $k - \mathfrak{b}$ are very small compared to $k$, (1) and (3) do not contribute much, so we can beat the $2$-approximation if we do for $\text{Laminar}cut{\mathfrak{b}}{\varepsilonilon_1}$.
\fi
\paragraph{Roadmap.} After some related work and preliminaries, we first
present the details of the reduction from $k$\textsc{-Cut}\xspace to \text{Laminar}kcut{k} in
Section~\ref{sec:reduction}. Then in Section~\ref{sec:laminar} we give
the algorithm for \text{Laminar}kcut{k} assuming an algorithm for \textsc{Partial VC}\xspace. Finally
we give our FPT-AS for \textsc{Partial VC}\xspace in Section~\ref{sec:partial-vc}.
\subsection{Other Related Work}
\langlebel{sec:related}
The $k$\textsc{-Cut}\xspace problem has been widely studied. Goldschmidt and Hochbaum gave
an $O(n^{(1/2- o(1))k^2})$-time algorithm~\cite{GH94}; they also showed
that the problem is NP-hard when $k$ is part of the input. Karger and
Stein improved this to an $O(n^{(2-o(1))k})$-time randomized Monte-Carlo
algorithm using the idea of random edge-contractions~\cite{KS96}.
After Kamidoi et al.~\cite{KYN06} gave an $O(n^{4k + o(1)})$-time
deterministic algorithm based on divide-and-conquer,
Thorup gave an $\tilde{O}(n^{2k})$-time deterministic algorithm based on
tree packings~\cite{Thorup08}.
Small values of $k \in [2, 6]$ also have been separately studied~\cite{NI92, HO92, BG97, Karger00, NI00, NKI00, Levine00}.
On the approximation algorithms front, a $2(1-1/k)$-approximation was
given by Saran and Vazirani~\cite{SV95}. Naor and Rabani~\cite{NR01},
and Ravi and Sinha~\cite{RS02} later gave $2$-approximation algorithms
using tree packing and network strength respectively. Xiao et
al.~\cite{XCY11} completed the work of Kapoor~\cite{Kapoor96} and Zhao
et al.~\cite{ZNI01} to generalize Saran and Vazirani to essentially give
an $(2 - h/k)$-approximation in time $n^{O(h)}$. Very recently,
Manurangsi~\cite{Manurangsi17} showed that for any $\varepsilon > 0$, it is
NP-hard to achieve a $(2 - \varepsilon)$-approximation algorithm in time
$\mathrm{poly}(n,k)$ assuming the Small Set Expansion Hypothesis.
\varepsilonmph{FPT algorithms:} Kawarabayashi and Thorup give an
$f(\varepsilonnsuremath{\mathsf{Opt}}\xspace) \cdot n^{2}$-time algorithm~\cite{KT11} for unweighted
graphs. Chitnis et al.~\cite{Chitnis} used a randomized color-coding
idea to give a better runtime, and to extend the algorithm to weighted
graphs. In both cases, the FPT algorithm is parameterized by the
cardinality of edges in the optimal $k$\textsc{-Cut}\xspace, not by $k$. For a
comprehensive treatment of FPT algorithms, see the excellent
book~\cite{FPT-book}, and for a survey on approximation and FPT
algorithms, see~\cite{Marx07}.
\varepsilonmph{Multiway Cut:} A problem very similar to $k$\textsc{-Cut}\xspace is the
\textsc{Multiway Cut} problem, where we are given $k$ terminals and want
to disconnect the graph into at least $k$ pieces such that all terminals
lie in distinct components. However, this problem behaves quite
differently: it is NP-hard even for $k=3$ (and hence an $n^{f(k)}$
algorithm is ruled out); on the other hand several algorithms are
known to approximate it to factors much smaller than~$2$ (see,
e.g.,~\cite{BuchbinderSW17} and references therein). FPT algorithms
parameterized by the size of $\varepsilonnsuremath{\mathsf{Opt}}\xspace$ are also known; see~\cite{CaoCF14}
for the best result currently known.
\section{Notation and Preliminaries}
\langlebel{sec:prelims}
For a graph $G = (V,E)$, and a subset $S \subseteq V$, we use $G[S]$ to
denote the subgraph induced by the vertex set $S$. For a collection of
disjoint sets $S_1, S_2, \ldots, S_t$, let $E(S_1, \ldots, S_t)$ be the
set of edges with endpoints in some $S_i, S_j$ for $i \neq j$. Let
$\partial S = E(S, V \setminus S)$. We say two cuts $(A, V\setminus A)$
and $(B,V\setminus B)$ \varepsilonmph{cross} if none of the four sets $A
\setminus B, B \setminus A, A \cap B$, and $V \setminus (A \cup B)$ is
empty.
$\mathsf{Mincut}$ and $\text{\sf{Min-4-cut}}$ denote the weight of the min-2-cut
and the min-4-cut respectively.
A cut $(A, V \setminus A)$ is called $(1 + \varepsilon)$-mincut if
$w(A, V \setminus A) \leq (1 + \varepsilon) \mathsf{Mincut}$.
\begin{restatable}[\textsc{Laminar $k$-Cut}$(\varepsilon_1)$]{definition}{LamDef}
\langlebel{def:laminarcut}
The input is a graph $G = (V,E)$ with edge weights, and two parameters
$k$ and $\varepsilon_1$, satisfying two promises: (i)~no two $(1+\varepsilon_1)$-mincuts
cross each other, and (ii)~there exists a $k$-cut ${\cal S}' = \{S_1',
\ldots, S_k'\}$ in $G$ with $w(\partial(S_i')) \le (1+\varepsilon_1)\mathsf{Mincut}(G)$
for all $i \in [1,k-1]$. Find a $k$-cut with the total weight.
The approximation ratio is defined as the ratio of
the weight of the returned cut to the weight of the $k$\textsc{-Cut}\xspace ${\cal S}'$
(which can be possibly less than $1$).
\varepsilonnd{restatable}
\begin{definition}[\textsc{Partial VC}\xspacelong]
\langlebel{def:pvc}
Given a graph $G = (V,E)$ with edge and vertex weights, and an integer
$k$, find a vertex set $S \subseteq V$ with $|S| = k$ nodes, minimizing the
weight of the edges hitting the set $S$ plus the weight of all
vertices in $S$.
\varepsilonnd{definition}
\section{Reduction to $\text{Laminar}cut{k}{\varepsilon_1}$}
\langlebel{sec:reduction}
In this section we give our reduction from $k$\textsc{-Cut}\xspace to
$\text{Laminar}cut{k}{\varepsilon_1}$, showing that if we can get a better-than-2
approximation for the latter, we can beat the factor of two for the
general $k$\textsc{-Cut}\xspace problem too. We assume the reader is familiar with the
overview in Section~\ref{sec:overview-redn}. Formally, the main theorem
is the following.
\begin{theorem}
\langlebel{thm:reduction1}
Suppose there exists a $(2 - \varepsilon_2)$-approximation algorithm for
$\text{Laminar}cut{k}{\varepsilon_1}$ for some $\varepsilon_1 \in (0, 1/4)$ and
$\varepsilon_2 \in (0, 1)$ that runs in time $f(k) \cdot g(n)$. Then
there exists a $(2 - \varepsilon_3)$-approximation algorithm for $k$\textsc{-Cut}\xspace
that runs in time $2^{O(k^2 \log k)} \cdot f(k) \cdot (n^4 \log^3 n + g(n))$ for some constant
$\varepsilon_3 > 0$.
\varepsilonnd{theorem}
\begin{algorithm}
\caption{$\text{Main}(G = (V, E, w), k)$}
\langlebel{alg:main}
\begin{algorithmic}[1]
\State $k' = 1$, $S_1 \gets V$
\While {$k' < k$ }
\mathbb For {$\boldsymbol{\mathrm{r}} \in [k]^{k'}$ } \langlebel{line:start-of-check} \Comment {Further partition each $S_i$ into $r_i$ components by Laminar}
\State $|\boldsymbol{\mathrm{r}}| \gets \sum_{j=1}^{k'} r_j$; $\{ C_1, \dots,
C_{|\boldsymbol{\mathrm{r}}|} \} \gets \cup_{i \in [k']}
\text{Laminar}(\inducedG{S_i}, r_i)$.
\If{$|\boldsymbol{\mathrm{r}}| \geq k$} $C_k \gets C_k \cup \dots \cup C_{|\boldsymbol{\mathrm{r}}|}$
\Else \State $\{C_1, \dots, C_{k}\} \gets \text{Complete}(G, k, C_1, \dots, C_{|\boldsymbol{\mathrm{r}}|})$ \langlebel{line:complete}
\EndIf
\State \mathbb Record($\text{Guess}(\{C_1, \dots, C_k \})$)
\EndFor \langlebel{line:end-of-check}
\State {} \Comment {Split some $S_i$ by a mincut or a min-4-cut}
\If{$k' > k - 3$ or $\min_{i \in [k']} \mathsf{Mincut} (\inducedG{S_{i}}) \leq
\min_{i \in [k']} \text{\sf{Min-4-cut}} (\inducedG{S_{i}}) / 3$} \langlebel{line:start-extend}
\State $i \gets \min_{i} \mathsf{Mincut} (\inducedG{S_{i}})$; $\{ T_1, T_2 \} \gets \text{Mincut}(\inducedG{S_{i}})$
\State $S_i \gets T_1$; $S_{k' + 1} \gets T_2$; $c_{k'} \gets \mathsf{Mincut} (\inducedG{S_{i}})$; $k' \gets k' + 1$
\Else
\State $i \gets \arg\min_{i} \text{\sf{Min-4-cut}} (\inducedG{S_{i}})$; $\{ T_1, \dots, T_4 \} \gets \text{Min-4-cut}(\inducedG{S_{i}})$;
$S_{i} \gets T_1$
\State $S_{k' + 1}, S_{k' + 2}, S_{k' + 3} \gets T_2, T_3, T_4$;
$c_{k'}, c_{k' + 1}, c_{k' + 2} \gets \text{\sf{Min-4-cut}} (\inducedG{S_{i}}) / 3$;
$k' \gets k' + 3$
\EndIf \langlebel{line:end-extend}
\EndWhile
\State let ${\cal S} = \{S_1, \ldots, S_k\}$ be the final reference $k$-partition.
\State \mathbb Record($\text{Guess}(G, k, {\cal S})$) \langlebel{line:lastupdate}
\State Return the best recorded $k$-partition.
\varepsilonnd{algorithmic}
\varepsilonnd{algorithm}
\begin{algorithm}
\caption{$\text{Complete}(G = (V, E, w), k, {\cal C} = \{C_1, \ldots, C_\varepsilonll\})$}
\langlebel{alg:complete}
\begin{algorithmic}[1]
\While {$\varepsilonll < k$}
\State $i \gets \min_{i \in [\varepsilonll]} \mathsf{Mincut}(\inducedG{C_i})$; $T_1, T_2 \gets \text{Mincut}(\inducedG{C_i})$
\State $C_i \gets T_1$; $C_{\varepsilonll + 1} \gets T_2$; $\varepsilonll \gets \varepsilonll + 1$
\EndWhile
\State Return ${\cal C} := \{C_1, \dots, C_k\}$.
\varepsilonnd{algorithmic}
\varepsilonnd{algorithm}
\begin{algorithm}
\caption{$\text{Guess}(G = (V, E, w), k, {\cal C} = \{C_1, \dots, C_k\})$}
\langlebel{alg:guess}
\begin{algorithmic}[1]
\State \mathbb Record($C_1, \dots, C_k$) \Comment{Returned
partition no worse than starting partition}
\mathbb For {$\varepsilonmptyset \neq J \subsetneq [k]$ }
\mathbb For {$k' = 1, 2, \dots, k - 1$ }
\State $L \gets \cup_{j \in J} C_j$; $R \gets V
\setminus L$ \Comment {Divide $S_i$ into two groups,
take union of each group}
\State $D_1, \dots, D_{k'} \gets \text{Main}(\inducedG{L},
k')$ \Comment{and recurse}
\State $D_{k'+1}, \dots, D_k \gets \text{Main}(\inducedG{R}, k - k')$
\State \mathbb Record($D_1, \dots, D_k$)
\EndFor
\EndFor
\State Return the best recorded $k$-partition among all these guesses.
\varepsilonnd{algorithmic}
\varepsilonnd{algorithm}
The main algorithm is shown in Algorithm~\ref{alg:main} (``\text{Main}'').
It maintains a ``reference'' partition ${\cal S}$, which is initially the
trivial partition where all vertices are in the same part. At each
point, it guesses how many pieces each part $S_i$ of this reference partition
${\cal S}$ should be split into using the ``Laminar'' procedure, and then
extends this to a $k$-cut using greedy cuts if necessary
(Lines~\ref{line:start-of-check}--\ref{line:end-of-check}).
It then
extends the reference partition by either taking the best min-cut or the
best min-4-cut among all the parts
(Lines~\ref{line:start-extend}--\ref{line:end-extend}).
Every time it
has a $k$-partition, it guesses (using ``Guess'') if the union of some
of the parts equals some part of the optimal partition, and uses that to
try get a better partition.
If one of the guesses is right, we strictly increase the number of
connected components by deleting edges in the optimal $k$-cut, so we can
recursively solve the two smaller parts. If none of our guesses was
right during the algorithm, our analysis in
Section~\ref{subsec:approx_factor} shows that there exist values of $k',
\boldsymbol{\mathrm{r}}$ such that ${\cal C} = \{C_1, \dots, C_k\}$ in
Line~\ref{line:complete}, obtained from the reference partition ${\cal S} =
\{ S_1, \dots, S_{k'} \}$ by running Laminar($G[S_{i}], r_i$) for each $i \in [k']$
and using Complete if necessary to get $k$ components, beats the $2$-approximation. Finally, a couple words about
each of the subroutines.
\begin{itemize}
\item Mincut$(G = (V, E, w))$ (resp.\ Min-4-cut$(G)$)
returns the minimum $2$-cut (resp.\ $4$-cut) as a partition of $V$
into $2$ (resp. $4$) subsets.
\item The subroutine ``Laminar'' returns a $(2-\varepsilon_2)$-approximation
for \text{Laminar}cut{$k$}{$\varepsilon_1$}, using the algorithm from
Theorem~\ref{thm:laminar}. Recall the definition of the problem in Definition~\ref{def:laminarcut}.
\item The operation ``\mathbb Record(${\cal P}$)'' in \text{Guess}\ and \text{Main}\ takes a
$k$-partition ${\cal P}$ and compares the weight of edges crossing this
partition to the least-weight $k$-partition recorded thus far (within
the current recursive call). If the current partition has less weight,
it updates the best partition accordingly.
\item Algorithm~\ref{alg:complete}(``\text{Complete}'') is a simple algorithm
that given an $\varepsilonll$-partition ${\cal P}$ for some $\varepsilonll \leq k$, outputs a
$k$-partition by iteratively taking the mincut in the current graph.
\item Algorithm~\ref{alg:guess}(``\text{Guess}''), when given an
$\varepsilonll$-partition ${\cal P}$ ``guesses'' if the vertices belonging to some
parts of this partition $\{ S_j \}_{j \in J}$ coincide with the union
of some $k'$ parts of the optimal partition. If so, we have made
tangible progress: it recursively finds a small $k'$-cut in the graph
induced by $\cup_{j \in J} S_j$, and a small $k-k'$ cut in the
remaining graph. It returns the best of all these guesses.
\varepsilonnd{itemize}
\subsection{The Approximation Factor}
\langlebel{subsec:approx_factor}
\begin{lemma}[Approximation Factor]
\langlebel{lem:apx-main}
$\text{Main}(G, k)$ achieves a $(2 - \varepsilon_3)$ approximation for some
$\varepsilon_3 > 0$ that depends on $\varepsilon_1, \varepsilon_2$ in
Theorem~\ref{thm:reduction1}.
\varepsilonnd{lemma}
\begin{proof}
We prove the lemma by induction on $k$. The value of $\varepsilon_3$ will be
determined later. The base case $k = 1$ is trivial. Fix some value
of $k$, and a graph $G$. Let ${\cal S} = \{S_1, \dots, S_k\}$ be the
final reference partition generated by the execution of $\text{Main}(G, k)$,
and let $c_1, \dots, c_{k - 1}$ be the values associated with it. From the
definition of the $c_i$'s in Procedure~\text{Main}, $\sum_{i=1}^{k - 1} c_i =
w(E(S_1, \dots, S_k))$. The $k$-partition returned by $\text{Main}(G, k)$ is
no worse than this partition ${\cal S}$ (because of the update on
line~\ref{line:lastupdate}), and hence has cost at most $\sum_{i=1}^{k-1}
c_i = w(E(S_1, \dots, S_k))$. Let us fix an optimal $k$-cut ${\cal S}^*
= \{ S^*_1, \dots, S^*_k\}$, and let $w(\partial{S^*_1}) \leq \dots
\leq w(\partial{S^*_k})$. Let $\varepsilonnsuremath{\mathsf{Opt}}\xspace := w(E(S^*_1, \dots, S^*_k)) =
\sum_{i=1}^k w(\partial{S^*_i}) / 2$.
\begin{definition}[Conformity]
\langlebel{def:conform}
For a subset $\varepsilonmptyset \neq U \subsetneq V$, we say that $U$ {\varepsilonm
conforms} to partition ${\cal S}$ if there exists a subset $J
\subsetneq [k]$ of parts such that $U = \cup_{j \in J} S_j$. (See Figure~\ref{fig:conform}.)
\varepsilonnd{definition}
The following claim shows that if there exists a subset $\varepsilonmptyset
\neq I \subsetneq [k]$ of indices such that $\cup_{i\in I} S^*_i$
conforms to ${\cal S}$, the induction hypothesis guarantees a $(2 -
\varepsilon_3)$-approximation.
\begin{claim}
Suppose there exists a subset $\varepsilonmptyset \neq I \subsetneq [k]$ such
that $\cup_{i \in I} S_i^*$ conforms to ${\cal S}$. Then $\text{Main}(G, k)$
achieves a $(2 - \varepsilon_3)$-approximation.
\langlebel{claim:good}
\varepsilonnd{claim}
\begin{proof}
Since $S^*_I := \cup_{i \in I} S^*_i$ conforms to ${\cal S}$, during
the run of $\text{Guess}(G, k, {\cal S})$ it will record the $k$-partition
$(\text{Main}(\inducedG{S^*_I}, |I|), \text{Main}(\inducedG{V \setminus S^*_I},
k - |I|) )$, and hence finally output a $k$-partition which cuts no
more edges than this starting partition. By the induction
hypothesis, $\text{Main}(\inducedG{S^*_I}, |I|)$ gives a $|I|$-cut of
$\inducedG{S^*_I}$ whose cost is at most $(2 - \varepsilon_3)$ times
$w(E(S^*_i)_{i \in I})$, and $\text{Main}(\inducedG{V \setminus S^*_I}, k
- |I|)$ outputs a $(k - |I|)$-cut of $\inducedG{V\setminus S^*_I}$
of cost at most $(2 - \varepsilon_3)$ times $w(E(S^*_i)_{i \notin
I})$. Thus, the value of the best $k$-partition returned by
$\text{Main}(G, k)$ is at most
\begin{align*}
& w(E(S^*_I, V \setminus S^*_I)) + (2 - \varepsilon_3) \left(
w(E(S^*_i)_{i \in I}) + w(E(S^*_i)_{i \notin I}) \right) \\
\leq & \ (2 - \varepsilon_3) w(E(S^*_1, \dots, S^*_k)) = (2 -
\varepsilon_3) \mathsf{Opt}. \qedhere
\varepsilonnd{align*}
\varepsilonnd{proof}
Therefore, to prove Lemma~\ref{lem:apx-main}, it suffices to assume
that no collection of parts in \varepsilonnsuremath{\mathsf{Opt}}\xspace conforms to our partition at any
point in the algorithm. I.e.,
\begin{leftbar}
\As{1}: for every subset $\varepsilonmptyset \neq I \subsetneq [k]$, $\cup_{i
\in I} S^*_i$ does not conform to ${\cal S} = \{ S_1, \dots, S_k\}$.
\varepsilonnd{leftbar}
Next, we study how $\varepsilonnsuremath{\mathsf{Opt}}\xspace$ is related to $w(\partial{S_1^*})$. Note
that $\varepsilonnsuremath{\mathsf{Opt}}\xspace \geq (k/2) \cdot w(\partial{S_1^*})$. The next claim shows
that we can strictly improve the $2$-approximation if $\varepsilonnsuremath{\mathsf{Opt}}\xspace$ is even
slightly bigger than that.
\begin{claim}
For every $i = 1, \dots, k-1$, $c_i \leq w(\partial{S^*_1})$. Moreover,
if $\varepsilonnsuremath{\mathsf{Opt}}\xspace \geq (k - 1)w(\partial{S_1^*}) / (2 - \varepsilon_3)$, $\text{Main}(G,
k)$ achieves a $(2 - \varepsilon_3)$-approximation.
\langlebel{claim:notgood}
\varepsilonnd{claim}
\begin{proof}
Consider the beginning of an arbitrary iteration of the while loop
of $\text{Main}(G, k)$. Let $k'$ and ${\cal S}' = \{ S_1, \dots, S_{k'} \}$
be the values at that iteration. By \As{1}, set $S_1^*$ does not
conform to ${\cal S}'$ (because ${\cal S}'$ only gets subdivided as the
algorithm proceeds, and $S_1^*$ does not conform to the final partition
${\cal S}$). So there exists some $i \in [k']$ such that $S_i$
intersects both $S^*_1$ and $V \setminus S^*_1$. If we consider
$\inducedG{S_i}$ and its mincut,
\[
\mathsf{Mincut}(\inducedG{S_i}) \leq
w(E(S_i \cap S^*_1, S_i \setminus S^*_1))
\leq w(\partial{S^*_1}).
\]
Now the new $c_j$ values created in this iteration of the while loop
are at most the smallest mincut value, so we have that each $c_j
\leq w(\partial{S_1^*})$. Therefore,
\[
w(E(S_1, \dots, S_k)) = \sum_{i=1}^{k - 1} c_i \leq (k - 1)\cdot
w(\partial{S_1^*}),
\]
and $\text{Main}(G, k)$ achieves a $(2 - \varepsilon_3)$-approximation if $(k
- 1) w(\partial{S^*_1}) \leq (2 - \varepsilon_3) \varepsilonnsuremath{\mathsf{Opt}}\xspace$.
\varepsilonnd{proof}
Consequently, it suffices to additionally assume that $\varepsilonnsuremath{\mathsf{Opt}}\xspace$ is
close to $(\nicefrac{k}{2}) \, w(\partial{S^*_1})$. Formally,
\begin{leftbar}
\As{2}: $ \varepsilonnsuremath{\mathsf{Opt}}\xspace < w(\partial{S^*_1}) \cdot \frac{k - 1}{2 - \varepsilon_3} $.
\varepsilonnd{leftbar}
Recall that $\varepsilon_1, \varepsilon_2 > 0$ are the parameters such that
there is a $(2 - \varepsilon_2)$-approximation algorithm for
$\text{Laminar}cut{k}{\varepsilon_1}$. Let $\mathfrak{a} \in [k]$ be the smallest
integer such that $c_{\mathfrak{a}} > w(\partial{S^*_1}) (1 -
\nicefrac{\varepsilon_1}{3})$ (set $\mathfrak{a} = k$ if there is no such integer).
(See Figure~\ref{fig:histo}.)
In other words, $\mathfrak{a}$ is the value of $k'$ in the while loop of
$\text{Main}(G, k)$ when both $\min_i \mathsf{Mincut}(\inducedG{S_i})$ and $\min_i
\text{\sf{Min-4-cut}}(\inducedG{S_i}) / 3$ are bigger than $w(\partial{S^*_1}) (1 -
\nicefrac{\varepsilon_1}{3})$ for the first time. Let $\varepsilon_4 > 0$ be a constant
satisfying
\begin{equation}
\langlebel{eq:para_1}
(2/3) \cdot \varepsilon_1 \varepsilon_4 \geq \varepsilon_3.
\varepsilonnd{equation}
The next claim shows that we are done if $\mathfrak{a}$ is large.
\begin{claim}
If $\mathfrak{a} \geq \varepsilon_4 k$, $\text{Main}(G, k)$ achieves a $(2 -
\varepsilon_3)$-approximation.
\langlebel{clm:left_tail}
\varepsilonnd{claim}
\begin{proof}
If $\mathfrak{a} \geq \varepsilon_4 k$, we have
\begin{align*}
\sum_{i=1}^{k-1} c_i &\leq
(\mathfrak{a} - 1) (1 - \nicefrac{\varepsilon_1}{3})\cdot w(\partial{S^*_1}) +
(k - \mathfrak{a})\cdot w(\partial{S^*_1}) \\
& \leq k \cdot w(\partial{S^*_1})\cdot (1 - \nicefrac{\varepsilon_1 \varepsilon_4}{3})
\leq (2 - (\nicefrac23) \varepsilon_1 \varepsilon_4) \varepsilonnsuremath{\mathsf{Opt}}\xspace \leq (2 - \varepsilon_3)
\varepsilonnsuremath{\mathsf{Opt}}\xspace. \qedhere
\varepsilonnd{align*}
\varepsilonnd{proof}
Thus, we can assume that our algorithm finds very few cuts appreciably
smaller than $w(\partial{S^*_1})$.
\begin{leftbar}
\As{3}: $\mathfrak{a} < \varepsilon_4 k$.
\varepsilonnd{leftbar}
Let $\mathfrak{b} \in [k]$ be the smallest number such that
$w(\partial{S^*_{\mathfrak{b}}}) > w(\partial{S^*_1}) (1 +
\nicefrac{\varepsilon_1}{3})$; let it be $k$ if there is no such
number. (Again, see Figure~\ref{fig:histo}.) Observe that $\mathfrak{a}$ is
defined based on our algorithm, whereas $\mathfrak{b}$ is defined based on the
optimal solution. Let $\varepsilon_5 > 0$ be a constant satisfying:
\begin{equation}
\langlebel{eq:para_2}
\frac{1}{2 - \varepsilon_3} \leq \frac{1 + \nf{\varepsilon_1 \varepsilon_5}{3}}{2}
~~~\Leftrightarrow~~~ (1 + \nf{\varepsilon_1 \varepsilon_5}{3})(2 -
\varepsilon_3) \geq 2.
\varepsilonnd{equation}
The next claim shows that $\mathfrak{b}$ should be close to $k$.
\begin{claim}
$\mathfrak{b} \geq (1 - \varepsilon_5)k$.
\varepsilonnd{claim}
\begin{proof}
Suppose that $\mathfrak{b} <(1 - \varepsilon_5) k$. We have
\begin{align*}
& \ \frac{ k \cdot w(\partial{S^*_1}) }{2 - \varepsilon_3} \stackrel{\As{2}}{>} \varepsilonnsuremath{\mathsf{Opt}}\xspace = \frac{1}{2} \sum_{i = 1}^k w(\partial{S^*_i}) \\
\geq& \ \frac{w(\partial{S^*_1})}{2} \left( (1 - \varepsilon_5)k +
\varepsilon_5 k (1 + \varepsilon_1 / 3) \right)
=
\frac{k\cdot w( \partial{S^*_1} )}{2} \left( 1 + \nf{\varepsilon_1 \varepsilon_5}{3} \right),
\varepsilonnd{align*}
which contradicts~\varepsilonqref{eq:para_2}.
\varepsilonnd{proof}
Therefore, we can also assume that very few cuts in \varepsilonnsuremath{\mathsf{Opt}}\xspace
are appreciably larger than $w( \partial{S^*_1})$.
\begin{leftbar}
\As{4}: $\mathfrak{b} \geq (1 - \varepsilon_5) k$.
\varepsilonnd{leftbar}
\textbf{Constructing an Instance of Laminar Cut:} In order to
construct the instance for the problem, let $S^*_{\geq \mathfrak{b}} = \cup_{i = \mathfrak{b}}^{k}
S^*_i$ be the union of these last few components from ${\cal S}^*$ which
have ``large'' boundary. Consider the iteration of the while loop
when $k' = \mathfrak{a}$ and consider $S_1, \dots, S_{\mathfrak{a}}$ in that
iteration. By its definition, $c_{\mathfrak{a}} > w(\partial{S_1^*}) (1 -
\nicefrac{\varepsilon_1}{3})$. Hence
\begin{gather}
\min_i \mathsf{Mincut}(\inducedG{S_i}) > w(\partial{S_1^*}) (1 -
\nicefrac{\varepsilon_1}{3}),
\langlebel{eq:stop1} \\
\min_i \text{\sf{Min-4-cut}}(G[S_i]) > 3 w(\partial{S_1^*}) (1 -
\nicefrac{\varepsilon_1}{3}).
\langlebel{eq:stop2}
\varepsilonnd{gather}
In particular, \varepsilonqref{eq:stop2} implies that no two near-min-cuts
cross, since two crossing near-min-cuts will result in a $4$-cut of
weight roughly at most $2 w(\partial{S_1^*})$.
However, we are not yet done, since we need to factor out the effects
of the $\mathfrak{a} - 1$ ``small'' cuts found by our algorithm. For this, we need
one further idea.
Let $\boldsymbol{\mathrm{r}} = (r_1, r_2, \ldots, r_{\mathfrak{a}}) \in [k]^{\mathfrak{a}}$ be such that
$r_i$ is the number of sets
$S^*_1, \dots, S^*_{\mathfrak{b} - 1}, S^*_{\geq \mathfrak{b}}$ that intersect with
$S_i$, and let $|\boldsymbol{\mathrm{r}}| := \sum_{i = 1}^{\mathfrak{a}} r_i$. If we consider the
bipartite graph where the left vertices are the algorithm's
components $S_1, \dots, S_{\mathfrak{a}}$, the right vertices are
$S^*_1, \dots, S^*_{\mathfrak{b} - 1}, S^*_{\geq \mathfrak{b}}$, and two sets have an
edge if they intersect, then $|\boldsymbol{\mathrm{r}}|$ is the number of edges. Since
there is no isolated vertex and the graph is connected (otherwise
there would exist $\varepsilonmptyset \neq I \subsetneq [k']$ and
$\varepsilonmptyset \neq J \subsetneq [k]$ with
$\cup_{i \in I}S_i = \cup_{j \in J} S^*_j$ contradicting~\As{1}),
the number of edges is $|\boldsymbol{\mathrm{r}}| \geq \mathfrak{a} + \mathfrak{b} - 1$.
\begin{claim}
\langlebel{clm:promises}
For each $i$ with $r_i \geq 2$,
the graph $\inducedG{S_i}$ satisfies the two promises of the problem
$\text{Laminar}cut{r_i}{\varepsilon_1}$.
\varepsilonnd{claim}
\begin{proof}
Fix $i$ with $r_i \geq 2$. Let
$J := \{ j \in [\mathfrak{b} - 1] \mid S_i \cap S^*_j \neq \varepsilonmptyset \}$ be
the sets $S_j^*$ among the first $\mathfrak{b} - 1$ sets in the optimal
partition that intersect $S_i$. Since $|J| \geq r_i - 1$ and $r_i \geq 2$, $|J| \geq 1$.
Note that
$(1 - \nf{\varepsilon_1}3)\cdot w(\partial{S^*_1}) <
\mathsf{Mincut}(\inducedG{S_i})$ by~\varepsilonqref{eq:stop1}.
For every $j \in J$,
\[
\mathsf{Mincut}(\inducedG{S_i}) \leq
w(E(S_i \cap S^*_j, S_i \setminus S^*_j)) \leq w(\partial{S^*_j})
\leq (1 + \varepsilon_1 / 3)\; w(\partial{S^*_1}) \leq (1 + \varepsilon_1)
\;\mathsf{Mincut}(\inducedG{S_i}).
\]
The first and second inequality hold since both parts
$S_i \cap S^*_j$ and $S_i \setminus S^*_j$ are nonempty, and hence
deleting all the edges in $\partial{S_j^*}$ would separate
$G[S_i]$.
The third inequality is by the choice of $\mathfrak{b}$, and the last
inequality uses~(\ref{eq:stop1}) and the fact that
$(1 + \nicefrac{\varepsilon_1}{3}) \leq (1 + \varepsilon_1)(1 -
\nicefrac{\varepsilon_1}{3})$ when $\varepsilon_1 < 1/4$.
This implies that in $\inducedG{S_i}$, for every $j \in J$,
$(S_i \cap S^*_j, S_i \setminus S^*_j)$ is a $(1 + \varepsilon_1)$-mincut.
Furthermore, in $\inducedG{S_i}$, no two $(1+\varepsilon_1)$-mincuts cross
because it will result a 4-cut of cost at most
\[
2(1+\varepsilon_1)\; \mathsf{Mincut}(\inducedG{S_i}) \leq 2(1+\varepsilon_1) (1 + \varepsilon_1 / 3)
\; w(\partial{S_1^*}),
\]
contradicting~\varepsilonqref{eq:stop2}. (Note that $2(1 +
\varepsilon_1) (1 + \nicefrac{\varepsilon_1}{3}) \leq 3(1 - \nicefrac{\varepsilon_1}{3})$ when
$\varepsilon_1 < 1/4$.) Hence, in $\inducedG{S_i}$, the two promises for
$\text{Laminar}cut{r_i}{\varepsilon_1}$ are satisfied.
\varepsilonnd{proof}
Our algorithm $\text{Main}(G, k$) runs $\text{Laminar}(\inducedG{S_i}, r_i)$ for
each $i \in [\mathfrak{a}]$ when it sets $k' = \mathfrak{a}$ and the vector $\boldsymbol{\mathrm{r}}$ as
defined above. As in the algorithm, let
${\cal C} = \{C_1, \dots, C_{k}\}$ be the partition obtained in
Line~\ref{line:complete}. In other words, to obtain the $k$ sets
$C_1, \dots, C_k$ from the set $V$, we take the reference partition
$S_1, \dots, S_{\mathfrak{a}}$ and further partition these sets using Laminar
to get $|\boldsymbol{\mathrm{r}}|$ parts $C_1, \dots, C_{|\boldsymbol{\mathrm{r}}|}$. If $|\boldsymbol{\mathrm{r}}| \geq k$, we
can merge the last $|\boldsymbol{\mathrm{r}}| - k + 1$ parts to get exactly $k$ parts if
we want (but we will not take any edge savings into account in this
calculation). If $|\boldsymbol{\mathrm{r}}| < k$, we get $k - |\boldsymbol{\mathrm{r}}|$ more parts using
the Complete procedure.
The total cost of this solution ${\cal C}$ is $w(E(C_1, \dots, C_k))$,
which is $\sum_{j=1}^{\mathfrak{a} - 1} c_j \leq (\mathfrak{a} - 1) w(\partial{S^*_1})$
plus the cost of $\text{Laminar}(\inducedG{S_i}, r_i)$ for all
$i \in [\mathfrak{a}]$ and the cost of $\text{Complete}$. Since
Claim~\ref{clm:promises} considers the partition of each
$\inducedG{S_i}$ obtained by cutting edges belonging to the optimal
$k$-partition,
the sum of the cost of the $r_i$-partition we compare to in
each \text{Laminar}kcut{r_i} is exactly $\varepsilonnsuremath{\mathsf{Opt}}\xspace$.
Hence the cost of the
solution given by $\text{Laminar}(\inducedG{S_i}, r_i)$ summed over
$i \in [\mathfrak{a}]$ is bounded by $(2 - \varepsilon_2) \varepsilonnsuremath{\mathsf{Opt}}\xspace$, by the approximation
assumption in Theorem~\ref{thm:reduction1}.
If $\cup_{i \in I} S^*_i$ for some $\varepsilonmptyset \neq I \subsetneq [k]$
conforms to ${\cal C}$, then since \text{Main}\ also records $\text{Guess}({\cal C})$,
the proof of Claim~\ref{claim:good} guarantees that $\text{Main}(G, k)$
gives a $(2 - \varepsilon_3)$ approximation using the induction hypothesis.
Otherwise, $S^*_1$ does not conform to ${\cal C}$, so the arguments used
in the proof of Claim~\ref{claim:notgood} show that the cost of
$\text{Complete}$ is at most $(k - |\boldsymbol{\mathrm{r}}|)\, w(\partial{S^*_1})$ if
$|\boldsymbol{\mathrm{r}}| \leq k$, and $0$ otherwise. Since $|\boldsymbol{\mathrm{r}}| \geq \mathfrak{a} + \mathfrak{b} - 1$,
the total cost $w(E(C_1, \dots, C_k))$ is then bounded
by
\begin{align*}
& (\mathfrak{a} - 1 ) w(\partial{S^*_1}) + (2 - \varepsilon_2) \varepsilonnsuremath{\mathsf{Opt}}\xspace +
(k - \mathfrak{a} - \mathfrak{b} + 1) w(\partial{S^*_1}) \\
= & \ (2 - \varepsilon_2) \varepsilonnsuremath{\mathsf{Opt}}\xspace + (k - \mathfrak{b}) w(\partial{S^*_1}) \\
\leq & \ (2 - \varepsilon_2) \varepsilonnsuremath{\mathsf{Opt}}\xspace + \varepsilon_5 k \cdot
w(\partial{S^*_1}) \tag{by \As{4}}\\
\leq & \ (2 - \varepsilon_2 + 2 \varepsilon_5) \varepsilonnsuremath{\mathsf{Opt}}\xspace.
\varepsilonnd{align*}
Therefore, if
\begin{equation}
\varepsilon_3 \leq \varepsilon_2 -2 \varepsilon_5,
\langlebel{eq:para_4}
\varepsilonnd{equation}
then $\text{Main}(G, k)$ gives a $(2 - \varepsilon_3)$ approximation in every
possible case. We set $\varepsilon_3, \varepsilon_4, \varepsilon_5 > 0$ so that they
satisfy the three conditions~\varepsilonqref{eq:para_1}, \varepsilonqref{eq:para_2},
and~\varepsilonqref{eq:para_4}, namely,
\[
(2/3) \cdot \varepsilon_1 \varepsilon_4 \geq \varepsilon_3, \quad (1 +
\varepsilon_1 \varepsilon_5 / 3)(2 - \varepsilon_3) \geq 2,
\quad \varepsilon_3 \leq \varepsilon_2 - 2 \varepsilon_5.
\]
(For instance, setting $\varepsilon_4 = \varepsilon_5 = \min(\varepsilon_1,
\varepsilon_2) / 3$ and $\varepsilon_3 = \varepsilon_4^2$ works.)
\varepsilonnd{proof}
\subsection{Running Time}
We prove that this algorithm also runs in FPT time, finishing the proof
of Theorem~\ref{thm:reduction1}.
\begin{lemma}
Suppose that $\text{Laminar}(G, k)$ runs in time $f(k) \cdot g(n)$.
Then Main$(G, k)$ runs in time $2^{O(k^2 \log k)} \cdot f(k) \cdot (g(n) + n^4 \log^3 n)$.
\varepsilonnd{lemma}
\begin{proof}
Let $\mathsf{Time}(\text{P})$ denote the running time of a procedure \text{P}.
Here each procedure is only parameterized by the number of sets it outputs (e.g., $\text{Main}(k), \text{Guess}(k), \text{Complete}(k), \text{Laminar}(k)$).
We use the fact that the global min-cut can be computed in time $O(n^2 \log^3 n)$~\cite{KS96} and the min-$4$-cut can be computed in $O(n^4 \log^3 n)$~\cite{Levine00}.
First, $\mathsf{Time}(\text{Complete}(k)) = O(kn^2 \log^3 n)$. For $\text{Guess}$ and $\text{Main}$,
\[
\mathsf{Time}(\text{Guess}(k)) \leq k \cdot 2^{k + 1} \cdot ( \mathsf{Time}(\text{Main}(k - 1)) + O(n) ),
\]
and
\begin{align*}
\mathsf{Time}(\text{Main}(k)) & \leq k^k \cdot (\mathsf{Time}(\text{Laminar}(k)) +
\mathsf{Time}(\text{Guess}(k)) + \mathsf{Time}(\text{Complete}(k))) + O(k n^4 \log^3 n) \\
& \leq 2^{O(k \log k)} \cdot f(k) \cdot (g(n) + O(n^4 \log^3 n)) +
2^{O(k \log k)} \cdot \mathsf{Time}(\text{Main}(k - 1)).
\varepsilonnd{align*}
We can conclude $\mathsf{Time}(\text{Main}(k)) \leq 2^{O(k^2 \log k)} \cdot f(k) \cdot (g(n) + n^4 \log^3 n)$.
\varepsilonnd{proof}
\section{An Algorithm for \text{Laminar}kcut{k}}
\langlebel{sec:laminar}
Recall the definition of the \text{Laminar}kcut{k} problem:
\LamDef*
Let $\mathcal O_{\varepsilon_1}$ contain all partitions $S_1,\ldots,S_k$ of $V$ with
the restriction that the boundaries of the first $k-1$ parts is
small---i.e., $w(\partial{S_i}) \le (1+\varepsilon_1)\mathsf{Mincut}(G)$ for all $i \in
[k-1]$. We emphasize that the weight of the last cut, i.e.,
$w(\partial{S_k})$, is unconstrained. In this section, we give an
algorithm to find a $k$-partition (possibly not in $\mathcal O_{\varepsilon_1}$) with
total weight
\[ w(E(S_1,\ldots,S_k)) \le (2 - \varepsilon_2) \min\limits_{\{S_i'\}\in
\mathcal O_{\varepsilon_1}} w(E(S_1',\ldots,S_k')). \]
Formally, the main theorem of this section is the following:
\begin{theorem}[Laminar Cut Algorithm]
\langlebel{thm:laminar}
Suppose there exists a $(1+\delta)$-approximation algorithm for
$\PartialVC{k}$ for some $\delta \in (0, 1/24)$ that runs in time
$f(k) \cdot g(n)$. Then, for any $\varepsilon_1\in(0,1/6-4\delta)$, there
exists a $(2 - \varepsilon_2)$-approximation algorithm for
$\text{Laminar}cut{k}{\varepsilon_1}$ that runs in time $2^{O(k)}f(k)(\tilde O(n^4) + g(n))$
for some constant $\varepsilon_2 > 0$.
\varepsilonnd{theorem}
In the rest of this section we present the algorithm and the analysis.
For a formal description, see the pseudocode in
Appendix~\ref{sec:pseudocode-laminar}.
\subsection{Mincut Tree}
The first idea in the algorithm is to consider the structure of a laminar family of cuts. Below, we introduce the concept of a \textit{mincut tree}. The vertices of the mincut tree are called \textit{nodes}, to distinguish them from the vertices of the original graph.
\begin{definition}[Mincut Tree]
A tree $\mathcal T = (V_{\mathcal T}, E_{\mathcal T}, w_{\mathcal T})$ is a \textbf{$(1+\varepsilon_1)$-mincut tree} on a graph $G=(V,E,w)$ with mapping $\phi : V \to V_{\mathcal T}$ if the following two sets are equivalent:
\begin{enumerate}
\item The set of all $(1+\varepsilon_1)$-mincuts of $G$.
\item Cut a single edge $e \in E_{\mathcal T}$ of the tree, and let $A_e
\subset V_{\mathcal T}$ be the nodes on one side of the cut. Define $S_e :=
\phi^{-1}(A_e) = \{v \mid \phi(v) \in A_e\}$ for each $e\in E_{\mathcal T}$,
and take the set of cuts $ \{ (S_e, V \setminus S_e) : e \in E_{\mathcal T}
\}$.
\varepsilonnd{enumerate}
Moreover, for every pair of corresponding $(1+\varepsilon_1)$-mincut $(S_e, V
\setminus S_e)$ and edge $e \in E_{\mathcal T}$, we have $w_{\mathcal T}(e) = w(E(S_e,
V\setminus S_e))$.
\varepsilonnd{definition}
We use the term \textit{mincut tree} without the $(1+\varepsilon_1)$ when the
value of $\varepsilon_1$ is either implicit or irrelevant.
For the rest of this section, let
\[ {\mu} := \mathsf{Mincut}(G) \] for brevity. Observe that the last condition
implies that ${\mu}\le w_{\mathcal T}(e)\le(1+\varepsilon_1){\mu}$ for all $e\in
E_{\mathcal T}$. The existence of a mincut tree (and the algorithm for it)
assuming laminarity, is standard, going back at least to Edmonds and
Giles~\cite{EG75}.
\begin{theorem}[Mincut Tree Existence/Construction]
\langlebel{thm:mincutTreeExistence}
If the set of $(1+\varepsilon_1)$-mincuts of a graph is laminar, then an
$O(n)$-sized $(1+\varepsilon_1)$-mincut tree always exists, and can
be found in $O(n^3)$ time.
\varepsilonnd{theorem}
\begin{proof}
We refer the reader to~\cite[Section~2.2]{KV12}. Fix a vertex $v \in
V$, and for each $(1+\varepsilon_1)$-mincut $(S,V\setminus S)$, pick the side
that contains $v$; this family of subsets of $V$ satisfies the laminar
condition in Proposition~2.12 of that book. Corollary~2.15 proves that
this family has size $O(n)$, and the construction of $T$ in
Proposition~2.14 gives the desired mincut tree. Furthermore, we can
compute the mincut tree in $O(n^3)$ time as follows: first precompute
whether $X \subset Y$ for every two sets $X$ and $Y$ in the family,
and then compute $T$ following the construction in the proof of
Proposition~2.14.
\varepsilonnd{proof}
\begin{definition}[Mincut Tree Terminology] Let $\mathcal T$ be a rooted mincut
tree. For $a \in V_{\mathcal T}$, define the following terms:
\begin{OneLiners}
\item[1.] $\ensuremath{\mathrm{children}}(a)$: the set of children of node $a$ in the rooted tree.
\item[2.] $\ensuremath{\mathrm{desc}}(a)$: the set of descendants of $a$, i.e., nodes $b \in V_{\mathcal T} \setminus a$ whose path to the root includes $a$.
\item[3.] $\ensuremath{\mathrm{anc}}(a)$: the set of ancestors of $a$, i.e., nodes $b \in V_{\mathcal T} \setminus a$ on the path from $a$ to the root.
\item[4.] $\ensuremath{\mathrm{subtree}}(a)$: vertices in the subtree rooted at $a$, i.e., $\{a\} \cup \ensuremath{\mathrm{desc}}(a)$.
\varepsilonnd{OneLiners}
\varepsilonnd{definition}
For the set of partitions $\mathcal O_{\varepsilon_1}$ (as defined at the beginning
of this section), we observe the following.
\begin{claim}[Representing Laminar Cuts in $\mathcal T$]
Let $\mathcal T = (V_{\mathcal T}, E_{\mathcal T}, w_{\mathcal T})$ be a $(1+\varepsilon_1)$-mincut tree of
$G=(V,E,w)$, and consider a partition $\{S_1, \ldots, S_k\} \in
\mathcal O_{\varepsilon_1}$. Then, there exists a root $r \in V_{\mathcal T}$ and nodes
$a_1, \ldots, a_{k-1} \in V_{\mathcal T} \setminus r$ such that if we root
the tree $\mathcal T$ at $r$,
\begin{enumerate}
\item For any two nodes in $\{a_1, \ldots, a_{k-1}\}$, neither is an
ancestor of the other. (We call two such nodes
\textbf{incomparable}).
\item For each $v_i$, let $A_i := \ensuremath{\mathrm{subtree}}(a_i)$, and let $A_k =
V_{\mathcal T} \setminus \bigcup_{i=1}^{k-1} A_i$ (so that $r \in A_k$). We
have the two equivalences $\{\phi^{-1}(A_i) \mid i \in [k-1]\} = \{S_1,
\ldots, S_{k-1}\}$ and $\phi^{-1}(A_k) = S_k$. In other words, the
components $A_i \subset V_{\mathcal T}$, when mapped back by $\phi^{-1}$,
correspond exactly to the sets $S_i \subset V$, with the additional
guarantee that $A_k$ and $S_k$ match.
\varepsilonnd{enumerate}
\varepsilonnd{claim}
\begin{proof}
Since $S_i$ is a $(1+\varepsilon_1)$-mincut for each $i \in [k-1]$, there
exists an edge $e_i \in E_{\mathcal T}$ such that the set $A_i'$ of nodes on
one side of $e_i$ satisfies $\phi^{-1}(A_i') = S_i $. The sets $A_i'$
for $i\in[k-1]$ are necessarily disjoint, and they cannot span all
nodes in $V_{\mathcal T}$, since $S_k$ is still unaccounted for. If we root
$\mathcal T$ at a node $r$ not in any $A_i'$, then each $A_i'$ is a subtree
of the rooted $\mathcal T$. Altogether, the roots of the subtrees $A_i'$
satisfy condition~(1) of the lemma, and the $A_i'$ themselves satisfy
condition~(2).
\varepsilonnd{proof}
For a graph $G=(V,E,w)$ and mincut tree $\mathcal T=(V_{\mathcal T},E_{\mathcal T},w_{\mathcal T})$
with mapping $\phi:V\to V_{\mathcal T}$, define $E_G(A,B)$ for $A,B \subset
V_{\mathcal T}$ as $E\left(\phi^{-1}(A), \phi^{-1}(B)\right)$, i.e., the total
weight of edges crossing the sets corresponding to $A$ and $B$ in $V$.
\begin{observation}
Given a root $r\in V_{\mathcal T}$ and incomparable nodes $a_1, \ldots,
a_{k-1} \in V_{\mathcal T} \setminus r$, we can bound the corresponding
partition $S_1,\ldots,S_k$ as follows:
\begin{align*}
w(E(S_1, \ldots, S_k)) &= \textstyle \sum_{i=1}^{k-1}
w(\partial(S_i)) - \sum_{i<j \le k-1} w(E(S_i, S_j)) \\ &=
\textstyle \sum_{i=1}^{k-1} w_{\mathcal T}(e_i) - \sum_{i<j\le k-1}
w(E_G(\ensuremath{\mathrm{subtree}}(a_i),\ensuremath{\mathrm{subtree}}(a_j))) ,
\varepsilonnd{align*}
where $e_i$ is the parent edge
of $v_i$ in the rooted tree.
\varepsilonnd{observation}
Note that ${\mu} \le w_{\mathcal T}(e) \le (1+\varepsilon_1){\mu}$ for all $e\in
E_{\mathcal T}$, so to approximately minimize the above expression for a fixed
root $r$, it suffices to approximately maximize
\begin{align*}
\textstyle \mathsf{Saved}(a_1,\ldots,a_{k-1}) := \sum\limits_{i<j\le k-1}
w(E_G(\ensuremath{\mathrm{subtree}}(a_i),\ensuremath{\mathrm{subtree}}(a_j))), \langlebel{eq:saved}
\varepsilonnd{align*}
which we think of as the edges \textit{saved} in the double counting of
$\sum_{i=1}^{k-1}w_{\mathcal T}(e_i)$. The actual approximation factor is made
precise in the proof of Theorem~\ref{thm:laminar}.
To maximize the number of saved edges over all partitions in
$\mathcal O_{\varepsilon_1}$, it suffices to try all possible roots $r$ and take the
best partition. Therefore, for the rest of this section, we focus on
maximizing $\mathsf{Saved}(a_1,\ldots,a_{k-1})$ for a fixed root
$r$.
Let $\varepsilonll^*(r)$ be that maximum value for root $r$, and let $\mathsf{Opt}(r) =
\{a_1^*, \ldots, a_{k-1}^*\} \subset V_{\mathcal T}$ be the solution that
attains it.
\subsection{Anchors}
Root the mincut tree $\mathcal T$ at $r$, and let $a_1^*,\ldots,a_{k-1}^*$ be incomparable nodes in the solution $\varepsilonnsuremath{\mathsf{Opt}}\xspace(r)$.
First, observe that we can assume w.l.o.g.\ that for each node $a_i^*$,
its parent node is an ancestor of some $a_j^* \neq a_i^*$: if not, we can replace $a_i^*$ with its parent, which can only increase $\mathsf{Saved}(a_1^*,\ldots,a_{k-1}^*)$.
\begin{observation}
Consider nodes $a_1^*,\ldots,a_s^* \in \mathsf{Opt}(r)$ which share the same parent $a \notin \mathsf{Opt}(r)$, and assume that $a$ has no other descendants. If we replace $a_1^*,\ldots,a_s^*$ in $\mathsf{Opt}(r)$ with $a$, then we lose at most $\mathsf{Saved}(a_1^*,\ldots,a_s^*)$ in our solution.\footnote{The new solution may no longer have $k-1$ nodes, but we will fix this problem in the proof of Theorem~\ref{thm:laminar}. For now, assume that we are allowed to choose any number up to $k-1$ nodes.}
\varepsilonnd{observation}
If $\mathsf{Saved}(a_1^*,\ldots,a_s^*)$ is small, i.e., compared to $(s-1){\mu}$, then we do not lose too much. This idea motivates the idea of anchors.
\begin{definition}[Anchors]
Let $\mathcal T=(V_{\mathcal T},E_{\mathcal T},w_{\mathcal T})$ be a rooted tree.
For a fixed constant $\varepsilon_3 > 0$, define an \textbf{$\varepsilon_3$-anchor} to be a node $a\in V_{\mathcal T}$ such that there exists $s \in [2,k-1]$ and $s$ children $a_1,\ldots,a_s$ such that $\mathsf{Saved}(a_1,\ldots,a_s) \ge \varepsilon_3(s-1){\mu}$. When the value of $\varepsilon_3$ is implicit, we use the term anchor, without the $\varepsilon_3$.
\varepsilonnd{definition}
We now claim that we can transform any solution to another
well-structured solution, with only a minimal loss.
\begin{lemma}[Shifting Lemma]
\langlebel{lem:anchor}
Let $a_1,\ldots,a_{k-1}$ be a set of incomparable nodes of a
$(1+\varepsilon_1)$-mincut tree $\mathcal T$. Then, there exists a set
$b_1,\ldots,b_s$ of incomparable nodes, for $1\le s\le k-1$, such that
\begin{enumerate}
\item The parent of every node $b_i$ is either an $\varepsilon_3$-anchor, or is
an ancestor of some node $b_j \neq b_i$ whose parent is an anchor.
\item $\mathsf{Saved}(b_1,\ldots,b_s) \ge \mathsf{Saved}(a_1,\ldots,a_{k-1}) -
\varepsilon_3(k-s){\mu}$.
\varepsilonnd{enumerate}
In particular, if $\{a_1, \ldots, a_{k-1}\} = \mathsf{Opt}(r)$, condition~(2)
implies $\mathsf{Saved}(b_1,\ldots,b_s) \ge \varepsilonll^*(r) - \varepsilon_3(k-1){\mu}$.
\varepsilonnd{lemma}
\begin{proof}
We begin with the solution $b_i=a_i$ for all $i$, and iteratively
shift non-anchors in the solution while maintaining the potential
function $\Phi:=\mathsf{Saved}(b_1,\ldots,b_s) - \mathsf{Saved}(a_1,\ldots,a_{k-1}) +
\varepsilon_3(k-s){\mu}$ nonnegative.
At the beginning, $\Phi=0$. Suppose there is a
node $b_i$ not satisfying condition (1). Choose one such $b_i$ of
maximum depth in the tree, and let $b'$ be its non-anchor parent. Then
the only descendants of $b'$ in the current solution are siblings of
$b_i$. Replace $b_i$ and its $s'$ siblings in the solution by
$b'$. Since $b'$ is not an anchor, $\mathsf{Saved}(b_1,\ldots,b_s)$ drops by
at most $\varepsilon_3(s'-1){\mu}$. This drop is compensated by the decrease
of the solution size from $s$ to $s-(s'-1)$.
\varepsilonnd{proof}
Hence, at a loss of $\varepsilon_3(k-1){\mu}$, it suffices to focus on a
solution $\mathsf{Opt}'(r)$ which fulfills condition (1) of
Lemma~\ref{lem:anchor} and has $\mathsf{Saved}$ value $\varepsilonll'(r) \geq \varepsilonll^*(r) -
\varepsilon_3(k-1){\mu}$.
The rest of the algorithm splits into two cases. At a high level, if
there are enough anchors in a mincut tree $\mathcal T$ that are incomparable
with each other, then we can take such a set and be done. Otherwise, the
set of anchors can be grouped into a small number of paths in $\mathcal T$, and
we can afford to try all possible arrangements of anchors. But first we
show how to find all the anchors in $\mathcal T$.
\subsection{Finding Near-Anchors}
\newcommand{\ensuremath{\mathrm{anc}}hors}{\mathcal{A}}
\begin{lemma}[Finding (Near-)Anchors]
\langlebel{lem:compute-anchors}
Assume access to a $(1+\delta)$-approximation algorithm for $\PartialVC k$
running in time $f(k) \cdot g(n)$. Then, there is an algorithm running
in time $O(n \cdot (n^2 + k \cdot f(k) \cdot g(n)))$ that computes a
set $\ensuremath{\mathrm{anc}}hors$ of ``near''-anchors in $\mathcal T$, i.e., vertices $a \in
V_{\mathcal T}$ for which there exists an integer $s \in [2,k-1]$ and $s$
children $b_1, \ldots, b_s$ such that $\mathsf{Saved}(b_1,\ldots,b_s) \geq
\varepsilon_3(s-1){\mu} - \delta(1+\varepsilon_1)s{\mu}$.
\varepsilonnd{lemma}
\begin{proof}
To determine if a node $a$ is an anchor or not, for each integer $s
\in [2, k-1]$ we wish to compute the maximum value of
$\mathsf{Saved}(b_1,\ldots,b_s)$ for $b_1,\ldots,b_s \in
\ensuremath{\mathrm{children}}(a)$. Consider the following weighted, complete graph with
vertex and edge weights: for each $b \in \ensuremath{\mathrm{children}}(a)$ create a vertex
$x_b$, and the edge $(x_{b_1}, x_{b_2})$ has weight
$\mathsf{Saved}(b_1,b_2)$. Each vertex $x_b$ also has weight $(1+\varepsilon_1){\mu}
- w(\partial x_b)$, where $w(\partial x_b)$ is the sum of the weights
of edges incident to $x_b$. Note that this graph is
$(1+\varepsilon_1){\mu}$-regular, if we include vertex weights in the
definition of vertex degree.
Observe that $w(\partial x_b)
\le \partial\left(\phi^{-1}(\ensuremath{\mathrm{subtree}}(b))\right) \le (1+\varepsilon_1){\mu}$,
since every edge in $G$ that contributes to $\mathsf{Saved}(b,b')$ for another
child $b'$ also contributes to the cut
$\partial\left(\phi^{-1}(\ensuremath{\mathrm{subtree}}(b))\right)$, which we know is $\le
(1+\varepsilon_1){\mu}$.
Therefore, each vertex has a nonnegative weight.
Also, a partial vertex cover on this graph with vertices $x_{b_1},
\ldots, x_{b_s}$ has weight exactly $(1+\varepsilon_1)s{\mu} - \mathsf{Saved} (b_1,
\ldots, b_s)$.
Let $b_1^*,\ldots,b_s^* \in \ensuremath{\mathrm{children}}(a)$ be the solution with maximum
$\mathsf{Saved}(b_1^*,\ldots,b_s^*)$. To compute this maximum, we can build
the above graph and run the $(1+\delta)$-approximate partial vertex
cover algorithm from Theorem~\ref{thm:pvc}. The solution
$b_1,\ldots,b_s$ satisfies
\[ (1+\varepsilon_1)s{\mu} - \mathsf{Saved}(b_1,\ldots,b_s) \le
(1+\delta)\left((1+\varepsilon_1)s{\mu} - \mathsf{Saved}(b_1^*,\ldots,b_s^*)\right),
\]
so that
\begin{align*}
\mathsf{Saved}(b_1,\ldots,b_s) & \ge (1+\delta)\,\mathsf{Saved}(b_1^*,\ldots,b_s^*)
- \delta(1+\varepsilon_1)s {\mu} \\
& \ge \mathsf{Saved}(b_1^*,\ldots,b_s^*) - \delta(1+\varepsilon_1)s{\mu}.
\varepsilonnd{align*}
We run this subprocedure for the vertex $a$ for each integer $2 \leq s
\le \min\{|\ensuremath{\mathrm{children}}(a)|, k-1\}$, and mark vertex $a$ if there exists
an integer $s$ such that the weight of saved edges is at least
$\varepsilon_3(s-1){\mu} - \delta(1+\varepsilon_1)s{\mu}$. The set $\ensuremath{\mathrm{anc}}hors$ of
near-anchors is exactly the set of marked vertices.
As for running time, for each node $a$, it takes $O(n^2)$ time to construct the $\textsc{Partial VC}\xspace$ graph and $O(k) \cdot f(k) \cdot g(n)$ time to solve $\PartialVC s$ for each $s \in [2,k-1]$. Repeating the above for each of the $O(n)$ nodes achieves the promised running time.
\varepsilonnd{proof}
\subsection{Many Incomparable Near-Anchors}
\begin{lemma}[Many Anchors]
\langlebel{lem:incomparable}
Suppose we have access to a $(1+\delta)$-approximation algorithm for
$\PartialVC k$ running in time $f(k) \cdot g(n)$. Suppose the set
$\ensuremath{\mathrm{anc}}hors$ of near-anchors contains $k-1$ incomparable nodes from the
mincut tree $\mathcal T$. Then, there is an algorithm computing a solution
with $\mathsf{Saved}$ value
$\ge \frac14\varepsilon_3(k-1){\mu} - \delta(1+\varepsilon_1)(k-1){\mu}$ for any
$\delta>0$, running in time
$O(n \cdot (n^2 + k \cdot f(k) \cdot g(n)))$.
\varepsilonnd{lemma}
\begin{proof}
First, we compute the set $\ensuremath{\mathrm{anc}}hors$ in $O(n \cdot (n^2 + k \cdot f(k)
\cdot g(n))$ time, according to Lemma~\ref{lem:compute-anchors}. If
$\ensuremath{\mathrm{anc}}hors$ contains $k-1$ incomparable nodes, we can \textit{find}
them in $O(n^2)$ time by greedily choosing nodes in a topological,
bottom-first order (see lines 4--11 in
Algorithm~\ref{alg:laminarRooted}). Each of these $k-1$ marked nodes
$a_1, \ldots, a_{k-1}$ has an associated value $s_i$, indicating that
$a_i$ has some $s_i$ children whose $\mathsf{Saved}$ value is at least
$\varepsilon_3(s_i-1){\mu} - \delta(1+\varepsilon_1)s_i{\mu}$. If we consider a
subset $A \subset [k-1]$ and choose the $s_i$ children for each $a_i$
with $i\in A$, then we get a set with $\sum_{i\in A} s_i$ nodes, whose
total $\mathsf{Saved}$ value at least \[\varepsilon_3\left( \sum_{i\in
A}(s_i-1)\right){\mu} - \delta(1+\varepsilon_1)\left( \sum_{i\in A}s_i
\right){\mu}.\] Assuming that $\sum_{i\in A}s_i \le k-1$, i.e., we
choose at most $k-1$ children, the second
$\delta(1+\varepsilon_1)\left(\sum_{i\in A}s_i\right){\mu}$ term is at most
$\delta(1+\varepsilon_1) (k-1){\mu}$. To optimize the $\varepsilon_3\left( \sum_{i\in
A}(s_i-1)\right){\mu}$ term, we reduce to the following knapsack
problem: we have $k-1$ items $i \in [k-1]$ where item $i$ has size
$s_i \in [2,k-1]$ and value $s_i-1$, and our bag size is $k-1$. A
knapsack solution of value $Z:=\sum_{i\in A}(s_i-1)$ translates to a
solution with $\mathsf{Saved}$ value $\ge \varepsilon_3{\mu} \cdot Z -
\delta(1+\varepsilon_1)(k-1){\mu}$. By Lemma~\ref{lemma:knapsack}, when $k
\ge 5$, we can compute a solution $A \subset [k-1]$ of value $\ge
(k-1)/4$ in $O(k)$ time. (If $k\le4$, we can use the exact $\tilde O(n^4)$
$k$\textsc{-Cut}\xspace algorithm from~\cite{Levine00}.) Selecting the children of each
$u_i$ with $i\in A$ gives a total $\mathsf{Saved}$ value of at least
$\frac14\varepsilon_3(k-1){\mu} - \delta(1+\varepsilon_1)(k-1){\mu}$.
\varepsilonnd{proof}
\subsection{Few Incomparable Near-Anchors}
\begin{figure}
\centering
\begin{tikzpicture} [xscale=.5, yscale=.3]
\tikzstyle{every node}=[circle, fill, scale=.3];
\node (v1) at (-1,4) {};
\node (v2) at (-2,2) {};
\node (v12) at (0.5,2) {};
\node (v3) at (-3,0) {};
\node (v10) at (-1.5,0) {};
\node (v11) at (0,0) {};
\node (v13) at (1,0) {};
\node (v14) at (2,0) {};
\node (v4) at (-4,-2) {};
\node (v7) at (-2,-2) {};
\node (v5) at (-5,-4) {};
\node (v6) at (-3.5,-4) {};
\node (v8) at (-2.5,-4) {};
\node (v9) at (-1,-4) {};
\node (v15) at (1,-2) {};
\node (v16) at (3,-2) {};
\draw [line width=.5pt] (v1) edge (v2);
\draw [line width=.5pt] (v2) edge (v3);
\draw [line width=.5pt] (v3) edge (v4);
\draw [line width=.5pt] (v4) edge (v5);
\draw [line width=.5pt] (v4) edge (v6);
\draw [line width=.5pt] (v7) edge (v3);
\draw [line width=.5pt] (v8) edge (v7);
\draw [line width=.5pt] (v9) edge (v7);
\draw [line width=.5pt] (v10) edge (v2);
\draw [line width=.5pt] (v11) edge (v12);
\draw [line width=.5pt] (v12) edge (v1);
\draw [line width=.5pt] (v13) edge (v12);
\draw [line width=.5pt] (v12) edge (v14);
\draw [line width=.5pt] (v15) edge (v14);
\draw [line width=.5pt] (v14) edge (v16);
\draw (v3) circle (.3);
\draw (v14) circle (.3);
\draw (v16) circle (.3);
\draw (v5) circle (.3);
\draw (v8) circle (.3);
\draw (v9) circle (.3);
\draw [line width=.5pt] (v5) edge (-5.5,-5);
\draw [line width=.5pt] (v5) edge (-4.5,-5);
\draw [line width=.5pt] (v8) edge (-3,-5);
\draw [line width=.5pt] (v8) edge (-2.25,-5);
\draw [line width=.5pt] (v9) edge (-1.25,-5);
\draw [line width=.5pt] (v9) edge (-0.5,-5);
\varepsilonnd{tikzpicture}
\qquad
\begin{tikzpicture} [xscale=.5, yscale=.3]
\tikzstyle{every node}=[circle, fill, scale=.5];
\node (v1) at (-1,4) {};
\node (v3) at (-3,0) {};
\node (v7) at (-2,-2) {};
\node (v5) at (-5,-4) {};
\node (v8) at (-2.5,-4) {};
\node (v9) at (-1,-4) {};
\node (v16) at (3,-2) {};
\draw [white,line width=.5pt] (v5) edge (-5.5,-5);
\draw [white,line width=.5pt] (v5) edge (-4.5,-5);
\draw [white,line width=.5pt] (v8) edge (-3,-5);
\draw [white,line width=.5pt] (v8) edge (-2.25,-5);
\draw [white,line width=.5pt] (v9) edge (-1.25,-5);
\draw [white,line width=.5pt] (v9) edge (-0.5,-5);
\draw [line width=1pt] (v1) edge (v3);
\draw [line width=1pt] (v3) edge (v5);
\draw [line width=1pt] (v3) edge (v7);
\draw [line width=1pt] (v7) edge (v8);
\draw [line width=1pt] (v9) edge (v7);
\draw [line width=1pt] (v1) edge (v16);
\varepsilonnd{tikzpicture}
\qquad
\begin{tikzpicture} [xscale=.5, yscale=.3]
\tikzstyle{every node}=[circle, fill, scale=.3];
\node[fill=brown] (v1) at (-1,4) {};
\node[fill=blue] (v2) at (-2,2) {};
\node[fill=green] (v12) at (0.5,2) {};
\node[fill=blue] (v3) at (-3,0) {};
\node[fill=] (v10) at (-1.5,0) {};
\node[fill=] (v11) at (0,0) {};
\node[fill=] (v13) at (1,0) {};
\node[fill=green] (v14) at (2,0) {};
\node[fill=red] (v4) at (-4,-2) {};
\node[fill=purple] (v7) at (-2,-2) {};
\node[fill=red] (v5) at (-5,-4) {};
\node[fill=] (v6) at (-3.5,-4) {};
\node[fill=orange] (v8) at (-2.5,-4) {};
\node[fill=yellow] (v9) at (-1,-4) {};
\node[fill=] (v15) at (1,-2) {};
\node[fill=green] (v16) at (3,-2) {};
\draw [blue,line width=1pt] (v1) edge (v2);
\draw [blue,line width=1pt] (v2) edge (v3);
\draw [red,line width=1pt] (v3) edge (v4);
\draw [red,line width=1pt] (v4) edge (v5);
\draw [line width=1pt] (v4) edge (v6);
\draw [purple,line width=1pt] (v7) edge (v3);
\draw [orange,line width=1pt] (v8) edge (v7);
\draw [yellow,line width=1pt] (v9) edge (v7);
\draw [line width=1pt] (v10) edge (v2);
\draw [line width=1pt] (v11) edge (v12);
\draw [green,line width=1pt] (v12) edge (v1);
\draw [line width=1pt] (v13) edge (v12);
\draw [green,line width=1pt] (v12) edge (v14);
\draw [line width=1pt] (v15) edge (v14);
\draw [green,line width=1pt] (v14) edge (v16);
\draw [white,line width=.5pt] (v5) edge (-5.5,-5);
\draw [white,line width=.5pt] (v5) edge (-4.5,-5);
\draw [white,line width=.5pt] (v8) edge (-3,-5);
\draw [white,line width=.5pt] (v8) edge (-2.25,-5);
\draw [white,line width=.5pt] (v9) edge (-1.25,-5);
\draw [white,line width=.5pt] (v9) edge (-0.5,-5);
\varepsilonnd{tikzpicture}
\caption{\langlebel{figure:branches}
Establishing the set of branches $\mathcal B$. The circled nodes on the left are the near-anchors. The middle graph is the tree $\mathcal T'$. On the right, each non-black color is an individual branch; actually, the branches only consist of nodes, but we connect the nodes for visibility. Also, note that the root is its own branch. The red, orange, yellow, and green branches form an incomparable set.}
\varepsilonnd{figure}
If the condition in Lemma~\ref{lem:incomparable} does not hold, then
there exist $\le k-2$ paths from the root in $\mathcal T$ such that every node
in the near-anchor set $\ensuremath{\mathrm{anc}}hors$ lies on one of these paths. If we view
the union of these paths as a tree $\mathcal T'$ with $\le k-2$ leaves, then we
can partition the nodes in tree $\mathcal T'$ into a collection $\mathcal B$ of at
most $2k-3$ \textit{branches}. Each branch $B$ is a collection of
vertices obtained by taking either a leaf of $\mathcal T'$ or a vertex of
degree more than two, and all its immediate degree-2 ancestors; see
Figure~\ref{figure:branches}. Note that it is possible that the root
node is its own branch. Hence, given two branches $B_1, B_2 \in \mathcal B$,
either every node from $B_1$ is an ancestor of every node from $B_2$ (or
vice versa), or else every node from $B_1$ is incomparable with every
node from $B_2$.
Let $A' \subseteq \ensuremath{\mathrm{anc}}hors$ be the set of anchors with at least one child in
$\mathsf{Opt}'(r)=\{a_1^*,\ldots,a_s^*\}$; recall that $\mathsf{Opt}'(r)$ was produced
by the shifting procedure in Lemma~\ref{lem:anchor}. Let $A^* \subseteq A'$
be
the \textit{minimal} anchors in $A'$, i.e., every anchor in $A'$ that is
not an ancestor of any other anchor in $A'$. We know that every anchor in
$A^*$ falls inside our set of branches, although the algorithm does not
know where. Moreover, by condition~(1) of Lemma~\ref{lem:anchor}, the
parent of every $a_i^* \in \mathsf{Opt}'(r)$ either lies in $A^*$, or is an
ancestor of an anchor in $A^*$.
As a warm-up, consider the case where all the anchors in $A'$ are
contained within a single branch.
\begin{figure}
\centering
\begin{tikzpicture}[scale=0.25]
\node [circle, fill=red, scale=.3] (v3) at (0,5) {};
\node [circle, fill=red, scale=.3] (v2) at (2.5,7.5) {};
\node [circle, fill=red, scale=.3] (v1) at (5,10) {};
\node [circle, fill=red, scale=.3] (v4) at (-2.5,2.5) {};
\node [circle, fill=red, scale=.3] (v5) at (-5,0) {};
\node [circle, fill=red, scale=.3, label=left:$a^*$] (v6) at (-7.5,-2.5) {};
\draw (v1) edge (v2);
\draw (v2) edge (v3);
\draw (v3) edge (v4);
\draw (v4) edge (v5);
\draw (v5) edge (v6);
\node [circle, fill=blue, scale=.3] (v7) at (5,8) {};
\node [circle, fill=blue, scale=.3] (v8) at (6.5,8) {};
\node [circle, fill=blue, scale=.3] (v9) at (2.5,5.5) {};
\node [circle, fill=blue, scale=.3] (v10) at (4,5.5) {};
\node [circle, fill=blue, scale=.3] (v11) at (0,3) {};
\node [circle, fill=blue, scale=.3] (v12) at (1.5,3) {};
\node [circle, fill=blue, scale=.3] (v13) at (-2.5,0.5) {};
\node [circle, fill=blue, scale=.3] (v14) at (-1,0.5) {};
\node [circle, fill=blue, scale=.3] (v15) at (-5,-2) {};
\node [circle, fill=blue, scale=.3] (v16) at (-3.5,-2) {};
\node [circle, fill=blue, scale=.3] (v17) at (-9,-4.5) {};
\node [circle, fill=blue, scale=.3] (v18) at (-7.5,-4.5) {};
\node [circle, fill=blue, scale=.3] (v19) at (-6,-4.5) {};
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\varepsilonnd{tikzpicture}
\qquad
\begin{tikzpicture}[scale=0.25]
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\draw (v28) -- (v24) -- (4,-6) -- (5,-6) -- (v24) (v28) -- (v26) -- (5.5,-6) -- (6.5,-6) -- (6,-4.5) (6,-2.5) -- (v27) -- (7,-6) -- (8,-6) -- (v27);
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\draw [green, line width=1pt] plot[smooth, tension=.7] coordinates {(-6,-5.5) (-4.5,-5.5) (-3.5,-3)};
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\varepsilonnd{tikzpicture}
\caption{\langlebel{figure:single-branch}Left (Claim~\ref{claim:singleBranch}): The red nodes form our branch $B$, and the blue nodes form the set $\ensuremath{\mathrm{children}}((\{a^*\}
\cup \ensuremath{\mathrm{anc}}(a^*)) \cap B)$. The triangles are the subtrees participating in the $\textsc{Partial VC}\xspace$ instance.
Right (Lemma~\ref{lem:chains}): The red nodes form our two incomparable branches. The green edges are internal edges, while the blue edges are external.}
\varepsilonnd{figure}
\begin{claim}[Warm-up]
\langlebel{claim:singleBranch}
Assume there exists a $(1+\delta)$-approximation algorithm for $\PartialVC k$ running
in time $f(k) \cdot g(n)$. Suppose the set of anchors $A'$ with at least one child in
$\mathsf{Opt}'(r)$ is contained within a single branch $B$. Then there is an
algorithm computing a solution with $\mathsf{Saved}$ value at least $\varepsilonll'(r)
- \delta(1+\varepsilon_1)(k-1){\mu}$, running in time $O(n \cdot (n^2 + f(k) \cdot g(n)))$.
\varepsilonnd{claim}
\begin{proof}
If all of $A'$ lies on $B$, the minimal anchor $a^* \in A^*$ must also
be in $B$. Moreover, for every $a_i^*\in\mathsf{Opt}'(r)$, its parent is
either $a^*$ or an ancestor of $a^*$, which means that $\mathsf{Opt}'(r) \subseteq
\ensuremath{\mathrm{children}}((\{a^*\} \cup \ensuremath{\mathrm{anc}}(a^*)) \cap B)$. Since the nodes in $\ensuremath{\mathrm{children}}((\{a^*\}
\cup \ensuremath{\mathrm{anc}}(a^*)) \cap B)$ are incomparable (see Figure~\ref{figure:single-branch}), we can construct the same
graph as the one in Lemma~\ref{lem:compute-anchors} on all these nodes
in $\ensuremath{\mathrm{children}}((\{a^*\} \cup \ensuremath{\mathrm{anc}}(a^*)) \cap B)$ and run the \textsc{Partial VC}\xspace-based
algorithm to get the same $\mathsf{Saved}$ guarantees (see
Algorithm~\ref{alg:subtreePVC}).
Therefore, the algorithm guesses the location of $a^*$ inside $B$ by
trying all possible $|B|=O(n)$ nodes, and for each choice of $a^*$,
runs the $(1-\delta)$-approximate \textsc{Partial VC}\xspace-based algorithm from
Lemma~\ref{lem:compute-anchors} on the corresponding graph (see
Algorithm~\ref{alg:singleBranch}).
\varepsilonnd{proof}
Now for the general case. Consider $\mathsf{Opt}'(r)$ and the set of all
branches $\mathcal B$. Let $\mathcal B^* \subseteq \mathcal B$ be the incomparable branches that
contain the minimal anchors, i.e., those in $A^*$. We classify the
$\varepsilonll(r')$ saved edges in $\mathsf{Opt}'(r)$ into two groups (see Figure~\ref{figure:single-branch}): if an edge is
saved between the subtrees below $a_i^*,a_j^* \in \mathsf{Opt}'(r)$ whose
parent(s) belong to the same branch in $\mathcal B^*$, then call this
\textit{an internal edge}. Otherwise, it is an \textit{external edge}:
these are saved edges in $\mathsf{Opt}'(r)$ that either go between two subtrees
in different branches, or between subtrees in the same branch in $\mathcal B
\setminus \mathcal B^*$. One of the two sets has $\ge \frac12\varepsilonll'(r)$ saved
edges, and we provide two separate algorithms, one to approximate each
group.
\begin{lemma}\langlebel{lem:chains}
Assume there exists a $(1+\delta)$-approximation algorithm for $\PartialVC k$ running
in time $f(k) \cdot g(n)$. Suppose that all anchors of $\mathsf{Opt}'(r)$ are contained in a set $\mathcal B$ of
$\le 2k-3$ branches. Then there is an algorithm that computes a
solution with $\mathsf{Saved}$ value $\ge \frac12\varepsilonll'(r) -
\delta(1+\varepsilon_1)(k-1){\mu}$, running in time $ 2^{O(k)} \cdot (n^2+f(k)\cdot g(n)) $.
\varepsilonnd{lemma}
\begin{proof}
\varepsilonmph{Case I: internal edges $\ge\frac12\varepsilonll'$.} For each branch
$B\in\mathcal B$ and each $s\in [k-1]$, compute a solution of $s$ nodes that
maximizes the number of internal edges \textit{within branch
$\mathcal B$}, in the same manner as in
Claim~\ref{claim:singleBranch}; this takes time $O(k^2n \cdot (n^2 + f(k) \cdot g(n)))$. Finally, guess all possible $\le
2^{2k-3}$ subsets of incomparable branches; for each subset
$\mathcal B'\subseteq\mathcal B$, try all vectors $\mathbf i \in [k-1]^{\mathcal B'}$ with
$\sum_{B\in\mathcal B'} i_B \le k-1$, look up the solution using $i_B$
vertices in branch $B$, and sum up the total number of internal
edges. Actually, trying all vectors $\mathbf i \in [k-1]^{\mathcal B'}$ takes $k^{O(k)}$ time, but we can speed up this step to $\mathrm{poly}(k)$ time using dynamic programming. Since one
of the guesses $\mathcal B'$ will be $\mathcal B^*$, the best solution will save at
$\ge\frac12\varepsilonll'(r)-\delta(1+\varepsilon_1)(k-1){\mu}$ edges. The total running time for this case is $O(k^2 \cdot f(k) \cdot g(n) + 2^{2k}\cdot\mathrm{poly}(k))$.
\varepsilonmph{Case II: external edges $\ge\frac12\varepsilonll'$.} Again, we guess the
set $\mathcal B^*\subset\mathcal B$ of incomparable branches containing minimal
anchors $A^*$. For a branch $B \in \mathcal B^*$, let $a_B:=(a \in B : B
\setminus a \subseteq \ensuremath{\mathrm{desc}}(a))$ be the ``highest'' node in $B$, that is an
ancestor of every other node in $B$. For each branch, we can replace
all nodes in $\mathsf{Opt}'(r)$ that are descendants of $a_B$ with just $a_B$;
doing can only increase the number of external edges. The new solution
has all nodes contained in the set
\[ \ensuremath{\mathrm{children}}\bigg(\ensuremath{\mathrm{anc}}\bigg(\bigcup_{B\in\mathcal B^*}\{a_B\}\bigg)\bigg), \]
which is a set of incomparable nodes. Therefore, we can construct the
graph of Lemma~\ref{lem:compute-anchors} and use the \textsc{Partial VC}\xspace-based
algorithm with this node set instead. This gives a solution with $\ge
\frac12\varepsilonll'(r)-\delta(1+\varepsilon_1)(k-1){\mu}$ saved edges. The total running time for this case is $O(2^{2k}\cdot (n^2 + f(k) \cdot g(n)))$.
\varepsilonnd{proof}
\subsection{Combining Things Together}
Putting things together, we conclude with Theorem~\ref{thm:laminar}. We
refer the reader to Algorithm~\ref{alg:laminar} for the pseudocode of
the entire algorithm.
\begin{proof}[Proof (Theorem~\ref{thm:laminar}).]
Let the original graph be $G=(V,E,w).$ We compute a $(1+\varepsilon_1)$-mincut
tree $\mathcal T=(V_{\mathcal T},E_{\mathcal T},w_{\mathcal T})$ with mapping $\phi:V\to V_{\mathcal T}$ in time $O(n^3)$, following Theorem~\ref{thm:mincutTreeExistence}.
Then, by running the two algorithms in Lemma~\ref{lem:incomparable} and
Lemma~\ref{lem:chains}, we compute a solution with $s \le k-1$
vertices with $\mathsf{Saved}$ value at least
\begin{align*}
& \max\left\{
\frac14\varepsilon_3(k-1){\mu} - \delta(1+\varepsilon_1)(k-1){\mu},\
\frac12\varepsilonll'(r) - \delta(1+\varepsilon_1)(k-1){\mu} \right\} \\
= & \max\left\{ \frac14\varepsilon_3(k-1){\mu}, \ \frac12\varepsilonll'(r) \right\} - \delta(1+\varepsilon_1)(k-1){\mu}
\varepsilonnd{align*}
for each root $r \in V_{\mathcal T}$ (see
Algorithm~\ref{alg:laminarRooted}). Using $\max\{p, q\} \geq
(4p+2q)/6$ and
$\varepsilonll'(r) \ge \varepsilonll^*(r) - \varepsilon_3(k-1){\mu}$
we get a solution with $\mathsf{Saved}$ value at least
\begin{align*}
& \frac16 \left( 4 \cdot \frac14\varepsilon_3(k-1){\mu} + 2 \cdot \frac12 \left[ \varepsilonll^*(r) - \varepsilon_3(k-1){\mu} \right] \right) - \delta(1+\varepsilon_1)(k-1){\mu} \\
\ge & \frac16\varepsilonll^*(r) - 2\delta(k-1){\mu},
\varepsilonnd{align*}
using that $\varepsilon_1 \leq 1$. In particular, the best solution
$v_1,\ldots,v_s \in V_{\mathcal T}$ over all $r$ satisfies
\[ \mathsf{Saved}(v_1,\ldots,v_s) \ge \frac16 \varepsilonll^* - 2\delta(k-1){\mu} ,\]
where $\varepsilonll^*(r)$ was replaced by $\varepsilonll^*$.
Let $v_1,\ldots,v_s \in V_{\mathcal T}$ be our solution with
$\mathsf{Saved}(v_1,\ldots,v_s) \ge \frac16 \varepsilonll^* - 2\delta(k-1){\mu}$. Let
$S_1,\ldots,S_s \subset V$ be the corresponding subsets in $V$, i.e.,
$S_i := \phi^{-1}(\ensuremath{\mathrm{subtree}}(v_i))$. Then, add the complement set
$S_{s+1}:=V \setminus \bigcup_{i\in[s]}S_i$ to the solution, so that
the sets $S_i$ partition $V$, and
\[w(E(S_1,\ldots,S_{s+1})) \le s(1+\varepsilon_1){\mu} - \left(\frac16 \varepsilonll^* -
2\delta(k-1){\mu}\right). \] Then, extend the solution to a
$k$-partition using Algorithm~\ref{alg:complete}. We now claim that every additional
cut that Algorithm~\ref{alg:complete} makes is a $(1+\varepsilon_1)$-mincut.
To see this, observe that $S_1^*, \ldots, S_{k-1}^*$ are all $(1+\varepsilon_1)$-mincuts and one
of them, say $S_j^*$, has to intersect some $S_i$. Then, the cut $(S_i \cap S_j^*, S_i \setminus S_j^*)$
is a $(1+\varepsilon_1)$-mincut in $S_i$. We can repeat this argument as long as we have $< k$ components $S_i$.
At the end, we have a
solution $S_1', \ldots, S_k'$ satisfying
\begin{align*}
w(E(S_1', \ldots, S_k')) & \le w(E(S_1,\ldots,S_s)) +
(k-1-s)(1+\varepsilon_1){\mu}
\\
& \le (k-1)(1+\varepsilon_1){\mu}- \left(\frac16 \varepsilonll^* -
2\delta(k-1){\mu}\right)
\varepsilonnd{align*}
Let $S^*_1,\ldots,S^*_k$ be the optimal partition in $\mathcal O_{\varepsilon_1}$
satisfying $\phi(r) \in S^*_k$, and let $\varepsilonll^*$ be the maximum of
$\mathsf{Saved}(v_1^*,\ldots,v_{k-1}^*)$ over incomparable
$v_1^*,\ldots,v_{k-1}^*$. Our solution has approximation ratio
\begin{align*}
\frac{w(E(S_1,\ldots,S_k))}{w(E(S_1^*,\ldots,S_k^*))} & \le
\frac{(k-1)(1+\varepsilon_1){\mu} - \frac16\varepsilonll^* +
2\delta(k-1){\mu}}{(k-1){\mu} - \varepsilonll^*} \\
& = \frac{(k-1)(1+\varepsilon_1){\mu} - \frac16\varepsilonll^* }{(k-1){\mu} -
\varepsilonll^*} + \frac{ 2\delta(k-1){\mu}}{(k-1){\mu} - \varepsilonll^*} \\
& \le 2(1+\varepsilon_1) - \frac16 + 4\delta,
\varepsilonnd{align*}
with the worst case achieved at $\varepsilonll^*=\frac12(k-1){\mu}$, which is
the highest $\varepsilonll^*$ can be. Setting $\varepsilon_2:=1/6 - 2\varepsilon_1-4\delta$
concludes the proof.
As for running time, we run the algorithms in Lemma~\ref{lem:incomparable} and Lemma~\ref{lem:chains} sequentially, and the final running time is $\tilde 2^{O(k)}f(k)(\tilde O(n^4) + g(n))$. (The $\tilde O(n^4)$ comes from the case when $k\le 4$, in which we solve the problem exactly in $\tilde O(n^4)$ time.)
\varepsilonnd{proof}
\newcommand{\mathsf{Wdeg}}{\mathsf{Wdeg}}
\section{An FPT-AS for \textsc{Partial VC}\xspacelong}
\langlebel{sec:partial-vc}
Recall the \textsc{Partial VC}\xspacelong (\textsc{Partial VC}\xspace) problem: the input is a graph $G = (V,E)$
with edge and vertex weights, and an integer $k$. For a set $S$, define
$E_S$ to be the set of edges with at least one endpoint in $S$. The goal
of the problem is to find a set $S$ with size $|S| = k$, minimizing the
weight $w(E_S) + w(S)$, i.e., the weight of all edges hitting $S$ plus
the weight of all vertices in $S$. Our main theorem is the following.
\begin{theorem}[\textsc{Partial VC}\xspacelong]
\langlebel{thm:pvc}
There is a randomized algorithm for \textsc{Partial VC}\xspace on weighted graphs that, for
any $\delta \in (0,1)$, runs in $O(2^{k^6/\delta^3} (m +
k^8/\delta^3)\,n \log n)$ time and outputs a
$(1+\delta)$-approximation to \textsc{Partial VC}\xspace with probability $1 - 1/\mathrm{poly}(n)$.
\varepsilonnd{theorem}
We first extend a result of Marx~\cite{Marx07} to give a
$(1+\delta)$-approximation algorithm for the case where $G$ has edge
weights being integers in $\{1, \ldots, M\}$ and no vertex weights, and
then show how to reduce the general case to this special case, losing
only another $(1+\delta)$-factor.
\subsection{Graphs with Bounded Weights}
\begin{lemma}
\langlebel{lem:pvc-simple}
Let $\delta \leq 1$. There is a randomized algorithm for the \textsc{Partial VC}\xspace
problem on simple graphs with edge weights in $\{1, \ldots, M\}$ (and
no vertex weights) that runs in $O(m+Mk^4/\delta)$ time, and outputs
a $(1+\delta)$-approximation with probability at least
$2^{-(Mk^2/\delta)}$.
\varepsilonnd{lemma}
\begin{proof}
This is a simple extension of a result for the
maximization case given by Marx~\cite{Marx07}. We give two algorithms: one for the case when the
optimal value is smaller than $\tau := Mk^2/\delta$ (which returns the
correct solution in time, but with probability $2^{-(Mk^2/\delta)}$),
and another for the case of the optimal value being at least $\tau$
(which deterministically returns a $(1+\delta)$-approximation in
linear time). We run both and return the better of the two solutions.
First, the case when the optimal value is at least $\tau$. Let the
\varepsilonmph{weighted degree} of a node $v$, denoted $w(\partial v)$ be
defined as $\sum_{e: v \in e} w(e)$. Observe that for
any set $S$ with $|S| \leq k$,
\[ 0 \leq \sum_{v \in S} w(\partial v) - w(E_S) \leq M\cdot
\binom{k}{2}. \] Hence, if $S^*$ is the optimal solution and
$w(E_{S^*}) \geq \tau$, then picking the set of $k$ vertices with the
least weighted degrees is a $(1+\delta)$-approximation.
Now for the case when the optimal value is at most $\tau$. In this case,
the optimal set $S^*$ can have at most $\tau$ edges incident to it, since
each edge must have weight at least $1$. Consider the color-coding
scheme where we independently and uniformly colors the vertices of $G$
with two colors (red and blue). With probability $2^{-(\tau+k)}$, all the
vertices in $S^*$ are colored red, and all the vertices in $N(S^*)
\setminus S^*$ are colored blue. Consider the ``red components'' in
the graph obtained by deleting the blue vertices. Then $S^*$ is the
union of one or more of these red components. To find it, define the
``size'' of a red component $C$ as the number of vertices in it, and
the ``cost'' as the total weight of edges in $G$ that are incident to
it (i.e., cost $= \sum_{e \in E: e \cap C \neq \varepsilonmptyset} w(e)$.)
Now we can use dynamic programming to find a collection of red
components with total size equal to $k$ and minimum total cost: this
gives us $S$ (or some other solution of equal cost). Indeed, if we
define the ``type'' of each component to be the tuple $(s,c)$ where $s
\in [1\ldots k]$ is the size (we can drop components of size greater
than $k$) and $c \in [1\ldots \tau]$ is the cost (we can drop all
components of greater cost). Let $T(s,c)$ be the number of copies of
type $(s,c)$, capped at $k$. Assume the types are numbered $\tau_1,
\tau_2, \ldots, \tau_{k\tau}$. Now if $C(i,j)$ is the minimum cost we can
have with components of type $\leq \tau_i = (s,c)$ whose total size is
$j$, then
\[ C(i,j) = \min_{0 \leq \varepsilonll \leq T(s,c)} C(i-1, j - \varepsilonll s) + \varepsilonll
c. \] Finally, we return the component achieving $C(k\tau,k)$. This can
all be done in $O(m + k^2\tau)$ time.
\varepsilonnd{proof}
Repeating the algorithm $O(2^{\tau + k} \log n) = O(2^{Mk^2/\delta + k} \log n)$ times and outputting
the best set found in these repetitions gives an algorithm that finds a
$(1+\delta)$-approximation with probability $1 - 1/\mathrm{poly}(n)$.
\subsection{Solving The General Case}
We now reduce the general \textsc{Partial VC}\xspace problem, where we have no bounds on the
edge weights (and we have vertex weights), to the special case from the
previous section.
The idea is simple: given a graph $G = (V,E)$ with edge and vertex
weights, we construct a collection of $|V|$ simple graphs $\{ H_v \}_{v
\in V}$, each defined on the vertex set $V$ plus a couple new nodes,
and having $O(|V|+|E|)$ edges, with each edge-weight $w'(e)$ being an
integer in $\{1, \ldots, M\}$ and $M = O(k/\delta)^2$, and with no vertex
weights. We find a $(1+\delta/2)$-approximate \textsc{Partial VC}\xspace solution on each $H_v$,
and then output the set $S$ which has the smallest weight (in $G$) among
these. We show how to ensure that $S \subseteq V$ and that it is a
$(1+\delta)$-approximation of the optimal solution in $G$.
\begin{proof}[Proof of Theorem~\ref{thm:pvc}]
Let $S^*$ be an optimal solution on $G$. Define the \varepsilonmph{extended
weighted degree} of a vertex $v$, denoted by $\mathsf{Wdeg}(v)$, to be its
vertex weight plus the weight of all edges adjacent to it. I.e.,
$\mathsf{Wdeg}(v) := w(v) + w(\partial v)$.
Firstly, assume we know a vertex $v^* \in S^*$ with the largest
$\mathsf{Wdeg}(v^*)$; we just enumerate over all vertices to find this vertex.
We now proceed to construct the graph $H_{v^*}$. Let $L =
\mathsf{Wdeg}(v^*)$, and delete all vertices $u$ with $\mathsf{Wdeg}(u) > L$. Note
that (a)~any solution containing $v^*$ has total weight at least $L$,
and (b)~each remaining edge and vertex has weight $\leq L$.
Assume that $G$ is simple, since we can combine parallel edges
together by summing their weights. Create two new vertices $p, q$,
and add an edge of weight $L k^2$ between them; this ensures that
neither of these vertices is ever chosen in any near-optimal
solution.
Let $\delta' > 0$ be a parameter to be fixed later; think of $\delta'
\approx \delta$. For each edge $e = (u,v)$ in the edge set $E$ that
has weight $w(e) < L\delta'/k^2$, remove this edge and add its weight
$w(e)$ to the weight of both its endpoints $u,v$. Finally, when there
are no more edges with $w(e) < L\delta'/k^2$, for each vertex $u$ in
$V$, create a new edge $\{u,p\}$ with weight being equal to the
current vertex weight $w(u)$, and zero out the vertex weight. Let the
new edge set be denoted by $E'$. We claim that for any set $S
\subseteq V$ of size $\le k$,
\[ \left( \sum_{e \in E': e \cap S} w(e)\right) - \left( \sum_{e \in
E: e \cap S} w(e) + \sum_{v \in S} w(v) \right) \leq \delta' L. \]
Indeed, the only change comes because of edges with weight $w(e) <
L\delta'/k^2$ and with both endpoints within $S$---these edges
contributed once earlier, but replacing them by the two edges means we
now count them twice. Since there are at most $\binom{k}{2}$ such
edges, they can add at most $\delta' L$.
At this point, all edges in the original edge set $E$ have weights in
$[L \delta'/k^2, Lk^2]$; the only edges potentially having weights $<
L \delta'/k^2$ are those between vertices and the new vertex $p$. For
any such edge with weight $< L \delta'/k$, we delete the edge. This
again changes the optimal solution by at most an additive $L \delta'$,
and ensure all edges in the new graph have weights in $[L \delta'/k^2,
Lk^2]$. Note that since the optimal solution has value at least $L$
by our guess, these additive changes of $L \delta'$ to the optimal
solution mean a multiplicative change of only $(1+\delta')$.
Finally, discretize the edge weights by rounding each edge weight to
the closest integer multiple of $L \delta'^2/k^2$. Since each edge
weight $\geq L \delta'/k^2$, each edge weight incurs a further
multiplicative error at most $1+\delta'$. Note that $M =
k^4/\delta'^2$. Now use Lemma~\ref{lem:pvc-simple} to get a
$(1+\delta')$-approximation for \textsc{Partial VC}\xspace on this instance with high
probability. Setting $\delta' = O(\delta)$ ensures that this solution
is within a factor $(1+\delta)$ of that in $G$.
\varepsilonnd{proof}
\section{Conclusion and Open Problems}
\langlebel{sec:conclusion}
Putting the sections together, we conclude with a proof of our main theorem.
\begin{proof}[Proof of Theorem~\ref{thm:kcut-main}]
Fix some $\delta \in (0,1/24)$. By Theorem~\ref{thm:pvc}, there is a
$(1+\delta)$-approximation algorithm for $\PartialVC k$ running in
time $O(2^{k^6/\delta^3} (m + k^8/\delta^3)\,n \log n) = 2^{O(k^6)}
n^4$ time. Plugging in $f(k) := 2^{O(k^6)}$ and $g(n) := n^4$ into
Theorem~\ref{thm:laminar}, we get a $(2-\varepsilon_2)$-approximation algorithm
to $\text{Laminar}cut k{\varepsilon_1}$ in time $2^{O(k)}f(k)(n^3 + g(n)) =
2^{O(k^6)} n^4$, for a fixed $\varepsilon_1 \in (0,1/6-4\delta)$. Plugging in
$f(k):=2^{O(k^6)}$ and $g(n):=n^4$ into Theorem~\ref{thm:reduction1}
gives a $(2-\varepsilon_3)$-approximation for $$k$\textsc{-Cut}\xspace$ in time $2^{O(k^2 \log k)} \cdot
f(k) \cdot (n^4 \log^3 n + g(n)) = 2^{O(k^6)} n^4 \log^3 n$.
Finally, for our approximation factor. Theorem~\ref{thm:laminar} sets
$\varepsilon_2:=1/6 - 2\varepsilon_1 - 4\delta$ for any small enough $\delta$. We can
take $\varepsilon_1$ and $\varepsilon_2$ to be equal, so that $\varepsilon_1 = \varepsilon_2 = 1/18 -
\nicefrac43\cdot\delta$. Finally, setting
$\varepsilon_4=\varepsilon_5=\min(\varepsilon_1,\varepsilon_2)/3$ and $\varepsilon_3:=\varepsilon_4^2$ in
Theorem~\ref{thm:reduction1} gives $\varepsilon_3 = 1/54^2 - \delta'$ for some
arbitrarily small $\delta'>0$. In other words, our approximation
factor is $2 - 1/54^2 + \delta'$, or $1.9997$ for an appropriately
small $\delta'$.
\varepsilonnd{proof}
Our result
combines ideas from approximation algorithms and FPT algorithms and
shows that considering both settings simultaneously can help bypass
lower bounds in each individual setting, namely the $W[1]$-hardness of
an exact FPT algorithm and the SSE-hardness of a polynomial-time
$(2-\varepsilon)$-approximation. While our improvement is quantitatively modest,
we hope it will prove qualitatively significant. Indeed, we hope these
and other ideas will help resolve whether an $(1+\varepsilon)$-approximation
algorithm exists in FPT time, and to show a matching lower and upper
bound.
\paragraph{Acknowledgments.} We thank Marek Cygan for generously giving
his time to many valuable discussions.
{\small
}
\appendix
\section{Pseudocode for \text{Laminar}cut{$k$}{$\varepsilon_1$}}
\langlebel{sec:pseudocode-laminar}
\begin{algorithm}
\caption{SubtreePartialVC$(G,\mathcal T,A,s,\delta)$}
\langlebel{alg:subtreePVC}
\begin{algorithmic}
\If {$|A| < s$}
\State \mathbb Return None
\EndIf
\mathbb For {$a \in A$}
\State $C_a \gets V(a) \cup \displaystyle\bigcup\limits_{a' \in \ensuremath{\mathrm{desc}}(a)}V(a')$
\EndFor \Comment{\textbf{Assert}: $C_a$ are all disjoint}
\
\State $\mathcal C \gets \{ C_a : a \in A\}$
\State $H \gets \text{Contract}(G, \mathcal C)$ \Comment{For each $C_a \in \mathcal C$, contract all vertices in $C_a$ into a single vertex in $H$}
\
\mathbb For {$i \in [k-1]$}
\State $P_{i} \gets \text{PartialVC}(H,i)$ \Comment{$P_{i} \in V(H)^i$}
\State $\mathcal S_i \gets \text{Expand}(H,P_i)$ \Comment { \parbox[t]{.5\linewidth}{ Map each $v \in P_i$ to the set of vertices in $V$ which contract to $v$ in $H$, and call the result $\mathcal S_i \in \left(2^V\right)^i$ } }
\EndFor
\State \mathbb Return $\{\mathcal S_{i} : i \in [s]\}$
\varepsilonnd{algorithmic}
\varepsilonnd{algorithm}
\begin{algorithm}
\caption{SingleBranch$(G,\mathcal T,B,k,\delta)$}
\langlebel{alg:singleBranch}
\begin{algorithmic}
\mathbb For{$a \in B$}
\State $\mathbb Record(\text{SubtreePartialVC}(G, \mathcal T, \ensuremath{\mathrm{children}} \left((\{a\} \cup \ensuremath{\mathrm{anc}}(a)) \cap B \right), k-1, \delta))$
\EndFor
\State Return the best recorded solution $\{v_1, \ldots, v_{k-1}\} \in V_{\mathcal T}$.
\varepsilonnd{algorithmic}
\varepsilonnd{algorithm}
\begin{algorithm}
\caption{Laminar$(G=(V,E,w),\mathcal T,k,\varepsilon_1,\delta)$}
\langlebel{alg:laminar}
\begin{algorithmic}
\State $\mathcal T=(V_{\mathcal T},E_{\mathcal T},w_{\mathcal T}) \gets \text{MincutTree}(G)$.
\mathbb For{$r \in V_{\mathcal T}$}
\State Root $\mathcal T$ at $r$.
\State $\mathbb Record(\text{LaminarRooted}(G,\mathcal T,r,k,\varepsilon_1,\delta))$
\EndFor
\State Return the best recorded $k$-partition.
\varepsilonnd{algorithmic}
\varepsilonnd{algorithm}
\begin{algorithm}
\caption{LaminarRooted$(G=(V,E,w),\mathcal T,r,k,\delta_1,\delta)$}
\langlebel{alg:laminarRooted}
\begin{algorithmic}
\mathbb For {$a \in V(\mathcal T)$}
\State $\{S_{a,i} : i \in [k-1]\} \gets \text{SubtreePartialVC}(G, \mathcal T, \ensuremath{\mathrm{children}}(a), k-1, \delta)$ \Comment{$S_{a,i} \in \left(2^V\right)^i$}
\EndFor
\
\State $A \gets \varepsilonmptyset$ \Comment{$A \subset V(\mathcal T) \times [k]$ is the set of \textit{anchors}}
\mathbb For {$a \in V(\mathcal T)$ in topological order from leaf to root}
\State $\varepsilon_3 \gets \frac{1-\delta}4-2\varepsilon_1$ \Comment {The optimal value of $\varepsilon_3$}
\State $I_a \gets \{i \in [k-1] : \text{Value}(P_{a,i}) \ge \varepsilon_3(1-\delta)(i-1){\mu} \}$
\If {$I_a \ne \varepsilonmptyset$ \textbf{and} $\nexists (a',i) \in A : a' \in \ensuremath{\mathrm{desc}}(a)$} \Comment{Only take \textit{minimal} anchors}
\State $A \gets A \cup \{(a, \max I_a)\}$
\EndIf
\EndFor
\
\If {$|A| \ge k-1$} \Comment{\textbf{Case (K)}: Knapsack}
\State $A' \gets \text{Knapsack}(A)$ \Comment{The Knapsack algorithm as described in Lemma~\ref{lem:incomparable}}
\State $\mathcal S \gets \displaystyle\bigcup\limits_{(a,i) \in A} \{S_{a,i}\}$ \Comment{The partition for Case (K), to be computed. \textbf{Assert}: $|S| \le k-1$}
\State $\mathbb Record(\text{Complete}(G,k,\mathcal S))$
\Else
\State $\mathcal B \gets \text{Branches}(A)$ \Comment{$\mathcal B \subset \left( 2^{V(\mathcal T)} \right)^r$ for some $k-1 \le r \le 2k-3$}
\
\mathbb For {$B \in \mathcal B$} \Comment {\textbf{Case (B1)}: Compute branches independently}
\State $\{P_{B,i} : i \in [k-1]\} \gets \text{SingleBranch}(G, \mathcal T, B, k-1, \delta)$ \Comment{$P_{B,i} \in V^i$}
\EndFor
\State $(\mathcal B^*,\mathbf i^*) \gets \operatornamewithlimits{argmin}\limits_{ \substack {\mathcal B' \subset \mathcal B \text{ incomparable},\\ \mathbf i \in [k-1]^{\mathcal B'} : \ \sum_{B} i_B\ =\ k-1} } \ \displaystyle\sum\limits_{B \in \mathcal B'} w(E(P_{B,i_B}))$ \Comment{Computed by brute force}
\State $\mathcal S_1 \gets \bigcup_{B \in \mathcal B^*} \{P_{B, i_B}\}$ \Comment{The partition in Case (B1)}
\State $\mathbb Record(\text{Complete}(G,k,\mathcal S_1))$
\
\mathbb For {$B \in \mathcal B$} \Comment{\textbf{Case (B2)}: Guess the branches with the anchors}
\State $a_B \gets (a \in B : B \setminus a \subset \ensuremath{\mathrm{desc}}(a))$ \Comment{$a_B$ is the common ancestor of branch $B$}
\EndFor
\mathbb For {$\mathcal B' \subset \mathcal B$ s.t.\ $\nexists B_1,B_2\in\mathcal B' : B_1 \subset \ensuremath{\mathrm{desc}}(B_2)$} \Comment{Subsets whose branches are incomparable}
\State $A_{\mathcal B'} \gets \ensuremath{\mathrm{children}}\left(\bigcup_{B \in \mathcal B'} \left( \{a_B\} \cup \ensuremath{\mathrm{anc}}(a_B) \right) \right)$
\State $\mathcal S_{2,\mathcal B'} \gets \text{SubtreePartialVC}(G,\mathcal T,A_{\mathcal B'},k-1,\delta)$ \Comment{The partition for $\mathcal B'$ in Case (B2)}
\State $\mathbb Record(\text{Complete}(G,k,\mathcal S_{2,\mathcal B'}))$
\EndFor
\EndIf
\
\State Return the best recorded $k$-partition.
\varepsilonnd{algorithmic}
\varepsilonnd{algorithm}
\section{Missing Proofs}
\langlebel{sec:missing-proofs}
\begin{lemma} \langlebel{lemma:knapsack}
Consider the knapsack instance of $k-1$ items $i \in [k-1]$ where item $i$ has size $s_i \in [2,k-1]$ and value $s_i-1$. There is an algorithm achieving value $\ge (k-1)/4$ for $k \ge 5$, running in $O(k)$ time.
\varepsilonnd{lemma}
\begin{proof}
Consider the greedy
knapsack solution where we always choose the heaviest item, if still
possible. Let $A \in [k-1]$ be our solution. If our total size
$\sum_{i\in A}s_i$ is at least $k - 1 - \sqrt k$, then our value is
at least $\sum_{i\in A}(s_i-1) \ge \sum_{i\in A}s_i/2 \ge (k-1-\sqrt
k)/2$. Otherwise, since we could not fit the next item of size at
least $\sqrt k$ into our solution, all of our items have size at
least $\sqrt k$. Furthermore, our total solution size is at least
$(k-1)/2$, so $\sum_{i \in A}(s_i-1) \ge \sum_{i\in A}(1-1/\sqrt
k)s_i \ge (1-1/\sqrt k)(k-1)/2$. When $k \ge 5$, the
value is $\ge (1-1/\sqrt 5)(k-1)/2 \ge (k-1)/4$.
\varepsilonnd{proof}
\varepsilonnd{document}
|
math
|
/**
* @license
* Copyright Google Inc. All Rights Reserved.
*
* Use of this source code is governed by an MIT-style license that can be
* found in the LICENSE file at https://angular.io/license
*/
import {Component} from '@angular/core';
import {Http, Response} from '@angular/http';
import 'rxjs/add/operator/map';
@Component({
selector: 'http-app',
template: `
<h1>people</h1>
<ul class="people">
<li *ngFor="let person of people">
hello, {{person['name']}}
</li>
</ul>
`
})
export class HttpCmp {
people: Object[];
constructor(http: Http) {
http.get('./people.json')
.map((res: Response) => res.json())
.subscribe((people: Array<Object>) => this.people = people);
}
}
|
code
|
Aliquippa water customers are being asked to conserve water after the second line break on Franklin Avenue in a day occurred overnight.
"This has been one of those nightmare things," said Aliquippa Water Authority General Manager Bob DiGiovine. "Anything that could have gone wrong, went wrong."
DiGiovine said he hoped service would be retored by late this afternoon.
Residents without water can bring their containers to the Aliquippa fire station on Kennedy Avenue or the city's filtration plant at 160 Hopewell Ave.
|
english
|
کرن راٹھوڑ چھِ اَکھ ہِندوستٲنؠ اَداکارہ یۄس فِلمَن مَنٛز چھِ کٲم کَران.
زٲتی زِندگی
فِلمی دور
== حَوالہٕ ==
|
kashmiri
|
کیا از چھا موسم اصل
|
kashmiri
|
\begin{document}
\title{Applying Brownian motion to the study of birth-death chains.}
\author{
\begin{tabular}{c}
\textit{Greg Markowsky} \\
[email protected] \\
(054)279-5828 \\
Pohang Mathematics Institute \\
POSTECH \\
Pohang, 790-784 \\
Republic of Korea
\end{tabular}}
\begin{abstract}
Basic properties of Brownian motion are used to derive two results concerning birth-death chains. First, the probability of extinction is calculated. Second, sufficient conditions on the transition probabilities of a birth-death chain are given to ensure that the expected value of the chain converges to a limit. The theory of Brownian motion local time figures prominently in the proof of the second result.
\end{abstract}
\begin{keyword}Birth-death chain, Markov chain, Brownian motion, local time.
\end{keyword}
\maketitle
\section{Introduction}
Let $X_m$ be a Markov chain taking values on the nonnegative integers with the following transition probabilities for $n \neq 0$
\be p_{nj} = \left \{ \begin{array}{ll}
r_{n} & \qquad \mbox{if } j=n+1 \\
l_n & \qquad \mbox{if } j=n-1 \\ 0 & \qquad \mbox{if } |n-j| \neq 1\;.
\end{array} \right. \ee
Implicit here is the fact that $r_n+l_n=1$. We suppose further for simplicity that $X_0 = k$ almost surely, for some $k \in \mathbb{N}$. $X_m$ is essentially a random walk on the nonnegative integers, moving to the right from state $n$ with probability $r_n$ and to the left with probability $l_n$. We refer to such a Markov chain as a {\it birth-death chain}. This name comes from considering $X_m$ as the number of members in a population, where at each step either a new member is born or an old member dies, causing the process to increase or decrease by 1. We can assume $p_{00}=1$ and $p_{0j}=0$ for any $j \neq 0$, as when the population reaches 0 it is considered to have gone extinct with no possibility of regeneration. The purpose of this paper is to introduce a method of using properties of Brownian motion to deduce two fundamental theorems concerning birth-death chains. The first theorem, presented in the next section, gives the probability that a birth-death chain goes extinct at some finite time. The second theorem, presented in Section 3, gives sufficient conditions for $E[X_m]$ to converge as $m \longrightarrow \ff$. The properties of Brownian motion which will be utilized are standard and can be found in many references on Brownian motion, such as \cite{rosmarc} or \cite{revyor}.
\vski
We will now introduce the basic setup. Let $t_0 := 1$ and
\be \label{pred}
t_n := \frac{l_1 l_2 \ldots l_n}{r_1 r_2 \ldots r_n}
\ee
for $n>0$. Define a sequence $\{x_n\}_{n=0}^\ff$ recursively by setting $x_0=0$, and having defined $x_n$ let $x_{n+1}=x_n+t_n$. Since the sequence $\{x_n\}$ is increasing it converges to a limit $x_\ff$, possibly infinite, as $n \longrightarrow \ff$. Let $B_t$ be a Brownian motion starting at $x_k$ and stopped at the first time $T_{\Delta}$ it hits $0$ or $x_\ff$. The recurrence properties of Brownian motion imply that $T_{\De} < \ff$ almost surely. We define a sequence of stopping times $T_m$ which are, roughly speaking, the successive hitting times of $\cal{A} :=$ $ \{ x_n \}_{n=0}^\ff$. More rigorously, $T_m$ is defined recursively by setting $T_0=0$, and having defined $T_m$ we let $T_{m+1}=\inf_{t>T_m}\{B_t \in \cal{A},$ $B_t \neq B_{T_m}\}$. We see that the variables $B_{T_0},B_{T_1},B_{T_2}, \ldots $ form a random process taking values in $\cal{A}$. The strong Markov property of Brownian motion, together with the standard exit distribution of Brownian motion from an interval, imply that
\bea \label{}
\nn && P(B_{T_{m+1}} = x_{n+1} | B_{T_{m}} = x_{n}) = \frac{x_{n}-x_{n-1}}{x_{n+1}-x_{n-1}} = \frac{t_{n-1}}{t_{n-1}+t_n} = \frac{1}{1+l_n/r_n} = r_n
\eea
and, likewise,
\bea \label{}
\nn && P(B_{T_{m+1}} = x_{n-1} | B_{T_{m}} = x_{n}) = l_n
\eea
If we define $\phi$ on $\{x_n\}_{n=0}^\ff$ by $\phi(x_n)=n$, we see that $\phi(B_{T_0}),\phi(B_{T_1}),\phi(B_{T_2}), \ldots $ is a realization of our original birth-death chain. The picture below gives an example, where we have oriented the time axis vertically and the space axis horizontally.
\vski
\hspace{.8cm} \includegraphics[width=110mm,height=80mm]{bdchainpic1.pdf}
{\small Figure 1: The Brownian path pictured realizes the birth-death path $k,k+1,k,k+1,k,k-1,k,k-1,k-2, \ldots$}
\vski
Given this framework, we are ready to prove several theorems. In the sequel, any reference to $X, B, x_n, T_{\Delta},\phi,$ etc. will refer to the definitions presented in this section.
\section{The extinction probability of a birth-death chain}
Perhaps the most fundamental question one can ask regarding a birth-death chain is whether the population must go extinct or not, that is, whether $P(X_m = 0$ for some $m)=1$ or $P(\lim_{m \longrightarrow \ff} X_m = +\ff)>0$. Let $P_k$ be the probability that the birth-death chain eventually hits 0 (recall $X_0=k$ a.s.). We then have the following.
\bt \label{surf}
\be \label{mass}
P_k = \frac{\sum_{j=k}^\ff t_j}{\sum_{j=0}^\ff t_j}
\ee
where this quotient is interpreted as being equal to 1 if the sums diverge.
\et
This elegant theorem has a straightforward proof using recurrence relations; see \cite{norr} or \cite{sysk}. A potentially pleasing aspect of the proof below, however, lies in giving a clear, visual intuition for the sums in \rrr{mass}.
\vski
{\bf Proof of Theorem \ref{surf}:}
Recall that $x_\ff = \lim_{n \longrightarrow \ff} x_n$ is given by
\be \label{}
x_\ff=\sum_{j=0}^\ff t_j
\ee
If $x_\ff=\ff$, so that both sums in \rrr{mass} diverge, then $B_{T_{\DD}}=0$ almost surely. This implies that the population dies out with probability $1$. On the other hand, if $x_\ff<\ff$ then $P(B_{T_{\De}}=0)$ is given by
\be \label{}
\frac{x_\ff-x_k}{x_\ff-0} = \frac{\sum_{j=k}^\ff t_j}{\sum_{j=0}^\ff t_j}
\ee
However, as in the first case, $P(B_{T_\De}=0)$ is precisely $P_k$, the probability of extinction. This is because the Brownian motion hitting $x_\ff$ before $0$ implies $B_{T_m}\longrightarrow x_\ff$, hence $\phi(B_{T_m}) \longrightarrow \ff$, whereas hitting $0$ before $x_\ff$ implies $\phi(B_{T_\De}) =0$ for some $m$. The two cases ($x_\ff=\ff$ and $x_\ff < \ff$) are illustrated in the following figure.
\includegraphics[width=140mm,height=110mm]{bdchain_compare.pdf}
{\small Figure 2: The left panel illustrates the situation in which $\sum_{j=0}^\ff t_j$ diverges. Eventually, $B_t$ hits $0$ and the population goes extinct. The right panel illustrates the other scenario, in which $\sum_{j=0}^\ff t_j = x_\ff < \ff$. In this case, there is a positive probability that $B_t$ hits $x_\ff$ before $0$, in which case the population never goes extinct.}
\vski
This completes the proof of Theorem 1. {
$\square$
}
\section{The long-term average of a birth-death chain}
Recall that $B$ and $X$ are stopped upon reaching $0$. It will therefore be convenient to let $X_m$ be defined to be $0$ for all $m>m_0$, where $m_0$ is the smallest integer, if it exists, for which $X_{m_0}=0$. Similarly, for convenience let $T_m=T_\DD$ for all $m > m_0$, where $m_0$ is the smallest integer, if it exists, for which $B_{T_{m_0}}=0$. In the case $r_i=l_i=\frac{1}{2}$ for all $i$, it is well known that $X_m$ is a martingale, and therefore $E[X_m]=E_0=k$ for all $m$. This occurs despite the fact that $P(X_m = 0) \longrightarrow 1$ as $m \longrightarrow \ff$, as the average value of $X_m$ on $\{X_m \neq 0\}$ grows at exactly the right speed to balance the set of large probability upon which $X_m=0$. Such behavior certainly does not hold for the general case, since we no longer have the martingale property, but we will see that the Brownian motion model presented above can shed light on the behavior of $E[X_m]$ as $m \longrightarrow \ff$.
\vski
Recall that $\cal{A}$ $=\{x_n\}_{n=0}^\ff$. Let $\phi: \cal{A}$ $\longrightarrow \mathbb{R}^+$ be extended to a continuous function from $\mathbb{R}^+$ to $\mathbb{R}^+$ by defining $\phi$ to be linear on each interval $(x_{n-1},x_n)$. Alternatively, we may think of $x_n=x(n)$ as a function from $\mathbb{N}$ to $\mathbb{R}$ which can be extended by linear interpolation to an increasing function from $\mathbb{R}^+$ to $\mathbb{R}^+$. In this case, $\phi$ is simply $x^{-1}$. $\phi$ is therefore a piecewise linear function, and $\phi'$ exists on $\mathbb{R}^+ - \cal{A}$. Let $\phi'_n$ be the value of $\phi'$ on $(x_{n-1},x_n)$. We will prove the following theorem.
\bt \label{bigguyii}
If $\phi'_\ff = \lim_{n \longrightarrow \ff} \phi'_n$ exists, then
\be \label{yokoneg1}
\lim_{m \longrightarrow \ff} E[X_m] = x_k \phi'_\ff
\ee
\et
Note that we are allowing $\phi'_\ff = +\ff$ or $0$. Writing $\phi'_\ff$ and $x_k$ in terms of the $l_n$'s and $r_n$'s shows that the following statement is equivalent to Theorem \ref{bigguyii}.
\vski
{\it If $t_\ff := \lim_{n \longrightarrow \ff} \frac{l_1\ldots l_{n}}{r_1\ldots r_{n}}$ exists then $\lim_{m \longrightarrow \ff}E[X_m]$ exists, and }
\be \label{}
\lim_{m \longrightarrow \ff}E[X_m] = \frac{1+ \frac{l_1}{r_1} + \ldots + \frac{l_1\ldots l_{k-1}}{r_1\ldots r_{k-1}}}{t_\ff}
\ee
\vski
The bulk of the rest of this section is devoted to the proof of this theorem. We will simplify initially by assuming $\sum_{n=1}^{\ff} |\phi'_{n+1}-\phi'_n| < \ff$; this condition will be removed at the end of the proof. For the case in which there is a positive probability that the population never goes extinct, it is easy to see that $E[X_m] \longrightarrow \ff$ as $m \longrightarrow \ff$, and that $\phi'_\ff$ exists and is equal to $+\ff$, so that \rrr{yokoneg1} is valid. We will therefore assume that $P(X_m = 0$ for some $m) = 1$. Note that $\phi'_{n+1} = \frac{1}{t_n}$, and $x_{n+1}-x_n = t_n$, so that $\phi_{n+1}'(x_{n+1}-x_n) = 1$. Note also that $x_1 \phi'_1 = 1$. This allows us to perform the following manipulations to obtain an expression which will be more convenient for the purposes of the proof.
\bea \label{yokoa}
&& x_k \phi'_\ff = x_k (\phi'_\ff - \phi'_{k}) + x_k \phi'_k + k - x_1\phi'_1 - \sum_{n=1}^{k-1} \phi_{n+1}(x_{n+1}-x_n)
\\ \nn && \hspace{1.1cm}= k + x_k (\phi'_\ff - \phi'_{k}) + \sum_{n=1}^{k-1} (\phi'_{n+1}-\phi'_n) x_n
\eea
The last equality uses summation by parts; see \cite{lang}. We see that the conclusion of the theorem is equivalent to showing
\be \label{yoko}
\lim_{m \longrightarrow \ff} E[X_m] = k + x_k (\phi'_\ff - \phi'_{k}) + \sum_{n=1}^{k-1} (\phi'_{n+1}-\phi'_n) x_n
\ee
This is what we will prove. We will proceed through several lemmas, and will need properties of Brownian motion {\it local time}, which is the density of the occupation measure of Brownian motion with respect to Lebesgue measure. That is, the local time $L_t^x$ satisfies
\be \label{}
L_t^x dx = \int_{0}^{t} 1_{B_s \in dx} ds
\ee
It is well known that $L_t^x$ exists and that
\be \label{}
L_t^x = \lim_{\varepsilon \longrightarrow 0} \frac{1}{2\varepsilon} \int_{0}^{t} 1_{|B_s-x|<\varepsilon}ds
\ee
almost surely. See \cite{rosmarc} for a comprehensive treatment of local time, or the more general reference \cite{revyor}. The following is an extension of Tanaka's formula, Theorem VI.1.2 in \cite{revyor}.
\bl \label{45}
Almost surely, for any stopping time $T$,
\be \label{maz}
\phi(B_T) = k + \int_{0}^{T} \phi'(B_s) dB_s + \sum_{n=1}^\ff \frac{(\phi'_{n+1}-\phi'_n)}{2} L^{x_n}_T
\ee
where $L^{x_n}_T$ denotes the local time of $B_t$ at $x_n$ at time $T$.
\el
{\bf Proof:} Note that $\phi''(x) = \sum_{n=1}^\ff (\phi'_{n+1}-\phi'_n) \delta_{x_n}(x)$ in the sense of distributions, where $\delta_{x_n}(x) = \delta_0(x-x_n)$ denotes the Dirac delta function at point $x_n$. Lemma \ref{45} is therefore seen to be a special case of the It\^{o}-Tanaka formula, Theorem VI.1.5 of \cite{revyor}, provided that $\phi$ can be realized as a difference of two convex functions. However, any piecewise-linear function can be realized as the difference of two convex functions, provided that the points of nondifferentiability do not accumulate. We may argue as follows. $\phi'$ is a piecewise constant function, which is therefore of bounded variation on bounded intervals, and as such we may write $\phi' = f-g$ where $f$ and $g$ are nondecreasing. Let $F$ and $G$ be antiderivatives of $f$ and $g$ chosen so that $\phi = F-G$. Then $F$ and $G$ are convex, and the result follows. {
$\square$
}
Applying this lemma to the stopping time $T_m$, we immediately obtain
\be \label{kok}
E[X_m] = k + \sum_{n=1}^\ff \frac{(\phi'_{n+1}-\phi'_n)}{2} E[L^{x_n}_{T_m}],
\ee
Note that the convergence of the sum at this point is not an issue, since $L^{x_n}_{T_m} = 0$ for $n>m+k$. Using the identity \rrr{kok} does not seem to be an effective way to calculate $E[X_m]$, due to the difficulty of obtaining information about $T_m$. Nonetheless, we do know that $T_m \nearrow T_\DD$ as $m \longrightarrow \ff$, and this implies
\be \label{oko}
\lim_{m \longrightarrow \ff} E[X_m] = k + \sum_{n=1}^\ff \frac{(\phi'_{n+1}-\phi'_n)}{2} E[L^{x_n}_\ff],
\ee
provided that $E[L^{x_n}_\ff]$ can be bounded uniformly, which we will show soon to be the case. We should mention that it was in obtaining \rrr{oko} that we used the assumption that $\sum_{n=1}^{\ff} |\phi'_{n+1}-\phi'_n| < \ff$. This is because a priori the quantities $E[L^{x_n}_{T_m}]$ may be growing in some strange way that causes problems if $\sum_{n=1}^{\ff} |\phi'_{n+1}-\phi'_n| = \ff$. We will return to this point at the end of the proof. In light of \rrr{oko}, we must compute $E[L^{x_n}_\ff]$.
\bl \label{sbtm}
\be \label{a3}
E[L^{x_n}_\ff] = 2 \min (x_k,x_n)
\ee
\el
{\bf Proof:} One may derive this through standard calculations involving the probability density function of $B_t$, but the following is a quicker and easier proof. Let us suppose first that $n=k$. From Tanaka's formula, $E[L^{x_k}_\ff] = \lim_{t \longrightarrow \ff} E[|B_t-x_k|]$. Furthermore, $B_t$ is a martingale, so $E[(B_t-x_k)]=0$ for all $t$. It follows from this that
\bea \label{}
&& E[L^{x_k}_\ff]=2\lim_{t \longrightarrow \ff} E[\max(-(B_t-x_k),0)]
\\ \nn && \hspace{1.3cm} = 2 \lim_{t \longrightarrow \ff} \Big( x_k P(B_t=0) + \int_{0}^{k} (x_k-x) P(B_t \in dx) \Big)
\eea
As $\lim_{t \longrightarrow \ff} P(B_t=0) = 1$, we can conclude that $E[L^{x_k}_\ff]=2x_k$. Now suppose $n \neq k$. Note that, if we let $T_{x_n} = \inf \{t: B_t=x_n\}$, then
\be \label{runy}
L^{x_n}_\ff = L_{T_{x_n}} + L_\ff(B \circ \theta_{T_{x_n}})1_{T_{x_n}<T_\DD} = L_\ff(B \circ \theta_{T_{x_n}})1_{T_{x_n}<T_\DD}
\ee
where $\theta$ denotes the standard shift operator and $L_t(B \circ \theta_{T_{x_n}})$ is the local time of the shifted process $B \circ \theta_{T_{x_n}}$.
Let $E_{x_j}$ denote expectation with respect to a Brownian motion $W$ which starts at $x_j$ and is stopped upon hitting $0$. The prior calculation together with \rrr{runy} and the strong Markov property of Brownian motion imply that
\bea \label{}
&& E[L^{x_n}_\ff] = P(T_{x_n}<T_\DD) E_{x_n} [L_\ff^{x_n}]
\\ \nn && \hspace{1.32cm} = P(T_{x_n}<T_\DD) 2x_n
\eea
The general result follows from noting that $P(T_{x_n}<T_\DD)$ is $1$ if $x_n < x_k$ and $\frac{x_k}{x_n}$ if $x_n > x_k$. {
$\square$
}
Combining \rrr{oko} and Lemma \ref{sbtm} gives
\be \label{yoko2}
\lim_{m \longrightarrow \ff} E[X_m] = k + \sum_{n=1}^{k-1} (\phi'_{n+1}-\phi'_n) x_n + x_k \sum_{n=k}^{\ff}(\phi'_{n+1} - \phi'_n)
\ee
Since $\sum_{n=k}^{\ff}(\phi'_{n+1} - \phi'_n) = (\phi'_\ff-\phi'_{k})$, we are done in this case. It remains only to remove the restriction that $\sum_{n=1}^{\ff} |\phi'_{n+1}-\phi'_n| < \ff$. The following lemma is key.
\bl \label{}
For any $m$, and any $n \geq k$, $E[L^{x_n}_{T_m}] \geq E[L^{x_{n+1}}_{T_m}]$.
\el
{\bf Proof:} In fact, we may prove somewhat more, namely that if $T$ is any stopping time with $B_T \in \cal{A}$ almost surely, then $E[L^{x_n}_{T}] \geq E[L^{x_{n+1}}_{T}]$. Let $E_{x_j}$ and $W$ be as in the proof of Lemma \ref{sbtm}.
Using Lemma \ref{sbtm} and the strong Markov property of Brownian motion, we obtain
\bea \label{}
&& E[L^{x_n}_{T}] = E[L^{x_n}_{\ff}] - E\Big[\lim_{\varepsilon \longrightarrow 0}\frac{1}{2\varepsilon}\int_{T}^{\ff} 1_{(-\varepsilon,\varepsilon)}(B_s - x_n) ds \Big]
\\ \nn && \hspace{1.33cm} = 2 x_k - \sum_{j=1}^{\ff} P(B_T = x_j)E_{x_j} \Big[ \lim_{\varepsilon \longrightarrow 0} \frac{1}{2\varepsilon} \int_{0}^{\ff} 1_{(-\varepsilon,\varepsilon)}(W_s- x_n) ds \Big]
\\ \nn && \hspace{1.33cm} = 2 x_k - \sum_{j=1}^{n-1} P(B_T = x_j)x_j - \sum_{j=n}^{\ff} P(B_T = x_j)x_n
\eea
Similarly,
\be \label{}
E[L^{x_{n+1}}_{T}] = 2 x_k - \sum_{j=1}^{n} P(B_T = x_j)x_j - \sum_{j=n+1}^{\ff} P(B_T = x_j)x_{n+1}
\ee
The conclusion of the lemma now follows from the fact that $x_{n+1} > x_n$. {
$\square$
}
We may now complete the proof of the theorem. Recall \rrr{kok}, and observe that $E[L^{x_n}_{T_m}] = 0$ for $n>k+m$, since $X_m \leq m+k$. This means that \rrr{kok} is in fact a finite sum.
\bea \label{kok2}
&& E[X_m] = k + \sum_{n=1}^{m+k} \frac{(\phi'_{n+1}-\phi'_n)}{2} E[L^{x_n}_{T_m}]
\\ \nn && \hspace{1.3cm} = k + \sum_{n=1}^{k-1} \frac{(\phi'_{n+1}-\phi'_n)}{2} E[L^{x_n}_{T_m}] + \sum_{n=k}^{k+m} \frac{(\phi'_{n+1}-\phi'_n)}{2} E[L^{x_n}_{T_m}]
\eea
The indices of the first sum in the final expression of \rrr{kok2} are independent of $m$. This implies that the sum converges as $m\longrightarrow \ff$, since $E[L^{x_n}_{T_m}] \longrightarrow 2x_n$ as $m \longrightarrow \ff$ for $n \leq k$. We must show that the second sum converges as $m\longrightarrow \ff$. We use summation by parts again, which gives
\bea \label{kok2}
&& \sum_{n=k}^{k+m} \frac{(\phi'_{n+1}-\phi'_n)}{2} E[L^{x_n}_{T_m}] = \frac{1}{2}\Big( \phi'_{k+m+1}E[L^{x_{k+m+1}}_{T_m}] - \phi'_{k}E[L^{x_{k}}_{T_m}]
\\ \nn && \hspace{1cm} - \sum_{n=k}^{k+m} \phi'_{n+1} (E[L^{x_{n+1}}_{T_m}] - E[L^{x_{n}}_{T_m}]) \Big)
\eea
Let us assume that $\phi'_\ff < \ff$, and let $\varepsilon>0$ be given. We may choose $N>k$ such that $\phi'_n \in (\phi_\ff-\varepsilon,\phi_\ff+\varepsilon)$ for all $n \geq N$. Having chosen this, we may choose $M > N-k$ such that $2x_k \geq E[L^{x_{n}}_{T_m}] > 2x_k - \varepsilon$ for all $n \in [k,N], m \geq M$. Using the fact that $E[L^{x_{k+m+1}}_{T_m}]=0$, and setting $\overline{\phi'} = \sup_{j > 0} \phi'_j$, we see that for $m>M$
\bea \label{}
&& \sum_{n=k+1}^{k+m} \frac{(\phi'_{n+1}-\phi'_n)}{2} E[L^{x_n}_{T_m}]
\\ \nn && \hspace{1cm} \leq \frac{1}{2}\Big( -\phi'_{k} (2x_k - \varepsilon) + \sum_{n=k}^{N} \phi'_{n+1} (E[L^{x_{n}}_{T_m}] - E[L^{x_{n+1}}_{T_m}])
\\ \nn && \hspace{2cm} + \sum_{n=N+1}^{k+m} \phi'_{n+1} (E[L^{x_{n}}_{T_m}] - E[L^{x_{n+1}}_{T_m}]) \Big)
\\ \nn && \hspace{1cm} \leq \frac{1}{2}\Big( -\phi'_{k} (2x_k - \varepsilon) + \overline{\phi'} (E[L^{x_{k}}_{T_m}] - E[L^{x_{N+1}}_{T_m}])
\\ \nn && \hspace{2cm} + (\phi'_\ff + \varepsilon) (E[L^{x_{N+1}}_{T_m}] - E[L^{x_{k+m+1}}_{T_m}]) \Big)
\\ \nn && \hspace{1cm} \leq \frac{1}{2}\Big( -\phi'_{k} (2x_k - \varepsilon) + \overline{\phi'} \varepsilon + (\phi'_\ff + \varepsilon) 2x_k \Big)
\eea
This shows that
\be \label{}
\limsup_{m \longrightarrow \ff} \sum_{n=k}^{k+m} \frac{(\phi'_{n+1}-\phi'_n)}{2} E[L^{x_n}_{T_m}] \leq x_k(\phi'_\ff - \phi'_{k})
\ee
Proceeding similarly, we can obtain
\be \label{}
\liminf_{m \longrightarrow \ff} \sum_{n=k+1}^{k+m} \frac{(\phi'_{n+1}-\phi'_n)}{2} E[L^{x_n}_{T_m}] \geq x_k(\phi'_\ff - \phi'_{k})
\ee
Together, these prove the desired convergence. The case $\phi'_\ff = +\ff$ is similar but easier and is omitted. This completes the proof of Theorem \ref{bigguyii}. {
$\square$
}
We conclude with a simple but counterintuitive example. Let $l_n=\frac{n}{2n+1}, r_n = \frac{n+1}{2n+1}$ for $n \geq 1$. Then $t_n = \frac{1}{n+1}$, so that $t_\ff = 0$. On the other hand, $x_\ff = 1 + \sum_{n=1}^{\ff}t_n = \ff$. We see that the birth-death chain $X_m$ built upon these transition probabilities has an extinction probability of 1, but $E[X_m] \longrightarrow \ff$ as $m \longrightarrow \ff$.
\end{document}
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math
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There are many reasons a company will hire a person, the top two are: (1) the candidate fills an employer’s immediate need and (2) they have the character that will thrive in the corporate culture.
When it comes to making a hiring decision, many factors are simply beyond your control. The interview is clearly not one of them.
You may be surprised how many people are unprepared when they interview. Never enter an interview without rehearsal; trying to “wing it” only makes you seem less competent than you may actually be. Practice with a friend, in the mirror or record yourself with a camcorder; any practice will help you perform better.
Have a list of the important issues about the job. You are trying to fill an organization’s need, so focus less on yourself and more on how your skills can benefit the organization. This is done with a combination of closely following the job description and a little research into the company.
This may be obvious, but it is so critical the process, it bears repeating. Being late to an interview—for any number of reasons—is one of the quickest ways to derail your chances. If you have no choice, then call the office and try to reschedule—being late will reflect poorly on you and your organizational skills. Wasting an interviewer’s time rarely results in a job offer.
Show, don’t sell, your communication skills.
You may talk a good game, but it is just that—talk. Prove yourself by communicating effectively, clearly and concisely. The best way to “sell” your skills is demonstrating them real-time, by being organized and professional. Never ramble or go off on tangents—stay focused on what you can do for the company once you are hired.
Look the interviewer directly in the eyes, but don’t make it a staring contest. Take notes and be agreeable. Have some questions prepared—at least three or four of them. If you do your homework, you will be far above the average job seeker.
Send an email to the interviewer. Better yet, go “old school” and post a handwritten thank you note. Remind them of the highlights of the interview, or some details you may have forgotten at the time.
Taking the time to hit the basics–the “simple stupid” interview tips–can make you appear confident, poised and professional, the type of person anyone wants to hire.
One last point: always make sure you have references lined up, so you can turn over a list immediately when asked.
|
english
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A joint programme developed by Civil Defence Academy (CDA) and funded by Ministry of Foreign Affairs, Singapore.
Dr Robert de Souza, Executive Director, invited to present "Collaborations in Humanitarian Supply Chain Management" at Disaster Risk Reduction and Management (DRRM), a joint programme developed by Civil Defence Academy (CDA) and funded by Ministry of Foreign Affairs, Singapore.
Organised by the Coordinating Ministry for Economic Affairs, Republic of Indonesia.
THINKLog: Train-the-Trainer Workshop, organised by the Coordinating Ministry for Economic Affairs, Republic of Indonesia.
TLI-AP was invited to conduct a logistics board game session for the participants of the ACE Programme by the ASEAN Coordinating Centre for Humanitarian Assistance in Jakarta.
*You can click NEXT if you do not wish to complete the whole survey till you reach Best Educational Course Provider page.
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english
|
I’ve been hard at work getting a Nebula 3 plugin for Maya to have all the features we’d want. Why you may ask, isn’t the pipeline awesome as it is? Well, the answer is no, no it isn’t. It works, but it is far from smooth, and the Nebula plugin aims to address that. The plugin is currently only for Maya, but there will be a Motion Builder version as well. The plugin basically just wrap FBX exporting, and makes sure it’s suitable for Nebula by using a preset included with the Nebula distribution. It also runs the FBX batcher which converts the fbx-file to the model files and mesh files. It can also preview the mesh if it’s exported, and will export it if it doesn’t exist already. This allows for immediate feedback how the model will look in Nebula. It also tries preserve the shader variables, but it’s impossible to make it keep the material. That’s because DirectX doesn’t support setting a vertex layout with a vertex shader with a smaller input than the layout. This is a problem because converting a skin from static to skinned will cause Nebula to crash, seeing as the material is preserved between exports. So the plugin offers a way to get meshes, including characters, directly to Nebula, which is very nice indeed.
I’ve also been working with getting a complete Motion Builder scene into Nebula, and I actually got Zombie to work with all features. This means the skin, along with more than one animation clip (yay) can be loaded into Nebula seamlessly by simply saving the Motion Builder file and running the exporter. I will probably make a Nebula 3 plugin for Motion Builder as well, so we can have the exact same export and preview capabilities as in Maya.
I know I have been promising a video showing some of the stuff we’ve done, but I just haven’t had the time! Right now, I will start working on the documentation for our applications so there are clear directions for anyone who wish to use them (mainly our graphics artists here at Campus) . The plugin already redirects from Maya to three different HTML docs, each of which will describe the different tools. That’s nice and all, except for the fact that the HTML docs are completely empty.
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english
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२९ सालों का अन्धविश्वाश तोड़कर नोएडा पहुंचे सीएम योग, कहा- विकास कार्यों से ऊंचाई पर पहुंचेगा उत्तर प्रदेश - ग्रेनोन्यूज
२९ सालों का अन्धविश्वाश तोड़कर नोएडा पहुंचे सीएम योग, कहा- विकास कार्यों से ऊंचाई पर पहुंचेगा उत्तर प्रदेश
नोएडा : आखिरकार उत्तर प्रदेश के मुख्यमंत्री योगी आदित्यनाथ २९ साल का अन्धविश्वाश तोड़कर आज नोएडा पहुंचे। उन्होंने यहां मेट्रो की नई मेजेंटा लाइन के उद्घाटन की तैयारियों का जायजा लिया। बता दें आगामी २५ दिसंबर को पूर्व प्रधानमंत्री अटल बिहारी वाजपेयी के जन्मदिवस के मौके पर नरेंद्र मोदी नोएडा से साउथ दिल्ली के बीच मेट्रो की शुरुआत करेंगे। योगी के नोएडा आने की चर्चा इसलिए जोरों पर है क्योंकि एक अंधविश्वास के चलते पिछले २९ सालों में यूपी के सीएम नोएडा जाने से परहेज करते आए हैं।
नोएडा में एक प्रेस वार्ता को संबोधित करते हुए सीएम ने कहा- नोएडा ने और ग्रेटर नोएडा का विकास करेंगे और प्रदेश को नई ऊंचाइयों पर ले जाएंगे। वायर्स की समस्या पर उन्होंने कहा कि पिछली सरकार में क्या हुआ मुझे नहीं पाते लेकिन हम आवास सर्टिफिकेट के माध्यम से दे रहे हैं।
क्या है नोएडा का अंधविश्वास?
नोएडा के बारे में कहा जाता है, जब भी प्रदेश का कोई सीएम यहां आया, वह दोबारा सत्ता में नहीं लौटा। पिछले २९ सालों से अंधविश्वास की यही धारणा चली आ रही है।
मायावती ने साल २०११ में यह अंधविश्वास तोड़ने की कोशिश की थी, लेकिन वह २०१२ के चुनाव में वापस सत्ता में नहीं आईं। मायावती १४ अगस्त, २०११ को दिल्ली और नोएडा के बॉर्डर पर एक रैली की थी पर नोएडा की सीमा में एंट्री नहीं की थी।
१९८८ में सीएम वीर बहादुर सिंह ने नोएडा आने के बाद ही अपनी कुर्सी खो दी थी। कल्याण सिंह, राजनाथ सिंह और मुलायम सिंह यादव भी नोएडा जाने से परहेज करते रहे।
प्रेरणा शोध संस्थान न्यास द्वारा आयोजित सिटीजन जर्नलिज्म वर्कशॉप का समापन
नोएडा: एक्सपोर्ट कम्पनी में लगी भीषण आग
सरकारी काम में बाधा पहुँचाने पर सपा नेता गिरफ्तार
वाणिज्य कर विभाग ने जीएसटी पर किया वर्कशाप का आयोजन
माँ अहिल्या सेवा संस्था द्वारा एक भव्य भंडारे का आयोजन
बेकाबू कार नाले में गिरी, रेडियो जॉकी की मौत
गुरु पूर्णिमा एवं दक्ष प्रजापति जयंती पर फल वितरण
बोरवेल में गिरे दो मजदूर
हरियाणा-पंजाब हिंसा के बाद नोएडा पुलिस सतर्क , एसएसपी समेत पुलिस के अधिकारी गश्त पर निकले
अमरनाथ श्रद्धालुओं पर हमला का विरोध, पाक पीएम नवाज शरीफ का पुतला फूंका
अवैध शराब के साथ युवक गिरफ्तार
ग्रेटर नोएडा में दो दिवसीय युवा संगीत सम्मेलन कल से
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hindi
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आगरा: दुनिया के सात अजूबों में शामिल देश के आगरा में स्थित ताजमहल को यूपी सरकार पर नज़र अंदाज करने का आरोप है।उत्तर प्रदेश सरकार की पर्यटन गंतव्यों की बुकलेट जारी होते ही विवादों में आई गई है। सबसे चौंकाने वाली बात तो यह है कि इस बुकलेट में ताजमहल को ही शुमार नहीं किया गया है जबकि सीएम योगी आदित्यनाथ के गोरखधाम मंदिर को जगह दी गई है।
टाइम्स नाउ की खबर के मुताबिक यूपी सरकार ने जो नई बुकलेट जारी की गई है उसमें दो पेज सिर्फ गोरखधाम मंदिर को दिए गए हैं। वहीं बुकलेट का पहला पेज वाराणसी की गंगा आरती पर है और दूसरे पेज में मुख्यमंत्री योगी आदित्यनाथ और पर्यटन मंत्री रीता बहुगुणा जोशी की तस्वीर है। इसके अलावा बुकलेट में पर्यटन विकास योजनाओं के बारे में भी बताया गया है।
बता दें कि ताजमहल को लेकर योगी सरकार अक्सर सुखिऱ्यों में रहती हैं। हाल ही में सरकार ने ताजमहल को यूपी की सांस्कृतिक विरासत लिस्ट में शुमार नहीं किया था। कहा जाता है कि सीएम योगी आदित्यनाथ के ताजमहल को लेकर अलग मत है। एक मीडिया रिपोर्ट के मुताबिकए बिहार की दरभंगा रैली में योगी ने यह तक कह दिया था कि उनके लिए ताजमहल एक इमारत के सिवा और कुछ नहीं है।
पर्यटन गंतव्यों की बुकलेट में ताजमहल का नाम न आने की खबर मीडिया में आते ही योगी सरकार जागी। देर शाम यूपी सरकार ने एक प्रेस नोट जारी करते हुए कहा कि प्रो. पुअर पर्यटन योजना के तहत ताजमहल और उससे जुड़े क्षेत्र के विकास के लिए १५६ करोड़ रूपए का प्रस्ताव वल्र्ड बैंक के पास प्रस्तावित हैं। इस योजना पर अगले तीन माह में स्वीकृति अपेक्षित है।
इसके अलावा ताजमहल और आगरा किले के बीच शाहजहां पार्क और वॉक वे के पुनरुद्धार के लिए २२ करोड़ ६६ लाख रूपए की परियोजना भी सम्मलित की गई है।
जानिए क्यों इन किसानों में ली जमीन समाधि, देखिए यह तस्वीर!
मुण्डन के लिए मंदिर जा रही ट्रैक्टर ट्राली पलटी तीन की मौत, ४० घायल
अप्रैल २, २017 टॉस्न्यूज कम्मंट ऑफ ऑन प्रशांत भूषण को एक ट्विट करना पड़ा महंगा, पुलिस ने दर्ज की एफआईआर !
|
hindi
|
हैलो दोस्तो ! दोस्तो जेसे हम सभी को पता हे की हमारे इस ब्रांहंड मे कही एसी रहस्यमय चिजे छुपी हुई हे जिसका पता लगाना एक सपने के बराबर हे। तो आज हम इस आर्टिक्ल साइंटिस्ट फाउंड ब्लैक होल् नियर तो अर्थ - वैज्ञानिको ने पृथ्वी से करीब ब्लैक होल ढूंढा मे एसे ही एक रहस्य की बात करेंगे।
आज पूरी दुनिया मे साइंटिस्ट को ब्रांहंड के रहस्यो की खोज और उसे सोल्व करने मे बहोत दिलचस्पी हे और इसके लिए कही सारी स्पेस एजेन्सी काम कर रही हे। इसमे से कही रहस्यो को साइंटिस्ट भेद चुके हे ओर कहिओ को समज ने मे बहोत टाइम लग रहा हे।
उसमे से ही एक रहस्य हे ब्लैक होल अभी अभी बहोत ही चर्चा का विषय बना ब्लैक होल क्या हे? ब्लैक होल से क्या होता हे? अगर उसमे हम चले जाए तो हमारे साथ क्या होता हे? कही सारे सवाल हमारे मन मे हो रहे होगे।
रीडर्स, हमने पौराणिक कथाओ मे सुना होगा की पहेले के जमाने मे पृथ्वी मानवभाकक्षी पर दैत्य हुआ करते जिनकी भूख इतनी हुआ करती थी की वो पूरे गाव को खा जा शकते ते। ब्रांहंड मे भी कुछ एसे खगोलीय पिंड हे, जिनकी भूख उन दत्यो के समान मनी जा रही हे।
यह पिंड का गुरुत्वाकर्षण बल इतना प्रबल हे की उसकी और आने वाली किसी भी खगोलीय पिंड को अपने मे समा लेता हे। साइंटिस्ट का कहेना हे की इसके प्रभाव से प्रकाश भी नहीं बच सकता हे वो भी उसे बच नहीं सकता।
ब्रांहंड मे मोजूद ब्लैक होल नाम का यह दानव एक अति घनत्व वाला पिंड होता हे। इतने से छोटे छेत्रों मे इतना द्वयमान होता हे की उस से उत्पन्न होने वाले गुरुत्वाकर्षण बल किसी अन्य बल से बहोत ही अधिक होता हे।
इस बल के प्रभाव से सबसे तेज़ कहेने वाला प्रकाश भी नहीं बच सकता हे, प्रकाश भी उसके अंदर से पास नहीं हो सकता। इतना ही नहीं ब्लैक होल के पास समय(टाइम) भी अपना प्रभावित होता हे।
दोस्तो अंतरिक्क्ष के कही सारे तारे(स्टार) होते हे, जिन मे से वो तारे जिनका मास हमारे सूर्य से भी २० गुना अधिक होता हे इसे बड़े तारे माने जाते हे , ओर जब एसे तारे मरते हे तब ब्लैक होल बन जाते हे। इन तारो मे ऊर्जा नूक्लियर फेशन की क्रिया से उत्पन्न होती हे।
यही तरी की कोर मे हऐरोजन फ्यूज़ हो के हिलियम बनते हे, ओर येही हिलियम फ्यूज़ होके कार्बन बनाते हे ओर यह क्रिया आगे बढ़ती रहेती हे एक समय एसा आता हे की आर्यन (फ़ें) बना लेती हे।ओर यह एक बहोत ही हेवी एलिमंट्स होता हे।
अब कोर इस आर्यन(फ़ें) को फुस नहीं कर सकता हे, ओर एसे नूक्लियर फेशन की क्रिया बन्ध हो जाती हे। जिसे तारो को ब ऐलैंस मे रखने वाला फ़ोर्स, अंदर की ओर लगने वाली ग्राविटी, ओर बाहर की ओर नूक्लियर फुशन के कारण लाग्ने वाले प्रेशर बराबर मे नहीं रहे पते हे।
फुशन क्रिया बन्ध होने के बाद कोर की ग्राविटी बाहर की ओर लगने वाले प्रेशर से बहोत से बहोत ज्यादा हो जाता हे ओर इस प्रकार कोर अपने ही ग्राविटी के कारण कोल्प हो कर सिकुडना सुरू हो जाता हे ओर येही कोर बहोत की कम समय मे अरबों क्ग गॅस को अपने अंदर सामने लगता हे। ओर इसके कारण एक बहोत बड़े धमाके के साथ तारा फट जाता हे। धमाके को सुपर नोवा एक्सप्लोशन कहेते हे जो की भयंकर ऊर्जा उत्सर्जन करता हे।
यह ऊर्जा इतनी अधिक होती हे की सूर्य उसके पूरे कल मे जितनी ऊर्जा उत्पन्न करता हे उसे करीब १०० गुना अधिक होती हे। ओर इस धमाके के बाद बचता हे एक जबरजस्त ग्राविटी वाला "ब्लैक होल"।
क्या होगा गर हम ब्लैक होल मे चले जाए तो?
फ्रेंडस, यह एक डरावना ओर रहस्यमय सवाल हे की अगर हेम ब्लैक होल के अंदर चले जाए या फिर गिर जाए तो हमारा क्या होता हे? तो आइए जानते हे अब इसके बारे मे ,
जेसे की हम जानते हे की ब्लैक होल उसके ग्रविटेशनल फब्रिक मे एक बहोत ही बड़ा खड्डा बनाते हे। जब यह ब्लैक होल हमे ग्रविटेशनल फ़ोर्स से उसकी ओर खिचता हे तब एसा लगता हे की हमारी बॉडी एक स्प्रिंग की तरह काम करती हे। एसा इसलिए लगता हे क्यू की ब्लैक होल जब किसी चीज़ को अपनी तरफ खिचता हे तब ज्यादा ग्रविटेशनल फ़ोर्स उस चीज़ के निचले हिस्से पर लगता हे ईएसए ही हमारी बॉडी के साथ होता हे।
ब्लैक होल एक एसी जगह हे जहा पर टिमर स्पेस कॉल्ब्से करता हे। लेकिन क्या यह बता ता हे की हम बाल्क होल मे जाके खतम हो जाते हे? इसके लिए हमे यह जानना जरूरी हे की इसके बारे मे स्टीफन हॉकिंग ने क्या कहा था?
दोस्तो स्टीफन हॉकिंग को तो आज सब जानते हे , इनहोने अपनी पुरू उम्र अवकासी खोज (स्पेस रिसर्च) करने मे व्यतीत किया हे। ओर उनहो ने बहोत एसी खोज भी की हे जिसके काही रहस्यो को सुलजाने मे मदद मिलेगी।
दोस्तो हमने देखा की ब्लैक होल केसे बनता हे , और उसमे क्या होता हे। तो ब्लैक होल की आजूबाजू मे एक रेशन होता हे , जिसका नामे ईवेंट होरिज़न हे। यह लेयर हम नीचे इमेज मे देख सकते हे। ईवेंट होरिज़न एक ब्लैक हाल की बाउँड्री हे। जब तक हम इस बाउँड्री के बार हे तब तक हम सुरक्षित हे, लेकिन अगर हम इसे क्रॉस कर दे तो हम कभी वापस बार नहीं आ सकते।
तो इसी कारण से स्टीफन हॉकिंग ने इस पर रिसर्च चालू किया की यह आखिर काम केसे करता हे? तो उनहो ने पाया की ईवेंट होरिज़न ब्लैक होल का एक हॉलोग्राम की तरह कम करता हे। ब्लैक होल के अन्दर क्या क्या चल रहा हे इसकी इंफोरमेशन ईवेंट होरिज़न के अन्दर ही छुपी होती हे। ओर हमे बस उसे समज ना हे ताकि हम इस रहस्य को सुलजा सके।
स्टीफन हॉकिंग का कहेना हे की इस इंफोरमेशन को रेड करना बहोत ही मुसकिल काम हे लेकिन उनहो ने इसे रेड कर के एक फार्मुला दिया जिसकी मदद से हम ब्लैक होल के अन्दर की एनर्जी का पता लगा सकते हे, हमे उसके लिए सिर्फ एरिया पता होना चाहिए।
जेसे हम सब जानते हे की ए = मै^२ फॉर्मूला से हम इस ब्लैक होल का मास भी पता लगा सकते हे। ओर इसी के साथ हमे यह भी पता चल सकता हे की उसका मास ओर एनर्जी बढ़ेगी साथ ही साथ ईवेंट होरिज़न की रेडियस की बढ़ेगी
रीडर्स, तो हमने सबसे रहस्यमय कहेने वाले ब्लैक होल् के बारे मे सारी जानकारी देखि , मे उम्मीद करता हु की आपको यह जानकारी अच्छी लगी होगी ।
आगर आपको भी स्पेस(अवकास) के बारे मे एसी ही जानकारी प्राप्त करने मे रुचि हे तो हमारे इस वेबसाइट पर विजिट करते रहे।
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hindi
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/*****************************************************************************
* Project: RooFit *
* Package: RooFitCore *
* File: $Id: RooAbsLValue.h,v 1.12 2007/05/11 09:11:30 verkerke Exp $
* Authors: *
* WV, Wouter Verkerke, UC Santa Barbara, [email protected] *
* DK, David Kirkby, UC Irvine, [email protected] *
* *
* Copyright (c) 2000-2005, Regents of the University of California *
* and Stanford University. All rights reserved. *
* *
* Redistribution and use in source and binary forms, *
* with or without modification, are permitted according to the terms *
* listed in LICENSE (http://roofit.sourceforge.net/license.txt) *
*****************************************************************************/
#ifndef ROO_ABS_LVALUE
#define ROO_ABS_LVALUE
#include <list>
#include <string>
#include "Rtypes.h"
class RooAbsBinning ;
class RooAbsLValue {
public:
// Constructors, cloning and assignment
RooAbsLValue() ;
virtual ~RooAbsLValue();
virtual void setBin(Int_t ibin, const char* rangeName=0) = 0 ;
virtual Int_t getBin(const char* rangeName=0) const = 0 ;
virtual Int_t numBins(const char* rangeName=0) const = 0 ;
virtual Double_t getBinWidth(Int_t i, const char* rangeName=0) const = 0 ;
virtual Double_t volume(const char* rangeName) const = 0 ;
virtual void randomize(const char* rangeName=0) = 0 ;
virtual const RooAbsBinning* getBinningPtr(const char* rangeName) const = 0 ;
virtual std::list<std::string> getBinningNames() const = 0;
virtual Int_t getBin(const RooAbsBinning*) const = 0 ;
protected:
ClassDef(RooAbsLValue,1) // Abstract variable
};
#endif
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code
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We are looking for exceptional people because our projects and products are exceptional. That is why our colleagues are talented and competitive. Both initiative and motivation are required to work in our dynamic and energetic team. When people can enjoy what they do, they find the personal motivation that leads to professional growth. Optimissa promotes integrity and transparency in the workplace. We work better when we work in teams and we support a culture of participation differentiated by problem-solving through collaboration.
We are part of a team where excellence is cultivated and rewarded. Advancement depends on talent and motivation, as well as on the quality of the service you provide. If you are able to add value, new doors will open and there will be new opportunities to boost your career progress. At Optimissa, effort and merit are rewarded.
We hire the best, and we help our employees to better themselves; that is why we empower them with training, both in-house and through external providers. We also use web-based tools, in order to leverage the benefits offered by distance learning. We know that our team’s success becomes our clients’ success. We firmly believe that our people’s skills and experience are the added-value that differentiates us.
You can read more about the opportunities we offer in a rapidly-growing company, where talent and performance are rewarded, on our Careers and vacancies website.
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english
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class Hash
# Standard in Ruby 1.8.8. See official documentation[http://ruby-doc.org/core-1.9/classes/Hash.html]
class << self
def try_convert(x)
return nil unless x.respond_to? :to_hash
x.to_hash
end unless method_defined? :try_convert
end
# Standard in Ruby 1.9. See official documentation[http://ruby-doc.org/core-1.9/classes/Hash.html]
def default_proc=(proc)
replace(Hash.new(&Backports.coerce_to(proc, Proc, :to_proc)).merge!(self))
end unless method_defined? :default_proc=
# Standard in Ruby 1.9. See official documentation[http://ruby-doc.org/core-1.9/classes/Hash.html]
def assoc(key)
val = fetch(key) do
return find do |k, v|
[k, v] if k == key
end
end
[key, val]
end unless method_defined? :assoc
# Standard in Ruby 1.9. See official documentation[http://ruby-doc.org/core-1.9/classes/Hash.html]
def rassoc(value)
k = key(value)
v = fetch(k){return nil}
[k, fetch(k)] if k || v == value
end unless method_defined? :rassoc
end
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code
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/* ==== index
##. ER Wait Time
##.
============= */
/* === ER Wait Time === */
nv.addGraph(function() {
var chart = nv.models.multiBarChart()
.transitionDuration(350)
.reduceXTicks(true) //If 'false', every single x-axis tick label will be rendered.
.rotateLabels(0) //Angle to rotate x-axis labels.
.showControls(false) //Allow user to switch between 'Grouped' and 'Stacked' mode.
.groupSpacing(0.5) //Distance between each group of bars.
;
chart.xAxis
.tickFormat(d3.format(',f'));
chart.yAxis
.axisLabel('Bookings $k')
.tickFormat(d3.format(',.1f'));
d3.select('#chart1 svg')
.datum(exampleData())
.call(chart);
nv.utils.windowResize(chart.update);
return chart;
});
//Generate some nice data.
function exampleData() {
return stream_layers(2,10+Math.random()*100,12).map(function(data, i) {
return {
key: 'Debt' + i+2,
values: data
};
});
}
/* Inspired by Lee Byron's test data generator. */
function stream_layers(n, m, o) {
if (arguments.length < 3) o = 0;
function bump(a) {
var x = 1 / (.1 + Math.random()),
y = 2 * Math.random() - .5,
z = 10 / (.1 + Math.random());
for (var i = 0; i < m; i++) {
var w = (i / m - y) * z;
a[i] += x * Math.exp(-w * w);
}
}
return d3.range(n).map(function() {
var a = [], i;
for (i = 0; i < m; i++) a[i] = o + o * Math.random();
for (i = 0; i < 5; i++) bump(a);
return a.map(stream_index);
});
}
/* Another layer generator using gamma distributions. */
function stream_waves(n, m) {
return d3.range(n).map(function(i) {
return d3.range(m).map(function(j) {
var x = 20 * j / m - i / 3;
return 2 * x * Math.exp(-.5 * x);
}).map(stream_index);
});
}
function stream_index(d, i) {
return {x: i, y: Math.max(0, d)};
}
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کیا ژٕ ہیٚککھا مےٚ ۲۰۱۳۲۰۱۴ تعلٟمی ورۍ یَس مَنٛز ۷ ساروی کھوتہٕ تھٔد منظور شدٕ اِدارن ہِنٛز لسٹ حٲصِل کٔرتھ
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kashmiri
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english
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\begin{equation}gin{document}
\tildetle{On a problem of Yau regarding a higher dimensional generalization of the
Cohn-Vossen inequality}
\author{Bo Yang}
\address{Department of Mathematics, University of California San Diego, La Jolla, CA 92093}
\email{{[email protected]}}
\begin{equation}gin{abstract}
We show that a problem by Yau in \cite{Yau} can not be true in
general. The counterexamples are constructed based on the recent
work of Wu and Zheng \cite{WZ}.
\end{abstract}
\title{On a problem of Yau regarding a higher dimensional generalization of the
Cohn-Vossen inequality}
\section{Introduction}
Shing-Tung Yau asked the following question in \cite{Yau}:
\begin{equation}gin{question} \label{problem 1.1} Given an n-dimensional complete manifold
with nonnegative Ricci curvature, let $B(r)$ be the geodesic ball
around some point $p$. Let $\sigmama_{k}$ be the $k$-th elementary symmetric
function of the Ricci tensor. Then is it true that $r^{-n+2k}
\int_{B(r)} \sigmama_{k}$ has an upper bound when $r$ tends to
infinity? This should be considered as a generalization of the
Cohn-Vossen inequality.
\end{question}
\vskip 1mm
In the K\"{a}hler category one would like to ask the following
similar question.
\begin{equation}gin{question} \label{Kahler category} On a complete K\"{a}hler manifold
with complex dimension $n$, if we denote $\omegaega$ and Ric the
K\"{a}hler form and the Ricci form respectively, one would like to
ask if $r^{-2n+2k} \int_{B(r)} Ric^{k} \wedge \omegaega^{n-k}$ is
bounded for any $1 \leq k \leq n$ when $r$ goes to infinity.
\end{question}
In this note we exhibit counterexamples to Question \mathrm{Re}f{problem
1.1} in the case of $1<k \leq n$ via the recent interesting work of
Wu and Zheng \cite{WZ}. We will show that for any complex dimension
$n \geq 2$ and any $2 \leq k < n$, there exists a $U(n)$ invariant
complete K\"{a}hler metrics on $\mathbb{C}^{n}$ with nonnegative
bisectional curvature such that $r^{-2n+2k} \int_{B(r)} \sigmama_{k}$
is unbounded when $r$ large (See Theorem \mathrm{Re}f{question 1.1 not
true}). We also prove that Question \mathrm{Re}f{Kahler category} is true
for all $U(n)$ invariant complete K\"{a}hler metrics on
$\mathbb{C}^{n}$ with nonnegative bisectional curvature (See Theorem
\mathrm{Re}f{question 1.2 true}).
\markright{On a problem of Yau}
\section{Results of Wu and Zheng}
In the section, we collect some of the results from the recent work of
Wu and Zheng \cite{WZ} since they will be used in our constructions of
counterexamples to Question \mathrm{Re}f{problem 1.1}. Unless stated otherwise all
results in this section are due to Wu and Zheng \cite{WZ}.
Wu and Zheng \cite{WZ} develops a systematic way
to construct $U(n)$ invariant complete K\"{a}hler metrics on
$\mathbb{C}^{n}$ with positive bisectional curvature. One of the
motivation behind their work is the uniformization conjecture by Yau
\cite{Yau2}. The conjecture states that a complete noncompact
K\"{a}hler manifold with positive bisectional curvature is
biholomorphic to the complex Euclidean space. See
\cite{ChenTangZhu},\cite{ChauTam1}, \cite{ChauTam2}, \cite{ChauTam3}
and reference therein for some recent progress towards Yau's
uniformization conjecture. See also \cite{K},\cite{Cao1}, and
\cite{Cao2} for some earlier works on the construction of
rotationally symmetric complete K\"{a}hler metrics with positive
curvature on $\mathbb{C}^{n}$.
We follow the notations in \cite{WZ}. Let $z=(z_1,\cdots,z_n)$ be the standard
coordinate on $\mathbb{C}^{n}$ and $r=|z|^{2}$. A $U(n)$ invariant
K\"{a}hler metric on $\mathbb{C}^{n}$ has the K\"{a}hler form
\begin{equation}gin{equation}
\omegaega=\frac{\sqrt{-1}}{2} \partial \overline{\partial} p(r)
\label{Kahler form}
\end{equation} where $p \in C^{\infty}
[0,+\infty)$. Under the local coordinates, the metric has
components:
\begin{equation}gin{equation}
g_{i\overline{j}}=f(r)\delta_{ij}+f'(r) \overline{z}_i z_j.
\end{equation}
We further denote:
\begin{equation}gin{equation}
f(r)=p'(r),\ \ \ \ h(r)=(rf)'. \label{def of f and h}
\end{equation}
Then the K\"{a}hler from $\omegaega$ will give a complete metric if and
only if
\begin{equation}gin{equation}
f>0, \ \ h>0, \ \ \int_{0}^{+\infty} \frac{\sqrt{h}}{\sqrt{r}}
dr=+\infty. \label{being complete}
\end{equation}
Now if we compute the components of the curvature tensor at
$(z_1,0,\cdots,0)$ under the orthonormal frame $\{
e_1=\frac{1}{\sqrt{h}} \partial_{z_1},e_2=\frac{1}{\sqrt{f}}
\partial_{z_2}, \cdots, e_n=\frac{1}{\sqrt{f}} \partial_{z_n} \}$,
then define $A, B, C$ respectively by:
\begin{equation}gin{equation}
A=R_{1\overline{1}1\overline{1}}=-\frac{1}{h} (\frac{rh'}{h})',\
B=R_{1\overline{1}i\overline{i}}=\frac{f'}{f^2}-\frac{h'}{hf},\
C=R_{i\overline{i}i\overline{i}}=2R_{i\overline{i}j\overline{j}}=-\frac{2f'}{f^2},
\label{ABC one form}
\end{equation} where we assume $2 \leq i \neq j \leq n$,
It is easy to check all other components of curvature tensor are
zero.
Let $\mathcal{M}_{n}$ denote the space of all $U(n)$ invariant
complete K\"{a}hler metrics on $\mathbb{C}^{n}$ with positive
bisectional curvature.
\begin{equation}gin{theorem}[\textbf{Characterization of $\mathcal{M}_{n}$ by the $ABC$
function}] Suppose $n \geq 2$ and h is a smooth positive function
on $[0,+\infty)$ satisfying (\mathrm{Re}f{being complete}), then the form
defined by (\mathrm{Re}f{Kahler form}) gives a complete K\"{a}hler metric
with positive (nonnegative) bisectional curvature if and only if
$A,B,C$ are positive (nonnegative).
\end{theorem}
If we define another function $\xi \in C^{\infty} [0,+\infty)$ by
\begin{equation}gin{equation}
\xi(r)=-\frac{r h'(r)}{h}, \label{def of xi}
\end{equation} then $h$ determines $\xi$ uniquely. On the other hand,
note that $\xi$ determines $h$ by $h(r)=h(0) e^{\int_{0}^{r}
\frac{\xi(t)}{t} dt}$, hence $\omegaega$ up to scaling. The following
interesting theorem in \cite{WZ} reveals that the space
$\mathcal{M}_{n}$ is in fact quite large.
\begin{equation}gin{theorem}[\textbf{Characterization of $\mathcal{M}_{n}$ by the $\xi$
function}] Suppose $n\geq 2$ and h is a smooth positive function on
$[0,+\infty)$, then the form defined by (\mathrm{Re}f{Kahler form}) gives a
complete K\"{a}hler metric with positive bisectional curvature on
$\mathbb{C}^n$ if and only if $\xi$ defined by (\mathrm{Re}f{def of xi})
satisfying \begin{equation}gin{equation} \xi(0)=0,\ \ \ \xi^{\prime}>0,\ \ \ \xi<1.
\end{equation}
\end{theorem}
Fix a metric $\omegaega$ in $\mathcal{M}_{n}$, the geodesic distance
between the origin and a point $z \in \mathbb{C}^{n}$ is:
\begin{equation}gin{equation}
s=\int_{0}^{r} \frac{\sqrt{h}}{2\sqrt{r}} dr. \label{s distance}
\end{equation} where $r=|z|^2$. We denote $B(s)$ the ball in
$\mathbb{C}^n$ centered at the origin and with the radius $s$ with
respect to $\omegaega$. It is further shown in \cite{WZ} that:
\begin{equation}gin{equation}
\operatorname{Vol} (B(s))=c_n (rf)^n. \label{vol formula}
\end{equation} where $c_n$ is the Euclidean volume of the Euclidean
unit ball in $\mathbb{C}^n$.
Using Theorem 2.2 Wu and Zheng further proved the following
estimates on volume growth of geodesics ball $B(s)$ and the first
Chern number for metrics in $\mathcal{M}_{n}$. Note that an estimate
on volume growth of geodesics ball in the general case has been
proved by Chen and Zhu \cite{ChenZhu2}.
\begin{equation}gin{proposition}[\textbf{Volume growth estimates for metrics in
$\mathcal{M}_{n}$}] $r f = f(1)+2\sqrt{h(1)} (s-s(1))$ for $r>1$ and
$rf \leq s^2$ for any $r \geq 0$. So there exists a constant $C$
such that:
\begin{equation}gin{equation} C s^{n} \leq \operatorname{Vol}(B(s)) \leq c_n s^{2n}.
\end{equation} for $s$ large enough.
\end{proposition}
\begin{equation}gin{proposition}[\textbf{Bounding the first Chern number for metrics in
$\mathcal{M}_{n}$}] Given any $\omegaega$ in $\mathcal{M}_{n}$ with $n
\geq 1$, we have
\begin{equation}gin{equation} \int_{C^{n}} (Ric)^n=c_n (\frac{n
\xi(+\infty)}{\pi})^n \leq c_n (\frac{n}{\pi})^{n}.
\end{equation} while $Vol(B(s))=c_n v^n$.
\end{proposition}
In order to construct more examples and compute the scalar curvature
curvature of metrics in $\mathcal{M}_{n}$ in a more convenient way,
Wu and Zheng \cite{WZ} introduced another function $F$ in the
following way: First we define $x=\sqrt{rh}$ and a nonnegative
function $y$ of $r$ by
\begin{equation}gin{equation}
y(0)=0, \ \ \ \ \ {x^{\prime}}^{2}+{y^{\prime}}^{2}=\frac{h}{4r}, \ \ \ \ y' > 0. \label{def of y}
\end{equation}
One can check that $x(r)$ is strictly increasing and then we may
define $F(x)$ a function on $[0,x_0)$ by $y=F(x)$, where
\begin{equation}gin{equation}
x_0^2=\lim_{r \rightarrow +\infty} rh=h(1)e^{\int_{1}^{+\infty}
\frac{1-\xi}{r} dr}. \label{def of x0}
\end{equation}
Extending $F$ to $(-x_0,x_0)$ by letting $F(x)=F(-x)$, one can check
that $F$ is a smooth, even function on $|x|<x_0$. Starting with such
a $F$ satisfying certain conditions, one can recover the metric
$\omegaega$ in a geometric way. See Section 5 in \cite{WZ} for details.
This result is summarized as the following theorem.
\begin{equation}gin{theorem} [\textbf{Characterization of
$\mathcal{M}_{n}$ by the $F$ function}] Suppose $n\geq 1$, there is
a one to one correspondence of between the set $\mathcal{M}_{n}$ and
the set of $\mathcal{F}$ of smooth, even function $F(x)$ defined on
$(-x_0,x_0)$ satisfying
\begin{equation}gin{equation} F(0)=0,\ \ \ \ F''>0,\ \ \ \ \lim_{x \rightarrow x_0} F(x)=+\infty.
\end{equation}
\end{theorem}
\vskip 1mm
Denote $v=rf$, one can rewrite $s$ and $\operatorname{Vol}(B(s))$ in terms of $F$:
\begin{equation}gin{equation}
s=\int_{0}^{x} \sqrt{1+(F'(\tau))^2} d \tau, \ \ \ \operatorname{Vol}(B(s))=c_n
v^n=c_n (\int_{0}^{x} 2\tau \sqrt{1+(F'(\tau))^2} d\tau)^n.
\label{distance and volume}
\end{equation}
Rewrite $A$, $B$, and $C$ defined in (\mathrm{Re}f{ABC one form}) in terms
of F:
\begin{equation}gin{equation}
A=\frac{F' F''}{2x(1+{F'}^2)^2}, \ \ \
B=\frac{x^2}{v^2}-\frac{1}{v\sqrt{1+{F'}^2}},\ \ \
C=\frac{2}{v}-\frac{2x^2}{v^2}. \label{ABC another form}
\end{equation}
Recall the scalar curvature at the point $z=(z_1,0,\cdots,0)$ is
given by
\begin{equation}gin{equation}
R=A+2(n-1)B+\frac{1}{2}n(n-1)C. \label{scalar curvature}
\end{equation}
Using (\mathrm{Re}f{scalar curvature}),(\mathrm{Re}f{ABC another form}),
(\mathrm{Re}f{distance and volume}) and a careful integration by parts, Wu
and Zheng \cite{WZ} proved the following relation between average
scalar curvature decay and volume growth of geodesic balls. See also
\cite{ChenZhu2} for a related result on any complete K\"{a}hler
manifold with positive bisectional curvature.
\vskip 1mm
\begin{equation}gin{proposition}
[\textbf{Estimates on average scalar
curvature for metrics in $\mathcal{M}_{n}$}]
Given any K\"{a}hler metric $\omegaega$ in $\mathcal{M}_{n}$ with $n
\geq 2$, there exists a constant $c>0$ such that \begin{equation}gin{equation}
\frac{1}{c(1+v)} \leq \frac{1}{\operatorname{Vol}(B(s))} \int_{B(s)} R(s) w^n \leq
\frac{c}{1+v}.
\end{equation} while $\operatorname{Vol}(B(s))=c_n v^n$.
\end{proposition}
\section{Counterexamples to Question 1.1}
Let $\overline{\mathcal{M}}_{n}$ denote the space of all $U(n)$
invariant complete K\"{a}hler metrics on $\mathbb{C}^{n}$ with
nonnegative bisectional curvature. First we state a generalization
of Theorem 2.2 to the space $\overline{\mathcal{M}}_{n}$.
\begin{equation}gin{proposition} [\textbf{Characterization of $\overline{\mathcal{M}}_{n}$ by
the $\xi$ function}] \label{xi function} Suppose $n\geq 2$ and h is
a smooth positive function on $[0,+\infty)$, then the form defined
by (\mathrm{Re}f{Kahler form}) gives a complete K\"{a}hler metric with
nonnegative bisectional curvature if and only if $\xi$ defined by
(\mathrm{Re}f{def of xi}) satisfying
\begin{equation}gin{equation} \xi(0)=0,\ \ \ \xi' \geq 0,\ \ \ \xi \leq 1.
\end{equation}
\end{proposition}
\begin{equation}gin{proof}[Proof of Proposition \mathrm{Re}f{xi function}]
The original proof of Theorem 2.2 due to Wu and Zheng \cite{WZ}
works here. Now we only sketch the necessary part. First from (\mathrm{Re}f{def of xi}) we know $\xi(0)=0$.
Note that (\mathrm{Re}f{def of xi}) and Theorem 2.1 imply
\begin{equation}gin{equation}
A=\frac{\xi^{\prime}}{h} \geq 0
\end{equation} which leads to $\xi^{\prime} \geq 0$.
To prove $\xi \leq 1$, argument by contradiction as in \cite{WZ}. Assume $\lim_{r \rightarrow +\infty}=b>1$, then take
$\delta_{0}>0$ such that $1+\delta_{0}<b$. It follows that there exists $r_0>0$ with $\xi(r_0) \geq 1+\delta_{0}$. Thus
integrating (\mathrm{Re}f{def of xi}) leads to
$h(r)=h(0) \exp{\int_{0}^{r} \frac{\xi}{r} dr} \leq \frac{c}{r^{1+\delta_0}} $ which contradicts to the completeness of the metric (\mathrm{Re}f{being complete}).
\end{proof}
It also follows from the original proof of Proposition 2.3 and 2.4 due to Wu and Zheng
that the same conclusion holds for the space
$\overline{\mathcal{M}}_{n}$. Namely, for any metric $\omegaega$ in
$\overline{\mathcal{M}}_{n}$, $C s^{n} \leq \operatorname{Vol}(B(s)) \leq c_n
s^{2n}$ holds for $s$ sufficiently large. and $\int_{C^{n}} (Ric)^n
\leq c_n (\frac{n}{\pi})^{n}$ is true. We remark here that the
estimate on lower bounds of the volume growth of $B(s)$ here can not
be true for an arbitrary complete noncompact K\"{a}hler manifolds
with nonnegative bisectional curvature. For example, take
$\Sigma_{1} \tildemes \mathbb{CP}^{1} \tildemes \cdots \tildemes
\mathbb{CP}^{1}$ where $\Sigma_{1}$ is a capped cylinder on one end
and $\mathbb{CP}^{1}$ is the complex projective plane with the
standard metric.
Next we state another generalization of Theorem 2.5 to
$\overline{\mathcal{M}}_{n}$.
\vskip 1mm
\begin{equation}gin{theorem} [\textbf{Characterization of $\overline{\mathcal{M}}_{n}$
by the $F$ function}] \label{F function} Suppose $n\geq 1$, there
is a partition of the set $\overline{\mathcal{M}}_{n} \setminus
{\{g_{e}\}}=S_1 \cup S_2 \cup S_3$ where $g_{e}$ is the standard
Euclidean metric on $\mathbb{C}^n$ such that:
(1) $S_1$ has a one to one correspondence with the set of
$\mathcal{F}$ of smooth, even function $F(x)$ on $(-\infty,+\infty)$
defined above satisfying
\begin{equation}gin{equation} F(0)=F'(0)=0,\ \ \ \ F'' \geq 0,\ \ \ \ F'(\infty)< +\infty, \ \ \ F(\infty) = +\infty.
\end{equation}
$S_1$ consists of nonflat K\"{a}hler metrics in
$\overline{\mathcal{M}}_{n}$ whose geodesic balls have Euclidean
volume growth.
(2) $S_2$ has a one to one correspondence with the set of
$\mathcal{F}$ of smooth, even function $F(x)$ on $(-x_0,x_0)$ (where
$x_0$ is either finite or $+\infty$) satisfying
\begin{equation}gin{equation} F(0)=F'(0)=0,\ \ \ \ F'' \geq 0,\ \ \ \ \ F'(x_0)=F(x_0)=+\infty.
\end{equation}
$S_2$ includes K\"{a}hler metrics in
$\overline{\mathcal{M}}_{n}$ whose geodesic balls
have strictly less than Euclidean volume growth and bisectional
curvatures in the radial direction strictly positive along a
sequence of points in $\mathbb{C}^n$ tending to infinity.
(3) For any metric $\omegaega \in S_3$, there exists a positive real
number $r_0$ such that $r_0=\inf \{ r: \xi(r)=1 \} $ and a
corresponding positive real number $x_0$ such that there exists a
smooth even function $F(x)$ defined on $(-x_0,x_0)$ such that
\begin{equation}gin{equation} F(0)=F'(0)=0,\ \ \ \ F'' \geq 0,\ \ \ \ \
F'(x_0)=+\infty,\ \ \ \ F(x_0)<\infty,
\end{equation}
$S_3$ is
the set of metrics with geodesic balls having half Euclidean volume
growth and whose bisectional curvatures in the radial direction
vanish outside a compact set. A standard example in complex dimension 1 is a capped cylinder on one end.
\end{theorem}
\begin{equation}gin{proof}[Proof of Theorem \mathrm{Re}f{F function}]
The proof of Theorem \mathrm{Re}f{F function} is based on a modification of
Theorem 2.5 due to Wu and Zheng. From Proposition \mathrm{Re}f{xi function},
we know for any K\"{a}hler metric in $\overline{\mathcal{M}}_{n}$,
there exists a corresponding $\xi(r)$ on $[0,+\infty)$ with
$\xi(0)=0, \xi' \geq 0$, and $\xi \leq 1$. Denote $r_0=\inf {\{ r:
\xi(r)=1 \}}$.
Recall the definition of $x$ and $y$ in (\mathrm{Re}f{def of y}), $x=\sqrt{rh}$ and ${x^{\prime}(r)}^2+{y^{\prime}(r)}^2=\frac{h}{4r}$ with $y(0)=0$ and $y^{\prime} \geq 0$. It is easy to check:
\begin{equation}gin{equation}
\frac{dx}{dr}=(1-\xi) \sqrt{\frac{h}{4r}}, \label{dx wrt dr}
\end{equation} then we know $x(r)$ and $y(r)$ are both nondecreasing with respect to $r$.
\textbf{(Case I)} $r_0=+\infty$. From the definition of $x_0$ in (\mathrm{Re}f{def of x0}) and (\mathrm{Re}f{dx wrt dr})
we know $x(r)$ is strictly increasing on $[0,+\infty)$, then we can define $F(x)$ by $y=F(x)$ on $x \in (-x_0,x_0)$ after an even extension by letting F(-x)=F(x). It is not hard to see that
\begin{equation}gin{equation}
F(0)=0,\ \ \ F'(x) \geq 0, \ \ \ 1+[F'(x)]^2=\frac{1}{(1-\xi)^2}. \label{about F}
\end{equation}
Recall that $0 \leq \xi(r) \leq 1$ is nondecreasing on
$(-\infty,+\infty)$, we conclude that $F^{\prime \prime} \geq 0$.
Moreover, (\mathrm{Re}f{dx wrt dr}) and (\mathrm{Re}f{about F}) implies:
\begin{equation}gin{eqnarray}
\lim_{x \rightarrow x_0} F(x) &=&\int_{0}^{x_0} \sqrt{\frac{1}{(1-\xi)^2}-1} dx \label{F(x0)}\\
&=& \int_{0}^{+\infty} \sqrt{1-(1-\xi)^2} \sqrt{\frac{h}{4r}} dr \nonumber \\
&\geq& \sqrt{1-(1-\xi(+\infty))^2} \int_{0}^{+\infty} \sqrt{\frac{h}{4r}} dr. \nonumber
\end{eqnarray}
Note that the integral in the last step of (\mathrm{Re}f{F(x0)}) is distance function (\mathrm{Re}f{s distance}). we conclude $F(x_0)=\infty$
if and only if $\xi(+\infty)>0$. Note that the latter condition is satisfied when $\omegaega$ is nonflat.
We further divide our discussion into two subcases:
\textbf{(Subcase Ia)} $0<\xi(+\infty)<1$. In this case we have $F^{\prime}$ is bounded on $(-x_0,x_0)$ and $x_0=+\infty$.
Moreover, we will prove that the geodesic balls of $(\mathbb{C}^n, \omegaega)$ has
Euclidean volume growth. We follow the method of Wu and Zheng (See P528 of \cite{WZ}). Note that (\mathrm{Re}f{s distance}),(\mathrm{Re}f{vol formula}), $(rf)'(r)=h$ and $(rh)'(r)=h(1-\xi)$, it follows from the
L'Hospital's rule that:
\begin{equation}gin{eqnarray}
\lim_{s \rightarrow +\infty} \frac{\operatorname{Vol} (B(s))}{s^{2n}} &=& \lim_{r \rightarrow +\infty} \frac{c_n (rf)^n}{s^{2n}} \label{volume growth} \\
&=&\lim_{r \rightarrow +\infty} c_n (\frac{ \sqrt{rf} }{s})^{2n} \nonumber \\
&=& c_n (1-\xi(+\infty))^{4n} \nonumber
\end{eqnarray}
\textbf{(Subcase Ib)} $\xi(+\infty)=1$, It follows from the (\mathrm{Re}f{volume growth}) that in this case the geodesic balls of $(\mathbb{C}^n, \omegaega)$ has strictly less than Euclidean volume growth. Since $A=\frac{\xi^{\prime}}{h}$, $\xi(0)=0$, and $\xi(+\infty)=1$ we also know that bisectional
curvatures in the radial direction strictly positive along at least a sequence of points in $\mathbb{C}^n$ tending to infinity.
\vskip 1mm
\textbf{(Case II)} $r_0>0$ is finite. Note that $(rh)'=h(1-\xi)$, we
conclude that $x_0=\lim_{r \rightarrow +\infty} \sqrt{rh}$ is finite
and $x_0^2=r_0 h(r_0)$. This implies that $F(x)$ is well defined on
$(-x_0,x_0)$ with $F(x_0)<+\infty$. Since $A=\frac{\xi^{\prime}}{h}$
we conclude that bisectional curvatures in the radial direction
vanishes outside a compact set in $\mathbb{C}^n$. Next we proceed to
show that the geodesic balls of $(\mathbb{C}^n, \omegaega)$ has half
Euclidean volume growth. Again the methods follows from Wu and Zheng
(See P528 of \cite{WZ}).
\begin{equation}gin{eqnarray}
\lim_{s \rightarrow +\infty} \frac{\operatorname{Vol} (B(s))}{s^{n}} &=& \lim_{r \rightarrow +\infty} c_n (\frac{ rf }{s})^{n} \label{half volume growth}\\
&=& \lim_{r \rightarrow +\infty} c_n (2 \sqrt{rh})^{n} \nonumber \\
&=& 2c_n x_0. \nonumber
\end{eqnarray}
Denote $S_1$, $S_2$, and $S_3$ the sets of metrics in the above three cases (Subcase Ia), (Subcase Ib), and (Case II) respectively, we have proved Theorem \mathrm{Re}f{F function}.
\end{proof}
Next we gives some more explicit description of $S_3$. Given any metric
$\omegaega$ in $S_3$, $\frac{d (rh)}{dr}=(1-\xi) h$, $h=(rf)'$ and $\xi(r)=1$ when
$r>r_0$, then:
\begin{equation}gin{equation} rf |_{r_0}^{r}=\int_{r_0}^{r} \frac{r_0 h(r_0)}{r}
dr,
\end{equation} which further implies:
\begin{equation}gin{equation} rf=x_0^2 \ln{\frac{r}{r_0}}+r_0 f(r_0).
\label{rf}
\end{equation}
Now we compute A, B, C with (\mathrm{Re}f{ABC one form}) in Section 2 when $r \geq r_0$:
\begin{equation}gin{equation}
A=-\frac{1}{h} (\frac{rh'}{h})'=\frac{{\xi}'}{h}=0, \label{A for S3}
\end{equation}
\begin{equation}gin{eqnarray}
B &=& \frac{f'}{f^2}-\frac{h'}{h f}=\frac{1}{r}
(\frac{(rf)'-f}{f^2}-\frac{r h'}{h f}) \label{B for S3} \\ &=& \frac{1}{r}
(\frac{h}{f^2}-\frac{1-\xi}{f})=\frac{h}{r f^2}=\frac{x_0^2}{r^2
f^2}. \nonumber
\end{eqnarray}
\begin{equation}gin{eqnarray}
C=-\frac{2f'}{f^2}=(-2) \frac{h-f}{r f^2} =2 \frac{r f-rh}{(r
f)^2}=2 \frac{x_0^2 (\ln{\frac{r}{r_0}}-1)+r_0 f(r_0) }{[x_0^2
\ln{\frac{r}{r_0}}+r_0 f(r_0)]^2} \label{C for S3}
\end{eqnarray}
We also see the distance function for metrics in $S_3$:
\begin{equation}gin{equation}
s(r)=\int_{0}^{r_0} \sqrt{\frac{h}{4r}} dr+ \frac{x_0}{2} \ln{\frac{r}{r_0}}. \label{s for S3}
\end{equation}
Now one can estimate the average of $A$, $B$, and $C$ inside $B(s)$
from (\mathrm{Re}f{scalar curvature}), (\mathrm{Re}f{A for S3}),(\mathrm{Re}f{B for
S3}),(\mathrm{Re}f{C for S3}), (\mathrm{Re}f{vol formula}), and (\mathrm{Re}f{s for S3}). Namely, if $n \geq 2$, for any metric in $S_3$ there
exists a constant $c$ such that
\begin{equation}gin{equation}
\frac{1}{c \, rf} \leq \frac{1}{\operatorname{Vol}(B(s))}
\int_{B(s)} R \, {\omegaega}^{n} \leq \frac{c}{rf},
\end{equation} where $\operatorname{Vol}(B(s))=c_n (rf)^n$.
If $\omegaega$ is a nonflat K\"{a}hler metric in $S_1 \cup S_2$, we see from Theorem
\mathrm{Re}f{F function} that $F$ must have $F'(x_0)>0$. Then the formula of
$A$, $B$, and $C$ in terms of $F$ is exactly the same as (\mathrm{Re}f{ABC
another form}) derived in \cite{WZ} (See P536 in \cite{WZ}). Follow the proof of Proposition
2.6 in \cite{WZ}, we get the same conclusion. We summarize the above discussion
as the following result. Note that $f$ is defined in (\mathrm{Re}f{def of f and h}).
\begin{equation}gin{proposition} \label{average scalar
curvature decay} When $n \geq 2 $, given any non flat metric in $
\overline{\mathcal{M}}_{n}$, there exists a constant $C>0$ such that
\begin{equation}gin{equation} \frac{1}{c(1+rf)} \leq \frac{1}{\operatorname{Vol}(B(s))}
\int_{B(s)} R \, w^n \leq \frac{c}{1+rf}.
\end{equation} where $\operatorname{Vol}(B(s))=c_n (rf)^n$.
\end{proposition}
Now we state the main theorem of this note.
\begin{equation}gin{theorem} \label{question 1.1 not
true}
Given any $n \geq 2 $, any nonflat K\"{a}hler metric in
$\overline{\mathcal{M}}_{n}$ has $\int_{B(s)} \sigmama_{n} {\omegaega}^n$
unbounded when $s$ goes to infinity. Moreover, if $2 \leq k<n$ one
can construct a complete K\"{a}hler metric $\omegaega$ from $S_1
\subset \overline{\mathcal{M}}_{n}$ with bounded curvature on
$\mathbb{C}^n$ such that $\frac{1}{s^{2n-2k}} \int_{B(s)} \sigmama_k
\, {\omegaega}^n$ is unbounded when $s$ tends to infinity.
\end{theorem}
\begin{equation}gin{proof}[Proof of Theorem \mathrm{Re}f{question 1.1 not
true}]
It follows from (\mathrm{Re}f{ABC one form}) that for any metric in
$\overline{\mathcal{M}}_{n}$ we have Ricci curvature at $z$ given
by:
\begin{equation}gin{equation}
\lambdabda=R_{1\overline{1}}=A+(n-1)B,\
\mu=R_{i\overline{i}}=B+\frac{n}{2}C \ \ \ 2 \leq i \leq n.
\end{equation} Note that we are now working on the K\"{a}hler manifolds
$\mathbb{C}^n$ and the Ricci tensor is $J$-invariant where $J$ is
the standard complex structure on $\mathbb{C}^n$. Therefore the
Ricci tensor in the real case has eigenvalue $\lambdabda$ of
multiplicity 2 and $\mu$ of multiplicity $2n-2$. From now on, let
$\sigmama_{k}$ denote the $k$-th elementary symmetric function of the
Ricci curvature tensor.
First note that Question 1.1 are true for any metric $\omegaega \in
\overline{\mathcal{M}}_{n}$ when $k=1$. Since $\sigmama_1=2 R$ where
$R$ is the scalar curvature in the K\"{a}hler case, it follows from
Proposition \mathrm{Re}f{average scalar curvature decay} and the upper bound
of the volume growth of $B(s)$ after Proposition \mathrm{Re}f{xi function}
that $\frac{1}{s^{2n-2}} \int_{B(s)} R\, {\omegaega}^n$ is bounded when
$r$ tending to infinity. Therefore, we focus on Question 1.1 in the
case of $2 \leq k \leq n$, If $2 \leq k \leq n$, $\sigmama_{k}$ of
the Ricci tensor is a linear combination of ${\lambdabda}^{2}
{\mu}^{k-2}$, $\lambdabda {\mu}^{k-1}$, and ${\mu}^{k}$. To sum up,
$\sigmama_{k}$ is a linear combination of three types of quantities:
\vskip 1mm
(Type I) $A^{2} B^{i} C^{j}$, $A B^{1+i} C^{j}$, and $ B^{2+i}
C^{j}$ when $i \geq 0$, $j \geq 0$, and $i+j=k-2$.
\vskip 1mm
(Type II) $A B^{i} C^{j}$ and $B^{1+i} C^{j}$ when $i \geq 0$, $j
\geq 0$, and $i+j=k-1$.
\vskip 1mm
(Type III) $B^{i} C^{j}$ when $i \geq 0$, $j \geq 0$, and $i+j=k$.
\vskip 1mm
We divide the proof of Theorem \mathrm{Re}f{question 1.1 not
true} into two cases.
(\textbf{Case I}) If $k=n$, we only need to look at the term $C^n$
contained in $\sigmama_{n}$. Recall that if for any non flat
K\"{a}hler metric $\omegaega$ in $S_1 \cup S_2$, we may assume that
there exists $0<M_1<x_0$ such that $F'(x) \geq C_0$ where
$C_0=F'(M_1)>0$ when $x \geq M_1$. we have the expression of $C$
from (\mathrm{Re}f {ABC another form}):
\begin{equation}gin{eqnarray}
C &=& \frac{2v-2x^2}{v^2} \\
& = & \frac{\int_{0}^{x} 2 \tau (\sqrt{1+{F'(\tau)}^2}-1) d\tau}{v^2} \nonumber\\
& \geq & \frac{\int_{M_1}^{x} 2 \tau
(\frac{F'(\tau)^2}{\sqrt{1+{F'(\tau)}^2}+1}) d\tau} {v
(\int_{M_1}^{x} 2 \tau \sqrt{1+{F'(\tau)}^2} d\tau +\int_{0}^{M_1} 2
\tau \sqrt{1+{F'(\tau)}^2} d\tau )}. \nonumber
\end{eqnarray}
Note that we have $1 \leq \frac{F'(x)}{C_0}$ when $x \geq M_1$.
\begin{equation}gin{eqnarray}
C \geq
\frac{\frac{C_0}{1+\sqrt{C_0^2+1}}I(x)}{v(\sqrt{\frac{1}{C_0^2}+1}I(x)+M_1^2
\sqrt{1+C_0^2})} \geq \frac{C_1}{v},
\end{eqnarray} where \begin{equation}gin{equation}
C_1=\frac{\frac{C_0}{1+\sqrt{C_0^2+1}}I(x)}{\sqrt{\frac{1}{C_0^2}+1}I(x)+M_1^2
\sqrt{1+C_0^2}}, \ \ \ \ \ \ \ \ \ I(x)=\int_{M_1}^{x} 2\tau
F'(\tau) d\tau.
\end{equation} Since $I(x)$ goes to $\infty$ and $C_1$ is bounded when $x$ tends to
$x_0$, we conclude that there exists a $C_2$ and $M_2$ such that
when $x > M_2$,
\begin{equation}gin{equation}
C \geq \frac{C_2}{v}. \label {lower bound C}
\end{equation} We remark that (\mathrm{Re}f {lower bound C}) is used in the proof
of Proposition 2.3 in \cite{WZ}.
There exists a constant $C_3$ only depending on $n$ such that:
\begin{equation}gin{eqnarray}
\int_{B(s)} \sigmama_{n} {\omegaega}^n &\geq& C_3 \int_{B(s)} C^n
{\omegaega}^n \label {unbounded} \\
&=& C_3 c_n \int_{0}^{x} C^n d v^n \nonumber\\
&\geq & C_3 c_n \int_{v(M_2)}^{v(x)} (\frac{C_2}{v})^n d v^n \nonumber\\
&=& n C_3 c_n C_2^{n} \ln{\frac{v(x)}{v(M_2)}} \nonumber
\end{eqnarray} Of course (\mathrm{Re}f{unbounded}) is unbounded when $x$ tends to $x_0$ since $v(x_0)=+\infty$.
To sum up, we show that for any non flat K\"{a}hler metric $\omegaega$
in $S_1 \cup S_2$, $\int_{\mathbb{C}^n} \sigmama_{n} {\omegaega}^n$ is
$\infty$. If $\omegaega \in S_3$, it follows from (\mathrm{Re}f{rf}) and
(\mathrm{Re}f{C for S3}) that $\lim_{s \rightarrow +\infty } \int_{B(s)}
\sigmama_{n} {\omegaega}^n$ is unbounded when $s$ goes to infinity.
Therefore, Question 1.1 is false when $k=n$ for any non flat
K\"{a}hler metric $\omegaega$ in $\overline{\mathcal{M}}_{n}$.
\vskip 1mm
(\textbf{Case II}) If $2 \leq k < n$, For any fixed nonflat
K\"{a}hler metric $\omegaega$ in $S_1$. we may assume that there exist
$C_4$ and $M_3$ such that $F'(x) \leq C_4$ for all $x \in
(-x_0,x_0)$ and $F'(x) \geq \frac{1}{C_4}$ when $x \geq M_3$. Then
it follows from a similar argument as in (\mathrm{Re}f {lower bound C}), we
may further assume that there exist $C_5$ and $M_3$ such that for
any $x \geq M_3$ \begin{equation}gin{equation} C \geq \frac{C_5}{v}.
\end{equation}
Since
$A=\frac{F' F''}{2x(1+{F'}^2)^2}$, we conclude that $A$ and $
\frac{F''(x)}{x}$ are equivalent. If we can construct an K\"{a}hler
metric $\omegaega$ in $S_1$ such that
\begin{equation}gin{equation}
\frac{1}{s^{2n-2k}} \int_{B(s)} (\frac{F''(x)}{x})^{2} (\frac{C_5}{v})^{k-2} {\omegaega}^n
\label{integral for AC}
\end{equation}
is unbounded when $s$ tends to $\infty$, then so is
$\frac{1}{s^{2n-2k}} \int_{B(s)} A^{2} C^{k-2} {\omegaega}^n$. Note
that the term $A^{2} C^{k-2}$ is contained in $\sigmama_k$, it follows
that $\frac{1}{s^{2n-2k}} \int_{B(s)} \sigmama_k {\omegaega}^n$ will be
unbounded when $s$ tends to $\infty$.
Let us rewrite (\mathrm{Re}f{integral for AC}):
\begin{equation}gin{eqnarray}
& & \int_{B(s)} (\frac{F''(x)}{x})^{2} (\frac{C_5}{v})^{k-2} {\omegaega}^n \label{an integral} \\
&=& c_n C_5^{k-2} \int_{0}^{x} (\frac{F''(\tau)}{\tau})^{2} (\frac{1}{v})^{k-2} d v^{n} \nonumber\\
&=& 2nc_n C_5^{k-2} \int_{0}^{x} \frac{1}{\tau} (F''(\tau))^2
v^{n-k+1} \sqrt{1+(F'(\tau))^2} d\tau. \nonumber
\end{eqnarray}
Since $s=\int_{0}^{x} \sqrt{1+(F'(\tau))^2} d\tau$, $v=\int_{0}^{x}
2\tau \sqrt{1+(F'(\tau))^2} d\tau$ and $F'(x) \leq C_4$ we know $s$
and $x$ are equivalent, $v$ and $x^2$ are equivalent. In order to
estimate (\mathrm{Re}f{an integral}), it suffices to estimate the following.
\begin{equation}gin{equation}
\int_{0}^{x} (F'')^2 {\tau}^{2(n-k)+1} \, d\tau.
\end{equation}
To sum up, if there exists a function $\delta(x) \in C^{\infty}
[0,+\infty)$, such that
\begin{equation}gin{equation}
\lim_{x \rightarrow x_0} \frac{1}{x^{2n-2k}} \int_{0}^{x} {\delta}^2
(\tau) \, {\tau}^{2(n-k)+1} \, d\tau=+\infty,\ \ \ \ \
\int_{0}^{+\infty} \delta (x) \, dx<+\infty. \label{looking for}
\end{equation} Then we can solve $F(x)$ with $F''(x)=\delta (x)$ with the initial
value $F(0)=F'(0)=0$, it will follow from Theorem \mathrm{Re}f{F function} that
we can construct a complete K\"{a}hler metric $\omegaega$ in $S_1$ such
that
\begin{equation}gin{equation} \frac{1}{s^{2n-2k}} \int_{B(s)} \sigmama_k \,
{\omegaega}^n
\end{equation} is unbounded when $s$ tends to infinity. Hence both
Question 1.1 and 1.2 can not be true when $2 \leq k<n$.
In fact such a $\delta(x)$ is not hard to construct. Consider
$\bar{\delta}(x)$ defined by the following with $q$ an integer to be determined.
\begin{equation}gin {equation}
\bar{\delta}=
\begin{equation}gin{cases}
2 & x \in [2,2+(\frac{1}{2})^q] \\
\vdots & \vdots \\
l & x \in [l,l+(\frac{1}{l})^q] \\
\vdots & \vdots \\
0 & x \in [0,+\infty) \setminus (\cup_{l \geq 2}
[l,l+(\frac{1}{l})^q] )
\end{cases}
\end{equation}
Now set $q=\frac{5}{2}$, it is easy to verify that $\psi(x)$
satisfies (\mathrm{Re}f{looking for}). Choose $\delta(x)$ as a suitable
smoothing of $\bar{\delta}(x)$ on $[0,+\infty)$ which also satisfies
(\mathrm{Re}f{looking for}), we will get the desired counterexample. It can
be checked that the result metric $\omegaega \in S_1$ has bounded
curvature on $\mathbb{C}^n$.
\end{proof}
It follows from Theorem \mathrm{Re}f{F function} and Proposition
\mathrm{Re}f{average scalar curvature decay} that for any K\"{a}hler
metric $\omegaega \in S_1$, $(\mathbb{C}^n, \omegaega)$ has quadratic
average scalar curvature decay. Note that the same result for any
complete K\"{a}hler manifolds with bounded nonnegative bisectional
curvature and Euclidean volume growth has been proved by Ni (See
\cite{Ni2} and \cite{Ni3}). Now we construct the following example
which implies that in general only assuming Euclidean volume
growth one can not expect the same rate of decay for
$L^{p}$ norm of curvature for any $p>1$.
\begin{equation}gin{proposition} \label{lp example} For any $n
\geq 2$ and any $p>1$, there exists a metric $\omegaega \in
\overline{\mathcal{M}}_{n}$ such that the geodesic balls in
$(\mathbb{C}^n, \omegaega)$ has Euclidean volume growth. Moreover,
\begin{equation}gin{equation} \frac{s^2}{\operatorname{Vol}(B(s))} \int_{B(s)}
(Rm(\frac{\partial}{\partial s}, J\frac{\partial}{\partial
s},\frac{\partial}{\partial s} J \frac{\partial}{\partial s}))^p
{\omegaega}^n
\end{equation} is unbounded as $s$ goes to infinity. Here we denote $\frac{\partial}{\partial s}$ to the
unit radial direction on $\mathbb{C}^n$.
\end{proposition}
\begin{equation}gin{proof}[Proof of Proposition \mathrm{Re}f{lp example}]
For a given metric $\omegaega$ in $S_1$, follow a similar argument in (Case II) of the proof of
Theorem \mathrm{Re}f{question 1.1 not true}, it suffices to show that we can find a smooth
function $\eta(x)$ on $[0,+\infty)$ such that
\begin{equation}gin{equation}
\lim_{x \rightarrow +\infty} \frac{1}{x^{2n-2}} \int_{0}^{x}
{\eta}^p (\tau) \, {\tau}^{2n-1-p} \, d\tau=+\infty,\ \ \ \ \
\int_{0}^{+\infty} \eta (x) \, dx<+\infty. \label{delta wanted}
\end{equation}
Consider $\bar{\eta}(x)$ defined by the following where $\alpha$ and $\begin{equation}ta$ are two
integers to be determined.
\begin{equation}gin {equation}
\bar{\eta}=
\begin{equation}gin{cases}
2^{\alpha} & x \in [2,2+(\frac{1}{2})^{\begin{equation}ta}] \\
\vdots & \vdots \\
l^{\alpha} & x \in [l,l+(\frac{1}{l})^{\begin{equation}ta}] \\
\vdots & \vdots \\
0 & x \in [0,+\infty) \setminus (\cup_{l \geq 2}
[l,l+(\frac{1}{l})^q] )
\end{cases}
\end{equation}
Pick any $\alpha>1$ and $1+\alpha < \begin{equation}ta < p(\alpha-1)+2$, then
$\bar{\eta}$ defined above satisfies (\mathrm{Re}f{delta wanted}). It is not hard to find
$\eta(x)$ from a suitable smoothing of $\bar{\eta}(x)$ which will result in
the desired metric $\omegaega$. Note that $(\mathbb{C}^n,\omegaega)$ we constructed has unbounded curvature
on $\mathbb{C}^n$. \end{proof}
We proceed to show that Question \mathrm{Re}f{Kahler category} is true for
$\overline{\mathcal{M}}_{n}$. It seems that Question \mathrm{Re}f{Kahler
category} should be a more suitable conjecture at least for complete
K\"{a}hler manifolds with nonnegative bisectional curvature.
\begin{equation}gin{theorem} \label{question 1.2 true} For any
metric $\omegaega \in \overline{\mathcal{M}}_{n}$, then $s^{-2n+2k} \int_{B(s)} Ric^{k} \wedge
\omegaega^{n-k}$ is bounded when $s$ goes to infinity.
\end{theorem}
\begin{equation}gin{proof}[Proof of Proposition \mathrm{Re}f{question 1.2 true}] \, First we remark that
it directly follows from analogues of Proposition 2.3, 2.4 and 2.6
in \cite{WZ} for the space $\overline{\mathcal{M}}_{n}$ (See
Proposition \mathrm{Re}f{average scalar curvature decay} and the paragraph
after Proposition \mathrm{Re}f{xi function}) that Question 1.3 is true for
$k=1$ and $k=n$. If suffices to show that $s^{-2n+2k} \int_{B(s)}
Ric^{k} \wedge \omegaega^{n-k}$ is bounded for any $2 \leq k <n$.
Note that for $2 \leq k <n$, $Ric^{k} \wedge \omegaega^{n-k}$ is a
linear combination of $\lambdabda {\mu}^{k-1}$ and ${\mu}^{k}$. It
turns out that we only needs to show that $\frac{1}{s^{2n-2k}}
\int_{B(s)} P(A,B,C) {\omegaega}^n$ is bounded when $s$ goes to
infinity where $P$ is a monomial of the following two types:
\vskip 1mm
(Type I) $A B^{i} C^{j}$, and $ B^{1+i} C^{j}$ when $i \geq 0$, $j
\geq 0$, and $i+j=k-1$.
\vskip 1mm
(Type II) $B^{p} C^{q}$ when $p \geq 0$, $q \geq 0$, and $p+q=k$.
\vskip 1mm
First we consider any K\"{a}hler metric $\omegaega$ in $S_1 \cup S_2$.
Note that (\mathrm{Re}f{ABC another form}) implies that
\begin{equation}gin{equation}
B \leq \frac{x^2}{v^2} \leq \frac{1}{v},\ \ \ C\leq \frac{2}{v}.
\end{equation}
Then we have the following estimate:
\begin{equation}gin{eqnarray}
& & \frac{1}{s^{2n-2k}} \int_{B(s)} B^{p} C^{q} {\omegaega}^n \label{hoho1} \\
&\leq & 2^{q} c_n \frac{1}{s^{2n-2k}} \int_{0}^{v(x)} \frac{1}{v^k}
n v^{n-1} dv \nonumber \\
& \leq & \frac{2^q n c_n (\int_{0}^{x} 2\tau \sqrt{1+(F'(\tau))^2}
d\tau)^{n-k}}{ (n-k)(\int_{0}^{x} \sqrt{1+(F'(\tau))^2} d\tau
)^{2n-2k}} \nonumber
\end{eqnarray}
According to the L'Hospital's rule, (\mathrm{Re}f{hoho1}) has the limit when
$x$ tends to $x_0$: \begin{equation}gin{equation} \lim_{x \rightarrow
x_0}\frac{(\int_{0}^{x} 2\tau \sqrt{1+(F'(\tau))^2}
d\tau)^{n-k}}{(\int_{0}^{x} \sqrt{1+(F'(\tau))^2} d\tau
)^{2n-2k}}=(\frac{2}{\sqrt{1+\lim_{x \rightarrow x_0}
F'(x)}})^{n-k}.
\end{equation} We conclude that $\frac{1}{s^{2n-2k}} \int_{B(s)} B^{p} C^{q}
{\omegaega}^n$ is bounded when $s$ goes to infinity.
Next we turn to the term $A B^{i} C^{j}$, integrate by parts as in
the original proof of Proposition 2.6 in \cite{WZ}.
\begin{equation}gin{eqnarray}
& & \int_{B(s)} A B^{i} C^{j} {\omegaega}^n \label{hoho2} \\
&=& c_n \int_{0}^{x} \frac{F'F''}{2\tau (1+(F')^2)^2}
\frac{1}{v^{k-1}}
n v^{n-1} 2\tau \sqrt{1+(F')^2} d\tau \nonumber \\
&=& c_n \int_{0}^{x} \frac{F'F''}{(1+(F')^2)^{\frac{3}{2}}} n v^{n-k} d\tau \nonumber \\
&=& c_n \int_{0}^{v} n v^{n-k} d(-\frac{1}{\sqrt{1+(F')^2}}) \nonumber \\
&=& (-c_n n v^{n-k}\frac{1}{\sqrt{1+(F')^2}})|_{0}^{v}+ c_n
n(n-k)\int_{0}^{v}
\frac{1}{\sqrt{1+(F')^2}} v^{n-k-1} dv \nonumber \\
&\leq & c_n n v^{n-k}. \nonumber
\end{eqnarray}
Using the L'Hospital's rule again, we conclude that
\begin{equation}gin{equation}
\frac{1}{s^{2n-2k}} \int_{B(s)} A B^{i} C^{j} {\omegaega}^n
\end{equation} is bounded when $s$ tends to infinity.
It remains to verify that Question \mathrm{Re}f{Kahler category} is true
when $2 \leq k<n$ for any metric $\omegaega \in S_3$. Note that in this case
we have (\mathrm{Re}f{B for S3}), (\mathrm{Re}f{C for S3}), (\mathrm{Re}f{s for S3}), and $A=0$ outside a compact set
for metrics in $S_3$, it follows from a
straightforward calculation that $\frac{1}{s^{2n-2k}} \int_{B(s)}
Ric^{k} \wedge \omegaega^{n-k}$ is bounded when $s$ goes to infinity.
Hence we finish the proof of Proposition \mathrm{Re}f{question 1.2 true}.
\end{proof}
\vskip 1mm
We also have the following result relating the growth of the
coordinate function $z_i$ to the volume growth of the geodesic
balls with respect to the metric $\omegaega$ in
$\overline{\mathcal{M}}_{n}$.
\begin{equation}gin{proposition}
\label{holomorphic function}
Given any metric $\omegaega \in \overline{\mathcal{M}}_{n}$, if some
coordinate function $z_i$ for some $1 \leq i \leq n$ has polynomial
growth with respect to $\omegaega$, then the geodesic balls of
$(\mathbb{C}^n, \omegaega)$ have Euclidean volume growth.
\end{proposition}
\begin{equation}gin{proof}[Proof of Proposition \mathrm{Re}f{holomorphic function}] Assume some
coordinate function $z_i$ for some $1 \leq i \leq n$ has polynomial
growth with respect to $\omegaega$ in $\overline{\mathcal{M}}_{n}$. it
follows from $\omegaega$ being rotationally symmetric that there exists
some integer $\alpha$ and constant $C_6>0$ such that:
\begin{equation}gin{equation}
r=|z|^2 \leq C_6 s^{\alpha}. \label {polynomial growth}
\end{equation}
From Theorem \mathrm{Re}f{F function} it suffices to show that $\omegaega \in S_1$, namely
$F'(x)$ bounded when $x$ goes to $x_0$. First we note that $\omegaega$
can not be from $S_3$ from the explicit formula (\mathrm{Re}f{s for S3}) on the
distance with respect to metrics in $S_3$ given in Theorem 3.1.
Plugging (\mathrm{Re}f{distance and volume}) into (\mathrm{Re}f{polynomial growth})
leads to:
\begin{equation}gin{equation}
r\leq C_6 (\int_{0}^{x} \sqrt{1+(F'(\tau))^2} d \tau)^{\alpha}.
\label {polynomial growth 2}
\end{equation}
Note that:
\begin{equation}gin{equation}
\frac{dr}{dx}=\frac{2r}{(1-\xi)x}=\frac{2r \sqrt{1+(F'(x))^2}}{x}.
\label{equation r}
\end{equation}
Solve $r$ in terms of $x$ from (\mathrm{Re}f{equation r}) and plug into
(\mathrm{Re}f {polynomial growth 2}):
\begin{equation}gin{equation}
e^{2\int_{C_7}^{x} \frac{\sqrt{1+(F'(\tau))^2}}{\tau} d\tau} \leq
C_6 (\int_{0}^{x} \sqrt{1+(F'(\tau))^2} d \tau)^{\alpha}
\label{crucial inequality}
\end{equation} for any $C_7 \leq x < x_0$. Here $C_7$ is the value of $x$
which corresponds to $r=1$.
It is not hard to show $F'(x)$ is bounded for all $x \in (0,x_0)$
from (\mathrm{Re}f{crucial inequality}). First we see that $x_0$ must be
infinity. Otherwise, the left hand side $e^{2\int_{C_7}^{x}
\frac{\sqrt{1+(F'(\tau))^2}}{\tau} d\tau} \geq e^{\frac{2}{x_0}
\int_{C_7}^{x} \sqrt{1+(F'(\tau))^2} d\tau}$. then (\mathrm{Re}f{crucial
inequality}) could not be true since exponential functions can not
be controlled by any polynomials when $\int_{C_7}^{x}
\sqrt{1+(F'(\tau))^2} d\tau$ goes to infinity. Next we show that
$F'(x)$ is bounded for all $x \in (0,+\infty)$. It follows from
(\mathrm{Re}f{crucial inequality}) that
\begin{equation}gin{equation}
\frac {2\int_{C_7}^{x} \frac{\sqrt{1+(F'(\tau))^2}}{\tau} d\tau}
{\alpha \ln{\int_{0}^{x} \sqrt{1+(F'(\tau))^2} d\tau } +\ln C_6 }
\label {quantity}
\end{equation} should be bounded when $x$ tends to infinity.
It is easy to see that (\mathrm{Re}f {quantity}) has a limit when $x$ goes
to infinity.
\begin{equation}gin{equation}
\lim_{x \rightarrow +\infty} \frac {2\int_{C_7}^{x}
\frac{\sqrt{1+(F'(\tau))^2}}{\tau} d\tau} {\alpha \ln{\int_{0}^{x}
\sqrt{1+(F'(\tau))^2} d\tau } +\ln C_6 } =\frac{2}{\alpha}
\sqrt{1+(\lim_{x \rightarrow +\infty} F'(\tau))^2},
\end{equation} which implies that $F'(x)$ is bounded for all $x$ in
$[0,+\infty)$. Therefore $\omegaega \in S_1$. \end{proof}
\vskip 1mm
\begin{equation}gin{remark}
See \cite{Ni1}, \cite{ChenZhu1}, \cite{ChenZhu2}, \cite{Ni2},
\cite{NiTam}, and \cite{Ni3} for some results on the geometry of general
complete noncompact K\"{a}hler manifolds with nonnegative
bisectional curvature. Some further generalizations will also appear
in a separate paper by the author.
\end{remark}
\noindent\textbf{Acknowledgments.} The author thanks Professor Lei
Ni for making Wu and Zheng's paper \cite{WZ} available to him,
bringing Question \mathrm{Re}f{Kahler category} to his attention, as well as
many helpful discussions during the preparation of this paper. The
author also thanks Professor Bennett Chow and Professor Xiaohua Zhu
for helpful comments and Professor Fangyang Zheng for his interest
in this work.
\begin{equation}gin{thebibliography}{99}
\bibitem {Cao1} Cao, Huai-Dong. \emph{Existence of gradient K\"{a}hler-Ricci
solitons.} Elliptic and parabolic methods in geometry (Minneapolis,
MN, 1994), 1-16, A K Peters, Wellesley, MA, 1996.
\bibitem {Cao2} Cao, Huai-Dong. \emph{Limits of solutions to the K\"{a}hler-Ricci
flow.} Journal of Differential Geometry, {\bf 65} (1997), 257-272.
\bibitem {ChauTam1} Chau, Albert; Tam, Luen-Fai. \emph{On the complex structure
of K\"{a}hler manifolds with nonegative curvature.} Journal of
Differential Geometry {\bf 73}, 2006, no.3, 491-530.
\bibitem {ChauTam2} Chau, Albert; Tam, Luen-Fai.. \emph{Non-negatively curved
K\"{a}hler manifolds with average quadratic curvature decay}
Communications in Analysis and Geometry {\bf 15}, 2007, no.1,
121-146.
\bibitem {ChauTam3} Chau, Albert; Tam, Luen-Fai. \emph{On the Steinness of a
class of K\"{a}hler manifolds} Journal of Differential Geometry {\bf
79}, 2008, no.2, 167-183.
\bibitem {ChenZhu1} Chen, Bing-Long; Zhu, Xi-Ping. \emph{On complete noncompact K\"{a}hler
manifolds with positive bisectional curvature.} Mathematische
Annalen {\bf 327}, 2003, no.1, 1-23.
\bibitem {ChenZhu2} Chen, Bing-Long; Zhu, Xi-Ping. \emph{Volume growth and curvature decay
of positively curved K\"{a}hler manifolds.} Quarterly Journal of
Pure and Applied Mathematics {\bf 1}, 2005, no.1, 68-108.
\bibitem {ChenTangZhu} Chen, Bing-Long; Tang, Siu-Hung; Zhu, Xi-Ping. \emph{A
uniformization theorem for complete non-compact K\"{a}hler surfaces
with positive bisectional curvature.} Journal of Differential
Geometry {\bf 67}, 2004, no.3, 519-570.
\bibitem {K} Klembeck, Paul F. \emph{A complete K\"{a}hler metric of positive
curvature on $\mathbb{C}^{n}$.} Proceedings of the American
Mathematical Society, {\bf 64}, 1977, no.2, 313-316.
\bibitem {Ni1} Ni, Lei. \emph{A monotonicity formula on complete
K\"{a}hler manifolds with nonnegative bisectional curvature.}
Journal of the American Mathematical Society {\bf 17}, 2004, no.4,
909-946.
\bibitem {Ni2} Ni, Lei. \emph{Ancient solutions to K\"{a}hler-Ricci flow.}
Mathematical Research Letters {\bf 12}, 2005, no.5-6, 633-653.
\bibitem {Ni3} Ni, Lei. \emph{Monotonicity and Holomorphic functions.}
Geometry and Analysis Volume I, Advanced Lecture in Mathematics {\bf
17}, Higher Education Press and International Press, Beijing and
Boston, 2010, 447-457.
\bibitem {NiTam} Ni, Lei; Tam, Luen-Fai. \emph{Plurisubharmonic functions
and the structure of complete K\"{a}hler manifolds with nonnegative
curvature.} Journal of Differential Geometry {\bf 64}, 2003, no.3,
457-524.
\bibitem {WZ} Wu, Hung-Hsi; Zheng, Fangyang. \emph{Examples of positively curved
complete K\"{a}hler manifolds.} Geometry and Analysis Volume I,
Advanced Lecture in Mathematics {\bf 17}, Higher Education Press and
International Press, Beijing and Boston, 2010, 517-542.
\bibitem {Yau} Yau, Shing-Tung. \emph{Open problems in geometry.} Chern--a great
geometer of the twentieth century, International Press, Hong Kong,
1992, 275-319.
\bibitem {Yau2} Yau, Shing-Tung. \emph{A review of complex differential geometry.}
Several complex variables and complex geometry, Part 2 (Santa Cruz,
CA, 1989), Proceedings of Symposia in Pure Mathematics, 619-625,
{Vol \bf52}, Part II, American Mathematical Society, 1991.
\end{thebibliography}
\end{document}
|
math
|
Located in Gillingham, is this mid-terraced family home. The house was built in the 1950s and offers spacious accommodation throughout. As you step inside, the first thing that you will notice is that it is a lot larger on the inside than it appears on the outside and offers plenty of scope for the new owner to put their stamp on it. The good size living room and separate dining room offers a perfect place for when you are entertaining guests. There is also a utility room and a downstairs toilet. Upstairs, there are three bedrooms, so there is ample space for a family to grow and room for friends when they come to stay. All of the bedrooms are big enough for a double bed and the bathroom has been converted into a wet room. Outside, there is a generous garden which is over 70' long and the aspect will ensure that you get the sun all day long. This creates a perfect place to host a summer barbeque for your family and friends. There are two outbuildings, which could quite easily become part of the property, or alternatively would make an ideal home/work office or hobbies room. Gillingham is very popular with families, as there are primary schools and a nursary opposite the house and buses that service all of the secondary schools. The town centre has been pedestrianised and you will find all of the shops that you would expect from a busy High Street. There is also a retail park nearby which offers larger stores and eateries. For those who commute, the A2 can be easily reached and there is a train station at Gillingham which, which connects up with the high speed link into London. This house will make a great home for a growing family.
Disclaimer - Property reference 527264-1. The information displayed about this property comprises a property advertisement. Rightmove.co.uk makes no warranty as to the accuracy or completeness of the advertisement or any linked or associated information, and Rightmove has no control over the content. This property advertisement does not constitute property particulars. The information is provided and maintained by Purplebricks, covering Meridian. Please contact the selling agent or developer directly to obtain any information which may be available under the terms of The Energy Performance of Buildings (Certificates and Inspections) (England and Wales) Regulations 2007 or the Home Report if in relation to a residential property in Scotland.
|
english
|
अम्बेडकर की सबसे ऊंची प्रतिमा बनाएंगे विश्वकर्मावंशी पद्मश्री राम वी० सुतार विश्वकर्मा किरण
अम्बेडकर की सबसे ऊंची प्रतिमा बनाएंगे विश्वकर्मावंशी पद्मश्री राम वी० सुतार
मुम्बई। मुम्बई के दादर में इंदू मिल परिसर में लगने वाली बाबा साहेब भीमराव अम्बेडकर की दुनिया की सबसे ऊंची प्रतिमा के निर्माण का काम महाराष्ट्र सरकार ने प्रसिद्ध मूर्तिकार विश्वकर्मावंशी पद्मश्री राम वी० सुतार को सौंपा है। 25० फीट ऊंची यह मूर्ति दुनिया में अब तक की अम्बेडकर की सबसे ऊंची प्रतिमा होगी, जिसे दो साल में तैयार किया जाएगा।
राम सुतार के नोएडा स्टूडियो में होगा मूर्ति का निर्माण
मूर्ति का निर्माण राम वी० सुतार के नोएडा स्टूडियो में किया जाएगा। उनके स्टूडियो में बने डिजाइन के आधार पर चीन में सांचा तैयार किया जाएगा, जिससे अम्बेडकर की कांसे की प्रतिमा तैयार होगी। इंदू मिल में इसके अलावा अम्बेडकर की २५ फीट की प्रतिमा भी लगेगी। जिसे राम वी० सुतार तैयार कर चुके हैं। मुम्बई में ही लगने वाली छत्रपति शिवाजी की ३९८ फीट ऊंची विश्व की सबसे ऊंची मूर्ति और गुजरात के भरूच में लगने वाली ५९७ फीट ऊंची सरदार पटेल की मूर्ति का निर्माण भी राम वी० सुतार कर रहे हैं।
इंदू मिल की जमीन पर बन रहा अम्बेडकर स्मारक
महाराष्ट्र सरकार मुम्बई के दादर में स्थित शिवाजी पार्क के इंदू मिल की जमीन पर १२.४ एकड़ जमीन पर ४25 करोड़ की लागत से अम्बेडकर स्मारक का निर्माण करा रही है। 1४ अप्रैल २०२० तक स्मारक का कार्य पूरा होना है। 1४ अप्रैल को बाबा साहेब की जयंती मनाई जाती है। २०१५ में प्रधानमन्त्री नरेन्द्र मोदी ने स्मारक का भूमि पूजन किया था। यह जमीन पहले केंद्र सरकार के पास थी। पिछले साल केंद्र सरकार ने जमीन महाराष्ट्र सरकार को हस्तांतरित कर दी थी।
इंदू मिल की जमीन पर क्यों बन रहा अम्बेडकर स्मारक
६ दिसंबर 195६ को अम्बेडकर का निधन दिल्ली में हुआ था, लेकिन उनका अंतिम संस्कार मुम्बई में समुद्र किनारे किया गया था। यहां एक स्मारक है, जिसे चैत्यभूमि कहते हैं। इसी के बगल में इंदू मिल है। यहां काफी समय से अम्बेडकर स्मारक बनाने की मांग हो रही थी। इसी कारण इंदू मिल और चैन्यभूमि की जमीन को मिलाकर अम्बेडकर स्मारक बनाने का निर्णय हुआ।
दलित वर्ग को होगा गौरव का बोध
मोदी सरकार बनने के बाद दलितों में गौरव बोध कराने के लिए किए गए कार्यों में से इसे एक बड़ा कदम माना जा रहा है।
अम्बेडकर स्मृतियां संजोने के लिए मोदी सरकार के अन्य कार्य
-अलीपुर रोड दिल्ली में अम्बेडकर स्मारक
-१५ जनपथ दिल्ली में अम्बेडकर इंटरनेशनल सेंटर
-१० किंग हेनरी रोड लंदन में अम्बेडकर स्मारक
अम्बेडकर प्रतिमा से जुड़ी कुछ अन्य जानकारी
-मूर्ति को सौ फीट ऊंचे बेस पर रखा जाएगा।
-बेस में बौद्ध धर्म की झलक होगी।
-जमीन तल से प्रतिमा की पूरी ऊंचाई ३५० फीट होगी।
-मूर्ति के बेस में कमल के फूल बने होंगे।
यह दुनिया की अब तक की अम्बेडकर की सबसे ऊंची प्रतिमा होगी। इसका डिजाइन संसद भवन में लगी अम्बेडकर प्रतिमा की तरह होगा।
-पद्मश्री राम वी० सुतार, प्रसिद्ध मूर्तिकार
प्रेवियस धर्मशाला जीर्णोद्धारक ओमप्रकाश धीमान की प्रतिमा का हुआ अनावरण
नेक्स्ट गरीब की झोपड़ी से निकला प्रकाश कर रहा समाज को गौरवान्वित
|
hindi
|
3 дни дизел - 2.11. Грешни данни.
The best petrol station in the area, which opened soon! Everything is new! Not crowded as the nearby old OMV. Very good coffee! And the girls who are working at the cash desk are young and always smiling! Greetings!
Местен клон на веригата, качествени горива на конкурентни цени, любезно обслужване!
|
english
|
Everquest is massively multiplayer online role-playing game (MMORPG) with a Tolkienesque fantasy setting.
In this role-playing game, you play on your computer across the Internet, and explore and adventure in a fantasy world. Meanwhile, other players also play in the same fantasy world, from their computers, and you may encounter them in the game.
Make 4,000 Platinum an Hour, Completely Legal. No Hacks or Cheats Needed.
|
english
|
أمۍ پرٛژھس موجی ژٕ کِتھہٕ پٲٹھۍ چھَکھ زِنٛدٕ روزان
|
kashmiri
|
import os
import unittest
from typing import Dict
from unittest.mock import patch
from ray_release.buildkite.concurrency import (
get_test_resources_from_cluster_compute,
get_concurrency_group,
CONCURRENY_GROUPS,
)
from ray_release.buildkite.filter import filter_tests, group_tests
from ray_release.buildkite.settings import (
split_ray_repo_str,
get_default_settings,
update_settings_from_environment,
Frequency,
update_settings_from_buildkite,
Priority,
)
from ray_release.buildkite.step import get_step
from ray_release.config import Test
from ray_release.exception import ReleaseTestConfigError
from ray_release.wheels import (
DEFAULT_BRANCH,
)
class MockBuildkite:
def __init__(self, return_dict: Dict):
self.return_dict = return_dict
def __call__(self, key: str):
return self.return_dict.get(key, None)
class BuildkiteSettingsTest(unittest.TestCase):
def setUp(self) -> None:
self.buildkite = {}
self.buildkite_mock = MockBuildkite(self.buildkite)
def testSplitRayRepoStr(self):
url, branch = split_ray_repo_str("https://github.com/ray-project/ray.git")
self.assertEqual(url, "https://github.com/ray-project/ray.git")
self.assertEqual(branch, DEFAULT_BRANCH)
url, branch = split_ray_repo_str(
"https://github.com/ray-project/ray/tree/branch/sub"
)
self.assertEqual(url, "https://github.com/ray-project/ray.git")
self.assertEqual(branch, "branch/sub")
url, branch = split_ray_repo_str("https://github.com/user/ray/tree/branch/sub")
self.assertEqual(url, "https://github.com/user/ray.git")
self.assertEqual(branch, "branch/sub")
url, branch = split_ray_repo_str("ray-project:branch/sub")
self.assertEqual(url, "https://github.com/ray-project/ray.git")
self.assertEqual(branch, "branch/sub")
url, branch = split_ray_repo_str("user:branch/sub")
self.assertEqual(url, "https://github.com/user/ray.git")
self.assertEqual(branch, "branch/sub")
url, branch = split_ray_repo_str("user")
self.assertEqual(url, "https://github.com/user/ray.git")
self.assertEqual(branch, DEFAULT_BRANCH)
def testSettingsOverrideEnv(self):
settings = get_default_settings()
# With no environment variables, default settings shouldn't be updated
updated_settings = settings.copy()
update_settings_from_environment(updated_settings)
self.assertDictEqual(settings, updated_settings)
# Invalid frequency
os.environ["RELEASE_FREQUENCY"] = "invalid"
updated_settings = settings.copy()
with self.assertRaises(ReleaseTestConfigError):
update_settings_from_environment(updated_settings)
# Invalid priority
os.environ["RELEASE_PRIORITY"] = "invalid"
updated_settings = settings.copy()
with self.assertRaises(ReleaseTestConfigError):
update_settings_from_environment(updated_settings)
os.environ["RELEASE_FREQUENCY"] = "nightly"
os.environ["RAY_TEST_REPO"] = "https://github.com/user/ray.git"
os.environ["RAY_TEST_BRANCH"] = "sub/branch"
os.environ["RAY_WHEELS"] = "custom-wheels"
os.environ["TEST_NAME"] = "name_filter"
os.environ["RELEASE_PRIORITY"] = "manual"
updated_settings = settings.copy()
update_settings_from_environment(updated_settings)
self.assertDictEqual(
updated_settings,
{
"frequency": Frequency.NIGHTLY,
"test_name_filter": "name_filter",
"ray_wheels": "custom-wheels",
"ray_test_repo": "https://github.com/user/ray.git",
"ray_test_branch": "sub/branch",
"priority": Priority.MANUAL,
"no_concurrency_limit": False,
},
)
def testSettingsOverrideBuildkite(self):
settings = get_default_settings()
with patch(
"ray_release.buildkite.settings.get_buildkite_prompt_value",
self.buildkite_mock,
):
# With no buildkite variables, default settings shouldn't be updated
updated_settings = settings.copy()
update_settings_from_buildkite(updated_settings)
self.assertDictEqual(settings, updated_settings)
# Invalid frequency
self.buildkite["release-frequency"] = "invalid"
updated_settings = settings.copy()
with self.assertRaises(ReleaseTestConfigError):
update_settings_from_buildkite(updated_settings)
# Invalid priority
self.buildkite["release-priority"] = "invalid"
updated_settings = settings.copy()
with self.assertRaises(ReleaseTestConfigError):
update_settings_from_buildkite(updated_settings)
self.buildkite["release-frequency"] = "nightly"
self.buildkite["release-ray-test-repo-branch"] = "user:sub/branch"
self.buildkite["release-ray-wheels"] = "custom-wheels"
self.buildkite["release-test-name"] = "name_filter"
self.buildkite["release-priority"] = "manual"
updated_settings = settings.copy()
update_settings_from_buildkite(updated_settings)
self.assertDictEqual(
updated_settings,
{
"frequency": Frequency.NIGHTLY,
"test_name_filter": "name_filter",
"ray_wheels": "custom-wheels",
"ray_test_repo": "https://github.com/user/ray.git",
"ray_test_branch": "sub/branch",
"priority": Priority.MANUAL,
"no_concurrency_limit": False,
},
)
def _filter_names_smoke(self, *args, **kwargs):
filtered = filter_tests(*args, **kwargs)
return [(t[0]["name"], t[1]) for t in filtered]
def testFilterTests(self):
tests = [
Test(
{
"name": "test_1",
"frequency": "nightly",
"smoke_test": {"frequency": "nightly"},
}
),
Test(
{
"name": "test_2",
"frequency": "weekly",
"smoke_test": {"frequency": "nightly"},
}
),
Test({"name": "other_1", "frequency": "weekly"}),
Test(
{
"name": "other_2",
"frequency": "nightly",
"smoke_test": {"frequency": "multi"},
}
),
Test({"name": "other_3", "frequency": "disabled"}),
Test({"name": "test_3", "frequency": "nightly"}),
]
filtered = self._filter_names_smoke(tests, frequency=Frequency.ANY)
self.assertSequenceEqual(
filtered,
[
("test_1", False),
("test_2", False),
("other_1", False),
("other_2", False),
("test_3", False),
],
)
filtered = self._filter_names_smoke(tests, frequency=Frequency.NIGHTLY)
self.assertSequenceEqual(
filtered,
[
("test_1", False),
("test_2", True),
("other_2", False),
("test_3", False),
],
)
filtered = self._filter_names_smoke(tests, frequency=Frequency.WEEKLY)
self.assertSequenceEqual(filtered, [("test_2", False), ("other_1", False)])
filtered = self._filter_names_smoke(
tests, frequency=Frequency.NIGHTLY, test_name_filter="other"
)
self.assertSequenceEqual(
filtered,
[
("other_2", False),
],
)
filtered = self._filter_names_smoke(
tests, frequency=Frequency.NIGHTLY, test_name_filter="test"
)
self.assertSequenceEqual(
filtered, [("test_1", False), ("test_2", True), ("test_3", False)]
)
def testGroupTests(self):
tests = [
(Test(name="x1", group="x"), False),
(Test(name="x2", group="x"), False),
(Test(name="y1", group="y"), False),
(Test(name="ungrouped"), False),
(Test(name="x3", group="x"), False),
]
grouped = group_tests(tests)
self.assertEqual(len(grouped), 3) # Three groups
self.assertEqual(len(grouped["x"]), 3)
self.assertSequenceEqual(
[t["name"] for t, _ in grouped["x"]], ["x1", "x2", "x3"]
)
self.assertEqual(len(grouped["y"]), 1)
def testGetStep(self):
test = Test(
{
"name": "test",
"frequency": "nightly",
"run": {"script": "test_script.py"},
"smoke_test": {"frequency": "multi"},
}
)
step = get_step(test, smoke_test=False)
self.assertNotIn("--smoke-test", step["command"])
step = get_step(test, smoke_test=True)
self.assertIn("--smoke-test", step["command"])
step = get_step(test, priority_val=20)
self.assertEqual(step["priority"], 20)
def testInstanceResources(self):
# AWS instances
cpus, gpus = get_test_resources_from_cluster_compute(
{
"head_node_type": {"instance_type": "m5.4xlarge"}, # 16 CPUs, 0 GPUs
"worker_node_types": [
{
"instance_type": "m5.8xlarge", # 32 CPUS, 0 GPUs
"max_workers": 4,
},
{
"instance_type": "g3.8xlarge", # 32 CPUs, 2 GPUs
"min_workers": 8,
},
],
}
)
self.assertEqual(cpus, 16 + 32 * 4 + 32 * 8)
self.assertEqual(gpus, 2 * 8)
cpus, gpus = get_test_resources_from_cluster_compute(
{
"head_node_type": {
"instance_type": "n1-standard-16" # 16 CPUs, 0 GPUs
},
"worker_node_types": [
{
"instance_type": "random-str-xxx-32", # 32 CPUS, 0 GPUs
"max_workers": 4,
},
{
"instance_type": "a2-highgpu-2g", # 24 CPUs, 2 GPUs
"min_workers": 8,
},
],
}
)
self.assertEqual(cpus, 16 + 32 * 4 + 24 * 8)
self.assertEqual(gpus, 2 * 8)
def testConcurrencyGroups(self):
def _return(ret):
def _inner(*args, **kwargs):
return ret
return _inner
test = Test(
{
"name": "test_1",
}
)
def test_concurrency(cpu, gpu, group):
with patch(
"ray_release.buildkite.concurrency.get_test_resources",
_return((cpu, gpu)),
):
group_name, limit = get_concurrency_group(test)
self.assertEqual(group_name, group)
self.assertEqual(limit, CONCURRENY_GROUPS[group_name])
test_concurrency(12800, 8, "large-gpu")
test_concurrency(12800, 7, "small-gpu")
test_concurrency(12800, 1, "small-gpu")
test_concurrency(12800, 0, "large")
test_concurrency(512, 0, "large")
test_concurrency(511, 0, "medium")
test_concurrency(128, 0, "medium")
test_concurrency(127, 0, "small")
test_concurrency(1, 0, "small")
|
code
|
لایٹَن ہٕنزۍ چمک ہُرراو
|
kashmiri
|
کیٛاہ مےٚ آیا ۴۷۷۴ پۄنٛسہٕ واپس سٹرک ٹرانزیکشن پؠٹھٕ
|
kashmiri
|
So for those curious, the Architecture Chat is still going - I know the last post was on February 22nd... apologies for the slackness.
Currently we have been using Google Docs (after the demise of google wave) to arrange the chats, which all the regular folk get invited to (the document that is) - and just announcing the doc on Twitter (follow myself @bittercoder or the hash tag #ArchitectureChat to see announcements of upcoming chats).
If your are a past regular/planning to come along more often in the future - then let me know your email address and I can start inviting you to the google doc as well.
The next chat is Thursday this week, hope to see you all there!
We're having another Architecture Chat tomorrow - 23rd February 2012, 11:30am onwards, Garrisons, Sylvia Park.
70x Higher Join Perf in MySql?
Ayende's blog post series "Limit your abstractions"
Topics this week were collected here.
|
english
|
I lost 15 pounds, 6 inches from my waist, and 5% body fat. I increased muscle mass in my legs, arms, back, and abs.
I work out three to four times a week, doing strength training classes or HIIT training. I eat high-protein, low-carbohydrate breakfasts, and I have reduced carbohydrate intake throughout the day. I also drink plenty of water. I look great on the outside, and I feel great on the inside. Social media makes it easy to share my results and start conversations with friends, family members, acquaintances, and more.
|
english
|
Welcome to the first instalment of "Weekly Wonders". Every week is full of various wonders of one kind or another for us, as I am sure it is for you too.
I am going to try and do a weekly wrap up of sorts for all 3 kiddos in the one post. So yes it will be a huge post but I will break it down into Tot School, the Preschool Corner, followed by Grade School stuff for B.
It definitely has been an effort to compile this but I really want to try and start breaking things down and writing things down more each week so fingers crossed for me here. Not only does it show what the kids have been doing and what they learn without me even intervening at times, but it also shows the weaknesses, the areas we need to cover more, this will be come very obvious over the next few weeks, so my hope is to use it in a positive way to bring more into our schooling weeks.
I am linking up to Tot School for C, he is currently 18 months old. It seems like forever since we have done a Tot School post.
C had the opportunity for lots of swimming practice due to us going away for the weekend. He is just about swimming between two adults (a couple of metres) just needs to remember to always kick his little legs. But considering he has had no formal swimming lessons (just lots of practice in various pools) he is doing really well and absolutely loves the water. Sorry no photo's I forgot to take my camera.
Lots of books, stories and songs. His favourite of the week is Toot Toot Beep Beep by Emma Garcia.
I purchased this book a little while ago and it got lost amongst a pile of stuff and totally forgotten, until I found it this week that was! C loves it, lots of simple repetitive language and noises from all sorts of things that move (what boy wouldn't love this book). The illustrations are adorable as well, they appear to be painted by children. I'd really like for us to have a go at doing some of our own Toot Toot Beep Beep type illustrations.
We've thoroughly enjoyed this book and I have decided to link it up to What My Child Is Reading over at Learning With Mouse.
C had a morning away from us during our first day back with our home school network classes program. Although he is never a problem I wanted to be able to have my full attention available for K this term to ensure that she was settling into her classes well. This is an 8 week term and C will spend a fun few hours with his Nan during that time slot. He was a little apprehensive when we were leaving this morning but by the time we had all gotten into the car he was happily waving goodbye. We’ll reassess the situation for next term.
We baked a batch of Chocolate Brownies during the week. This was the first time C has been involved with the whole cooking process, he was so keen to get in and give things a stir, but licking the spoon at the end was definitely the most enjoyable part of the process.
C had his first shaving cream experience this week. I covered the table with just plain shaving cream, as it was his first introduction I wanted to keep it simple. He was a little unsure at first and just poked and touched it with only a couple of fingers, but once he saw his brother and sister jump right on he was ready to go. It only took a few minutes before he had shaving cream right up to his arm pits. Shaving cream is cool, squishy and smelly and is quite invigorating on the senses. As long as you there with your toddler (especially for those that put everything in their mouths) this is a great first introduction to things like finger painting.
He also spent quite a bit of time with these Fisher Price blocks. We have had them for years and he has had access to them for some time now but this was the first time he actually sat down and went through most of them, touching, feeling and listening to what was inside.
C got to join in with a few live productions that B and K had put on during the week. It was very difficult to grab a snap of him as he didn’t keep still for any longer than one second. However in this one he is the Prince from Sleeping Beauty. He loves it when the big kids include him in their games and activities, just like he is one of them. It’s these times that remind me what he would be missing out on if they were away at school every day.
As is every week he spent a considerable amount of time building blocks with his brother. Or should I say B spends a considerable amount of time building and C spends the bulk of his time destroying B’s creations!!
That's our Tot School for the week, to see what other Tots got up to visit 1+1+1=1.
Preschool Corner is hosted over at Homeschool Creations and I am happy to finally be joining back in with this Linky.
Since we are getting into Montessori in a bigger way this year I have decided to link up to One Hook Wonder and Montessori Monday. Our Montessori related posts will increase as we (I) either gain/present or make more materials.
K continued to work with the Reading Eggs software, she really does enjoy this program and is working on Level 1 discs 1 and 2, although currently I am unsure if she is actually learning anything. I’ll reassess this in a few weeks time. I’m still working on Montessori inspired materials for her for letter/sound recognition. It really is time consuimg work, but at least when it comes time for C all the work will be done!
I also introduced her to the Metal Insets (aids in pre writing skills and pencil grip) this week (one of our many new Montessori Materials that arrived over the summer holidays). At first she didn’t like this material, said it was too hard and that she didn’t want to do it, but once I left her alone she went on to complete a few insets.
She has also been grabbing various picture books and writing out words from them. This has come completely from her, definitely a sensitive period happening here at the moment.
We always have lots of read alouds and I am trying to add them to the reading log as we go along. I saw somewhere in the blogging world a family keeping 'book journals' and I love this idea. I hope to introduce them soon.
K completed more of her Singapore Preschool Math, this week was filled with counting, individual items and counting groups of items, number recognition and number writing practice. She also got an introduction to picture graphs, even numbers and making pairs.
She enjoyed all of the sensorial materials presented to her this week.
Colour Sorting Task, at first she found slightly difficult but she trudged on and got there in the end.
The Knobbed Cylinders, as this was the first time she has ever seen these I wasn't sure how many to present her with. I decided on 2 knowing that 1 would be far too simple. She breezed through the 2 of them and when I suggested we try and add a 3rd she jumped at the opportunity. For a moment or two she wasn't so sure that she had made the right decision, but this was perfect for her, still a challenge but one she could complete.
Fabric Sorting, was fun, slightly beneath her level but she wasn't up for trying the blindfold out just yet. I'll leave it on the shelf and see I guess.
This is a Charlie and Lola Balancing Game, made of timber. You have various sized wooden cylinders that you need to balance on the U shaped 'holder'. The kids used this indpendently this week, which is what I liked about this material, you don't have to play it as a game.
K is continuing to be very eager in the kitchen and almost daily comes and asks to help prepare dinner. We made a quick batch of brownies and some easy Mango Ice Blocks this week.
She has also been doing really well with her daily chores, making sure her room is tidy and her bed is made. In fact completely out of the blue this week she tidied an entire room that was pretty much covered with toys, all without me knowing she was doing it. Those little things bring tears to my eyes to see my children growing into beautiful, thoughtful, caring, people.
K watched two documentaries, one on Tornadoes and one about Whales.
To celebrate dad’s birthday and Australia Day we spent the weekend at Grandma’s and Grandad’s Unit at Mudjimba on the Sunshine Coast. We actually did the same thing last year, do I see a new family tradition emerging!!
The unit is situated whereby you open their gate and your toes are touching the water so when we stay there the kids are water babies and spend hours upon hours in the pool. I do attribute these extended times swimming to greatly improving their skills in the water. Katelyn had a fabulous weekend and has very nearly perfected her diving skills from being able to practice so much.
K had her first Gymnastics Class this week with our home school network. What a success that was for both her and I! She was very keen to go and super excited the night before (generally she is walking around the house saying she doesn’t want to go to activities). I reminded her that she wouldn’t be able to wear a dress (K is a dress girl) and was prepared for the argument that would follow but she simply chose a pair of shorts and shirt to match.
When the time came to head into the Gym class she just went off with everyone else as if she had been participating for weeks. Mummy had a grin from ear to ear, I was so very proud of her, she really has grown up so much and her confidence is growing in leaps and bounds.
She also mastered the Moneky Bars this week and was so proud of her achievement with that. She really doesn't like it when things seem to be beating her so that was a big notch on her belt.
We began history this week, I have already written about that here >>> I was pleasantly surprised to see that K was happy to participate in this. I never expected her to so that was a bonus.
Our first Art Lesson for the year with Miss Liz (our teacher we used last year) focused on colour theory, using colours, how colours make us feel, blending colours, making patterns with colours and using various art materials (paintbrush, rollers, tips) to paint and create with. The class also touched on the artist Kadinsky as their inspiration for this week’s art class.
The following was created from her collage tray and felt pens. She has really put a lot of thought into the use of the stickers and papers in this picture.
K and B also prepared a couple of 5 minute skits for me during the week. These things are generally always K’s idea but B is ever so obliging and most of the time goes along with it. First up was Star Wars. One guess why she chose that as her topic? She knew full well that B would never refuse her! No pics of that one but as you can imagine, lots of light sabers and ‘forcing’ each other, as they call it.
Then she turned on the charm and wanted to do Sleeping Beauty, B wasn’t impressed and I did have to remind him that she chose Star Wars earlier for him. He of course was the prince, complete with his homemade sword to chop down all of the vines to the castle tower. He wasn’t so impressed with the kissing, getting married and living happily ever after part. That was when C decided to join in and so he was quickly given the role of prince, after all he is generally more than willing to give plenty of kisses and hugs!
If I took pictures of K every time she danced I would have enough to fill entire blog just on dancing alone. However dancing in the rain is a little special. Our outdoor tiled area is generally always the stage for her dance productions. Currently it is not under a roof so when it rains it is the perfect spot.
You can see in the backgound our (in the process) pool and deck area. I keep forgetting to do a post on this, maybe this week I'll find some time.
That brings us to the end of our Preschool Corner for the week.
Following is B's week in review.
Loads of swimming practice over the weekend at the unit for B. His current love is seeing how far he can swim underwater so he’s building on his lung capacity at the moment and has also decided that he’d give Butterfly (the stroke) a go. I was pretty impressed as to how well he did, he hasn’t been shown butterfly at lessons and has really only seen the stroke from watching his dad in the pool.
B is taking a Mixed Sports Class this term with the homeschool network. This class was so much fun, he had a grin from ear to ear the entire time. One of the best things about it was the ages of the children. (7 – 14) normally B is one of the oldest but in the class he had quite a few that were considerably older than he was. The class included ball handling skills, trampoline, using the gym equipment, tennis skills etc definitely a hit.
He also chose to take Gymnastics this term and is the age group above K. I wasn’t sure how we would find gym but loved it, he concentrated really very well and enjoyed the balance beam a lot.
|
english
|
<?php namespace Altenia\Ecofy\Dao;
use Altenia\Ecofy\Dao\Schema\Schema;
use Altenia\Ecofy\Util\UuidUtil;
/**
* Helper class that provides HTML rendering functionalites.
*/
abstract class BaseDao {
protected $modelFqn;
protected $schema;
public function __construct($modelFqn)
{
$this->modelFqn = $modelFqn;
$this->schema = new Schema;
}
public function setSchema($schema)
{
$this->schema = $schema;
}
public function getSchema()
{
return $this->schema;
}
/**
* Returns model's classname
*/
public function modelClassName()
{
return '\\' . $this->modelFqn;
}
public function newModel()
{
$modelClassName = $this->modelClassName();
$model = new $modelClassName();
}
public function genUuid()
{
return UuidUtil::generate();
}
// Template method called prior insertion
public function beforeInsert(&$record)
{
// Make proper conversion of value representation, e.g. ISO date to MySQL's
/* Seems like Laravel getDates() already provide this functianality
$fields = $this->schema->getFields();
foreach ($fields as $fieldName => $dataType) {
if ($dataType->getName() === 'DateTime') {
if (isset($record->$fieldName)) {
$record->$fieldName = $this->toDbDateTime($record->$fieldName)
}
}
}*/
}
// Template method called prior update
public function beforeUpdate(&$record)
{}
abstract public function query($criteria, $sortParams = array(), $offset = 0, $limit=100);
/**
* Returns the count of records satisfying the critieria.
*
* @param array $queryParams Parameters used for querying
* @return int number of records that satisfied the criteria
*/
abstract public function count($criteria);
/**
* Inserts record.
* Mostly wrapper around insert with pre and post processing.
*
* @param array $record Model tha has already validated
* @return mixed null if successful, validation object validation fails
*/
abstract public function insert($record);
/**
* Retrieves a single record.
*
* @param int $pk The primary key for the search
* @return Transaction
*/
abstract public function find($criteria);
/**
* Retrieves a single record.
*
* @param int $pk The primary key for the search
* @return AccessControl
*/
abstract public function findByPK($pk);
/**
* Update the specified resource.
*
* @param int $pk The primary key of the record to update
* @param array $data The data of the update
* @return mixed null if successful, validation if validation error
*/
abstract public function update($record);
/**
* Update the fields.
*
* @param int $pk The primary key of the record to update
* @param array $data Fields to update
* @return mixed Returns the newely updated record
*/
abstract public function updateFields($pk, $data);
/**
* Remove the specified resource from storage.
*
* @param int $pk
* @return Object the object that was deleted, null if not found
*/
abstract public function delete($pk);
/**
* @param $date Either null or string in iso format
*/
abstract public function toDbDateTime($time);
}
|
code
|
जिया को लिफ्ट वाले घर की तलाश
जिया को लिफ्ट वाले घर की तलाश
जिया खान आजकल मुंबई में अपने लिये घर की तलाश में जुटी हुई है। ऐसा नहीं है कि वह बेघर है और फुटपाथ में रह रही है। दरअसल वह जिस इमारत में है, वहां लिफ्ट का इंतजाम नहीं है और उसका घर छठी मंजिल पर है। अब वह इतनी नाजुक भी नहीं है कि छह मंजिल तक सीढियों में चढाई ही नहीं कर सके।
असल में उसके साथ कुछ महिने से उसकी मां भी रहने आ गई है और मां को सीढियां चढ़ने उतरने में बहुत परेशानी का सामना करना पडता है। यही वजह है कि जिया को नये घर की तलाश है जिसमें कि लिफ्ट लगी हो ताकि उसकी मां को परेशानी का सामना ना करना पडे़।
जिहा चूंकि शूटिंग में लगातार व्यस्त रहती है, इसलिए उसे घर ढूंढने के लिए समय नहीं मिल पाता है। वह आमिर खान की फिल्म गजिनी कर रही है। उसने अपनी समस्या जब आमिर को बताई तो आमिर ने जिया को घर ढूंढने में पूरा सहयोग करने का वादा किया और उसकी मुलाकात स्टेट एजेंट से करवाकर दी। संभव है शीघ्र ही जिहा को ऐसा घर मिल जाये, जहां लिफ्ट लगी हो।
|
hindi
|
<!DOCTYPE html>
<html data-require="math">
<head>
<title>Exchange Sort Question: Inversions</title>
<script src="//cdnjs.cloudflare.com/ajax/libs/require.js/2.1.14/require.min.js"></script>
<script src="../../khan-exercises/local-only/main.js" ></script>
</head>
<body>
<div class="exercise">
<div class="problems">
<div id="ExchangeMCQinverp">
<p class="question">An inversion is:</p>
<div class="solution">When a record with key value greater than
the current record's key appears before it in the array</div>
<ul class="choices" >
<li>A swap</li>
<li>A type of sort</li>
</ul>
<div class="hints">
<p>We have not used "inversion" to refer to a type of sort.</p>
<p>While "inversions" are related to swaps (since ultimately a
swap is needed to undo an inversion), it is not a swap.</p>
<p>Inversion refers to an instance of a record being out of order.</p>
</div>
</div>
</div>
</div>
</body>
</html>
|
code
|
होम > उत्पादों > जल उपचार बहाव एलिमिनेटर
(जल उपचार बहाव एलिमिनेटर के लिए कुल २४ उत्पादों)
थोक चीन से जल उपचार बहाव एलिमिनेटर , लेकिन कम कीमत के अग्रणी निर्माताओं के रूप में सस्ते जल उपचार बहाव एलिमिनेटर खोजने की आवश्यकता है। बस जल उपचार बहाव एलिमिनेटर पर उच्च गुणवत्ता वाले ब्रांडों पा कारखाना उत्पादन, आप आप क्या चाहते हैं, बचत शुरू करते हैं और हमारे जल उपचार बहाव एलिमिनेटर का पता लगाने के बारे में भी राय, आप में सबसे तेजी से उत्तर हम करूँगा कर सकते हैं।
|
hindi
|
The complete mitochondrial genome of the Chinstrap penguin (Pygoscelis antarcticus) was sequenced and compared with other penguin mitogenomes. The genome is 15,972bp in length with the number and order of protein coding genes and RNAs being very similar to that of other known penguin mitogenomes. Comparative nucleotide analysis showed the Chinstrap mitogenome shares 94% homology with the mitogenome of its sister species, Pygoscelis adelie (Adélie penguin). Divergence at nonsynonymous nucleotide positions was found to be up to 23 times less than that observed in synonymous positions of protein coding genes, suggesting high selection constraints. The complete mitogenome data will be useful for genetic and evolutionary studies of penguins. © 2014 Informa UK Ltd. All rights reserved: reproduction in whole or part not permitted.
|
english
|
//------------------------------------------------------------------------------
// <auto-generated>
// This code was generated by a tool.
// Runtime Version:4.0.30319.0
//
// Changes to this file may cause incorrect behavior and will be lost if
// the code is regenerated.
// </auto-generated>
//------------------------------------------------------------------------------
namespace Crate.Core {
using System;
/// <summary>
/// A strongly-typed resource class, for looking up localized strings, etc.
/// </summary>
// This class was auto-generated by the StronglyTypedResourceBuilder
// class via a tool like ResGen or Visual Studio.
// To add or remove a member, edit your .ResX file then rerun ResGen
// with the /str option, or rebuild your VS project.
[global::System.CodeDom.Compiler.GeneratedCodeAttribute("System.Resources.Tools.StronglyTypedResourceBuilder", "4.0.0.0")]
[global::System.Diagnostics.DebuggerNonUserCodeAttribute()]
[global::System.Runtime.CompilerServices.CompilerGeneratedAttribute()]
internal class Resource {
private static global::System.Resources.ResourceManager resourceMan;
private static global::System.Globalization.CultureInfo resourceCulture;
[global::System.Diagnostics.CodeAnalysis.SuppressMessageAttribute("Microsoft.Performance", "CA1811:AvoidUncalledPrivateCode")]
internal Resource() {
}
/// <summary>
/// Returns the cached ResourceManager instance used by this class.
/// </summary>
[global::System.ComponentModel.EditorBrowsableAttribute(global::System.ComponentModel.EditorBrowsableState.Advanced)]
internal static global::System.Resources.ResourceManager ResourceManager {
get {
if (object.ReferenceEquals(resourceMan, null)) {
global::System.Resources.ResourceManager temp = new global::System.Resources.ResourceManager("Crate.Core.Resource", typeof(Resource).Assembly);
resourceMan = temp;
}
return resourceMan;
}
}
/// <summary>
/// Overrides the current thread's CurrentUICulture property for all
/// resource lookups using this strongly typed resource class.
/// </summary>
[global::System.ComponentModel.EditorBrowsableAttribute(global::System.ComponentModel.EditorBrowsableState.Advanced)]
internal static global::System.Globalization.CultureInfo Culture {
get {
return resourceCulture;
}
set {
resourceCulture = value;
}
}
/// <summary>
/// Looks up a localized string similar to Object can't be an IEnumerable!.
/// </summary>
internal static string IEnumerableError {
get {
return ResourceManager.GetString("IEnumerableError", resourceCulture);
}
}
}
}
|
code
|
होम झाँसी कारागार में लगाया साक्षरता शिविर
कारागार में लगाया साक्षरता शिविर
झांसी। राष्ट्रीय विधिक सेवा प्राधिकरण की योजना एवं उ.प्र राज्य विधिक सेवा प्राधिकरण के निर्देशानुसार १० दिवसीय अभियान के अन्तर्गत कारागार में निरूद्ध महिला बन्दियो एवं उनके साथ रह रहे बच्चों को विधिक सेवायें, स्वास्थ्य शिक्षा आदि सुविधायें उपलब्ध कराने एवं उनकी समस्याओं के विधिसम्मत निदान हेतु जनपद न्यायाधीश, अध्यक्ष जिला विधिक सेवा प्राधिकरण निसामद्दीन के मार्ग दर्शन में १० दिवसीय अभियान के अष्टम दिवस पर कारागार में निरूद्ध महिला बंदियो एवं उनके साथ निवास कर रहे बच्चों के लिये विधिक सेवा, साक्षरता शिविर का आयोजन किया गया।
मनोज कुमार तिवारी, सचिव जिला विधिक सेवा प्राधिकरण द्वारा अपने वक्तव्य में महिला बन्दियो एवं उसके साथ निवास कर रहे बच्चों के सम्बन्ध में जेल अधिकारियों, कर्मचारियों के कर्तव्यों के सम्बन्ध मंे विस्तारपूर्वक जानकारी दी गयी। महिला बन्दियों को साक्षर बनाये जाने हेतु शिक्षा विभाग की ओर से उपलब्ध करायी गयी। शिक्षण सामग्री भी वितरित की गयी। आर्ट आफ लिविंग संस्था के सहयोग से महिला बन्दियो को योग शिविर के माध्यम से योगाभ्यास भी कराया गया।
इस मौके पर राजीव शुक्ला, वरिष्ठ अधीक्षक जिला कारागार कैलाश चन्द्र जेलर, धर्मेन्द्र कुमार, ए.बी.आर.सी आर्ट आॅफ लिविंग से श्रीमती कंचन आहूजा, समाजसेविका श्रीमती रजनी वर्मा, श्रीमती वसुधा प्रेमानी आदि मौजूद रहे।
नेक्स्ट आर्टियलराशन कार्ड व गैस कनेक्शन में गड़बड़ी को लेकर प्रदर्शन कर ज्ञापन दिया
सनसाइन लेडीज क्लब ने वैवाहिक कन्या को दिये उपहार
विश्व मानवाधिकार दिवस पर दिया रेडियो बुंदेलखंड ९०.४ से सन्देश
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hindi
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Royal Mail honours Tommy Flowers with a first class Colossus stamp.
Article via The National Museum of Computing Web site.
* A recording of the sounds of the Colossus Rebuild have recently been made by sound artist and composer Matt Parker and will be deposited in the British Library sound archives. Get a link via the TNMOC article.
Click the "Return to List" link below to visit the Cryptologic Bytes Archives.
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साइकिल अभिलेखागार - एक गुस्सा गामेर
टैग अभिलेखागार: चक्र
महाकाव्य खेलों की दुकान से पता चलता है कि नए खेल समयबद्ध विशिष्टता के रूप में खरीदे जाते हैं
बाय एतान ऑन अगस्त २०, २०19
एपिक गेम्स किसी भी चीज को स्वीप करने की कोशिश कर रहे हैं और हर चीज पर उनका हाथ हो सकता है, जिसका मतलब है टिम स्वीनी ...
साइकल, को-ऑप साइंस-फाई शूटर एक और एपिक गेम्स स्टोर एक्सक्लूसिव है
बाय बिली डी ऑन मार्च २२, २०१९
हमने इन दिनों यागर से अब तक बहुत कुछ नहीं सुना है। स्टूडियो के बाद छः फुट और ... के साथ ड्रेडनॉट बाहर लुढ़का
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hindi
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# Porocladium aquaticum Descals, 1976 SPECIES
#### Status
ACCEPTED
#### According to
The Catalogue of Life, 3rd January 2011
#### Published in
in Descals, Nawawi & Webster, Trans. Br. mycol. Soc. 67(2): 211 (1976)
#### Original name
Porocladium aquaticum Descals, 1976
### Remarks
null
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code
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مگر پٔنٛزِس یِتھُے أمۍ سُنٛد اِرادٕ فِکرِ توٚر سُہ ژوٚل تیز تہٕ ترٛٲوٕن ناگس منٛز وۄٹھ
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kashmiri
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I know its inevitable that China will hack stuff. I guess my position is that we developers have a first movers advantage and can at least try to get in early and make a few bucks in China before our stuff gets hacked, rather than do nothing, make no money, and still get hacked.
My game now exists legally in both the Lenovo store and the Samsung store. The Samsung store is totally out-performing the Lenovo store, but still not nearly as good as the Google Play store. The Samsung store has a strict 10 day review period, before your app can exist in their store. There are no hacked copies of anything in the Samsung store and so that is why I think my app is performing better there. In the Lenovo store, everybody has already played the hacked copies of my game and have no interest in playing the legal copy.
I happened to be watching a tv show the other day called "Jungle Gold", where these two Americans pre-arrange with a local tribe chief to do some gold mining on his land in Ghana, Africa. When they arrive in Ghana, the mining location is already overrun by Chinese gold miners who are toting guns. The Americans say to the tribe chief "you promised us that we could do gold mining on your land, and we paid you money already to reserve your land, why are all these Chinese guys here instead?". The chief said "there was nothing I could do, the Chinese just showed up with guns and told me they are going to start mining my land!".
Everyone can do gold mining. All you need is a metal detector for gold.
Last edited by ShayneThill on Sat Sep 14, 2013 3:42 am, edited 1 time in total.
Well, started receiving some crash report on google analytics and found out my game was on some chinese stores.
Here's the interesting part, seems they can't play the game. As far as I can tell they are stuck without getting to the menu. The apk seems not altered, same byte count as the original.
I've put a check for google play services when the game starts, maybe this is preventing the game to continue?
Has anyone got any experience on this?
Traktor Digger 2 finally got accepted into the Samsung App Store after a 10 day review process. It was instantly placed in the Puzzle Top New chart at position #6 !!! ...even though it had zero downloads. It must be a human decision as to what's the initial position on the chart.
Anyway I suggest you all make sure you get your games into the Samsung App Store for extra visibility.
So by now we have pinpointed most of china's markets, but for the past few days, I am getting downloads+crash reports(missing GP services) from INDIA.
Google Analytics shows all downloads coming form india, and the devices are GIONEE a made in India manufacturer.
Does anybody knows about a third party market or webpage/facebook sharing apks in India?
This is very helpful topic. Thanks to the author.
I've been trying to get my game into chinese app stores, but have been having some difficulty. Googling the problem has lead me here. This thread has been extremely helpful, but there's one thing I just can't get round.
Just use this fake phone number….
But don't they use the number to send a confirmation code?
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english
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Today is my grandma's birthday; she's 91. (I'm sure she's thrilled I just told the whole internet that!) Anyway, I made this card for her by scanning a portrait of her as a young woman. I added the greeting (font is Richard Murray), printed it out, and attached it to a cream-coloured card using black and gold photo squares. It came out really nice, if I do say so myself!
Yeah, I know, it's been a while. I keep thinking about doing the Positives, then I forget. Then it's Tuesday, and I've missed it completely. But, not today: today I remembered, and I'm doing my Participation Positives right now!!
I used the dill on the salmon fillet I cooked for dinner and it was tres yummy!
We successfully moved my eight-year-old recliner from our house to my grandparent's house, where it's is now my grandpa's chair.
I am now hunting for a new chair. I have found two I like so far: the Andre leather recliner w/ ottoman, and the Courtney chair and matching ottoman. I just can't decide which I like best.
I'm leaning towards the Andre right now, because I've always wanted a leather chair, but the Courtney was soft and snuggly. I need to decide soon, however, as I'm having to use my office chair to watch TV , and it's definitely not very comfy!
See these shoes? Cute, aren't they? Yes, they are; they are adorable. However, they killed my feet. The left foot is covered in band-aids because there are no fewer than three blisteres on it; the right foot was not as badly damaged, but I covered with moleskin to protect it from its mate's fate. My left foot is still healing, and when it does, I'm going right back to those dang cute shoes. Why? Am I nuts? Well, yes, but that's a story for another time. No, the reason is that I LOVE THESE DANG CUTE SHOES!!! Plus, next time, I'll remember to put on the moleskin first!
"Enjoy Life!" said this mural on the side of Max and Erma's restaurant, and that is exactly what I did during my all-too-brief trip to Ohio to visit my friend Jewel. You can read all about it and see all the pictures here. Hope you enjoy it! Cheers!
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/*
* Copyright 2010-2015 Amazon.com, Inc. or its affiliates. All Rights Reserved.
*
* Licensed under the Apache License, Version 2.0 (the "License").
* You may not use this file except in compliance with the License.
* A copy of the License is located at
*
* http://aws.amazon.com/apache2.0
*
* or in the "license" file accompanying this file. This file is distributed
* on an "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either
* express or implied. See the License for the specific language governing
* permissions and limitations under the License.
*/
package com.amazonaws.services.cloudwatch.model.transform;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import com.amazonaws.AmazonClientException;
import com.amazonaws.Request;
import com.amazonaws.DefaultRequest;
import com.amazonaws.http.HttpMethodName;
import com.amazonaws.services.cloudwatch.model.*;
import com.amazonaws.transform.Marshaller;
import com.amazonaws.util.StringUtils;
/**
* EnableAlarmActionsRequest Marshaller
*/
public class EnableAlarmActionsRequestMarshaller
implements
Marshaller<Request<EnableAlarmActionsRequest>, EnableAlarmActionsRequest> {
public Request<EnableAlarmActionsRequest> marshall(
EnableAlarmActionsRequest enableAlarmActionsRequest) {
if (enableAlarmActionsRequest == null) {
throw new AmazonClientException(
"Invalid argument passed to marshall(...)");
}
Request<EnableAlarmActionsRequest> request = new DefaultRequest<EnableAlarmActionsRequest>(
enableAlarmActionsRequest, "AmazonCloudWatch");
request.addParameter("Action", "EnableAlarmActions");
request.addParameter("Version", "2010-08-01");
request.setHttpMethod(HttpMethodName.POST);
com.amazonaws.internal.SdkInternalList<String> alarmNamesList = (com.amazonaws.internal.SdkInternalList<String>) enableAlarmActionsRequest
.getAlarmNames();
if (!alarmNamesList.isEmpty() || !alarmNamesList.isAutoConstruct()) {
int alarmNamesListIndex = 1;
for (String alarmNamesListValue : alarmNamesList) {
if (alarmNamesListValue != null) {
request.addParameter("AlarmNames.member."
+ alarmNamesListIndex,
StringUtils.fromString(alarmNamesListValue));
}
alarmNamesListIndex++;
}
}
return request;
}
}
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code
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राकेश सिंह का कांग्रेस पर गंभीर आरोप, बोले- "मोटी रकम लेकर किए जा रहे तबादले'
जबलपुर। मध्यप्रदेश में हुए संपन्न हुए विधानसभा चुनाव के बाद से ही अधिकारियों के लगातार तबादले हो रहे थे। लेकिन लोकसभा चुनाव के मद्देनजर लागू आचार सहिंता के कारण इस पर अंकुश लग गया था। अब आचार सहिंता हटाई जाने के बाद फिर से अधिकारियों के तबादले होना शुरू हो गए हैं।
इसी बात को लेकर भाजपा के प्रदेशाध्यक्ष व जबलपुर सांसद राकेश सिंह ने प्रदेश सरकार पर तबादला उद्योग चलाने का आरोप लगा दिया है। पत्रकारों से चर्चा करते हुए राकेश सिंह ने प्रदेश सरकार पर आरोप लगाते हुए कहा है कि विधानसभा चुनावों में किसानों को राहत देने, बेरोजगारों को रोजगार देने और बिजली के बिल आधा करने जैसी बाते कांग्रेस ने कही थी। लेकिन सभी बातें झूठी निकली हैं। जिसको लेकर प्रदेश में कांग्रेस की सरकार के खिलाफ जनता में आक्रोश है।
राकेश सिंह ने मध्यप्रदेश सरकार पर धावा बोलते हुए कहा कि चुनाव खत्म होने के साथ ही प्रदेश में एक बार फिर तबादला उद्योग शुरू कर दिया है। कांग्रेस अपने आर्थिक हितों के लिए तबादला उद्योग जमकर चला रही है। राकेश सिंह ने कहा कि मध्यप्रदेश में अधिकारी यदि कांग्रेस के हित में काम नहीं कर रहे हैं, तो उनका तबादला कर दिया जा रहा है । कांग्रेस जनता के हितों में नहीं बल्कि पार्टी के हित में काम करने वाले अधिकारियों को तवज्जो दे रही है। साथ ही राकेश सिंह ने मोटी रकम लेकर तबादले किए जाने का आरोप भी प्रदेश सरकार पर लगाया है।
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hindi
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Situated on Leonard St., this spacious 3 br.,2 bath condo features an open concept concerning living room, dining rm., and kitchen with cathedral ceilings. The fully equipped oak kitchen also has laminate flooring and an island with breakfast bar. This condominium has a split bedroom design with the master BR., bath and walk-in closet on one side of the home, and 2 bedrooms and bath on the other side. At the rear of the home is really nice 10 x 16 family room/den giving the owners a separate space for relaxation. 2 car garage holds the electric, F.A.furnace, and central air, and electrical water heater. Wes Gar, Inc. has 1st Right of Refusal. $85 monthly association fee for snow removal, lawn maintained and 1x year bush trimming.
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english
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सुल्तानपुर। यूपी के सुल्तानपुर में शुक्रवार की सुबह कार और ट्रक की आमने-सामने की टक्कर हो गई। हादसे में कार सवार दो दरोगा की मौत हो गई है। वहीं, ३ लोग गंभीर रूप से घायल बताए जा रहे हैं। घायलों को अस्पताल में भर्ती करवाया गया है। सीएम योगी आदित्यनाथ ने दो दरोगा के निधन पर गहरा शोक व्यक्त किया है। वहीं, जिला प्रशासन के अधिकारियों को हादसे में घायल हुए लोगों का समुचित उपचार कराने के निर्देश दिए हैं।
प्रयागराज हाईकोर्ट जा रहे थे दोनों दरोगा
घटना सुल्तानपुर जिले केकूरेभार थाना इलाके की बताई जा रही है। जानकारी के मुताबिक, दोनों दरोगा कार से प्रयागराज हाईकोर्ट जा रहे थे। अयोध्या-प्रयागराज हाईवे पर रास्ते में जमौली के पास एक ट्रक ने उनकी कार को टक्कर मार दी। हादसे में कार के परखच्चे उड़ गए। कार सवार एक दरोगा की मौके पर ही मौत हो गई, जबकि दूसरे दरोगा ने अस्पताल में दम तोड़ दिया। वहीं, तीन अन्य घायल हैं जिनका अस्पताल में इलाज चल रहा है। मृतक दरोगाओं की पहचान नित्यानंद और राजकुमार यादव के रूप में हुई है। दोनों सुल्तानपुर जिले के चील्हपुर थाने में तैनात थे।
सीएम योगी आदित्यनाथ ने दो दरोगा के निधन पर गहरा शोक व्यक्त किया है। उन्होंने दिवंगत लोगों की शांति की कामना करते हुए मृतकों के शोक संतप्त परिजनों के प्रति अपनी संवेदना व्यक्त की है। वहीं, जिला प्रशासन के अधिकारियों को हादसे में घायल हुए लोगों का समुचित उपचार कराने के निर्देश दिए हैं।
ये भी पढ़ें: करीबियों पर हमले के बाद जेल से बाहर नहीं निकल रहा मुख्तार अंसारी, स्पेशल कोर्ट ने खारिज की जमानत
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hindi
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\begin{document}
\thispagestyle{empty}
\author{
{{\bf B. S. El-Desouky\footnote{Corresponding author: b\[email protected]} \; and Abdelfattah Mustafa}
\newline{\it{{} }}
} { }
\\
\small \it Department of Mathematics, Faculty of Science, Mansoura University,Mansoura 35516, Egypt
}
\title{New Results on Higher-Order Daehee and Bernoulli Numbers and Polynomials}
\date{}
\maketitle
\small \pagestyle{myheadings}
\markboth{{\scriptsize New Results on Higher-Order Daehee and Bernoulli Numbers and Polynomials }}
{{\scriptsize {El-Desouky and Mustafa}}}
\hrule \vskip 8pt
\begin{abstract}
We derive new matrix representation for higher order Daehee numbers and polynomials, the higher order $\lambda$-Daehee numbers and polynomials and the twisted $\lambda$-Daehee numbers and polynomials of order $k$. This helps us to obtain simple and short proofs of many previous results on higher order Daehee numbers and polynomials. Moreover, we obtained recurrence relation, explicit formulas and some new results for these numbers and polynomials. Furthermore, we investigated the relation between these numbers and polynomials and Stirling numbers, N\"{o}rlund and Bernoulli numbers of higher order. The results of this article gives a generalization of the results derived very recently by El-Desouky and Mustafa \cite{El-DesoukyMustafa2014}.
\end{document}nd{abstract}
\noindent
{\bf Keywords:}
{\it Daehee numbers, Daehee polynomials, Higher order Daehee numbers, Higher order Daehee polynomials, Higher order Bernoulli polynomials, Matrix representation.}
\noindent
{\bf \it 2010 MSC:}
{\end{document}m 05A19, 11C20 , 11B73, 11T06}
\section{Introduction}
For $\alpha \in \mathbb{N}$, the Bernoulli polynomials of order $\alpha$ are defined by, see \cite{Carlitz1961}-\cite{Ozdenetal2009}
\begin{equation}\label{1}
\left(\frac{t}{e^t-1}\right)^{\alpha} e^{xt}= \sum_{n=0}^\infty B_n^{(\alpha) } (x) \frac{ t^n}{n!}.
\end{document}nd{equation}
\noindent
When $x=0,B_n^{(\alpha) }=B_n^{(\alpha)} (0)$ are the Bernoulli numbers of order $\alpha$, defined by
\begin{equation}\label{2}
\left( \frac{t}{(e^t-1} \right)^{\alpha}=\sum_{n=0}^\infty B_n^{(\alpha) } \frac{ t^n}{n!}.
\end{document}nd{equation}
\noindent
The Daehee polynomials are defined by, see \cite{KimKim2013, KimSimsek2008} and \cite{Ozdenetal2009}.
\begin{equation} \label{3}
\left( \frac{\log(1+t)}{t} \right) (1+t)^x = \sum_{n=0}^\infty D_n (x) \frac{ t^n}{n!}.
\end{document}nd{equation}
\noindent
In the special case, $x=0,D_n=D_n (0)$ are called the Daehee numbers, defined by
\begin{equation} \label{4}
\left(\frac{\log(1+t)}{t} \right)=\sum_{n=0}^\infty D_n \frac{ t^n}{n!}.
\end{document}nd{equation}
\noindent
The Stirling numbers of the first and second kind are defined, respectively, by
\begin{equation} \label{5}
(x)_n = \prod_{i=0}^n (x-i) = \sum_{l=0}^n s_1 (n,l) x^l,
\end{document}nd{equation}
where $s_1(n,0)=\delta_{n,0}, \; s_1(n,k)=0, \; \mbox{for} \; k>n,$, and
\begin{equation} \label{5a}
x^n = \sum_{k=0}^n s_2 (n,k) (x)_k,
\end{document}nd{equation}
where $s_2(n,0)=\delta_{n,0}, \; s_2(n,k)=0, \; \mbox{for} \; k>n$, and $\delta_{n,k}$ is the kronecker delta.
\noindent
The Stirling numbers of the second kind have the generating function, see \cite{Carlitz1961,Comtet1974,El-Desouky1994,El-Desoukyetal2010} and \cite{Gould1972}.
\begin{equation} \label{6}
\left(e^t-1 \right)^m = m! \sum_{l=m}^\infty s_2 (l,m) \frac{ t^l}{l!}.
\end{document}nd{equation}
\section{Higher order Daehee Numbers and Polynomials}
In this section, we derive an explicit formulas and recurrence relations for the higher order Daehee numbers and polynomials of the first and second kinds. Also the relation between these numbers and N\"{o}rlund numbers are given. Furthermore, we introduce the matrix representation of some results for higher order Daehee numbers and polynomial obtained by Kim et al. \cite{Kimetal2014} in terms of Stirling numbers, N\"{o}rlund numbers and Bernoulli numbers of higher order and investigate a simple and short proofs of these results.
\noindent
Kim et al. \cite{Kimetal2014} defined the Daehee numbers of the first kind of order $k$, by the following generating function
\begin{equation} \label{7}
\sum_{n=0}^\infty D_n^{(k)} \frac{t^n}{n!}=\left(\frac{\log(1+t)}{t} \right)^k.
\end{document}nd{equation}
\noindent
Next, an explicit formula for $D_n^{(k) }$ is given by the following theorem.
\begin{theorem}
For $n\in \mathbb{Z},\; k \in \mathbb{N}$, we have
\begin{equation}\label{8}
D_n^{(k) } = n! \sum_{l_1+l_2+ \cdots +l_k=n+k } \frac{(-1)^n}{l_1 l_2 \cdots l_k }.
\end{document}nd{equation}
\end{document}nd{theorem}
\begin{proof}
From Eq. (\ref{7}), we have
\begin{equation*}
\sum_{n=0}^\infty D_n^{(k)} \frac{t^{n+k}}{n!}=\left(\log(1+t) \right)^k=\left( \sum_{l=1}^\infty \frac{(-1)^{l-1}}{l} t^l \right)^k.
\end{document}nd{equation*}
\noindent
Using Cauchy rule of product of series, we obtain
\begin{equation*}
\sum_{n=0}^\infty D_n^{(k)} \frac{t^{n+k}}{n!} = \sum_{r=k}^\infty \sum_{l_1+l_2+ \cdots +l_k=r}^\infty \frac{(-1)^{r-k}}{l_1 l_2 \cdots l_k } t^r,
\end{document}nd{equation*}
let $r-k=n$, in the right hand side, we have
\begin{equation*}
\sum_{n=0}^\infty D_n^{(k)} \frac{t^{n+k}}{n!}= \sum_{n=0}^\infty \sum_{l_1+l_2+ \cdots +l_k=n+k}^\infty \frac{(-1)^n}{l_1 l_2 \cdots l_k } t^{n+k}.
\end{document}nd{equation*}
Equating the coefficients of $t^{n+k}$ on both sides yields (\ref{8}). This completes the proof.
\end{document}nd{proof}
\begin{remark}
It is worth noting that setting $k=1$ in (\ref{8}), we get \cite[ Eq. (2.2)]{El-DesoukyMustafa2014} as a special case.
\end{document}nd{remark}
\noindent
Kim et al. \cite[2014, Theorem 1]{Kimetal2014} proved that, see \cite{Wang2010}, for $n \in \mathbb{Z}, \; k\in \mathbb{N}$, we have
\begin{equation} \label{9}
D_n^{(k) }=\frac{s_1 (n+k,k)}{\left(^{n+k}_{\; \; \; k} \right)}.
\end{document}nd{equation}
\noindent
We can represent the Daehee numbers of the first kind of order $k$, by $(n+1)\times(k+1)$ matrix , $0\leq k \leq n$, as follows
\begin{equation*}
{\bf D}^{(k) }=
\left(
\begin{array}{ccccc}
D_0^{(0)} & D_0^{(1)} & D_0^{(2)} & \cdots & D_0^{(k)} \\
D_1^{(0)} & D_1^{(1)} & D_1^{(2)} & \cdots & D_1^{(k)} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
D_n^{(0) }& D_n^{(1) }& D_n^{(2) } & \cdots & D_n^{(k) }
\end{document}nd{array}
\right).
\end{document}nd{equation*}
For example if $0\leq n \leq 3,\; 0 \leq k \leq n$, we have
\begin{equation*}
{\bf D}^{(k)}=
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
0 &-1/2 & -1 &-3/2 \\
0 & 2/3 & 11/6 & 7/2 \\
0 & -3/2 & -5 & -45/4 \\
\end{document}nd{array}
\right).
\end{document}nd{equation*}
\noindent
Kim et al. \cite[Theorem 4]{Kimetal2014}, proved the following result.
For $n \in \mathbb{Z}, \; k\in \mathbb{N}$, we have
\begin{equation}\label{10}
B_n^{(k)}=\sum_{m=0}^n D_m^{(k)} s_2 (n,m).
\end{document}nd{equation}
\noindent
\begin{remark}
We can write this relation in the matrix form as follows.
\begin{equation}\label{11}
{\bf B}^{(k)}= {\bf S}_2 \, {\bf D}^{(k)},
\end{document}nd{equation}
\end{document}nd{remark}
\noindent
where ${\bf D}^{(k) }$ is $(n+1)\times (k+1), \, 0\leq k\leq n,$ matrix for the Daehee numbers of the first kind of order $k$ and ${\bf S}_2$ is $(n+1)\times(n+1)$ lower triangular matrix for the Strirling numbers of the second kind and ${\bf B}^{(k) }$ is $(n+1)\times(k+1), \, 0\leq k \leq n,$ matrix for the Bernoulli numbers of order $k$.\\
For example, if setting $0\leq n \leq 3, \; 0 \leq k \leq n$ , in (\ref{11}), we have
\begin{equation*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 1 & 3 & 1
\end{document}nd{array}
\right)\left(
\begin{array}{ccccc}
1 & 1 & 1 & 1 \\
0 & -1/2 & -1 & -3/2 \\
0 & 2/3 & 11/6 & 7/2 \\
0 & -3/2 & -5 & -45/4
\end{document}nd{array}
\right)=\left(
\begin{array}{ccccc}
1 & 1 & 1 & 1 \\
0 & -1/2 & -1 &-3/2 \\
0 & 1/6 & 5/6 & 2 \\
0 & 0 & -1/2 & -9/4
\end{document}nd{array}
\right).
\end{document}nd{equation*}
\noindent
Kim et al. \cite[Theorem 3]{Kimetal2014} introduced the following result.
For $n\in \mathbb{Z},\; k\in \mathbb{N}$, we have
\begin{equation} \label{12}
D_n^{(k)}= \sum_{m=0}^n s_1 (n,m) B_m^{(k)}.
\end{document}nd{equation}
\noindent
We can write this relation in the matrix form as follows
\begin{equation} \label{13}
{\bf D}^{(k)}={\bf S}_1 \, {\bf B}^{(k)},
\end{document}nd{equation}
\noindent
where ${\bf S}_1$ is $(n+1)\times(n+1)$ lower triangular matrix for the Strirling numbers of the first kind.
\noindent
For example, if setting $0\leq n \leq 3, \; 0 \leq k \leq n$ , in (\ref{13}), we have
\begin{equation*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 2 & -3 & 1
\end{document}nd{array}
\right)\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
0 & -1/2 & -1 & -3/2 \\
0 & 1/6 & 5/6 & 2 \\
0 & 0 & -1/2 & -9/4
\end{document}nd{array}
\right)=
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
0 & -1/2 & -1 & -3/2 \\
0 & 2/3 & 11/6 & 7/2 \\
0 & -3/2 & -5 & -45/4
\end{document}nd{array}
\right).
\end{document}nd{equation*}
\begin{remark}
Using the matrix form (\ref{13}), we easily derive a short proof of Theorem 4 in Kim et al. \cite{Kimetal2014}. Multiplying both sides by the Striling number of second kind as follows.
\begin{equation*}
{\bf S}_2 \, {\bf D}^{(k) }= {\bf S}_2 \, {\bf S}_1 \, {\bf B}^{(k)}={\bf I\, B}^{(k)}={\bf B}^{(k) },
\end{document}nd{equation*}
\noindent
where {\bf I} is the identity matrix of order $(n+1)$.
\end{document}nd{remark}
\noindent
Kim et al. \cite{Kimetal2014} defined the Daehee polynomials of order $k$ by the generating function as follows.
\begin{equation} \label{14}
\sum_{n=0}^\infty D_n^{(k) } (x) \frac{t^n}{n!}=\left(\frac{\log(1+t)}{t}\right)^k (1+t)^x.
\end{document}nd{equation}
\noindent
Liu and Srivastava \cite{LiuSrivastava2006} define the N\"{o}rlund numbers of the second kind $b_n^{(x) }$ as follows.
\begin{equation} \label{15}
\left(\frac{t}{\log(1+t)} \right)^x = \sum_{n=0}^\infty b_n^{(x) } t^n.
\end{document}nd{equation}
\noindent
Next, we find the relation between the Daehee polynomials of order $k$ and the N\"{o}rlund numbers of the second kind $b_n^{(x)}$ by the following theorem.
\begin{theorem}
For $m\in \mathbb{Z},\; k\in \mathbb{N}$, we have
\begin{equation} \label{16}
D_m^{(k) } (z)=m! \sum_{n=0}^m \left(^{\; \; \;z}_{m-n} \right) b_n^{(-k)}.
\end{document}nd{equation}
\end{document}nd{theorem}
\begin{proof}
From Eq. (\ref{15}), by multiplying both sides by $(1+t)^z$, we have
\begin{eqnarray}\label{17}
\left(\frac{t}{\log(1+t)}\right)^x (1+t)^z & = & \sum_{n=0}^\infty b_n^{(x)} t^n (1+t)^z = \sum_{n=0}^\infty b_n^{(x) } t^n \sum_{i=0}^\infty \left(^z_i \right) t^i
\nonumber\\
&= & \sum_{n=0}^\infty \sum_{m=n}^\infty b_n^{(x)} \left(^{\;\; \;z}_{m-n} \right) t^m
= \sum_{m=0}^\infty \sum_{n=0}^m \left(^{\; \;\; z}_{m-n}\right) b_n^{(x) } t^m.
\end{document}nd{eqnarray}
\noindent
Replacing $x$ by $-k$ in (\ref{17}), we have
\begin{eqnarray} \label{18}
\left(\frac{\log(1+t)}{t}\right)^k (1+t)^z & = & \sum_{m=0}^\infty \sum_{n=0}^m \left(^{\; \;\;z}_{m-n} \right) b_n^{(-k)} t^m.
\nonumber\\
& = & \sum_{m=0}^\infty m! \sum_{n=0}^m \left(^{\;\; \; z}_{m-n} \right) b_n^{(-k) } \frac{ t^m}{m!}.
\end{document}nd{eqnarray}
\noindent
From (\ref{14}) and (\ref{18}), we have (\ref{16}). This completes the proof.
\end{document}nd{proof}
\noindent
\begin{corollary}
Setting $k=1$ in (\ref{16}) we have
\begin{equation} \label{19}
D_m (z) = m! \sum_{n=0}^m \left(^{\; \; \; z}_{m-n} \right) b_n^{(-1)}.
\end{document}nd{equation}
\end{document}nd{corollary}
\noindent
Setting $z=0$, in (\ref{16}), we have the following relation between Daehee numbers of higher order and N\"{o}rlund numbers of the second kind.
\begin{corollary}.
For $k\in \mathbb{N}$, by setting $z=0$ in (\ref{16}) we obtain
\begin{equation} \label{20}
D_m^{(k)}=m!\, b_m^{(-k)}.
\end{document}nd{equation}
\end{document}nd{corollary}
\noindent
The relation between the Bernoulli numbers and Bernoulli polynomials of order k are given by Kimura \cite{Kimura2003}, as follows.
\begin{equation} \label{21}
B_n^{(k)} (x) = \sum_{j=0}^n \left(^n_j\right) B_j^{(k)} x^{n-j}.
\end{document}nd{equation}
\noindent
Therefore, we can represent (\ref{21}) in the matrix form
\begin{equation} \label{22}
{\bf B}^{(k)} (x)= {\bf P}(x) \, {\bf B}^{(k) },
\end{document}nd{equation}
\noindent
where ${\bf B}^{(k)} (x)$ is $(n+1)\times(k+1)$ matrix, $0\leq k \leq n$ for Bernoulli polynomials of order $k$ as follows
\begin{equation*}
{\bf B}^{(k) } (x)=
\left(
\begin{array}{ccccc}
B_0^{(0)}(x) & B_0^{(1)}(x) & B_0^{(2)}(x) & \cdots & B_0^{(k)}(x)\\
B_1^{(0)}(x) & B_1^{(1)}(x) & B_1^{(2)}(x) & \cdots & B_1^{(k)}(x)\\
\vdots & \vdots & \vdots &\ddots &\vdots\\
B_n^{(0)}(x) & B_n^{(1)}(x) & B_n^{(2)}(x) & \cdots & B_n^{(k)}(x))
\end{document}nd{array}
\right),
\end{document}nd{equation*}
where the column $k$ represents the Bernoulli polynomials of order $k$, ${\bf B}^{(k)}$ is $(n+1)\times(k+1)$ matrix, $0\leq k \leq n$ for Bernoulli numbers of order $k$ and the matrix ${\bf P}(x)$, the Pascal matrix, is $(n+1)\times(n+1)$ lower triangular matrix defined by
\begin{equation*}
({\bf P}(x))_{ij}=
\left\{
\begin{array}{cl}
\left(^i_j \right) x^{i-j}, &i\geq j,\\
0, & \mbox{otherwise}
\end{document}nd{array}
\right.
, \; i,j=0,1,\cdots,n.
\end{document}nd{equation*}
\noindent
For example if setting $0 \leq n \leq 3, \; 0 \leq k \leq n$ in (\ref{22}), we have
\begin{eqnarray*}
& &
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
x & 1 & 0 & 0 \\
x^2 & 2x & 1 & 0 \\
x^3 & 3x^2 & 3x & 1
\end{document}nd{array}
\right)\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
0 & -1/2 & -1 &-3/2 \\
0 & 1/6 & 5/6 & 2 \\
0 & 0 & -1/2 & -9/4
\end{document}nd{array}
\right)=
\\
&&
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\ x & x- \frac{1}{2} & x-1 & x-\frac{3}{2} \\ x^2 & x^2-x+\frac{1}{6} & x^2-2x+\frac{5}{6} & x^2-3x+2 \\ x^3 & x^3-\frac{3}{2}x^2 + \frac{1}{2}x & x^3-3x^2+\frac{5}{2}x-\frac{1}{2} & x^3-\frac{9}{2}x^2+6x-\frac{9}{4}
\end{document}nd{array}
\right).
\end{document}nd{eqnarray*}
\noindent
Kim et al. \cite[Theorem 5]{Kimetal2014} introduced the following result. For $n\in \mathbb{Z}, \; k\in \mathbb{N}$,
\begin{equation} \label{23}
D_n^{(k)} (x) = \sum_{m=0}^n s_1 (n,m) B_m^{(k)} (x).
\end{document}nd{equation}
\noindent
We can write this relation in the matrix form as follows
\begin{equation} \label{24}
{\bf D}^{(k)} (x) = {\bf S}_1 \, {\bf B}^{(k)} (x),
\end{document}nd{equation}
\noindent
where ${\bf D}^{(k) } (x)$ is $(n+1)\times (k+1)$ matrix for the Daehee polynomials of the first kind with order $k$ and ${\bf B}^{(k) } (x)$ is $(n+1)\times(k+1)$ matrix for the Bernoulli polynomials of order $k$.
\noindent
For example, if setting $0 \leq n \leq 3, \; 0 \leq k \leq n$ , in (\ref{24}), we have
\begin{eqnarray*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 2 & -3 & 1
\end{document}nd{array}
\right)\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
x & x-\frac{1}{2} & x-1 & x-\frac{3}{2} \\
x^2 & x^2-x+\frac{1}{6} & x^2-2x+\frac{5}{6} & x^2-3x+2 \\
x^3 & x^3-\frac{3}{2}x^2+\frac{1}{2} x & x^3-3x^2+\frac{5}{2} x-\frac{1}{2} & x^3-\frac{9}{2}x^2+6x-\frac{9}{4}
\end{document}nd{array}
\right)=
\\
\; \; \; \; \; \; \; \;
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
x & x-\frac{1}{2} & x-1 & x-\frac{3}{2} \\
x^2-x & x^2-2x+\frac{2}{3} & x^2-3x+\frac{11}{6} & x^2-4x+\frac{7}{2} \\
x^3-3x^2+2x & x^3-\frac{9}{2}x^2+\frac{11}{2}x-\frac{3}{2} & x^3-6x^2+\frac{21}{2}x-5 & x^3-\frac{15}{2}x^2 + 17x-\frac{45}{4}\\
\end{document}nd{array}
\right)
\end{document}nd{eqnarray*}
\noindent
Kim et al. \cite[Theorem 7]{Kimetal2014} introduced the following result. For $n\in \mathbb{Z}, \; k\in \mathbb{N}$,
\begin{equation} \label{25}
B_n^{(k) } (x) = \sum_{m=0}^n D_m^{(k) } (x) s_2 (n,m).
\end{document}nd{equation}
\noindent
We can write Eq. (\ref{25}) in the matrix form as follows
\begin{equation}\label{26}
{\bf B}^{(k) } (x)={\bf S}_2 \, {\bf D}^{(k) } (x).
\end{document}nd{equation}
\noindent
For example, if setting $0≤\leq n \leq 3, \; 0 \leq k \leq n$ , in (\ref{26}), we have
{\footnotesize
\begin{eqnarray*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 1 & 3 & 1
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
x & x-\frac{1}{2} & x-1 & x- \frac{3}{2} \\
x^2-x & x^2-2x+\frac{2}{3} & x^2-3x+\frac{11}{6} & x^2-4x+\frac{7}{2} \\
x^3-3x^2+2x & x^3-\frac{9}{2} x^2+\frac{11}{2}x-\frac{3}{2} & x^3-6x^2+\frac{21}{2}x-5 & x^3-\frac{15}{2}x^2+17x-\frac{45}{4}
\end{document}nd{array}
\right)
\\
= \left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
x & x-\frac{1}{2} & x-1 & x-\frac{3}{2} \\
x^2 & x^2-x + \frac{1}{6} & x^2-2x+\frac{5}{6} &x^2-3x+2\\
x^3 & x^3-\frac{3}{2}x^2+\frac{1}{2} x & x^3-3x^2+\frac{5}{2}x-\frac{1}{2} & x^3-\frac{9}{2}x^2+6x-\frac{9}{4}
\end{document}nd{array}
\right).
\end{document}nd{eqnarray*}
}
\begin{remark} We can prove Theorem 7 in Kim et al. \cite{Kimetal2014} by using the matrix form (\ref{24}) as follows . Multiplying both sides of (\ref{24}) by the Striling number of second kind, we have
\begin{equation*}
{\bf S}_2 \, {\bf D}^{(k)} (x) ={\bf S}_2 \, {\bf S}_1 \, {\bf B}^{(k)} (x)={\bf I\, B}^{(k)} (x)={\bf B}^{(k)} (x).
\end{document}nd{equation*}
\end{document}nd{remark}
\noindent
Kim et al. \cite{Kimetal2014} defined the Daehee numbers of the second kind of order $k$ by the generating function as follows.
\begin{equation} \label{27}
\sum_{n=0}^\infty \hat{D}_n^k [(k)] \frac{ t^n}{n!}=\left(\frac{(1-t) \log(1-t)}{-t} \right)^k.
\end{document}nd{equation}
\noindent
Kim et al. \cite[Theorem 8]{Kimetal2014} introduced the following result. For $n\in \mathbb{Z}, \; k \in \mathbb{N}$,
\begin{equation} \label{28}
\hat{D}_n^k [(k)]=\sum_{l=0}^n \Big[^{\; n}_{\; l}\Big] B_l^{(k) },
\end{document}nd{equation}
\noindent
where $\Big[^{\; n}_{\; l} \Big]=(-1)^{n-l} s_1 (n,l)=|s_1(n,k)|=\mathfrak{s}(n,k)$, where $\mathfrak{s}(n,k)$ is the signless Stirling numbers of the first kind, see \cite{Comtet1974} and \cite{El-Desouky1994, El-Desoukyetal2010}.
\noindent
We can write this theorem in the matrix form as follows
\begin{equation}\label{29}
\hat{{\bf D}}^{(k)}= {\bf \mathfrak{S}} \, {\bf B}^{(k) },
\end{document}nd{equation}
\noindent
where $\hat{{\bf D}}^{(k) }$ is $(n+1)\times(k+1)$ matrix of Daehee numbers of the second kind with order $k$ and ${\bf \mathfrak{S}}$ is $(n+1)\times(n+1)$ lower triangular matrix for the signless Stirling numbers of first kind.
\noindent
For example, if setting $0 \leq n \leq 3,\; 0 \leq k \leq n$ in (\ref{29}), we have
\begin{equation*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 2 & 3 & 1
\end{document}nd{array}
\right)\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
0 & -1/2 & -1 & -3/2 \\
0 & 1/6 & 5/6 & 2 \\
0 & 0 & -1/2 & -9/4
\end{document}nd{array}
\right)=\left(
\begin{array}{cccc}
1 & 1 & 1 & 1\\
0 & -1/2 & -1 & -3/2\\
0 & -1/3 & -1/6 & 1/2\\
0 & -1/2 & 0 & 3/4
\end{document}nd{array}
\right).
\end{document}nd{equation*}
\noindent
Kim et al. \cite[Theorem 9]{Kimetal2014} introduced the following result.
For $n\in \mathbb{Z},\; k\in \mathbb{N}$, we have
\begin{equation} \label{30}
B_n^{(k)}= \sum_{m=0}^n (-1)^{n-m} s_2 (n,m) \hat{D}_m^k [(k)].
\end{document}nd{equation}
\noindent
We can write Eq. (\ref{30}) in the matrix form as follows
\begin{equation} \label{31}
{\bf B}^{(k)}= \tilde{\bf S}_2 \, \hat{{\bf D}}^{(k)}.
\end{document}nd{equation}
\noindent
where $\tilde{\bf S}_2$ is $(n+1)\times(n+1)$ lower triangular matrix for signed Stirling numbers of the second kind defined by
\begin{equation*}
({\tilde{\bf S}}_2)_{ij}=
\left\{
\begin{array}{cl}
(-1)^{i-j} s_2(i,j), & i\geq j, \\
0, & \mbox{otherwise}.
\end{document}nd{array}
\right.,
\; i,j=0,1,\cdots,n.
\end{document}nd{equation*}
\noindent
For example, if setting $0\leq n \leq 3,\; 0 \leq k \leq n$ in (\ref{31}), we have
\begin{equation*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 1 & -3 & 1
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1\\
0 & -1/2 & -1 & -3/2\\
0 & -1/3 & -1/6 & 1/2\\
0 & -1/2 & 0 & 3/4
\end{document}nd{array}
\right)=
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
0 & -1/2 & -1 & - 3/2 \\
0 & 1/6 & 5/6 & 2 \\
0 & 0 & - 1/2 & - 9/4
\end{document}nd{array}
\right).
\end{document}nd{equation*}
\noindent
\begin{remark}
We can prove Theorem 9 in Kim et al. \cite{Kimetal2014} by using the matrix form (\ref{29}) as follows.\\
Multiplying both sides of (\ref{29}) by the matrix of sign Striling numbers of second kind $\tilde{\bf S}_2$ we have
\begin{equation*}
\tilde{\bf S}_2 \, \hat{{\bf D}}^{(k) }=\tilde{\bf S}_2 \, {\bf \mathfrak{S}} \, {\bf B}^{(k)}={\bf I} \, {\bf B}^{(k)}= {\bf B}^{(k)},
\end{document}nd{equation*}
\noindent
we obtain Eq. (\ref{31}), where we used the identity, $\tilde{\bf S}_2 \, {\bf \mathfrak{S}}={\bf I}$.
\end{document}nd{remark}
\noindent
Kim et al. \cite{Kimetal2014} defined the Daehee polynomials of the second kind of order $k$ by the generating function as follows.
\begin{equation} \label{32}
\sum_{n=0}^\infty \hat{D}_n^k [(k)] (x) \frac{t^n}{n!}=\left(\frac{(1-t) \log(1-t)}{-t} \right)^k (1-t)^x.
\end{document}nd{equation}
\noindent
Kim et al. \cite[Eq. (31)]{Kimetal2014} introduced the following result. For $n\in \mathbb{Z},\; k\in \mathbb{N}$,
\begin{equation} \label{33a}
\hat{D}_n^k [(k)](x)= \sum_{m=0}^n (-1)^{n-m} s_1 (n,m) B_m^{(k) } (-x).
\end{document}nd{equation}
\noindent
Eq. (\ref{33a}) is equivalent to
\begin{equation} \label{33}
\hat{D}_n^k [(k)](x)= \sum_{m=0}^n \mathfrak{s} (n,m) B_m^{(k) } (-x).
\end{document}nd{equation}
\noindent
We can write Eq. (\ref{33}) in the matrix form as follows
\begin{equation} \label{34}
\hat{{\bf D}}^{(k) } (x) = {\bf \mathfrak{S}} \, {\bf B}^{(k)} (-x),
\end{document}nd{equation}
\noindent
where $\hat{{\bf D}}^{(k) } (x) $ is $(n+1)\times(k+1)$ matrix of the Daehee polynomials of the second kind with order $k$ and ${\bf B}^{(k) } (x)$ is $(n+1)\times(k+1)$, the matrix of the Bernoulli polynomials when $x\rightarrow -x $ numbers.
\noindent
For example, if setting $0 \leq n \leq 3,\; 0 \leq k \leq n,$ in (\ref{34}), we have
\begin{eqnarray*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 2 & 3 & 1
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
-x & -x-\frac{1}{2} & -x-1 & -x-\frac{3}{2} \\
x^2 & x^2+x+\frac{1}{6} & x^2+2x+\frac{5}{6} & x^2+3x+2 \\
-x^3 & -x^3-\frac{3}{2}x^2-\frac{1}{2}x & -x^3-3x^2-\frac{5}{2}x-\frac{1}{2} & -x^3-\frac{9}{2}x^2-6x-\frac{9}{4}
\end{document}nd{array}
\right)
\\
=
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
-x & -x-\frac{1}{2} & -x-1 & -x-\frac{3}{2} \\
x^2-x & x^2-\frac{1}{3} & x^2+x-\frac{1}{6} & x^2+2x+\frac{1}{2} \\
3x^2-x^3-2x & \frac{3x^2}{2}-x^3+\frac{x}{2}-\frac{1}{2} & \frac{3x}{2}-x^3 & x-\frac{3x^2}{2}-x^3+ \frac{3}{4}
\end{document}nd{array}
\right).
\end{document}nd{eqnarray*}
\noindent
Kim et al. \cite[Theorem 11]{Kimetal2014} introduced the following result. For $n\in \mathbb{Z}, \;k\in \mathbb{N}$,
\begin{equation} \label{35}
B_n^{(k)} (-x) = \sum_{m=0}^n (-1)^{n-m} s_2 (n,m) \hat{D}_m^k [(k)](x).
\end{document}nd{equation}
\noindent
We can write Eq. (\ref{35}) in the matrix form as follows
\begin{equation}\label{36}
{\bf B}^{(k) } (-x)= \tilde{\bf S}_2 \, \hat{{\bf D}}^{(k)} (x).
\end{document}nd{equation}
\noindent
For example, if setting $0\leq n \leq 3, \; 0 \leq k \leq n$ in (\ref{36}), we have
\begin{eqnarray*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 1 & -3 & 1 \\
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
-x & -x-\frac{1}{2} & -x-1 & -x-\frac{3}{2} \\
x^2-x & x^2-\frac{1}{3} & x^2+x-\frac{1}{6} & x^2+2x+\frac{1}{2} \\
3x^2-x^3-2x & \frac{3x^2}{2}-x^3+\frac{x}{2}-\frac{1}{2} & \frac{3x}{2}-x^3 & x-\frac{3x^2}{2}-x^3+ \frac{3}{4}
\end{document}nd{array}
\right)
\\
= \left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
-x & -x-\frac{1}{2} & -x-1 & -x-\frac{3}{2} \\
x^2 & x^2+x+\frac{1}{6} & x^2+2x+\frac{5}{6} & x^2+3x+2 \\
-x^3 & -x^3-\frac{3x^2}{2}-\frac{x}{2} & -x^3-3x^2-\frac{5x}{2}-\frac{1}{2} & -x^3-\frac{9x^2}{2}-6x-\frac{9}{4}
\end{document}nd{array}
\right).
\end{document}nd{eqnarray*}
\begin{remark}
We can prove Eq. (\ref{36}), \cite[Theorem 11]{ Kimetal2014}, directly by using the matrix form (\ref{34}) as follows. Multiplying both sides of (\ref{34}) by $\tilde{\bf S}_2$ as follows.
\begin{equation*}
\tilde{\bf S}_2 \, \hat{{\bf D}}^{(k)} (x)=\tilde{\bf S}_2 \, {\bf \mathfrak{S}} \, {\bf B}^{(k)} (-x)={\bf I} \, {\bf B}^{(k)} (-x)={\bf B}^{(k)} (-x),
\end{document}nd{equation*}
thus, we have Eq. (\ref{36}).
\end{document}nd{remark}
\section{The $\lambda$- Daehee Numbers and Polynomials of Higher Order}
In this section we introduce the matrix representation for the $\lambda$-Daehee numbers and polynomials of higher order given by Kim et al. \cite{Kimetal2013}. Hence, we can derive these results in matrix representation and prove these results simply by using the given matrix forms.
\noindent
The $\lambda$-Daehee polynomials of the first kind with order $k$ can be defined by the generating function
\begin{equation} \label{37}
\left(\frac{\lambda \log(1+t)}{(1+t)^{\lambda}-1} \right)^k (1+t)^x = \sum_{n=0}^\infty D_{n,\lambda}^{(k)} (x) \frac{ t^n}{n!}.
\end{document}nd{equation}
\noindent
When $x=0,\; D_{n,\lambda}^{(k)}=D_{n, \lambda}^{(k)} (0)$ are called the $\lambda$-Daehee numbers of order $k$.
\begin{equation} \label{ln}
\left(\frac{\lambda \log(1+t)}{(1+t)^{\lambda}-1} \right)^k = \sum_{n=0}^\infty D_{n,\lambda}^{(k)} \frac{t^n}{n!}.
\end{document}nd{equation}
\noindent
It is easy to see that $D_n^{(k) } (x) = D_{n,1}^{(k) } (x)$ and $D_{n,\lambda} (x)=D_{n,\lambda}^{(1) } (x)$.
\noindent
Kim et al. \cite[Theorem 3]{Kimetal2013} obtained the following results. For $n\geq 0, \;k \in \mathbb{N}$,
\begin{equation} \label{38}
D_{n,\lambda}^{(k)} (x)= \sum_{m=0}^n s_1 (n,m) \lambda^m B_m^{(k)} \left(\frac{x}{\lambda} \right),
\end{document}nd{equation}
and
\begin{equation} \label{39}
\lambda^n B_n^{(k)} \left(\frac{x}{\lambda}\right)=\sum_{m=0}^n s_2 (n,m) D_{m,\lambda}^{(k) } (x),
\end{document}nd{equation}
\noindent
we can write these results in the following matrix forms
\begin{equation} \label{40}
{\bf D}_{\lambda}^{(k)} (x)= {\bf S}_1 \, {\bf \Lambda \, B}^{(k) } \left(\frac{x}{\lambda} \right),
\end{document}nd{equation}
and
\begin{equation} \label{41}
{\bf \Lambda}\, {\bf B}^{(k)} \left(\frac{x}{\lambda} \right)= {\bf S}_2\, {\bf D}_{\lambda}^{(k) } (x),
\end{document}nd{equation}
\noindent
where, ${\bf D}_{\lambda}^{(k)} (x)$ is $(n+1)\times (k+1)$ matrix for the $\lambda$-Daehee polynomials of the first kind with order $k$, ${\bf B}^{(k)} (x/\lambda)$ is $(n+1)\times(k+1)$ matrix for the Bernoulli polynomials of order $k$, when $x\rightarrow x/\lambda$ and ${\bf \Lambda}$ is $(n+1)\times(n+1)$ diagonal matrix with elements, $({\bf \Lambda})_{ii}=\lambda^i,\; i=j=0,1,\cdots,n$.
\noindent
For example, if setting $0 \leq n \leq 3,\; 0\leq k \leq n$, in (\ref{40}), we have
{\footnotesize
\begin{eqnarray*}
&&
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 2 & -3 & 1
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \lambda & 0 & 0 \\
0 & 0 & \lambda^2 & 0 \\
0 & 0 & 0 & \lambda^3
\end{document}nd{array}
\right) \times
\\
&&
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
\frac{x}{\lambda} & \frac{x}{\lambda}-\frac{1}{2} & \frac{x}{\lambda}-1 & \frac{x}{\lambda}-\frac{3}{2} \\
\frac{x^2}{\lambda^2} & \frac{x^2}{\lambda^2} -\frac{x}{\lambda} + \frac{1}{6} & \frac{x^2}{\lambda^2} -\frac{2x}{\lambda} + \frac{5}{6} & \frac{x^2}{\lambda^2} -\frac{3 x}{\lambda} + 2 \\
\frac{x^3}{\lambda^3} & \frac{x}{2\lambda}-\frac{3x^2}{2\lambda^2} + \frac{x^3}{\lambda^3} & \frac{5x}{2\lambda}- \frac{3x^2}{\lambda^2} + \frac{x^3}{\lambda^3} -\frac{1}{2} & \frac{6x}{\lambda}-\frac{9x^2}{2\lambda^2} + \frac{x^3}{\lambda^3} -\frac{9}{4}
\end{document}nd{array}
\right)=
\\
&&
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1\\
x & x-\frac{\lambda}{2} & x-\lambda & x-\frac{3\lambda}{2} \\
x(x-1)& \frac{\lambda^2}{6} - \lambda x+ \frac{\lambda}{2} + x^2-x & \frac{5\lambda^2}{6}- 2\lambda x + \lambda + x^2-x & 2\lambda^2 - 3 \lambda x + \frac{3\lambda}{2} + x^2-x \\
D_{3,\lambda}^{(0)}(x)& D_{3,\lambda}^{(1)}(x) & D_{3,\lambda}^{(2)}( x) & D_{3,\lambda}^{(3)}(x)
\end{document}nd{array}
\right).
\end{document}nd{eqnarray*}
}
where
\begin{eqnarray*}
D_{3,\lambda}^{(0)}(x)& = &x(x-1)(x-2), \;
D_{3,\lambda}^{(1)}(x)=- \frac{1}{2}(\lambda -2x+2)(\lambda-2x + x^2-\lambda x),
\\
D_{3,\lambda}^{(2)}(x) & = &-\frac{1}{2} (\lambda -x+1)(\lambda^2- 4\lambda x + 4\lambda + 2x^2-4x),
\\
D_{3,\lambda}^{(3)}(x)& = & -\frac{1}{4} (3\lambda-2x+2)(3\lambda^2-6\lambda x+6 \lambda + 2x^2-4x).
\end{document}nd{eqnarray*}
\begin{remark} In fact, we can prove Eq. (\ref{41}), simply by multiplying Eq. (\ref{40}) by ${\bf S}_2$ as follows.
\begin{equation*}
{\bf S}_2 \, {\bf D}_{\lambda}^{(k)} (x)={\bf S}_2 \, {\bf S}_1 \, {\bf \Lambda B}^{(k)} \left(\frac{x}{\lambda} \right)={\bf I \, \Lambda B}^{(k) } \left(\frac{x}{\lambda} \right)={\bf \Lambda \, B}^{(k)} \left(\frac{x}{\lambda} \right).
\end{document}nd{equation*}
\end{document}nd{remark}
\noindent
The following Theorem gives the relation between the Daehee polynomials of higher order and $\lambda$-Daehee polynomials of higher order.
\begin{theorem}
For $m \geq 0$, we have
\begin{equation} \label{42}
D_{m,\lambda}^{(k) } (\lambda x)=m! \sum_{n=0}^m \sum_{ i_1+i_2+ \cdots+i_n=m} \frac{D_n^{(k)}(x)}{n!} \left(^{\lambda}_{i_1} \right) \left(^{\lambda}_{i_2}\right) \cdots \left(^{\lambda}_{i_n} \right).
\end{document}nd{equation}
\end{document}nd{theorem}
\begin{proof}
From (\ref{14}), replacing $(1+t)$ by $(1+t)^\lambda$, we have
\begin{equation*}
\left(\frac{\lambda \log(1+t)}{(1+t)^\lambda-1}\right)^k (1+t)^{\lambda x}= \sum_{n=0}^\infty D_n^{(k)} (x) \frac{((1+t)^\lambda-1 )^n}{n!},
\end{document}nd{equation*}
thus from (\ref{37}), we get
\begin{eqnarray*}
\sum_{m=0}^\infty D_{m,\lambda}^{(k)} (\lambda x) \frac{ t^m}{m!} &= & \sum_{n=0}^\infty \frac{D_n^{(k)} (x)}{n!} \left(
\sum_{i=0}^\lambda \left(^{\lambda}_i \right) t^i -1 \right)^n
\\
&= &
\sum_{n=0}^\infty \frac{D_n^{(k)} (x)}{n!} \left(
\sum_{i=1}^\lambda \left(^{\lambda}_i \right) t^i \right)^n.
\end{document}nd{eqnarray*}
\noindent
Using Cauchy rule of product of series, we obtain
\begin{eqnarray*}
\sum_{m=0}^\infty D_{m,\lambda}^{(k)} (\lambda x) \frac{t^m}{m!} & = &
\sum_{n=0}^\infty \frac{D_n^{(k)} (x))}{n!} \sum_{m=n}^\infty \sum_{i_1+i_2+ \cdots +i_n=m} \left(^{\lambda}_{i_1} \right) \cdots \left(^{\lambda}_{i_n} \right) t^m
\\
&= &
\sum_{m=0}^\infty m! \sum_{n=0}^m \sum_{i_1+i_2+ \cdots +i_n=m}
\frac{D_n^{(k)} (x))}{n!} \left(^{\lambda}_{i_1} \right) \cdots \left(^{\lambda}_{i_n} \right) \frac{t^m}{m!}.
\end{document}nd{eqnarray*}
\noindent
Equating the coefficients of $t^m$ on both sides yields (\ref{42}). This completes the proof.
\end{document}nd{proof}
\noindent
Setting $x=0$, in (\ref{42}), we have the following corollary as a special case.
\begin{corollary}
For $m \geq 0$, we have
\begin{equation} \label{43}
D_{m,\lambda}^{(k)} = m! \sum_{n=0}^m \sum_{i_1+i_2+ \cdots+i_n=m} \frac{D_n^{(k) }}{n!} \left(^{\lambda}_{i_1} \right)\left(^{\lambda}_{i_2} \right) \cdots \left(^{\lambda}_{i_n} \right).
\end{document}nd{equation}
\end{document}nd{corollary}
\noindent
Kim et al. \cite{Kimetal2013}, defined the $\lambda$-Daehee polynomials of the second kind with order $k$ as follows.
\begin{equation} \label{44}
\left( \frac{\lambda \log(1+t)}{(1+t)^\lambda-1} \right)^k (1+t)^{\lambda k+x}) = \sum_{n=0}^\infty \hat{D}_{n,\lambda}^{(k)} (x)\frac{ t^n}{n!}.
\end{document}nd{equation}
\noindent
Kim et al. \cite[Theorem 5]{Kimetal2013} proved that
\begin{equation} \label{45}
\hat{D}_{m, \lambda}^{(k)} (x) = \sum_{l=0}^m s_1 (m,l) \lambda^l B_l^{(k)} \left(k+\frac{x}{\lambda} \right),
\end{document}nd{equation}
and
\begin{equation} \label{46}
\lambda^m B_m^{(k) } \left(k+\frac{x}{\lambda}\right)= \sum_{n=0}^m s_2 (m,n) \hat{D}_{n,\lambda}^{(k)} (x).
\end{document}nd{equation}
\noindent
Also, Kim et al. \cite[Eq. (35)]{Kimetal2013} introduced the following result
\begin{equation} \label{47}
B_n^{(k) } (k-x)=(-1)^n B_n^{(k)}(x).
\end{document}nd{equation}
\noindent
\begin{remark}
We can write (\ref{45}) and (\ref{46}), respectively, in the following matrix forms
\begin{equation} \label{48}
\hat{\bf D}_{\lambda}^{(k)} (x)={\bf S}_1 \, {\bf \Lambda}_1 \, {\bf B}^{(k)} \left(-\frac{x}{\lambda}\right),
\end{document}nd{equation}
and
\begin{equation} \label{49}
{\bf \Lambda}_1 \, {\bf B}^{(k)} \left(-\frac{x}{\lambda}\right)= {\bf S}_2 \, \hat{\bf D}_{\lambda}^{(k)} (x),
\end{document}nd{equation}
\end{document}nd{remark}
\noindent
where $\hat{\bf D}_\lambda (x)$ is $(n+1)\times(n+1)$ matrix for the $\lambda$-Daehee polynomials of the second kind of order $k$ and ${\bf \Lambda}_1$ is $(n+1)\times(n+1)$ diagonal matrix with elements $({\bf \Lambda}_1 )_{ii}=(-\lambda)^i$, for $i=j=0,1,\cdots ,n$.
\noindent
For example, if setting $0\leq n \leq 3, \;0 \leq k \leq n$, in (\ref{48}), we have
{\footnotesize
\begin{eqnarray*}
& &
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 2 & -3 & 1
\end{document}nd{array}
\right)\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & -\lambda & 0 & 0 \\
0 & 0 & \lambda^2 & 0 \\
0 & 0 & 0 & -\lambda^3
\end{document}nd{array}
\right)
\times
\\
& &
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
-\frac{x}{\lambda} & -\frac{x}{\lambda}-\frac{1}{2} &- \frac{x}{\lambda}-1 &-\frac{x}{\lambda}-\frac{3}{2} \\
\frac{x^2}{\lambda^2} & \frac{1}{\lambda^2}(\frac{\lambda^2}{6}+\lambda x+x^2 )& \frac{1}{\lambda^2}( \frac{5}{6} \lambda^2+2\lambda x+x^2 ) & \frac{1}{\lambda^2}(2\lambda + x)(\lambda + x) \\
-\frac{x^3}{\lambda^3} & \frac{x(\lambda + 2x)( \lambda + x)}{2\lambda^3 } &-\frac{(\lambda + x)(\lambda^2+4\lambda x+2x^2 )}{2\lambda^3} &-\frac{(3\lambda +2x)(3\lambda^2+6\lambda x+2x^2 )}{4\lambda^3}
\end{document}nd{array}
\right)=
\\
& &
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1\\
x & \frac{\lambda}{2}+x & \lambda+x &\frac{3\lambda}{2} + x \\
x(x-1) & \frac{\lambda^2 }{6}+\lambda x-\frac{\lambda}{2}+x^2-x & \frac{5\lambda^2}{6} +2\lambda x-\lambda+x^2-x &2\lambda^2+3\lambda x-\frac{3\lambda}{2} +x^2-x \\
\hat{D}_{3,\lambda}^{(0)}&\hat{D}_{3,\lambda}^{(1)}(x)& \hat{D}_{3,\lambda}^{(2)}(x) & \hat{D}_{3,\lambda}^{(3)}(x)
\end{document}nd{array}
\right).
\end{document}nd{eqnarray*}
}
where
\begin{eqnarray*}
\hat{D}_{3,\lambda}^{(0)}(x) & =& x(x-1)(x-2), \; \hat{D}_{3,\lambda}^{(1)}(x)=-\frac{1}{2}(\lambda+2x-2)(\lambda+2x-x^2-\lambda x)
\\
\hat{D}_{3,\lambda}^{(2)}(x)& = & \frac{1}{2} (\lambda + x-1)(\lambda^2 + 4\lambda x-4\lambda + 2x^2-4x),
\\
\hat{D}_{3,\lambda}^{(3)}(x)&=& \frac{1}{4} (3\lambda+2x-2)(3\lambda^2+6\lambda x-6\lambda+2x^2-4x))).
\end{document}nd{eqnarray*}
\begin{remark}
We can prove Eq. (\ref{46}) easily by using the matrix form, multiplying Eq.(\ref{48}) by ${\bf S}_2$ as follows.
\begin{equation*}
{\bf S}_2 \, \hat{\bf D}_\lambda^{(k)} (x)={ \bf S}_2 \, {\bf S}_1 \, { \bf \Lambda}_1 \, {\bf B}^{(k)} \left(-\frac{x}{\lambda} \right)={\bf I} \, {\bf \Lambda}_1 \, {\bf B}^{(k)} \left(- \frac{x}{\lambda} \right)={\bf \Lambda}_1 \, {\bf B}^{(k)} \left(- \frac{x}{\lambda} \right).
\end{document}nd{equation*}
\end{document}nd{remark}
\section{The Twisted $\lambda$-Daehee Numbers and Polynomials of Higher Order}
Kim et al. \cite{Kimetal2013b} defined the twisted $\lambda$-Daehee polynomials of the first kind of order $k$ by the generating function
\begin{equation} \label{50}
\left(\frac{\lambda \log(1+\xi t)}{(1+\xi t)^\lambda-1} \right)^k (1+\xi t)^x = \sum_{n=0}^\infty D_{n,\xi }^{(k)} (x|\lambda ) \frac{ t^n}{n!}.
\end{document}nd{equation}
\noindent
In the special case, $x=0,\; D_{n,\xi, \lambda}^{(k) } = D_{n,\xi }^{(k)} (0|\lambda)$ are called the twisted $\lambda$-Daehee numbers of the first kind of order $k$.
\begin{equation} \label{51}
\left( \frac{ \lambda \log(1+\xi t)}{(1+\xi t)^\lambda-1} \right)^k = \sum_{n=0}^\infty D_{n,\xi,\lambda}^{(k)} \frac{ t^n}{n!}.
\end{document}nd{equation}
\noindent
The twisted Bernoulli polynomials of order $r\in \mathbb{N}$ are defined by the generating function, see \cite{Dolgyetal2013}
\begin{equation} \label{52}
\left(\frac{t}{\xi e^t-1} \right)^r e^{xt}=\sum_{n=0}^\infty B_{n,\xi}^{(r)} (x) \frac{t^n}{n!}.
\end{document}nd{equation}
\noindent
The relation between the twisted $\lambda$-Daehee polynomials and $\lambda$-Daehee polynomials of order $k$, can be obtained by the following corollary.
\begin{corollary}
For $n \geq 0, \, k\in \mathbb{N}$, we have
\begin{equation} \label{a1}
D_{n,\xi}^{(k) } (x|\lambda)=\xi^n D_{n,\lambda}^{(k) } (x).
\end{document}nd{equation}
\end{document}nd{corollary}
\begin{proof}
Replacing $t$ with $\xi t$ in (\ref{37}), we have
\begin{equation} \label{a2}
\left( \frac{\lambda \log(1+\xi t)}{(1+\xi t)^\lambda-1} \right)^k (1+\xi t)^x= \sum_{n=0}^\infty D_{n,\lambda}^{(k)} (x)
\frac{(\xi t)^n}{n!} = \sum_{n=0}^\infty \xi^n D_{n,\lambda}^{(k) } (x) \frac{t^n}{n!},
\end{document}nd{equation}
\noindent
hence by virtue of (\ref{50}) and (\ref{a2}), we get (\ref{a1}). This completes the proof.
\end{document}nd{proof}
\noindent
If we put in (\ref{a1}), $x=0$ and $\lambda=1$, respectively, we have
\begin{equation*}
D_{n,\xi,\lambda}^{(k)}=\xi^n D_{n,\lambda}^{(k)}.
\end{document}nd{equation*}
and
\begin{equation*}
D_{n,\xi}^{(k)} (x)=\xi^n D_n^{(k)}(x).
\end{document}nd{equation*}
\noindent
Kim et al. \cite[Theorem 1]{Kimetal2013b} proved the following relation. For $m\in \mathbb{Z}, \; k\in \mathbb{N}$,
\begin{equation} \label{53}
D_{m,ξ}^{(k)} (x│ \lambda )=\xi^m \sum_{l=0}^m S_1 (m,l) \lambda^l B_l^{(k) } \left(\frac{x}{\lambda} \right),
\end{document}nd{equation}
and
\begin{equation} \label{54}
\lambda^m B_{m,\xi^\lambda}^{(k)} \left(\frac{x}{\lambda} \right)=\sum_{n=0}^m D_{n,\xi}^{(k)} (x│ \lambda ) \xi^{-n-x} s_2 (m,n).
\end{document}nd{equation}
\noindent
where $B_{m,\xi^\lambda }^{(k) } \left(\frac{x}{\lambda}\right)$ is defined by Kim et al. \cite[Eq. 15]{Kimetal2013b}, as follows
\begin{equation} \label{55}
\left(\frac{\lambda t}{\xi^\lambda e^{\lambda t}-1}\right)^k (\xi e^t )^x = \xi^x \sum_{m=0}^\infty \lambda^m B_{m,\xi^\lambda}^{(k)} \left(\frac{x}{\lambda}\right) \frac{ t^m}{m!}.
\end{document}nd{equation}
\begin{remark}
We can write (\ref{53}) in the following matrix form
\begin{equation} \label{56}
{\bf D}_\xi^{(k)} (x|\lambda)={\bf \Xi S}_1 \, {\bf \Lambda B}^{(k)} \left(\frac{x}{\lambda}\right),
\end{document}nd{equation}
\end{document}nd{remark}
\noindent
where ${\bf D}_\xi^{(k)} (x|\lambda)$ is $(n+1)\times(k+1)$ matrix for the twisted Daehee numbers of the first kind of the order $k$ and ${\bf \Xi}$ is $(n+1)\times(n+1)$ diagonal matrix with elements $({\bf \Xi})_{ii}=\xi^i$ for $i=j=0,1,\cdots,n$.
\noindent
For example, if setting $0\leq n \leq 3,\; 0\leq k \leq n$, in (\ref{56}), we have
{\footnotesize
\begin{eqnarray*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & \xi & 0 & 0\\
0 & 0 & \xi^2 & 0\\
0 & 0 & 0 & \xi^3
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 2 & -3 &1
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \lambda & 0 & 0 \\
0 & 0 & \lambda^2 & 0 \\
0 & 0 & 0 & \lambda^3
\end{document}nd{array}
\right)\times
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
\\
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1\\
\frac{x}{\lambda} & \frac{x}{\lambda}-\frac{1}{2} & \frac{x}{\lambda}-1 & \frac{x}{\lambda}-\frac{3}{2} \\
\frac{x^2}{\lambda^2} & \frac{x^2}{\lambda^2} -\frac{x}{\lambda}+\frac{1}{6} & \frac{x^2}{\lambda^2} -\frac{2x}{\lambda}+ \frac{5}{6} & \frac{x^2}{\lambda^2} -\frac{3x}{\lambda}+2 \\
\frac{x^3}{\lambda^3} & \frac{x}{2\lambda}-\frac{3x^2}{2\lambda^2}+ \frac{x^3}{\lambda^3} & \frac{5x}{2\lambda}-\frac{3x^2}{\lambda^2} + \frac{x^3}{\lambda^3} -\frac{1}{2} & \frac{6x}{\lambda}-\frac{9x^2}{2\lambda^2}+ \frac{x^3}{\lambda^3} -\frac{9}{4}
\end{document}nd{array}
\right)=
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
\\
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
\xi x & -\frac{\xi}{2} (\lambda-2x)&-\xi(\lambda-x)&-\frac{\xi}{2}(3\xi-2x) \\
\xi^2 x(x-1) & \xi^2 (\frac{\lambda^2}{6}-\lambda x+ \frac{\lambda}{2}+x^2-x)& \xi^2 (\frac{5}{6} \lambda^2-2\lambda x+\lambda+x^2-x)& \xi^2 (2\lambda^2-3\lambda x+\frac{3}{2} \lambda+x^2-x)\\
D_{3,\xi}^{(0)}(x|\lambda) & D_{3,\xi}^{(1)}(x|\lambda) & D_{3,\xi}^{(2)}(x|\lambda) & D_{3,\xi}^{(3)}(x|\lambda)
\end{document}nd{array}
\right).
\end{document}nd{eqnarray*}
}
\noindent
where
\begin{eqnarray*}
D_{3,\xi}^{(0)}(x|\lambda) &= &\xi^3 x(x-1)(x-2)\; D_{3,\xi}^{(1)}(x|\lambda)= -\frac{\xi^3}{2} (\lambda-2x+2)(\lambda-2x+x^2-\lambda x),
\\
D_{3,\xi}^{(2)}(x|\lambda) &=& -\frac{\xi^3}{2}(\lambda-x+1)(\lambda^2-4\lambda x+4\lambda+2x^2-4x),
\\
D_{3,\xi}^{(0)}(x|\lambda) &=& -\frac{\xi^3}{4}(3\lambda-2x+2)(3\lambda^2-6\lambda x+6\lambda+2x^2-4x).
\end{document}nd{eqnarray*}
\begin{remark}
In fact, it seems that the statement in (\ref{54}) is not correct, the second equation of, Kim et al. \cite[Theorem 1]{Kimetal2013b}.
From (\ref{56}), multiplying both sides by ${\bf \Xi}^{-1}$, we have,
\begin{equation*}
{\bf \Xi}^{-1} \, {\bf D}_\xi^{(k)} (x|\lambda)={\bf \Xi}^{-1} \, {\bf \Xi S}_1 \, {\bf \Lambda B}^{(k)} \left(\frac{x}{\lambda} \right)=
{\bf S}_1 \, {\bf \Lambda B}^{(k)} \left( \frac{x}{\lambda} \right),
\end{document}nd{equation*}
\noindent
then multiplying both sides by ${\bf S}_2$, we have
\begin{equation} \label{58}
{\bf S}_2 \, {\bf \Xi}^{-1} \, {\bf D}_\xi^{(k)} (x|\lambda)= {\bf S}_2 \, {\bf S}_1 \, {\bf \Lambda B}^{(k)} \left(\frac{x}{\lambda} \right)={\bf \Lambda B}^{(k)} \left(\frac{x}{\lambda}\right).
\end{document}nd{equation}
\noindent
From (\ref{54}) and (\ref{58}), it is clear that there is a contradiction.
\end{document}nd{remark}
\noindent
In the following theorem we obtained the corrected relation as follows.
\begin{theorem}
For $m\in \mathbb{Z},\; k\in \mathbb{N}$, we have
\begin{equation} \label{59}
\lambda^m B_m^{(k)} \left(\frac{x}{\lambda} \right)= \sum_{n=0}^m D_{n,\xi}^{(k)} (x| \lambda) \xi^{-n} s_2 (m,n).
\end{document}nd{equation}
\end{document}nd{theorem}
\begin{proof}
From Eq. (\ref{50}), replacing $t$ by $(e^t-1)/ \xi $ , we have
\begin{eqnarray} \label{60}
\left( \frac{ \lambda \log\left( 1 + \frac{\xi(e^t-1)}{\xi}\right)}{\left(1+ \frac{\xi(e^t-1)}{\xi} \right)^\lambda-1}\right)^k
\left( 1 + \frac{\xi(e^t-1)}{\xi} \right)^x & = & \sum_{n=0}^\infty D_{n,\xi}^{(k)} (x| \lambda) \frac{ (e^t-1)^n}{n! \xi^n}
\nonumber\\
\left( \frac{\lambda t}{e^{\lambda t}-1} \right)^k e^{tx} & = & \sum_{n=0}^\infty D_{n,\xi}^{(k)} (x|\lambda) \frac{(e^t-1)^n}{n!\xi^n}.
\end{document}nd{eqnarray}
\noindent
Substituting from (\ref{6}) into (\ref{60}), we have
\begin{eqnarray} \label{61}
\left( \frac{\lambda t}{e^{\lambda t}-1} \right)^k e^{\lambda t(\frac{x}{\lambda})} & = & \sum_{n=0}^\infty D_{n,\xi }^{(k)} (x|\lambda) \xi^{-n} \sum_{m=n}^\infty s_2 (m,n)\frac{t^m}{m!}
\nonumber\\
& = &\sum_{m=0}^ \infty \sum_{n=0}^m D_{n,\xi}^{(k) } (x|\lambda) \xi^{-n} s_2 (m,n)\frac{t^m}{m!}.
\end{document}nd{eqnarray}
\noindent
From (\ref{1}) and (\ref{61}), we have
\begin{equation} \label{62}
\sum_{m=0}^\infty \lambda^m B_m^{(k)} \left(\frac{x}{\lambda} \right) \frac{ t^m}{m!} = \sum_{m=0}^\infty \sum_{n=0}^m D_{n,\xi }^{(k)} (x|\lambda) \xi^{-n} s_2 (m,n) \frac{t^m}{m!}.
\end{document}nd{equation}
\noindent
Equating the coefficients of $t^m$ on both sides gives (\ref{59}). This completes the proof.
\end{document}nd{proof}
\noindent
Moreover, we can represent Equation (\ref{59}), in the following matrix form as (\ref{58}).
\begin{equation} \label{63}
{\bf B}^{(k)} \left(\frac{x}{\lambda} \right)= {\bf \Lambda}^{-1} \, {\bf S}_2 \, {\bf \Xi }^{-1} \, {\bf D}_\xi^{(k)} (x|\lambda). \end{document}nd{equation}
\noindent
For example, if setting $0 \leq n \leq 3,\; 0 \leq k \leq n$, in (\ref{63}), we have
{\footnotesize
\begin{eqnarray*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \frac{1}{\lambda} & 0 & 0 \\
0 & 0 & \frac{1}{\lambda^2} & 0 \\
0 & 0 & 0 & \frac{1}{\lambda^3}
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 1 & 3 & 1
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \frac{1}{\xi} & 0 & 0 \\
0 & 0 & \frac{1}{\xi^2} & 0 \\
0 & 0 & 0 & \frac{1}{\xi^3}
\end{document}nd{array}
\right)\times
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
\; \; \; \;
\\
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
\xi x & -\frac{\xi}{2} (\lambda-2x)&-\xi(\lambda-x)&-\frac{\xi}{2}(3\xi-2x)\\
\xi^2 x(x-1) & \xi^2 (\frac{\lambda^2}{6}-\lambda x+ \frac{\lambda}{2}+x^2-x) &\xi^2 (\frac{5}{6} \lambda^2-2\lambda x+\lambda+x^2-x) &\xi^2 (2\lambda^2-3\lambda x+\frac{3}{2} \lambda+x^2-x)\\
D_{3,\xi}^{(0)}(x|\lambda) & D_{3,\xi}^{(1)}(x|\lambda) & D_{3,\xi}^{(2)}(x|\lambda) & D_{3,\xi}^{(3)}(x|\lambda)
\end{document}nd{array}
\right)
\\
= \left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
\frac{x}{\lambda} & \frac{x}{\lambda}-\frac{1}{2} & \frac{x}{\lambda}-1 & \frac{x}{\lambda}-\frac{3}{2} \\
\frac{x^2}{\lambda^2} & \frac{x^2}{\lambda^2} -\frac{x}{\lambda}+ \frac{1}{6} & \frac{x^2}{\lambda^2} -\frac{2x}{\lambda}+\frac{5}{6} & \frac{x^2}{\lambda^2} -\frac{3x}{\lambda}+2 \\
\frac{x^3}{\lambda^3} & \frac{x}{2\lambda}-\frac{3x^2}{2\lambda^2}+ \frac{x^3}{\lambda^3} & \frac{5x}{2\lambda}-\frac{3x^2}{\lambda^2} + \frac{x^3}{\lambda^3} - \frac{1}{2} & \frac{6x}{\lambda}-\frac{9x^2}{2\lambda^2}+ \frac{x^3}{\lambda^3} -\frac{9}{4}
\end{document}nd{array}
\right)
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
\end{document}nd{eqnarray*}
}
\noindent
where
\begin{eqnarray*}
D_{3,\xi}^{(0)}(x|\lambda)& =& \xi^3 x(x-1)(x-2), \; D_{3,\xi}^{(1)}(x|\lambda)=-\frac{\xi^3}{2} (\lambda-2x+2)(\lambda-2x+x^2-\lambda x)
\\
D_{3,\xi}^{(2)}(x|\lambda)&=&-\frac{\xi^3}{2}(\lambda-x+1)(\lambda^2-4\lambda x+4\lambda+2x^2-4x),
\\
D_{3,\xi}^{(3)}(x|\lambda)&=&-\frac{\xi^3}{4}(3\lambda-2x+2)(3\lambda^2-6\lambda x+6\lambda+2x^2-4x).
\end{document}nd{eqnarray*}
\noindent
Kim et al. \cite{Kimetal2013b} introduced the twisted $\lambda$-Daehee polynomials of the second kind of order $k$ as follows:
\begin{equation} \label{64}
\left( \frac{\lambda \log(1+\xi t) (1+ \xi t)^\lambda}{(1+\xi t)^\lambda-1} \right)^k (1+ \xi t)^x = \sum_{n=0}^\infty \hat{D}_{n,\xi }^{(k) } (x|\lambda) \frac{ t^n}{n!}.
\end{document}nd{equation}
\noindent
Setting $x=0, \; \hat{D}_{n,\xi,\lambda}^{(k) }= \hat{D}_{n,\xi }^{(k)} (0|\lambda)$, we have the twisted Daehee numbers of second kind of order $k$.
\begin{equation} \label{65}
\left( \frac{\lambda \log(1+\xi t) (1+ \xi t)^\lambda}{(1+\xi t)^\lambda-1} \right)^k = \sum_{n=0}^\infty \hat{D}_{n,\xi , \lambda}^{(k) } \frac{ t^n}{n!}.
\end{document}nd{equation}
\noindent
Kim et al. \cite[Theorem 2]{Kimetal2013b}, proved that. For $m\in \mathbb{Z}, \; k\in \mathbb{N}$, we have
\begin{equation} \label{66}
\xi^{-m} \hat{D}_{n,\xi} (x│ \lambda )= \sum_{l=0}^m s_1 (m,l) \lambda^l B_l^{(k)} \left(k+\frac{x}{\lambda} \right),
\end{document}nd{equation}
and
\begin{equation} \label{67}
\lambda^m B_{m,\xi^\lambda}^{(k)} \left(k+\frac{x}{\lambda}\right)=\sum_{n=0}^m \hat{D}_{n,\xi}^{(k) } (x│ \lambda) s_2 (m,n) \xi^{-n-\lambda k-x}.
\end{document}nd{equation}
\noindent
Using Eq. (\ref{47}), we can write (\ref{66}) in the following matrix form.
\begin{equation} \label{68}
\hat{\bf D}_\xi^{(k) } (x|\lambda)= {\bf \Xi \, S}_1 \, {\bf \Lambda }_1 \, {\bf B}^{(k)} \left(-\frac{x}{\lambda} \right),
\end{document}nd{equation}
\noindent
where $\hat{\bf D}_\xi^{(k)} (x| \lambda)$ is $(n+1)\times(k+1)$ matrix for the twisted Daehee numbers of the second kind of the order $k$.
\noindent
For example, if setting $0\leq n \leq 3,\; 0\le k \leq n$, in (\ref{68}), we have
{\footnotesize
\begin{eqnarray*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \xi & 0 & 0 \\
0 & 0 & \xi^2 & 0 \\
0 & 0 & 0 & \xi^3
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 2 & -3 & 1
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & -\lambda & 0 & 0 \\
0 & 0 & \lambda^2 & 0 \\
0 & 0 & 0 & -\lambda^3
\end{document}nd{array}
\right)
\times
\; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \;
\; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\;
\\
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
-\frac{x}{\lambda} & - \frac{x}{\lambda}-\frac{1}{2} & -\frac{x}{\lambda}-1 & -\frac{x}{\lambda}- \frac{3}{2} \\
\frac{x^2}{\lambda^2} & \frac{x^2}{\lambda^2} + \frac{x}{\lambda}+\frac{1}{6} & \frac{x^2}{\lambda^2} + \frac{2x}{\lambda}+ \frac{5}{6} & \frac{x^2}{\lambda^2} + \frac{3x}{\lambda}+2 \\
-\frac{x^3}{\lambda^3} & -\frac{x}{2\lambda}-\frac{3x^2}{2\lambda^2}- \frac{x^3}{\lambda^3} & -\frac{5x}{2\lambda}-\frac{3x^2}{\lambda^2} -\frac{x^3}{\lambda^3} -\frac{1}{2} &-\frac{6x}{\lambda}-\frac{9x^2}{2\lambda^2}-\frac{x^3}{\lambda^3} -\frac{9}{4}
\end{document}nd{array}
\right)
=
\; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \;
\; \; \; \; \; \; \; \; \;\; \; \;
\\
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
\xi x & \frac{\xi}{2} (\lambda+2x)& \xi(\lambda+x) & \frac{\xi}{2}(3\xi+2x)\\
\xi^2 x(x-1)& \xi^2 (\frac{\lambda^2}{6} + \lambda x-\frac{\lambda}{2} + x^2-x)& \xi^2 (\frac{5}{6} \lambda^2+2\lambda x-\lambda+x^2-x)&\xi^2 (2\lambda^2+3\lambda x-\frac{3}{2} \lambda+x^2-x)\\
\hat{D}_{3,\xi}^{(0)}(x| \lambda) & \hat{D}_{3,\xi}^{(1)}(x| \lambda) & \hat{D}_{3,\xi}^{(2)}(x| \lambda) & \hat{D}_{3,\xi}^{(3)}(x| \lambda)
\end{document}nd{array}
\right).
\end{document}nd{eqnarray*}
}
where
\begin{eqnarray*}
\hat{D}_{3,\xi}^{(0)}(x| \lambda) &= & \xi^3 x(x-1)(x-2), \; \hat{D}_{3,\xi}^{(1)}(x| \lambda)= -\frac{\xi^3}{2} (\lambda+2x-2)(\lambda+2x-x^2-\lambda x),
\\
\hat{D}_{3,\xi}^{(2)}(x| \lambda)& = & \frac{\xi^3}{2}(\lambda+x-1)(\lambda^2+4\lambda x-4\lambda +2x^2-4x),
\\
\hat{D}_{3,\xi}^{(3)}(x| \lambda) & = &\frac{\xi^3}{4}(3\lambda+2x-2)(3\lambda^2+6\lambda x-6\lambda+2x^2-4x).
\end{document}nd{eqnarray*}
\begin{remark}
In fact, it seems that there is something not correct in (\ref{67}), the second equation of Kim et al. \cite[Theorem 2]{Kimetal2013b}.\\
From (\ref{68}), multiplying both sides by ${\bf \Xi}^{-1}$, we have,
\begin{equation*}
{\bf \Xi}^{-1} \, \hat{\bf D}_\xi^{(k)} (x|\lambda)= {\bf \Xi}^{-1} \, {\bf \Xi \, S}_1 {\bf \Lambda}_1 \, { \bf B}^{(k)} \left(-\frac{x}{\lambda} \right)={\bf S}_1 \, { \bf \Lambda}_1 \, {\bf B}^{(k) } \left(-\frac{x}{\lambda} \right)
\end{document}nd{equation*}
\noindent
multiplying both sides by ${\bf S}_2$, we have
\begin{equation} \label{70}
{\bf S}_2 \, {\bf \Xi}^{-1} \, \hat{\bf D}_\xi^{(k)} (x|\lambda)= {\bf S}_2 \, {\bf S}_1 \, { \bf \Lambda }_1 \, {\bf B}^{(k) } \left(-\frac{x}{\lambda} \right)= {\bf I} \, { \bf \Lambda }_1 \, {\bf B}^{(k) } \left(-\frac{x}{\lambda} \right)={ \bf \Lambda}_1 \, {\bf B}^{(k) } \left(-\frac{x}{\lambda} \right).
\end{document}nd{equation}
\noindent
From (\ref{67}) and (\ref{70}) there is a contradiction.
\end{document}nd{remark}
\noindent
We obtained the corrected relation in the following theorem as follows.
\begin{theorem}
For $m\in \mathbb{Z}, \; k\in \mathbb{N}$, we have
\begin{equation} \label{71}
\lambda^m B_m^{(k)} \left(k+ \frac{x}{\lambda} \right)= \sum_{n=0}^m \hat{D}_{n,\xi}^{(k)} (x|\lambda) \xi^{-n} s_2 (m,n).
\end{document}nd{equation}
\end{document}nd{theorem}
\begin{proof}
From Eq. (\ref{64}), replacing $t$ by $(e^t-1)/\xi$ , we have
\begin{equation*}
\left(\frac{\lambda \log\left(1+ \frac{\xi(e^t-1)}{\xi}\right) \left(1+\frac{\xi(e^t-1)}{\xi}\right)^\lambda}{\left(1+\frac{\xi(e^t-1)}{\xi} \right)^\lambda-1} \right)^k \left(1+\frac{\xi(e^t-1)}{\xi} \right)^x =
\sum_{n=0}^\infty \hat{D}_{n,\xi}^{(k)} (x|\lambda) \frac{ (e^t-1)^n}{n! \xi^n},
\end{document}nd{equation*}
\begin{eqnarray} \label{72}
\left( \frac{\lambda t}{e^{\lambda t}-1} \right)^k e^{(k \lambda+x)t} & = & \sum_{n=0}^\infty \hat{D}_{n,\xi}^{(k)} (x| \lambda) \frac{(e^t-1)^n}{n! \xi^n},
\nonumber\\
\left(\frac{\lambda t}{e^{\lambda t}-1} \right)^k e^{\lambda t\left(k+ \frac{x}{\lambda} \right)} & = & \sum_{n=0}^\infty \hat{D}_{n,\xi}^{(k)} (x| \lambda) \xi^{-n} \frac{(e^t-1)^n}{n!}.
\end{document}nd{eqnarray}
\noindent
Substituting from Eq. (\ref{6}) into (\ref{72}), we have
\begin{eqnarray} \label{73}
\left(\frac{\lambda t}{(e^{\lambda t}-1} \right)^k e^{\lambda t\left(k+\frac{x}{\lambda} \right)} & = &
\sum_{n=0}^\infty \hat{D}_{n,\xi}^{(k)} (x|\lambda) \xi^{-n} \sum_{m=n}^\infty s_2 (m,n) \frac{ t^m}{m!}
\nonumber\\
&= & \sum_{m=0}^\infty \sum_{n=0}^m \hat{D}_{n,\xi}^{(k)} (x|\lambda) \xi^{-n} s_2 (m,n) \frac{t^m}{m!}.
\end{document}nd{eqnarray}
\noindent
From Eq. (\ref{1}) and (\ref{73}), we have
\begin{equation} \label{74}
\sum_{m=0}^\infty \lambda^m B_m^{(k)} \left(k+\frac{x}{\lambda} \right) \frac{ t^m}{m!}=
\sum_{m=0}^\infty \sum_{n=0}^m \hat{D}_{n,\xi}^{(k)} (x|\lambda) \xi^{-n} s_2 (m,n) \frac{t^m}{m!}.
\end{document}nd{equation}
Equating the coefficients of $t^m$ on both sides gives (\ref{71}). This completes the proof.
\end{document}nd{proof}
\noindent
Moreover, by using Eq. (\ref{47}), we can represent Equation (\ref{71}), in the following matrix form.
\begin{equation} \label{75}
{\bf B}^{(k)} \left(-\frac{x}{\lambda}\right)= {\bf \Lambda}_1^{-1} \, {\bf S}_2 \, {\bf \Xi }^{-1} \, \hat{ \bf D}_\xi^{(k)} (x|\lambda),
\end{document}nd{equation}
where ${\bf \Lambda \, B}^{(k) } \left(k+\frac{x}{\lambda}\right)={\bf \Lambda}_1 \, {\bf B}^{(k)} \left(-\frac{x}{\lambda} \right)$.
\noindent
For example, if setting $0\leq n \leq 3,\, 0 \leq k \leq n$, in (\ref{75}), we have
{\footnotesize
\begin{eqnarray*}
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & -\frac{1}{\lambda} & 0 & 0 \\
0 & 0 & \frac{1}{\lambda^2} & 0 \\
0 & 0 & 0 & -\frac{1}{\lambda^3}
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 1 & 3 & 1
\end{document}nd{array}
\right)
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \frac{1}{\xi} & 0 & 0 \\
0 & 0 & \frac{1}{\xi^2} & 0 \\
0 & 0 & 0 & \frac{1}{\xi^3}
\end{document}nd{array}
\right)\times
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
\\
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
\xi x & \frac{\xi}{2} (\lambda+2x) & \xi(\lambda+x)& \frac{\xi}{2}(3\xi+2x)\\
\xi^2 x(x-1)& \xi^2 (\frac{\lambda^2}{6} + \lambda x- \frac{\lambda}{2} + x^2-x)& \xi^2 (\frac{5}{6} \lambda^2+2\lambda x-\lambda+x^2-x)& \xi^2 (2\lambda^2+3\lambda x-\frac{3}{2} \lambda + x^2-x)\\
\hat{D}_{3,\xi}^{(0)}(x|\lambda) & \hat{D}_{3,\xi}^{(1)}(x|\lambda) & \hat{D}_{3,\xi}^{(2)}(x|\lambda) & \hat{D}_{3,\xi}^{(3)}(x|\lambda)
\end{document}nd{array}
\right)
\\
=
\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
- \frac{x}{\lambda} &- \frac{x}{\lambda}- \frac{1}{2} & - \frac{x}{\lambda}-1 &- \frac{x}{\lambda}-\frac{3}{2} \\
\frac{x^2}{\lambda^2} & \frac{x^2}{\lambda^2} + \frac{x}{\lambda}+ \frac{1}{6} & \frac{x^2}{\lambda^2} + \frac{2x}{\lambda}+ \frac{5}{6} & \frac{x^2}{\lambda^2} + \frac{3x}{\lambda}+2\\
-\frac{x^3}{\lambda^3} & -\frac{x}{2\lambda}-\frac{3x^2}{2\lambda^2}-\frac{x^3}{\lambda^3} & -\frac{5x}{2\lambda}-\frac{3x^2}{\lambda^2} -\frac{x^3}{\lambda^3} -\frac{1}{2} &- \frac{6x}{\lambda}-\frac{9x^2}{2\lambda^2}-\frac{x^3}{\lambda^3} -\frac{9}{4}.
\end{document}nd{array}
\right)
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
\end{document}nd{eqnarray*}
}
where
\begin{eqnarray*}
\hat{D}_{3,\xi}^{(0)}(x|\lambda)
&= &\xi^3 x(x-1)(x-2), \; \hat{D}_{3,\xi}^{(1)}(x|\lambda)= -\frac{\xi^3}{2} (\lambda+2x-2)(\lambda+2x-x^2-\lambda x),
\\
\hat{D}_{3,\xi}^{(2)}(x|\lambda)&= &\frac{\xi^3}{2}(\lambda+x-1)(\lambda^2+4\lambda x-4\lambda +2x^2-4x),
\\
\hat{D}_{3,\xi}^{(3)}(x|\lambda)&= & \frac{\xi^3}{4}(3\lambda+2x-2)(3\lambda^2+6\lambda x-6\lambda+2x^2-4x).
\end{document}nd{eqnarray*}
\begin{thebibliography}{99}
\bibitem{Carlitz1961}
Carlitz, L., A note on Bernoulli and Euler polynomials of the second kind, {\sl Scripta Math}. {\bf 25} (1961), 323–-330.
\bibitem{Comtet1974}
Comtet, L. , Advanced Combinatorics, {\sl Reidel, Dordrecht}, 1974.
\bibitem{Dolgyetal2013}
Dolgy, D.V., Kim, T., Lee, B. and Lee, S.-H., Some new identities on the twisted Bernoulli and Euler polynomials, {\sl J. Comput. Anal. Appl.} {\bf 14} (2013), No. 3, 441--451.
\bibitem{El-Desouky1994}
El-Desouky, B.S., The multiparameter non-central Stirling numbers,{\sl Fibonacci Quart.} {\bf 32} (1994), 3, 218--225
\bibitem{El-Desoukyetal2010}
El-Desouky, B.S., Caki\'{c}, N.P. and Mansour, T., Modified approach to generalized Stirling numbers via differential operators, {\sl Appl. Math. Lett.} {\bf 23} (2010), 115–-120.
\bibitem{El-DesoukyMustafa2014}
El-Desouky, B. S. and Mustafa, A., New Results and Matrix Representation for Daehee and Bernoulli Numbers and Polynomials, {\sl arXiv:Submit/114836 [math.Co]} 29 Dec 2014, http://arxiv.org/pdf/1412.8259v1.pdf
\bibitem{Gould1972}
Gould, H. W., Explicit formulas for Bernoulli numbers, {\sl Amer. Math. Monthly,} {\bf 79} (1972), 44–-51.
\bibitem{Kimetal2014}
Kim, D.S., Kim, T., Lee, S-H. and J-J. Seo, Higher-Order Daehee Numbers and Polynomials, {\sl Int. J. Math. Anal. (HIKARI Ltd)}, {\bf 8} (2014), No. 6, 273-–283: http://dx.doi.org/10.12988/ijma.2014.4118
\bibitem{Kimetal2013}
Kim, D.S., Kim, T., Lee, S-H. and J-J. Seo, A Note on the lambda Daehee Polynomials, {\sl Int. J. Math. Anal. (HIKARI Ltd)}, {\bf 7} (2013), No. 62, 3069–-3080: http://dx.doi.org/10.12988/ijma.2013.311264
\bibitem{Kimetal2013b}
Kim, D.S., Kim, T., Lee, S-H. and J-J. Seo, A Note on Twisted $\lambda$- Daehee Polynomials, {\sl Appl. Math. Sci. (HIKARI Ltd)}, {\bf 7} (2013), No. 141, 7005–-7014: http://dx.doi.org/10.12988/ams.2013.311635
\bibitem{KimKim2013}
Kim, D.S. and Kim, T., Daehee Numbers and polynomials, {\sl Appl. Math. Sci. (HIKARI Ltd.)}, {\bf 7} (2013), No. 120, 5969-–5976, http://dx.doi.org/10.12988/ams.2013.39535
\bibitem{KimSimsek2008}
Kim, T. and Simsek, Y., Analytic continuation of the multiple Daehee q-l-functions associated with Daehee numbers, {\sl Russ. J. Math. Phys.} {\bf 15} (2008), 58--65.
\bibitem{Kimura2003}
Kimura, N., On universal higher order Bernoulli numbers and polynomials, {\sl Report of the Research Institute of Industrial Technology, Nihon University}, Number {\bf 70}, 2003: ISSN 0386-1678
\bibitem{LiuSrivastava2006}
Liu, G-D. and Srivastava, H.M. Explicit Formulas for the N\"{o}rlund Polynomials $B_n^{(x) }$ and $b_n^{(x) }$, {\sl Computers and Mathematics with Applicatios}, {\bf 51 } (2006), 1377--1384.
\bibitem{Ozdenetal2009}
Ozden, H. and Cangul, N. and Simsek, Y., Remarks on q-Bernoulli numbers associated with Daehee numbers, {\sl Adv. Stud. Contemp. Math. (Kyungshang)}, {\bf 18} (2009), 41--48.
\bibitem{Wang2010}
Wang, W., Generalized higher order Bernoulli number pairs and generalized Stirling number pairs,
{\sl J. Math. Anal. Appl.}, {\bf 364} (2010), 255-–274.
\end{document}nd{thebibliography}
\end{document}nd{document}
|
math
|
What a great review! I really enjoyed reading how your family used this, we struggled a bit; mostly because my kids already use a video curriculum for math and weren’t excited to help me out too much. Look forward to reading more of your reviews. Thanks for stopping by!
Thanks for commenting on my blog; wanted to hop over and read your math review–love your food theme (great analogy for those who find math hard to digest :o). I think his tutorials will really help me stay on top of the math my kids are doing.
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english
|
#pragma ident "$Id$"
/**
* @file SolarSystem.cpp
* Implement JPL solar system ephemeris handling, including reading and writing of
* ASCII and binary files and the computation of position and velocity of the sun,
* the moon and the 9 planets, as well as nutation and lunar libration and their rate.
* JPL ephemeris files may be obtained from ftp://ssd.jpl.nasa.gov/pub/eph/planets.
* Generally you should download the ASCII files and use tools based on this code
* to convert to your own binary files; this avoids compiler- and platform-dependent
* differences in the binary files.
*/
// ======================================================================
// This software was developed by Applied Research Laboratories, The
// University of Texas at Austin under contract to an agency or agencies
// within the U.S. Department of Defense. The U.S. Government retains all
// rights to use, duplicate, distribute, disclose, or release this
// software.
//
// Copyright 2008 The University of Texas at Austin
// ======================================================================
//------------------------------------------------------------------------------------
#include "CommonTime.hpp"
#include "SolarSystem.hpp"
#include "Matrix.hpp" // only for WGS84Position()
#include "GeodeticFrames.hpp" // only for WGS84Position()
#include "logstream.hpp"
#include "TimeString.hpp"
#include "JulianDate.hpp"
//------------------------------------------------------------------------------------
using namespace std;
using namespace gpstk::StringUtils;
namespace gpstk {
//------------------------------------------------------------------------------------
void SolarSystem::readASCIIheader(string filename) throw(Exception)
{
try {
// open the file
ifstream strm;
strm.open(filename.c_str());
if(!strm) {
Exception e("Failed to open input file " + filename + ". Abort.");
GPSTK_THROW(e);
}
// clear existing data
constants.clear();
// read the file one line at a time, process depending on the value of group
int group=0,n=0; // n will count lines/items within a group
string line,word;
vector<string> const_names; // Temporary - name,value stored in map constants
while(1) {
getline(strm,line);
stripTrailing(line,'\r');
// catch new groups
if(line.substr(0,5) == "GROUP") {
word = stripFirstWord(line);
group = asInt(stripFirstWord(line));
n = 0; // n will count lines/items within a group
continue;
}
// skip blank lines
stripLeading(line," ");
if(line.empty()) {
if(strm.eof() || !strm.good()) break; // if the last line is blank
else continue;
}
// process entire line at once
// first line (no GROUP)
if(group == 0) {
word = stripFirstWord(line);
word = stripFirstWord(line); // ignore KSIZE
word = stripFirstWord(line);
if(word == "NCOEFF=") {
Ncoeff = asInt(stripFirstWord(line));
continue;
}
else {
Exception e("Confused on the first line - 3rd word is not NCOEFF=");
GPSTK_THROW(e);
}
}
// GROUP 1010
else if(group == 1010) {
if(n > 2) { // this should not happen
Exception e("Too many labels under GROUP 1010");
GPSTK_THROW(e);
}
else {
stripTrailing(line," ");
label[n++] = line;
continue;
}
}
// GROUP 1030
else if(group == 1030) {
// start and stop times. These are meaningless here, because they will be
// determined by the data that follows this header, and so are meaningful
// only in the binary file.
startJD = for2doub(stripFirstWord(line));
endJD = for2doub(stripFirstWord(line));
// interval in days covered by each block of coefficients
interval = for2doub(stripFirstWord(line));
}
// GROUP 1070 - end-of-header
else if(group == 1070) {
break;
}
// process the line one (whitespace-separated) word at a time
while(!line.empty()) {
word = stripFirstWord(line);
if(group == 1040) {
if(n++ == 0) {
Nconst = asInt(word);
}
else {
const_names.push_back(word);
}
}
else if(group == 1041) {
if(n++ == 0) {
if(Nconst != asInt(word)) {
Exception e("Nconst does not match N in GROUP 1041 : " +
asString(Nconst) + " != " + word);
GPSTK_THROW(e);
}
}
else
constants[const_names[n-2]] = for2doub(word);
}
else if(group == 1050) {
if(n < 13) {
c_offset[n] = asInt(word);
}
else if(n < 26) {
c_ncoeff[n-13] = asInt(word);
}
else {
c_nsets[n-26] = asInt(word);
}
n++;
}
else {
Exception e("Confused about GROUP : " + asString(group));
GPSTK_THROW(e);
}
} // end loop over words
if(strm.eof() || !strm.good()) break; // if the last line is not blank
} // end read loop over lines
strm.clear();
strm.close();
// test that we got a full header
if(group != 1070) {
Exception e("Premature end of header");
GPSTK_THROW(e);
}
// EphemerisNumber != -1 means the header is complete
EphemerisNumber = int(constants["DENUM"]);
// clear the data arrays
store.clear();
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
int SolarSystem::readASCIIdata(vector<string>& filenames) throw(Exception)
{
try {
size_t i;
if(filenames.size() == 0) return 0;
// read the files, in any order; the data map will be sorted on time
for(i=0; i<filenames.size(); i++) {
int iret = readASCIIdata(filenames[i]);
if(iret) return iret;
}
// set the start and stop times in the header
map<double,vector<double> >::iterator it = store.begin();
startJD = it->second[0];
it = store.end(); it--;
endJD = it->second[1];
// Mod the header labels to reflect the new time limits
ostringstream oss;
CommonTime tt;
tt = JulianDate(startJD);
oss << "Start Epoch: JED= " << fixed << setw(10) << setprecision(1) << startJD
<< printTime(tt," %4Y %b %2d %02H:%02M:%02S");
label[1] = leftJustify(oss.str(),81);
oss.seekp(ios_base::beg);
tt = JulianDate(endJD);
oss << "Final Epoch: JED= " << fixed << setw(10) << setprecision(1) << endJD
<< printTime(tt," %4Y %b %2d %02H:%02M:%02S");
label[2] = leftJustify(oss.str(),81);
return 0;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
int SolarSystem::readASCIIdata(string filename) throw(Exception)
{
try {
if(EphemerisNumber < 0) {
Exception e("readASCIIdata called before header read");
GPSTK_THROW(e);
}
int iret=0;
string line,word;
ifstream strm;
// open the file
strm.open(filename.c_str());
if(!strm) {
Exception e("Could not open file " + filename);
GPSTK_THROW(e);
}
// expect this many lines per record
int nmax = Ncoeff/3 + (Ncoeff % 3 ? 1 : 0);
// loop over lines in the file
int ntot=0; // counts the total number of lines
int n=0; // counts the lines within a set of coefficients
int nc=0; // count coefficients within a record
int rec;
vector<double> data_vector;
while(1) {
getline(strm,line);
stripTrailing(line,'\r');
if(line.empty()) {
if(strm.eof()) break;
if(!strm.good()) { iret = -1; break; }
continue;
}
if(n == 0) {
rec = asInt(stripFirstWord(line)); // 1st word is the record number
int ncc = asInt(stripFirstWord(line)); // 2nd word is ncoeff
if(ncc != Ncoeff) {
Exception e("readASCIIdata finds conflicting sizes in header ("
+ asString(Ncoeff) + ") and data (" + asString(ncc) + ") in file "
+ filename + " at line #" + asString(ntot));
GPSTK_THROW(e);
}
nc = 0;
}
else {
for(int j=0; j<3; j++) {
double coeff = for2doub(stripFirstWord(line));
nc++;
data_vector.push_back(coeff);
if(nc >= Ncoeff) {
vector<double> dtemp=data_vector;
store[data_vector[0]] = dtemp;
data_vector.clear();
break;
}
}
}
if(strm.eof()) break;
if(!strm.good()) { iret = -1; break; }
if(n == nmax) n=0; else n++;
ntot++;
}
strm.close();
return iret;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
int SolarSystem::writeASCIIheader(ostream& os) throw(Exception)
{
try {
if(EphemerisNumber < 0) return -4;
int i;
string str;
string blank(81,' '); blank += string("\n");
ostringstream oss;
oss << "KSIZE= 0000 NSIZE=" << setw(5) << Ncoeff << blank;
os << leftJustify(oss.str(),81) << endl << blank; oss.seekp(ios_base::beg);
os << leftJustify("GROUP 1010",81) << endl << blank;
for(i=0; i<3; i++) {
str = label[i];
os << leftJustify(str,81) << endl;
}
os << blank;
os << leftJustify("GROUP 1030",81) << endl << blank;
oss << fixed << setprecision(2) << setw(12) << startJD
<< setw(12) << endJD << setw(12) << interval << blank;
os << leftJustify(oss.str(),81) << endl << blank; oss.seekp(ios_base::beg);
os << leftJustify("GROUP 1040",81) << endl << blank;
oss << setw(6) << Nconst << blank;
os << leftJustify(oss.str(),81) << endl; oss.seekp(ios_base::beg);
map<string,double>::const_iterator it=constants.begin();
for(i=0; it != constants.end(); it++,i++) {
oss << leftJustify(" " + it->first,8);
if((i+1)%10 == 0) {
oss << blank;
os << leftJustify(oss.str(),81) << endl;
oss.seekp(ios_base::beg);
}
}
if(Nconst%10 != 0) {
oss << blank;
os << leftJustify(oss.str(),81) << endl;
oss.seekp(ios_base::beg);
}
os << blank;
os << leftJustify("GROUP 1041",81) << endl << blank;
oss << setw(6) << Nconst << blank;
os << leftJustify(oss.str(),81) << endl; oss.seekp(ios_base::beg);
for(i=0,it=constants.begin(); it != constants.end(); it++,i++) {
oss << leftJustify(" " + doub2for(it->second,24,2),26);
if((i+1)%3 == 0) {
oss << blank;
os << leftJustify(oss.str(),81) << endl;
oss.seekp(ios_base::beg);
}
}
if(Nconst%3 != 0) {
i--;
while((i+1)%3 != 0) { oss << leftJustify(" " + doub2for(0.0,24,2),26); i++; }
oss << blank;
os << leftJustify(oss.str(),81) << endl;
oss.seekp(ios_base::beg);
}
os << blank;
os << leftJustify("GROUP 1050",81) << endl << blank;
for(i=0; i<13; i++) oss << rightJustify(asString(c_offset[i]),6);
oss << blank; os << leftJustify(oss.str(),81) << endl; oss.seekp(ios_base::beg);
for(i=0; i<13; i++) oss << rightJustify(asString(c_ncoeff[i]),6);
oss << blank; os << leftJustify(oss.str(),81) << endl; oss.seekp(ios_base::beg);
for(i=0; i<13; i++) oss << rightJustify(asString(c_nsets[i]),6);
oss << blank; os << leftJustify(oss.str(),81) << endl; oss.seekp(ios_base::beg);
os << blank;
os << leftJustify("GROUP 1070",81) << endl << blank;
os << blank;
return 0;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
int SolarSystem::writeASCIIdata(ostream& os) throw(Exception)
{
try {
if(EphemerisNumber < 0) return -4;
string blank(81,' '); blank += string("\n");
int i,nrec;
ostringstream oss;
map<double,vector<double> >::iterator jt;
for(nrec=1,jt=store.begin(); jt != store.end(); jt++,nrec++) {
os << setw(6) << nrec << setw(6) << Ncoeff << " " << endl;
for(i=0; i<Ncoeff; i++) {
oss << leftJustify(" " + doub2for(jt->second[i],24,2),26);
if((i+1)%3 == 0) {
oss << blank;
os << leftJustify(oss.str(),81) << endl;
oss.seekp(ios_base::beg);
}
}
if(Ncoeff % 3 != 0) {
i--;
while((i+1)%3 != 0) {
oss << leftJustify(" " + doub2for(0.0,24,2),26);
i++;
}
oss << blank;
os << leftJustify(oss.str(),81) << endl;
oss.seekp(ios_base::beg);
}
}
// TD clear the array after writing
//store.clear();
return 0;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
int SolarSystem::writeBinaryFile(string filename) throw(Exception)
{
try {
int recLength;
size_t i;
string str;
if(EphemerisNumber <= 0) return -4;
// open the output file
ofstream strm;
strm.open(filename.c_str(),ios::out | ios::binary);
if(!strm) {
Exception e("Failed to open output file " + filename + ". Abort.");
GPSTK_THROW(e);
}
// write the header records: two of them, both of length Ncoeff*sizeof(double)
// the structure and ordering taken from the JPL code
recLength = 0;
// first header record -------------------------------------------------
// 1. 3 labels each of length 84
for(i=0; i<3; i++) {
str = label[i];
writeBinary(strm,leftJustify(str,84).c_str(),84);
recLength += 84;
}
// 2. 400 keys from the const array, each of length 6
map<string,double>::const_iterator it=constants.begin();
for(i=0; i<400; i++) {
if(it != constants.end()) {
str = it->first;
writeBinary(strm,leftJustify(str,6).c_str(),6);
it++;
}
else
writeBinary(strm," ",6);
recLength += 6;
}
// 3. the three times
writeBinary(strm,(char *)&startJD,sizeof(double));
writeBinary(strm,(char *)&endJD,sizeof(double));
writeBinary(strm,(char *)&interval,sizeof(double));
recLength += 3*sizeof(double);
// 4. Ncoeff
writeBinary(strm,(char *)&Ncoeff,sizeof(int));
recLength += sizeof(int);
// 5. AU and EMRAT
writeBinary(strm,(char *)&constants["AU"],sizeof(double));
writeBinary(strm,(char *)&constants["EMRAT"],sizeof(double));
recLength += 2*sizeof(double);
// 6. c_arrays for the first 12 planets
for(i=0; i<12; i++) {
writeBinary(strm,(char *)&c_offset[i],sizeof(int));
writeBinary(strm,(char *)&c_ncoeff[i],sizeof(int));
writeBinary(strm,(char *)&c_nsets[i],sizeof(int));
recLength += 3*sizeof(int);
}
// 7. DENUM
writeBinary(strm,(char *)&constants["DENUM"],sizeof(double));
recLength += sizeof(double);
// 8. c_arrays for libration
writeBinary(strm,(char *)&c_offset[12],sizeof(int));
writeBinary(strm,(char *)&c_ncoeff[12],sizeof(int));
writeBinary(strm,(char *)&c_nsets[12],sizeof(int));
recLength += 3*sizeof(int);
// 9. pad
char c=' ';
for(i=0; i < Ncoeff*sizeof(double) - recLength; i++)
writeBinary(strm,&c,1);
// second header record -------------------------------------------------
// the second header record: 400 values from the const array
double z=0.0;
it = constants.begin();
for(i=0; i<400; i++) {
if(it != constants.end()) {
writeBinary(strm,(char *)&(it->second),sizeof(double));
it++;
}
else
writeBinary(strm,(char *)&z,sizeof(double));
}
for(i=0; i < (400-Nconst)*sizeof(double); i++)
writeBinary(strm,&c,1);
// data records ---------------------------------------------------------
// the data, in time order
int nrec=1;
map<double,vector<double> >::iterator jt;
for(jt=store.begin(); jt != store.end(); jt++) {
for(i=0; i<jt->second.size(); i++)
writeBinary(strm,(char *)&jt->second[i],sizeof(double));
nrec++;
}
// TD after writing it out, clear the store array
//store.clear();
strm.clear();
strm.close();
return 0;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
int SolarSystem::readBinaryFile(string filename) throw(Exception)
{
try {
int iret;
readBinaryHeader(filename);
iret = readBinaryData(true); // true: store ALL the data in map
istrm.clear();
istrm.close();
return iret;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
int SolarSystem::initializeWithBinaryFile(string filename) throw(Exception)
{
try {
int iret;
readBinaryHeader(filename);
iret = readBinaryData(false); // false: don't store data in map
if(iret == 0) {
// EphemerisNumber == -1 means the header has not been read
// EphemerisNumber == 0 means the fileposMap has not been read (binary)
// EphemerisNumber == constants["DENUM"] means object has been initialized
// (binary file), or header read (ASCII file)
EphemerisNumber = int(constants["DENUM"]);
}
return iret;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
// return 0 ok, or (from seekToJD)
// -1 out of range : input time is before the first time in file
// -2 out of range : input time is after the last time in file, or in a gap
// -3 stream is not open or not good, or EOF was found prematurely
// -4 EphemerisNumber is not defined
// For -3,-4 : initializeWithBinaryFile() has not been called, or reading failed.
int SolarSystem::computeState(double tt,
SolarSystem::Planet target,
SolarSystem::Planet center,
double PV[6],
bool kilometers) throw(Exception)
{
try {
int iret,i;
// initialize
for(i=0; i<6; i++) PV[i] = 0.0;
// trivial; return
if(target == center) return 0;
// get the right record from the file
iret = seekToJD(tt);
if(iret) return iret;
// compute Nutations or Librations
if(target == Nutations || target == Librations) {
computeState(tt, target==Nutations ? NUTATIONS : LIBRATIONS, PV);
return 0;
}
// define computeID's for target and center
computeID TARGET,CENTER;
if(target <= Sun) TARGET = computeID(target-1);
else if(target == SolarSystemBarycenter) TARGET = NONE;
else if(target == EarthMoonBarycenter) TARGET = EMBARY;
// (Nutations and Librations are done above)
if(center <= Sun) CENTER = computeID(center-1);
else if(center == SolarSystemBarycenter) CENTER = NONE;
else if(center == EarthMoonBarycenter) CENTER = EMBARY;
// Earth and Moon need special treatment
double PVMOON[6],PVEMBARY[6],Eratio,Mratio;
// special cases of Earth AND Moon: Moon result is always geocentric
if(target == Earth && center == Moon) TARGET = NONE;
if(center == Earth && target == Moon) CENTER = NONE;
// special cases of Earth OR Moon, but not both:
if((target == Earth && center != Moon) || (center == Earth && target != Moon)) {
Eratio = 1.0/(1.0 + constants["EMRAT"]);
computeState(tt, MOON, PVMOON);
}
if((target == Moon && center != Earth) || (center == Moon && target != Earth)) {
Mratio = constants["EMRAT"]/(1.0 + constants["EMRAT"]);
computeState(tt, EMBARY, PVEMBARY);
}
// compute states for target and center
double PVTARGET[6],PVCENTER[6];
computeState(tt, TARGET, PVTARGET);
computeState(tt, CENTER, PVCENTER);
// handle the Earth/Moon special cases
// convert from E-M barycenter to Earth
if(target == Earth && center != Moon)
for(i=0; i<6; i++) PVTARGET[i] -= PVMOON[i]*Eratio;
if(center == Earth && target != Moon)
for(i=0; i<6; i++) PVCENTER[i] -= PVMOON[i]*Eratio;
if(target == Moon && center != Earth)
for(i=0; i<6; i++) PVTARGET[i] = PVEMBARY[i] + PVTARGET[i]*Mratio;
if(center == Moon && target != Earth)
for(i=0; i<6; i++) PVCENTER[i] = PVEMBARY[i] + PVCENTER[i]*Mratio;
// final result
for(i=0; i<6; i++) PV[i] = PVTARGET[i] - PVCENTER[i];
if(!kilometers) {
double AU = constants["AU"];
for(i=0; i<6; i++) PV[i] /= AU;
}
return 0;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
// Return the geocentric (relative to Earth's center) position of the Sun at the
// input time, in WGS84 coordinates.
Position SolarSystem::WGS84Position(Planet body, const CommonTime time,
const EarthOrientation& eo) throw(Exception)
{
try {
int iret;
double PV[6];
double JD = static_cast<JulianDate>(time).jd;
iret = computeState(JD, body, Earth, PV); // result in km, km/day
Matrix<double> Rot;
Rot = GeodeticFrames::ECEFtoInertial(time, eo.xp, eo.yp, eo.UT1mUTC);
//, bool reduced = false);
Vector<double> Inertial(3),Geodetic(3);
for(int i=0; i<3; i++) Inertial(i) = PV[i];
Geodetic = transpose(Rot) * Inertial;
Geodetic *= 1000.0; // convert km to meters
Position result;
result.setECEF(Geodetic(0),Geodetic(1),Geodetic(2));
return result;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
//------------------------------------------------------------------------------------
// private
void SolarSystem::writeBinary(ofstream& strm, const char *ptr, size_t size)
throw(Exception)
{
try {
strm.write(ptr,size);
if(!strm.good()) {
Exception e("Stream error");
GPSTK_THROW(e);
}
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
// private
void SolarSystem::readBinary(char *ptr, size_t size) throw(Exception)
{
try {
istrm.read(ptr,size);
if(istrm.eof() || !istrm.good()) {
Exception e("Stream error or premature EOF");
GPSTK_THROW(e);
}
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
// private
void SolarSystem::readBinaryHeader(std::string filename) throw(Exception)
{
try {
int recLength;
size_t i;
char buffer[100];
double AU,EMRAT;
string word;
// open the input binary file
istrm.open(filename.c_str(), ios::in | ios::binary);
if(!istrm) {
Exception e("Failed to open input binary file " + filename + ". Abort.");
GPSTK_THROW(e);
}
// initialize
EphemerisNumber = -1;
constants.clear();
store.clear();
recLength = 0;
// ----------------------------------------------------------------
// read the first header record
// 1. 3 labels each of length 84
for(i=0; i<3; i++) {
readBinary(buffer,84); //istrm.read(buffer,84);
recLength += 84;
buffer[84] = '\0';
label[i] = stripTrailing(stripLeading(buffer," ")," ");
}
// 2. 400 keys from the const array, each of length 6
vector<string> consts_names;
buffer[6] = '\0';
for(i=0; i<400; i++) {
readBinary(buffer,6);
recLength += 6;
word = stripLeading(string(buffer));
if(!word.empty()) {
consts_names.push_back(word);
}
}
Nconst = consts_names.size();
// 3. the three times
readBinary((char *)&startJD,sizeof(double));
readBinary((char *)&endJD,sizeof(double));
readBinary((char *)&interval,sizeof(double));
recLength += 3*sizeof(double);
// 4. Ncoeff
readBinary((char *)&Ncoeff,sizeof(int));
recLength += sizeof(int);
// 5. AU and EMRAT
buffer[sizeof(double)] = '\0';
readBinary((char *)&AU,sizeof(double));
recLength += sizeof(double);
readBinary((char *)&EMRAT,sizeof(double));
recLength += sizeof(double);
// 6. c_arrays for the first 12 planets
for(i=0; i<12; i++) {
readBinary((char *)&c_offset[i],sizeof(int));
readBinary((char *)&c_ncoeff[i],sizeof(int));
readBinary((char *)&c_nsets[i],sizeof(int));
recLength += 3*sizeof(int);
}
// 7. DENUM
double denum;
readBinary((char *)&denum,sizeof(double));
recLength += sizeof(double);
// 8. c_arrays for libration
readBinary((char *)&c_offset[12],sizeof(int));
readBinary((char *)&c_ncoeff[12],sizeof(int));
readBinary((char *)&c_nsets[12],sizeof(int));
recLength += 3*sizeof(int);
// 9. pad - records are padded to be the same length as the data records b/c
// JPL does it (for Fortran reasons) - not necessary
for(i=0; i < Ncoeff*sizeof(double)-recLength; i++)
readBinary(buffer,1);
// ----------------------------------------------------------------
// the second header record: 400 values from the const array
double d;
for(i=0; i<400; i++) {
readBinary((char *)&d,sizeof(double));
if(i < Nconst) {
constants[stripTrailing(consts_names[i])] = d;
}
}
// pad
for(i=0; i < (400-Nconst)*sizeof(double); i++)
readBinary(buffer,1);
// ----------------------------------------------------------------
// test the header
if(denum == constants["DENUM"]) {
// EphemerisNumber == -1 means the header has not been read
// EphemerisNumber == 0 means the fileposMap has not been read (binary)
// EphemerisNumber == constants["DENUM"] means object has been initialized
// (binary file), or header read (ASCII file)
EphemerisNumber = 0;
// clear the data arrays
store.clear();
}
else {
LOG(WARNING) << "DENUM (" << denum << ") does not equal the array value ("
<< constants["DENUM"] << ")";
}
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
// private
// return 0 ok, or -4 EphemerisNumber is not defined
int SolarSystem::readBinaryData(bool save) throw(Exception)
{
try {
// has the header been read?
if(EphemerisNumber == -1) return -4;
// read the data, optionally storing it all; fill the file position map
int iret=-1,nrec=1;
double prev=0.0;
vector<double> data_vector;
while(!istrm.eof() && istrm.good()) {
long filepos = istrm.tellg();
iret = readBinaryRecord(data_vector);
if(iret == -2) { iret = 0; break; } // EOF
if(iret) break;
// if saving all in store, add it here
if(save)
store[data_vector[0]] = data_vector;
// put the first record in coefficients array
if(nrec == 1)
coefficients = data_vector;
// build the positions map
fileposMap[data_vector[0]] = filepos;
if(nrec > 1 && data_vector[0] != prev) {
ostringstream oss;
oss << "ERROR: found gap in data at " << nrec << fixed << setprecision(6)
<< " : prev end = " << prev << " != new beg = " << data_vector[0];
Exception e(oss.str());
GPSTK_THROW(e);
}
// remember the end time for next record
prev = data_vector[1];
// count records
nrec++;
}
istrm.clear();
return iret;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
// private
// return 0 ok, or
// -3 stream is not open or not good
// -4 EphemerisNumber is not defined
// For -3,-4 : initializeWithBinaryFile() has not been called, or reading failed.
int SolarSystem::readBinaryRecord(vector<double>& data_vector) throw(Exception)
{
try {
if(!istrm) return -3;
if(istrm.eof() || !istrm.good()) return -3;
if(EphemerisNumber <= -1) return -4;
data_vector.clear();
double d;
for(int i=0; i<Ncoeff; i++) {
istrm.read((char *)&d,sizeof(double)); // don't use readBinary(), to catch EOF
if(istrm.eof()) return -2;
if(!istrm.good()) return -3;
data_vector.push_back(d);
}
return 0;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
// private
// return 0 ok, or
// -1 out of range : input time is before the first time in file
// -2 out of range : input time is after the last time in file, or in a gap
// -3 stream is not open or not good, or EOF was found prematurely
// -4 EphemerisNumber is not defined
// For -3,-4 : initializeWithBinaryFile() has not been called, or reading failed.
int SolarSystem::seekToJD(double JD) throw(Exception)
{
try {
if(!istrm) return -3;
if(istrm.eof() || !istrm.good()) return -3;
if(EphemerisNumber != int(constants["DENUM"])) return -4;
if(coefficients[0] <= JD && JD <= coefficients[1]) return 0;
map<double,long>::const_iterator it; // key >= input
it = fileposMap.lower_bound(JD); // it points to first element with JD <= time
if(it == fileposMap.begin() && JD < it->first)
return -1; // failure: JD is before the first record
if(it == fileposMap.end() // if beyond the found record, go to previous;
|| JD < it->first) it--; // but beware the "lower_bound found the =" case
istrm.seekg(it->second,ios_base::beg); // get the record
int iret = readBinaryRecord(coefficients);
if(iret == -2) iret = -3; // this means EOF during data read
if(iret) return iret; // reading failed
if(JD > coefficients[1])
return -2; // failure: JD is after the last record, or
// JD is in a gap between records
return 0;
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
// private
void SolarSystem::computeState(double tt, SolarSystem::computeID which, double PV[6])
throw(Exception)
{
try {
int i,i0,ncomp;
size_t j;
for(i=0; i<6; i++) PV[i]=0.0;
if(which == NONE) return;
double T,Tbeg,Tspan,Tspan0;
Tbeg = coefficients[0];
Tspan0 = Tspan = coefficients[1] - coefficients[0];
i0 = c_offset[which]-1; // index of first coefficient in array
ncomp = (which == NUTATIONS ? 2 : 3); // number of components returned
// if more than one set, find the right set
if(c_nsets[which] > 1) {
Tspan /= double(c_nsets[which]);
for(j=c_nsets[which]; j>0; j--) {
Tbeg = coefficients[0] + double(j-1)*Tspan;
if(tt > Tbeg) { // == with j==1 is the default
i0 += (j-1)*ncomp*c_ncoeff[which];
break;
}
}
}
// normalized time
T = 2.0*(tt-Tbeg)/Tspan - 1.0;
// interpolate
unsigned int N=c_ncoeff[which];
vector<double> C(N,0.0); // Chebyshev
vector<double> U(N,0.0); // derivative of Chebyshev
for(i=0; i<ncomp; i++) { // loop over components
// seed the Chebyshev recursions
C[0] = 1; C[1] = T; //C[2] = 2*T*T-1;
U[0] = 0; U[1] = 1; //U[2] = 4*T;
// generate the Chebyshevs
for(j=2; j<N; j++) {
C[j] = 2*T*C[j-1] - C[j-2];
U[j] = 2*T*U[j-1] + 2*C[j-1] - U[j-2];
}
// compute P and V
// done above PV[i] = PV[i+3] = 0.0;
for(j=N-1; j>-1; j--) // POS
PV[i] += coefficients[i0+j+i*N] * C[j];
for(j=N-1; j>0; j--) // j>0 b/c U[0]=0 // VEL
PV[i+ncomp] += coefficients[i0+j+i*N] * U[j];
// convert velocity to 'per day'
PV[i+ncomp] *= 2*double(c_nsets[which])/Tspan0;
}
}
catch(Exception& e) { GPSTK_RETHROW(e); }
catch(exception& e) { Exception E("std except: "+string(e.what())); GPSTK_THROW(E); }
catch(...) { Exception e("Unknown exception"); GPSTK_THROW(e); }
}
//------------------------------------------------------------------------------------
} // end namespace gpstk
//------------------------------------------------------------------------------------
|
code
|
Hey friends, we are very thankful that you come to our website. Below you will find all the Sheep-related crossword clue answers and solutions. This is a very popular game and you can enjoy it for a very long time. Get all the solutions and answers of Crossword Clues Solver and we also recommend to find out all the solutions of Word Stacks which you will also like.
Everything about this awesome Crossword Clues Solver game is very interesting to see. All you have to do is see our posted solutions and answers for your everyday use to make easy. If you still require more relevant answers then ++PR1++, ++PR2++, ++NX1++ and ++NX2++ can provide you perfect solutions. We thank you all of you to for visiting our website and try to visiting for more information and solutions.
|
english
|
"""
For CommonsCloud copyright information please see the LICENSE document
(the "License") included with this software package. This file may not
be used in any manner except in compliance with the License
Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
"""
"""
Import Python/System Dependencies
"""
import csv
import json
import urllib2
"""
Import Flask Dependencies
"""
from flask.ext.rq import job
"""
Import Commons Cloud Dependencies
"""
from CommonsCloudAPI.extensions import rq
"""
Imports features from a CSV file based on user defined content
@requires
import csv
from .forfmat import FormatContent
@param (object) self
The object we are acting on behalf of
"""
@job
def import_csv(filename, storage, template_fields, activity_id):
"""
Ensure we have a valid URL, the nominal case is
"""
filename_ = validate_url(filename)
"""
List of new features that has been created
"""
features = []
headers = []
print filename_
"""
Open the CSV from a remote server (AmazonS3)
"""
try:
response = urllib2.urlopen(filename_)
reader = csv.reader(response)
except urllib2.HTTPError as error:
reader = error.read()
"""
Process each row of the CSV and save each row as a separate Feature
"""
for index, row in enumerate(reader):
if not index:
headers = process_import_headers(row, template_fields)
continue
feature_object = build_feature_object(row, headers)
features.append(feature_object)
"""
Send list of Features to our batch import function
"""
batch_url = ('http://api.commonscloud.org/v2/type_%s/batch.json') % (storage)
# batch_url = ('http://127.0.0.1:5000/v2/type_%s/batch.json') % (storage)
data = {
'features': features,
'activity_id': activity_id
}
content_length = len(data)
timeout = 3600
req = urllib2.Request(batch_url)
req.add_header('Content-Type', 'application/json')
# req.add_header('Content-Length', content_length)
response = urllib2.urlopen(req, json.dumps(data), timeout)
"""
Return a list of newly created Features
"""
return response
def build_feature_object(data, columns):
feature = {}
for index, column in enumerate(columns):
if column.startswith('type_'):
feature[column] = [
{
'id': data[index]
}
]
elif data[index] is None or data[index] == '':
feature[column] = None
else:
feature[column] = data[index]
return feature
def process_import_headers(headers, template_fields):
fields = []
for index, field in enumerate(headers):
if field.endswith('__id'):
field_name = field.replace('__id', '')
relationship_field = get_relationship_field(field_name, template_fields)
fields.append(relationship_field)
continue
fields.append(field)
return fields
def get_relationship_field(field_name, fields):
for field in fields:
if field.get('name') == field_name:
return field.get('relationship')
"""
Ensures that the URL we're opening is prepended with an appropriate
http:// or https://
@requires
@param (string) url
The URL to check for proper structure
"""
def validate_url(url):
if url.startswith('http://') or url.startswith('https://'):
return url
return ('http://%s' % url)
|
code
|
Judy Gladstone is the new Executive Director and National Board for the Documentary Organization of Canada (DOC). DOC announced the news today in a release that Gladstone will take over the post from outgoing Executive Director Pepita Ferrari.
Gladstone’s previous experience includes serving as Executive Director of Bravo!FACT and MaxFACT from 1997 to 2012, fostering the BellMedia Foundation for short film, and, since 2013, serving as a consultant for E & E Productions. She also brings to DOC experience in the non-profit sector as Chair of the Square Circle Circus, Treasurer of Tangled Art + Disability, and membership on the Park People Board of Directors, the Advisory Council of Akimbo, and the Caribbean Tales Film Festival Advisory Board.
“I look forward to working with Judy and my fellow board members to build upon the renewed vision for a stronger, more cohesive organization, which Pepita, Connie, and Anne helped to shepherd at DOC,” said Nadine Pequeneza, Chair of DOC’s Board of Directors. Former Chair Connie Edwards and Vice-Chair Anne Pick stepped down from DOC’s National Board to make room for new institutional directors.
One additional Director position remains vacant at this time.
Board members elected the Executive Committee comprised of Nadine Pequeneza as Board Chair, Isabelle Couture as Vice-Chair, David Vaisbord as Treasurer, and Ariella Pahlke as Secretary.
POV was founded by DOC in 1990. In 2010, POV became separately incorporated and enjoys its ongoing relationship with DOC and its members across the country.
|
english
|
# Chlorophytum gramineum var. typicum VARIETY
#### Status
SYNONYM
#### According to
The Catalogue of Life, 3rd January 2011
#### Published in
null
#### Original name
null
### Remarks
null
|
code
|
<article id="post-<?php the_ID(); ?>" <?php post_class("article"); ?>>
<header class="entry-header">
<?php if( ! has_post_format('quote') ) : ?><h4 class="entry-title"><a href="<?php the_permalink(); ?>" rel="bookmark"><?php the_title(); ?></a></h4><?php endif; ?>
<?php dw_minion_entry_meta(); ?>
</header>
<?php if( has_post_thumbnail() ) : ?>
<div class="entry-thumbnail"><a href="<?php the_permalink(); ?>" rel="bookmark"><?php the_post_thumbnail(); ?></a></div>
<?php endif; ?>
<div class="entry-content">
<?php the_content( __( '<span class="btn btn-small">Continue reading</span>', 'dw-minion' ) ); ?>
<?php
wp_link_pages( array(
'before' => '<div class="page-links">',
'after' => '</div>',
'link_before' => '<span class="btn btn-small">',
'link_after' => '</span>',
) );
?>
</div>
</article>
|
code
|
होम > अपराध > इंदौर में युवती ने बदमाशो से तंग आकर पिया एसिड
इंदौर में बदमाशो से तंग आकर एक युवती ने एसिड पि लिया. जानकारी के मुताबिक कुछ लड़के उसके साथ रोज शारब पीकर छेड़-छाड़ करते थे. जिससे परेशान होकर उस लड़की ने यह कदम उठाया था. पुलिस इस मामले में हर तरफ से जाँच कर रही है. पुलिस के मुताबिक आरोपियों को ढूंढा जा रहा है. उनकी हर जगह तलाश जारी है.
इस मामले में युवती के परिजनों का कहना है की युवती इन्ही बदमाशों से परेशान थी और इन्ही के कारण उसने ऐसा किया है. युवती को गंभीर हालत में एमवाय हॉस्पिटल में भर्ती कराया गया है जहा उसका इलाज जारी है. डॉक्टरों के मुताबिक युवती की हालत गंभीर बनी हुई है. युवती के परिवार वालों का अस्पताल में जमावड़ा लगा हुआ है.
आपको बता दें की रोज आए दिन इस तरह के अपराध होते जा रह है. शाशन और प्रशाशन के इतने सख्त होने के बाद भी साथ ही कानून के इतने कड़े होने के बाद भी अपराधियों के हौसले बुलंद है. और वह इस तरह की घटनाओं को अंजामं देने से पीछे नहीं हट रहे है. सरकार को इस तरह के अपराध रोकने के लिए और भी कड़े कदम उठाने होंगे क्योकि आए दिन हर जगह इस तरह की घटनाए हो रही है.
|
hindi
|
from django.contrib import admin # noqa
from .models import ScheduleItem
@admin.register(ScheduleItem)
class SchduleItemAdmin(admin.ModelAdmin):
list_filter = ('type', 'room')
|
code
|
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|
english
|
/// Creates a `Stream` from a seed and a closure returning a `Future`.
/// `Stream` from a seed value.
/// value `a`, and use `b` as the next internal state.
/// `Ok(Async::Ready(None))` in future calls to `poll()`.
/// will turn it into a `Stream` by repeating the operation.
|
english
|
वर्ल्ड कप २०१९ : धोनी के अलावा विराट कोहली की मदद कोई नहीं करेगा, जानिए किसने कहा ऐसा
नई दिल्ली। विकेटकीपर महेंद्र सिंह धोनी(महेन्द्र सिंह धोनी) भारतीय क्रिकेट टीम के लिए कितने महत्तव रखते हैं, शायद इसकी व्याख्या करना भी आसान नहीं। विकेटकीपिंग में महानता हासिल करने वाले इस दिग्गज खिलाड़ी ने कप्तानी के मामले में भी खूब नाम कमाया। यही कारण है कि इंग्लैंड एंड वेल्स में होने जा रहे विश्व कप(वर्ल्ड कप) के दाैरान भी सभी को उन्हीं से बड़ी उम्मीद लगी है। वह भले ही टीम के कप्तान ना हों लेकिन मैदान के अंदर जिस तरह से उनका रवैया रहता है, उसे देख ऐसा लगता है कि टीम वही चला रहे हैं। विराट कोहली(विरत कोहली) कप्तानी के मामले में इतने पक्के नहीं हैं। धोनी के बिना वह सटीक फैसला नहीं ले सकते, ऐसा सोशल मीडिया पर क्रिकेट फैंस का जबकि कई दिग्गजों का भी मानना है। वहीं धोनी के बचपन के कोच केशव रंजन बनर्जी का भी मानना है कि कप्तानी के मामले में कोहली के पास ऐसी कला नहीं है जो धोनी के पास है।
धोनी के अलावा कोहली की मदद कोई नहीं करेगा
धोनी को तैयार करने वाले इस कोच ने एमएस धोनी क्रिकेट अकादमी क्लीनिक लांच के मौके पर समर कैंप' के दौरान पत्रकारों से बातचीत करते हुए कहा कि जहां तक मैच की स्थितियों को पढ़ने और रणनीति बनाने की बात है तो धोनी की तुलना नहीं की जा सकती। यहां तक कि कोहली में भी यह कला नहीं है। इसलिए कोहली को कुछ सलाह लेने की जरूरत है। अगर धोनी भारतीय टीम का हिस्सा नहीं हों तो उनकी मदद के लिए कोई नहीं होगा।' उन्होंने कहा, विराट को अब भी कप्तान के तौर पर कुछ समय की जरूरत है और धोनी से सलाह से उसकी मदद ही होगी।
नंबर ४ के लिए धोनी ही सही
नंबर-४ पर काैन बल्लेबाजी करेगा, इसको लेकर कई दिग्गजों मेंं अभी भी आपसी बहस देखने को मिलती है। लेकिन बनर्जी के कोच का मानना है कि चाैथे नंबर की पहले अगर कोई सुलझा सकता है तो वह धोनी ही हैं। उन्होंने कहा, 'मुझे लगता है कि विश्व कप में धोनी को चौथे नंबर पर बल्लेबाजी करनी चाहिए। यह फैसला टीम प्रबंधन का है लेकिन यह मेरी राय है। अगर वह चौथे नंबर पर बल्लेबाजी करता है तो उसके बाद बल्लेबाज खुलकर खेल सकते हैं।' धोनी के पास नंबर-४ पर ३० मैच खेलने का अनुभव है। इसपर नंबर पर वह खरे भी उतरे हैं। इसका अंदाजा यह देखकर लगाया जा सकता है कि विश्व कप २०११ का खिताब दिलाने वाले धोनी ने नंबर-४ पर आकर ५६.५८ की एवरेज के साथ 13५८ रन बनाए हैं, जिसमें १२ अर्धशतक व एक नाबाद शतकीय पारी भी शामिल है।
संन्यास पर भी बोले
वहीं अब यह भी सवाल उठने लग पड़े हैं कि क्या धोनी का यह आखिरी विश्व कप होगा। जब उनके कोच से यह सवाल किया गया तो उन्होंने हंसते हुए इसका जवाह दिया। बनर्जी ने कहा कि आप उन्हें संन्यास लेते देखना चाहते हैं। आपको देखना चाहिए कि वह कितने फिट हैं। यह अहम बात है। धोनी कब संन्यास लेंगे यह उनकी पत्नी और पिता को भी नहीं पता होगा। इसके अलावा उन्होंने ऋषभ पंत को टीम में नहीं चुनने पर कहा कि उन्हें मौका देना जल्दबाजी होगा। भारत के पास अच्छी बेंच स्ट्रैंग्थ है। उन्हें विश्व कप के बाद मौका मिल सकता है।
|
hindi
|
using Entitas;
using System;
public class CallOnFrameEndSystem : IExecuteSystem, ISetPool {
Group group;
public void SetPool(Pool pool) {
group = pool.GetGroup(Matcher.CallOnFrameEnd);
}
public void Execute() {
foreach (Entity e in group.GetEntities()) {
Action<Entity> action = e.callOnFrameEnd.callback;
e.RemoveCallOnFrameEnd();
action.Invoke(e);
}
}
}
|
code
|
It’s not yet the primest of prime seasons for luxury beach house rentals SC, but it’s pretty close.
We’re past Easter and the Cooper River Bridge Run and not yet to Spoleto or Memorial Day, but it’s a fine time to be here, indeed. You’ll save on rates for all our SC vacation rentals, be they Isle of Palms vacation rentals or a Folly Beach house Charleston. The weather’s nice and there’s lots to do.
The stadium’s an easy drive from all beach rentals Charleston, especially our Isle of Palms luxury rentals. Just head over the IOP Connector into Mount Pleasant, take I-526 and get off at the Daniel Island exit. Follow the signs to parking for the big stadium, you can’t miss it. Click here for more on the tournament.
If you’re in a Charleston beach house this weekend, especially a luxury house on Isle of Palms, there’s a cool event for a good cause going on Saturday night, and it’s not far away. It’s a fundraiser for a young man named Drew Shelton, who was the passenger in a car accident in which the driver was killed, and he suffered unimaginable injuries.
It’s called MIRACLES FOR DREW.
The event involves a live auction with incredible stuff… a 10-day trip to Figi, an African Safari, a week in a luxury home in Palmetto Bluff, five private polo lessons with Thomas Ravenel (star of Bravo’s Southern Charm, who will be there), a trip to the Cook Islands, four tickets to see any Carolina Panthers game with behind the scenes access and a private tour by former pro and now coach Ricky Proehl, a Thanksgiving meal for 18-20 prepared in your home by an Expedition Chef who is starring in the Food Network’s show “Chopped”….a three-hour concert by the band Dead 27s…. backstage passes to Hootie & the Blowfish… a guitar signed by all four members… a Gulf Stream deep sea fishing or a private harbor cruise for six aboard the Caramba, the three-time winner of the Governor’s Cup. Oh, and there are over 100 silent auction items that are seriously way cool.
There will be live music, GREAT food and libations, and all the proceeds go toward Drew’s medical expenses.
Just cross the Connector from our IOP beach house rentals Charleston SC and turn left onto Rifle Range Road, and turn left into Cassina Plantation (just before the second roundabout at Bowman Road). Hope to see you there!
And remember, as always, to call EP for the best in rental homes South Carolina on the Charleston beaches.
Drew was born June 2, 1986, the middle of three boys in Charleston. He has an older brother, James, who is 35 and a younger brother, Keegan, who is 23. He was one of those rare babies that you could put in the crib and he would entertain himself until he fell asleep. Although he was a very content baby, he was also strong willed and independent. He kicked the slats out of his crib when he was a little over a year old and we had to get him into a bed early. I never knew how well that strong will would serve him until his accident 7 years ago.
Drew started playing soccer when he was almost 7. It became apparent early that he was a natural athlete and was good at anything he set his mind to. In the 7th grade, he was a Junior Scholar. He took the SAT in 7th grade and made a 940. He was very self-motivated and I never had to get on him about homework. He came to know the Lord when he was 10 years old and went on a missions trip at 12 with our church to a work camp to help people in an underprivileged neighborhood with house repairs. He also used to go downtown with his small group leader and talk to people about Jesus. He continued playing soccer until he was in 9th grade. He also surfed during his middle and high school years and played football in 7th, 8th and 9th grades. He was passionate about movies and all things French.
In high school he continued to do well in school. He got into the normal teenage scrapes and his grades fell in his junior year, however, he managed to pull them up and graduated with a Life Scholarship and a Presidential Scholarship. He went to the College of Charleston after graduation. Unfortunately, he lost his scholarship in his junior year because he was having a little too much fun.
Shortly after his 21st birthday, I asked him if he wanted to come back home to live for a while. He had been working in restaurants at night while in college and the party life was affecting him in a negative way. I did not expect him to accept my offer, but he did and in November of 2007, he moved home.
Drew had studied French for 2 years in middle school, 3 years in high school and 3 years in college and wanted to attend his last year of college in France or at least spent a summer there. He worked a couple of restaurant jobs but had an interview scheduled for February 21st at Blackbaud and he was very excited about that. Unfortunately, the accident happened in the early morning of February 21st and he never made that interview.
When we got to the hospital after that terrible phone call, we did not know what to expect. All we knew was that Drew had been in an accident, that he was a passenger, and that the driver did not make it. It was a one-car accident. We were finally told within the first few hours that if he survived, he would never breathe on his own and he would spend what was left of his life in a persistent vegetative state. Apparently, God had a different plan for Drew.
Several times during the first two weeks, we almost lost Drew. His injury was very serious. It seemed we were praying almost constantly, which is, I am sure, what saved him. His body temperature was lowered to 93 degrees on two occasions to try to relieve the pressure in his brain. Finally, a piece of skull on the right side of his head was removed and placed inside his abdomen and after the swelling was under control, the piece was replaced.
After three weeks or so, he began to stabilize. One day at about the 5 week point, the doctor came in Drew’s room and told me that he would need to be moved to a vent hospital in the next few weeks. When I asked what a vent hospital was, she told me that it was for people who would never breathe on their own. I was devastated, but before I even had a chance to tell anyone what I had been told, I received a text message from someone indicating that in their quiet time that morning, the Lord had said that a family member would be told discouraging news in Drew’s hospital room, but that God had the final word. He started breathing three days later – on Good Friday.
We almost lost Drew again two weeks later when he was moved from ICU to a step down unit, but again, God intervened.
Drew was at MUSC for almost 8 weeks, then was transferred to a hospital in Greenville. He needed time to continue healing, but did not need the ICU any more. We knew about 7 weeks after the accident that he was understanding what we were saying. Although Drew began mouthing words about 9 weeks after the accident, the doctors did not believe us. When he stopped mouthing words because of a fluid build-up (unbeknownst to us), no one really looked into it because they never believed that he had been doing that. About 6 weeks after arriving in Greenville, Drew was transferred to the Shepherd Center in Atlanta.
Their words may have been different, but all of the doctors had the same message – they all told us that Drew needed a miracle. Since God is in the miracle business, and the only qualification for getting one was needing one, I felt we were in the right camp.
Drew finally returned home from the hospital on July 19, 2008, about 5 months after the accident. We had a hard time finding good nurses the first year but in spite of that, looking back, it was probably the easiest year Drew had. The second year began with Drew having storms where his blood pressure, heart rate and temperature would skyrocket and the oxygen level in his blood would plummet. It is sort of like his “governor” was broken. Again, we almost lost him a couple of times, and again, God had a different plan.
About two years ago, we met a doctor who told us about neurofeedback, and it changed everything. Drew no longer has the storms he once had and no longer has seizures. He has made great strides in his level of alertness, although he is still not completely “awake”. We continue to see little things here and there as he continues the neurofeedback treatment.
Unfortunately, Drew still requires around-the-clock care. He has to be tilted in the wheelchair every 30 minutes while he is up, and he has to be turned every 3 ½ hours at night to keep from getting bedsores (which he has never had!) He has a baclofen pump implanted to keep his muscles from being stiff and drawn. He has a VP shunt to keep the fluid level down in his brain. And his temperature control, although somewhat improved, is still not functioning as it should. He is stable for the most part, but still has episodes of dysautonomia (which are similar to seizures but without the damaging effect on his brain). We make the formula that we feed him because the available formulas have a lot of sugar and other things that are not good for a healing body. As I mentioned before, he has understood what was going on and being said to him from early on, and for the last 3 years he has been blinking once to say yes and twice to say no. His comprehension does not seem to lack in any area, and he has a French lesson once a week.
Drew’s continued need for around-the-clock care has necessitated a dramatic change in the priorities and activities of his family. There is no “down” time for them – they are basically on call 24/7. Please assist Drew and his family by providing financial assistance. This fundraiser is to help with Drew’s ongoing and future medical expenses.
|
english
|
var expect = require('chai').expect;
var DateFormat = require('../lib/format');
describe("date-parser, formats", function () {
var formats;
beforeEach(function () {
formats = new DateFormat();
});
function getRegExp(format) {
return new RegExp(formats.buildLocaleRegExpStringByFormat(format));
}
describe("localizeRegExps", function () {
describe("en", function () {
it("(shortTime) should match to short time format, e.g.: 3:45 am", function () {
var regExp = getRegExp("shortTime");
expect(regExp.test("3:45 am")).to.be.true;
expect(regExp.test("03:45 am")).to.be.false;
expect(regExp.test("3:45:24 am")).to.be.false;
});
it("(mediumTime) should match to medium time format, e.g.: 3:45:23 am", function () {
var regExp = getRegExp("mediumTime");
expect(regExp.test("3:45 am")).to.be.false;
expect(regExp.test("03:45 am")).to.be.false;
expect(regExp.test("3:45:23 am")).to.be.true;
expect(regExp.test("03:45:23 am")).to.be.false;
});
it("(shortDate) should match to short date format, e.g.: 2/3/16", function () {
var regExp = getRegExp("shortDate");
expect(regExp.test("2/3/16")).to.be.true;
expect(regExp.test("02/03/2016")).to.be.false;
});
it("(mediumDate) should match to medium date format, e.g.: Jan 2, 2016", function () {
var regExp = getRegExp("mediumDate");
expect(regExp.test("Jan 2, 2016")).to.be.true;
expect(regExp.test("January 02, 2016")).to.be.false;
});
it("(longDate) should match to long date format, e.g.: January 2, 2016", function () {
var regExp = getRegExp("longDate");
expect(regExp.test("January 2, 2016")).to.be.true;
expect(regExp.test("Janu 02, 2016")).to.be.false;
});
it("(fullDate) should match to full date format, e.g.: Monday, January 2, 2016", function () {
var regExp = getRegExp("fullDate");
expect(regExp.test("Monday, January 2, 2016")).to.be.true;
expect(regExp.test("Mon, Jan 02, 2016")).to.be.false;
});
it("(short) should match to short date time format, e.g.: 2/3/16 3:45 am", function () {
var regExp = getRegExp("short");
expect(regExp.test("2/3/16 3:45 am")).to.be.true;
expect(regExp.test("02/03/2016 03:45")).to.be.false;
});
it("(medium) should match to medium date time format, e.g.: Jan 2, 2016 03:45:23 am", function () {
var regExp = getRegExp("medium");
expect(regExp.test("Jan 2, 2016 03:45:23 am")).to.be.true;
expect(regExp.test("January 02, 2016 3:45 am")).to.be.false;
});
});
describe("hu", function () {
beforeEach(function() {
formats = new DateFormat('hu');
});
it("(shortTime) should match to short time format, e.g.: 03:45", function () {
var regExp = getRegExp("shortTime");
expect(regExp.test("03:45")).to.be.true;
expect(regExp.test("03:45 am")).to.be.false;
expect(regExp.test("03:45:24")).to.be.false;
});
it("(mediumTime) should match to medium time format, e.g.: 03:45:23", function () {
var regExp = getRegExp("mediumTime");
expect(regExp.test("03:45")).to.be.false;
expect(regExp.test("03:45 am")).to.be.false;
expect(regExp.test("03:45:23")).to.be.true;
expect(regExp.test("03:45:23 am")).to.be.false;
});
it("(shortDate) should match to short date format, e.g.: 2016. 2. 3.", function () {
var regExp = getRegExp("shortDate");
expect(regExp.test("2016. 2. 3.")).to.be.true;
expect(regExp.test("2016. 02. 03.")).to.be.false;
});
it("(mediumDate) should match to medium date format, e.g.: 2016. Jan. 2.", function () {
var regExp = getRegExp("mediumDate");
expect(regExp.test("2016. Jan. 2.")).to.be.true;
expect(regExp.test("2016. Január 2.")).to.be.false;
});
it("(longDate) should match to long date format, e.g.: 2016. Január 2.", function () {
var regExp = getRegExp("longDate");
expect(regExp.test("2016. Január 2.")).to.be.true;
expect(regExp.test("2016. Jan. 2.")).to.be.false;
});
it("(fullDate) should match to full date format, e.g.: 2016. Január 2., Hétfő", function () {
var regExp = getRegExp("fullDate");
expect(regExp.test("2016. Január 2., Hétfő")).to.be.true;
expect(regExp.test("2016. Jan. 2. Hét.")).to.be.false;
});
it("(short) should match to short date time format, e.g.: 2016. 2. 3. 03:45", function () {
var regExp = getRegExp("short");
expect(regExp.test("2016. 2. 3. 03:45")).to.be.true;
expect(regExp.test("2016. 02. 03. 03:45:52")).to.be.false;
});
it("(medium) should match to medium date time format, e.g.: 2016. Jan. 3. 03:45:23", function () {
var regExp = getRegExp("medium");
expect(regExp.test("2016. Jan. 3. 03:45:23")).to.be.true;
expect(regExp.test("2016. Január 3. 3:45:23")).to.be.false;
});
});
});
describe("tokenRegExps", function () {
describe("months", function () {
it("(LLLL) should match to full month names", function () {
var regExp = getRegExp("LLLL");
expect(regExp.test("January")).to.be.true;
expect(regExp.test("january")).to.be.false;
expect(regExp.test("Január")).to.be.false;
});
it("(MMMM) should match to full month names", function () {
var regExp = getRegExp("MMMM");
expect(regExp.test("January")).to.be.true;
expect(regExp.test("january")).to.be.false;
expect(regExp.test("Január")).to.be.false;
});
it("(MMM) should match to short month names", function () {
var regExp = getRegExp("MMM");
expect(regExp.test("Jan")).to.be.true;
expect(regExp.test("jan")).to.be.false;
expect(regExp.test("01")).to.be.false;
});
it("(MM) should match to padded month numbers", function () {
var regExp = getRegExp("MM");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("00")).to.be.false;
expect(regExp.test("01")).to.be.true;
expect(regExp.test("10")).to.be.true;
expect(regExp.test("12")).to.be.true;
expect(regExp.test("13")).to.be.false;
});
it("(M) should match to normal month numbers", function () {
var regExp = getRegExp("M");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("00")).to.be.false;
expect(regExp.test("01")).to.be.false;
expect(regExp.test("1")).to.be.true;
expect(regExp.test("10")).to.be.true;
expect(regExp.test("12")).to.be.true;
expect(regExp.test("13")).to.be.false;
});
});
describe("years", function () {
it("(yyyy) should match to full years", function () {
var regExp = getRegExp("yyyy");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("11")).to.be.false;
expect(regExp.test("222")).to.be.false;
expect(regExp.test("1986")).to.be.true;
expect(regExp.test("2016")).to.be.true;
expect(regExp.test("3333")).to.be.false;
expect(regExp.test("44444")).to.be.false;
});
it("(yy) should match to two digit years", function () {
var regExp = getRegExp("yy");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("11")).to.be.true;
expect(regExp.test("86")).to.be.true;
expect(regExp.test("222")).to.be.false;
expect(regExp.test("3333")).to.be.false;
});
it("(y) should match to one digit years", function () {
var regExp = getRegExp("y");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("11")).to.be.true;
expect(regExp.test("222")).to.be.true;
expect(regExp.test("3333")).to.be.true;
});
});
describe("days", function () {
it("(EEEE) should match to full day names", function () {
var regExp = getRegExp("EEEE");
expect(regExp.test("Monday")).to.be.true;
expect(regExp.test("monday")).to.be.false;
expect(regExp.test("Hétfő")).to.be.false;
});
it("(EEE) should match to short day names", function () {
var regExp = getRegExp("EEE");
expect(regExp.test("Mon")).to.be.true;
expect(regExp.test("Monday")).to.be.false;
expect(regExp.test("mon")).to.be.false;
expect(regExp.test("hét")).to.be.false;
});
it("(RRRR) should match to relative day names", function () {
var regExp = getRegExp("RRRR");
expect(regExp.test("Yesterday")).to.be.true;
expect(regExp.test("yesterday")).to.be.false;
expect(regExp.test("Monday")).to.be.false;
});
it("(dd) should match to padded day of the month", function () {
var regExp = getRegExp("dd");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("00")).to.be.false;
expect(regExp.test("01")).to.be.true;
expect(regExp.test("11")).to.be.true;
expect(regExp.test("22")).to.be.true;
expect(regExp.test("30")).to.be.true;
expect(regExp.test("31")).to.be.true;
expect(regExp.test("32")).to.be.false;
});
it("(d) should match to any day of the month", function () {
var regExp = getRegExp("d");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("00")).to.be.false;
expect(regExp.test("1")).to.be.true;
expect(regExp.test("01")).to.be.false;
expect(regExp.test("11")).to.be.true;
expect(regExp.test("22")).to.be.true;
expect(regExp.test("30")).to.be.true;
expect(regExp.test("31")).to.be.true;
expect(regExp.test("32")).to.be.false;
});
});
describe("week", function () {
it("(ww) should match to padded number of week of year", function () {
var regExp = getRegExp("ww");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("00")).to.be.true;
expect(regExp.test("1")).to.be.false;
expect(regExp.test("01")).to.be.true;
expect(regExp.test("25")).to.be.true;
expect(regExp.test("52")).to.be.true;
expect(regExp.test("53")).to.be.true;
expect(regExp.test("54")).to.be.false;
});
it("(w) should match to the number of week of year", function () {
var regExp = getRegExp("w");
expect(regExp.test("0")).to.be.true;
expect(regExp.test("00")).to.be.false;
expect(regExp.test("1")).to.be.true;
expect(regExp.test("01")).to.be.false;
expect(regExp.test("25")).to.be.true;
expect(regExp.test("52")).to.be.true;
expect(regExp.test("53")).to.be.true;
expect(regExp.test("54")).to.be.false;
});
});
describe("hours", function () {
it("(HH) should match to padded 24h format", function () {
var regExp = getRegExp("HH");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("00")).to.be.true;
expect(regExp.test("1")).to.be.false;
expect(regExp.test("01")).to.be.true;
expect(regExp.test("13")).to.be.true;
expect(regExp.test("23")).to.be.true;
expect(regExp.test("24")).to.be.false;
});
it("(H) should match to 24h format", function () {
var regExp = getRegExp("H");
expect(regExp.test("0")).to.be.true;
expect(regExp.test("00")).to.be.false;
expect(regExp.test("1")).to.be.true;
expect(regExp.test("01")).to.be.false;
expect(regExp.test("13")).to.be.true;
expect(regExp.test("23")).to.be.true;
expect(regExp.test("24")).to.be.false;
});
it("(hh) should match to padded 12h format", function () {
var regExp = getRegExp("hh");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("00")).to.be.false;
expect(regExp.test("1")).to.be.false;
expect(regExp.test("01")).to.be.true;
expect(regExp.test("12")).to.be.true;
expect(regExp.test("13")).to.be.false;
expect(regExp.test("23")).to.be.false;
expect(regExp.test("24")).to.be.false;
});
it("(h) should match to 12h format", function () {
var regExp = getRegExp("h");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("00")).to.be.false;
expect(regExp.test("1")).to.be.true;
expect(regExp.test("01")).to.be.false;
expect(regExp.test("12")).to.be.true
expect(regExp.test("13")).to.be.false;
expect(regExp.test("23")).to.be.false;
expect(regExp.test("24")).to.be.false;
});
});
describe("minutes", function () {
it("(mm) should match to padded minutes in hours", function () {
var regExp = getRegExp("mm");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("00")).to.be.true;
expect(regExp.test("1")).to.be.false;
expect(regExp.test("01")).to.be.true;
expect(regExp.test("59")).to.be.true;
expect(regExp.test("60")).to.be.false;
});
it("(m) should match to minutes in hours", function () {
var regExp = getRegExp("m");
expect(regExp.test("0")).to.be.true;
expect(regExp.test("00")).to.be.false;
expect(regExp.test("1")).to.be.true;
expect(regExp.test("01")).to.be.false;
expect(regExp.test("59")).to.be.true;
expect(regExp.test("60")).to.be.false;
});
});
describe("seconds, milliseconds", function () {
it("(sss) should match to milliseconds", function () {
var regExp = getRegExp("sss");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("00")).to.be.false;
expect(regExp.test("000")).to.be.true;
expect(regExp.test("999")).to.be.true;
expect(regExp.test("0000")).to.be.false;
});
it("(ss) should match to padded seconds in hours", function () {
var regExp = getRegExp("ss");
expect(regExp.test("0")).to.be.false;
expect(regExp.test("00")).to.be.true;
expect(regExp.test("1")).to.be.false;
expect(regExp.test("01")).to.be.true;
expect(regExp.test("59")).to.be.true;
expect(regExp.test("60")).to.be.false;
});
it("(s) should match to seconds in hours", function () {
var regExp = getRegExp("s");
expect(regExp.test("0")).to.be.true;
expect(regExp.test("00")).to.be.false;
expect(regExp.test("1")).to.be.true;
expect(regExp.test("01")).to.be.false;
expect(regExp.test("59")).to.be.true;
expect(regExp.test("60")).to.be.false;
});
});
describe("other", function () {
it("(a) should match to am/pm", function () {
var regExp = getRegExp("a");
expect(regExp.test("am")).to.be.true;
expect(regExp.test("AM")).to.be.true;
expect(regExp.test("pm")).to.be.true;
expect(regExp.test("PM")).to.be.true;
expect(regExp.test("AP")).to.be.false;
});
it("(Z) should match to any timezone offset", function () {
var regExp = getRegExp("Z");
expect(regExp.test("+00:00")).to.be.true;
expect(regExp.test("-00:00")).to.be.true;
expect(regExp.test("00:00")).to.be.false;
expect(regExp.test("+0000")).to.be.true;
expect(regExp.test("-0000")).to.be.true;
expect(regExp.test("0000")).to.be.false;
expect(regExp.test("+12:00")).to.be.true;
expect(regExp.test("+13:00")).to.be.false;
expect(regExp.test("+00:40")).to.be.false;
});
});
describe("literals", function () {
it("(!) should handle character as literal", () => {
var regExp = getRegExp("!a");
expect(regExp.test("a")).to.be.true;
expect(regExp.test("AM")).to.be.false;
});
it("(!!) should handle ! as a character literal", () => {
var regExp = getRegExp("!!");
expect(regExp.test("!")).to.be.true;
expect(regExp.test("!a")).to.be.false;
});
});
describe("complex", function () {
it("should handle complex expression", () => {
var regExp = getRegExp("MMMM d, yyyy H a");
expect(regExp.test("March 3, 2020 11 AM")).to.be.true;
expect(regExp.test("Mar 03, 20 1a")).to.be.false;
});
it("should handle complex expression with literals", () => {
var regExp = getRegExp("MMMM d, yyyy !a!t H a CEST");
expect(regExp.test("March 3, 2020 at 11 AM CEST")).to.be.true;
expect(regExp.test("March 3, 2020 AMt 11 AM CEST")).to.be.false;
});
});
});
describe("_groupRegExpString()", function () {
it("should group with not matching group", function () {
expect(formats._groupRegExpString('string')).to.equal('(?:string)');
});
it("should group with matching group", function () {
expect(formats._groupRegExpString('string', true)).to.equal('(string)');
});
it("should not group if prohibited", function () {
expect(formats._groupRegExpString('string', false)).to.equal('string');
});
});
describe("buildRegExpStringByFormat()", function () {
it("should replace all possible tokens", function () {
var format = Object.keys(formats.tokenRegExps).join(':');
var expected = "^" + Object.keys(formats.tokenRegExps).map(function (token) {
return '(?:' + formats.tokenRegExps[token] + ')';
}).join(':') + "$";
expect(formats.buildRegExpStringByFormat(format)).to.equal(expected);
});
});
describe("buildLocaleRegExpStringByFormat()", function () {
it("should replace all possible locale tokens", function () {
var format = Object.keys(formats.localizeReqExps).join(':');
var expected = "^" + formats._replaceTokens(Object.keys(formats.localizeReqExps).map(function (token) {
return formats.localizeReqExps[token];
}).join(':')) + "$";
expect(formats.buildLocaleRegExpStringByFormat(format)).to.equal(expected);
});
});
});
|
code
|
Buy products related to cattle fence products and see what customers say about cattle fence products on Amazon.com ✓ FREE DELIVERY possible on eligible .
As with cattle-working pens, fences around swine confinement units are likely to receive heavy usage. Use heavy materials and sturdy construction for long life .
Livestock fencing is dependent on the type of livestock you intend to keep. are willing to spend to buy the supplies, and how long the fence-line needs to be (or, .
Tractor Supply Co. is your destination for garden fencing, livestock fencing, feed. Wood fences are the most common type of fence we build today, because of .
|
english
|
प्रणब मुखर्जी की बेटी शर्मिष्ठा, मीरा कुमार के बेटे अंशुल को बनाया कांग्रेस का राष्ट्रीय प्रवक्ता
नई दिल्ली। कांग्रेस ने पूर्व राष्ट्रपति प्रणब मुखर्जी की बेटी शर्मिष्ठा मुखर्जी (शर्मीस्था मुखर्जी) और पूर्व लोकसभा अध्यक्ष मीरा कुमार (मैरा कुमार) के बेटे अंशुल मीरा कुमार को राष्ट्रीय प्रवक्ता बनाया गया है। कांग्रेस की अंतरिम अध्यक्ष सोनिया गांधी ने शर्मिष्ठा और अंशुल की नियुक्ति को हरी झंडी दे दी है। इससे पहले शर्मिष्ठा दिल्ली कांग्रेस में प्रदेश प्रवक्ता थीं। साभार-खस्खबर.कॉम शर्फासबुक्तविटरव्हत्सप्प
साराधा स्कम : राजीव कुमार का पता लगाने के लिए सीबीआई गठित कर रही है एक विशेष टीम
|
hindi
|
एसपी के आदेश का असर, सहतवार में लाल पर्ची बंटने की प्रक्रिया शुरू - अखंड भारत न्यूज
होम उत्तर प्रदेश बलिया एसपी के आदेश का असर, सहतवार में लाल पर्ची बंटने की प्रक्रिया...
एसपी के आदेश का असर, सहतवार में लाल पर्ची बंटने की प्रक्रिया शुरू
थाने द्वारा कोई प्रूफ नही मिलने की शिकायत फरियादियो ने पुलिस कप्तान से पहले कई बार कर चुके ऐसे मामलों की गंभीरता को देखते हुए नवागत कप्तान ने सहतवार थाने को लाल परची की मुहैया कराया। यही नही इस लाल पर्ची पाकर फरियादियो में पुलिस द्वारा कार्रवाही होने की बात कहीं इससे फरियादियो में खुशी की लहर दौड़ गई है।
प्रेवियस आर्टियलशराब की दूकान में तोड़-फोड़ मचाने के लिए महिलाओं के खिलाफ मुकदमा दर्ज
नेक्स्ट आर्टियलसीडीओ ने विकास भवन स्थित कार्यालयों का किया औचक निरीक्षण
|
hindi
|
What would you like to consult?
The information obtained is totally anonymous, and in no circumstances could be associated to a specific, identified user.
The personal data provided in the course of your treatment request (the “Request”) will be collected, processed and used in order to manage your request for the subsequent execution of the Contract entered into, and for the purpose of maintaining the contractual relationship, the management, administration, provision, extension and improvement of the services, as well as the sending of technical and operational information related to the Treatment, by any means, including electronic mail and/or equivalent means. In this regard, we hereby inform you that your personal data together with the result of possible user satisfaction surveys will be used to evaluate your opinion and the study of your particular profile, with the sole purpose of improving the services we provide and thus adapt and design our commercial offerings.
To exercise these rights you can write to EUGIN using the reference “GDPR Rights”, to EUGIN Clinic, Patient Care Department, C/Travessera de les Corts, 322 and C/ Entença, 293 – 08029 or by email to the EUGIN DPO ([email protected]) by attaching a copy of an official identification document or by directed email.
Sending of commercial communications. Sending commercial communications, whether on paper or by electronic or telematic means, and direct contact regarding the services which, at any given time, Eugin Clinic commercialises.
Exercise of rights by the user. The user has the right to access their personal data, as well as to request the rectification of inaccurate data or to request its deletion when, among other reasons, the data is no longer necessary for the purposes for which it was collected. The user will be able to access their information, rectify it, request cancellation or oppose its processing, as well as withdraw consent at any time, without affecting the legality of the treatment based on the consent prior to its withdrawal by writing to [email protected].
Claims before the data protection authority. The user can direct their claims arising from the processing of their personal data to the Spanish Agency for Data Protection (www.agpd.es).
The information provided on this website serves to support, but in no case replace the relationship between patient and doctor.
Its content is intended for a wide audience, therefore, the language used is intended to be accessible and includes graphical illustrations designed to facilitate understanding.
As one would expect, the information contained here is general in nature. The peculiarities of each case are developed through direct contact with our experts. To this end, all EUGIN Clinic patients have a secure private area that allows for an individualised form of contact.
This website is the property of Clínica EUGIN and does not contain any advertising whatsoever. Thus, there is no conflict of interest that may influence the information described herein.
The artwork appearing on this website is owned by Eugin and its reproduction or dissemination, either totally or partially, in other media or platforms is strictly forbidden.
Analytical cookies: This type of cookies informs us how many users visit our site, allowing us to carry out subsequent statistical analysis of how users access our services. We analyze how you browse our web in order to improve the range of services and products we offer.
Advertising cookies: These cookies store information about users’ browsing habits through ongoing observation. They provide information on internet browsing habits and we can then show advertising that relates to your browser profile.
|
english
|
'use strict';
angular.module('webActivitiesApp.toolbar', [])
.controller('ToolbarCtrl', [ '$rootScope', '$scope', 'framework', '$q', function($rootScope, $scope, framework, $q ) {
$scope.liveActivity = null;
var updateActions = function() {
setTimeout(function() {
if ($scope.liveActivity && $scope.liveActivity.context.actions) {
$scope.activityActions = $scope.liveActivity.context.actions;
} else {
$scope.activityActions = [];
}
$scope.$apply();
},0);
};
var tracker = new LiveActivityTracker(framework.eventBus,
function(oldActivity) {
$scope.liveActivity = null;
updateActions();
},
function(newActivity) {
$scope.liveActivity = newActivity;
$scope.liveActivity.context.eventBus.on("actionsChanged",updateActions);
}
);
$scope.activityStack = function() {
if (framework.getActivityStack() == null) {
return [];
}
var clone = framework.getActivityStack().slice(0);
return clone.reverse();
};
$scope.toolbarActions = [];
$scope.activityActions = [];
$scope.maxBreadcrumbSize = 3;
framework.internalBus().syncTopic("com.newt.system.toolbar.actions",$scope.toolbarActions,function() {
$scope.toolbarActions.sort(function(a, b){
var aOrder = a.order || 10,
bOrder = b.order || 10;
return aOrder-bOrder;
});
$scope.$apply();
});
$scope.executeAction = function(action) {
if (action.handler) {
action.handler(action);
}
};
}])
//END
;
|
code
|
Background Skins, which are often referred to as Background Advertising or Wallpaper Advertising, are very effective branding unit for advertisers running branding campaigns with performance metrics similar to those of a TV media buy.
Background Skins are typically a HTML5 or is a static background visual (JPG) that serves on the Left and Right of a website. They can also occupy the space above the website also. Generally speaking, the entire skin area is clickable.
Publishers can select to run Background Skins on their website via the High Impact Page of our Publisher Console.
Background Skins are a high demand unit by Advertisers, particularly around Holidays and Events.
If you’re an approved Gourmet Ads Publisher and would like to include background skins on your site, please contact us.
|
english
|
अर्थ सभी यंत्रों में ड्वेलर
भगवान शनि देव और हनुमान जी को कुंभ राशि का आराध्य देव माना जाता है। कुम्भ राशि के सर्वायंत्रात्मका नाम के लड़कों की उत्तेजना और परिसंचरण को यूरेनस ग्रह द्वारा नियंत्रित किया जाता है। जिस समय वृक्ष फलों और फूलों से भर जाते हैं उस मौसम में सर्वायंत्रात्मका नाम के लड़के जन्म लेते हैं। कुंभ राशि के सर्वायंत्रात्मका नाम के लड़कों को गुस्सा अधिक आता है। इन सर्वायंत्रात्मका नाम के लड़कों में एलर्जी, सूजन और हृदय सम्बन्धी समस्याओं और गठिया होने की सम्भावना रहती है। इन सर्वायंत्रात्मका नाम के लड़कों में दूसरों की मदद करने का गुण, ऊर्जा और बुद्धि कूट कूट कर भरी होती है। इन्हें दोस्ती करना पसंद होता है।
सर्वायंत्रात्मका नाम बहुत सुंदर और आकर्षक माना जाता है। इतना ही नहीं इसका मतलब भी बहुत अच्छा होता है। आपको बता दें कि सर्वायंत्रात्मका नाम का अर्थ सभी यंत्रों में ड्वेलर होता है। सभी यंत्रों में ड्वेलर होना बहुत अच्छा माना जाता है और इसकी झलक सर्वायंत्रात्मका नाम के लोगों में भी दिखती है। शास्त्रों में सर्वायंत्रात्मका नाम को काफी अच्छा माना गया है और इसका मतलब यानी सभी यंत्रों में ड्वेलर भी लोगों को बहुत पसंद आता है। जैसा कि हमने बताया कि सर्वायंत्रात्मका का अर्थ सभी यंत्रों में ड्वेलर होता है और इस अर्थ का प्रभाव आप सर्वायंत्रात्मका नाम के व्यक्ति के व्यव्हार में भी देख सकते हैं। सर्वायंत्रात्मका नाम के अर्थ यानी सभी यंत्रों में ड्वेलर का असर आप इनके स्वभाव में साफ़ देख सकते हैं। आगे पढ़ें सर्वायंत्रात्मका नाम की राशि, इसका लकी नंबर क्या है, सर्वायंत्रात्मका नाम के सभी यंत्रों में ड्वेलर मतलब के बारे में विस्तार से जानें।
सर्वायंत्रात्मका नाम का स्वामी ग्रह शनि और शुभ अंक ८ है। जिन लोगों का नाम सर्वायंत्रात्मका और शुभ अंक ८ होता है, उन्हें धन की कोई कमी नहीं होती। सर्वायंत्रात्मका नाम वाले लोग अपनी शर्तों पर जीते हैं। ये अपने नियम खुद बनाते हैं। इस अंक के लोग संगीत प्रेमी होते हैं। सर्वायंत्रात्मका नाम के लोग मेहनत और लगन से सफल होते हैं। ये किस्मत पर निर्भर नहीं रहते। ८ अंक वाले लोगों को सफलता प्राप्त अवश्य होती है मगर देरी से और ये स्वभाव में दयावान होते हैं।
सर्वायंत्रात्मका नाम के लोगों की कुंभ राशि होती है। ये लोग आत्मनियंत्रित, प्रतिभाशाली व नरम दिल के होते हैं। इस नाम के लोगों के पास बुद्धि की कमी नहीं होती और इन्हें अपने ज्ञान पर काफी गर्व होता है। अक्सर इन लोगों को समझ पाना मुश्किल होता है। कुंभ राशि के लोग समाज में अच्छा व्यवहार करते है, लेकिन मित्र बहुत ध्यानपूर्वक चुनते हैं। सर्वायंत्रात्मका नाम के लोग दूसरों के प्रति काफी सहानुभूति रखते हैं और लोगों की मदद करना इन्हें अच्छा लगता है।
सर्वायंत्रात्मका की कुंभ राशि के हिसाब से और नाम
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hindi
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Will you be the next early career scientist representative for the Geodesy division?
In the run-up to the general assembly in 2017, The geodesy division is looking for a fresh early career scientist (ECS) to take over the role of the ECS-representative. But what comprises being an ECS-representative? And where can you sign up?
Early career scientists represent a significant share of the EGU general assembly attendees. It is therefore desirable to involve this group not only as participants, but also on the division and EGU level. To enable this, every division has assigned an early career scientist representative (ECS-rep), who have contact with the representatives from the other divisions. This allows to gather ideas, issues and thoughts relevant to the early career scientist group and communicate these at an EGU-wide level.
Since April 2014, I’ve been the early career scientist representative of the geodesy division. However, from April 2017, I will lay down this task and take over the role as the union-wide ECS-rep. It is therefore time to find a enthusiastic candidate for the next 2 years, who can bring her/his own ideas to the geodesy division. A prerequisite is that you fulfill the conditions of being an early career scientist, i.e. the date of obtaining your highest degree (BSc, MSc, or PhD) lies within the past 7 years.
Communicate with the Geodesy division president and its deputies. This may cover ECS involvement but also programmatic issues, topics such as the division business meeting, and website content.
Participate with the EGU-wide ECS-rep discussions. We all meet in person during the EGU but several times over the course of the year we set up skype meetings to discuss new ideas, recommendations, questionnaire results, short-course organization and much more. These meetings are essential for the workings of the ECS representation at EGU level.
How did you like being the representative?
I’m obviously biased now, but initially I was a bit skeptical about being involved. I’m now looking back at some valuable years adding to my skillset. I got to know people from other divisions, learned about the workings of the EGU, and got more involved in organizing short courses. Furthermore, contact with the Geodesy division program committee has been very fruitful and I appreciate their openness.
Outside the ECS-lounge. During the general assembly, you can see ample mentions of early career scientists related events and facilities.
Where and when can I sign up?
Please send a brief motivation letter and a CV to the current geodesy division president, Michael Schmidt, at [email protected]. Deadline is the 31st of March 2017. If you have any further questions or if you want a first hand report, you can also contact me under [email protected] or on twitter.
Your scientific talk: mental breakdown or conference highlight?
After last years success, we’re again organizing a short course on presentation techniques. EGU GA 2016 participants who are interested in rehearsing their talk and getting feedback can sign up of for a rehearsal here (deadline 31 March 2016). Of course we welcome and encourage contributions from all divisions.
You can feel it coming, sometimes it kicks in days before your talk, at other times just moments before you climb the podium. When it is at its peak, speech anxiety or, in scientific terms, glossophobia, may even have physical ramifications. Your heart rate raises, your breathing is irregular and your armpits are spraying sweat, or at least you think they are. In this state, your body is in an excellent shape to the one thing it considers sane: flee.
The problem is you can’t. You are a scientific speaker at a conference and there is an audience eagerly waiting to hear about your research. Some of us may be tempted to opt for something in between fleeing and presenting, but this usually results in someone hiding and whispering behind a lectern. Alternatively, you have the option to look your speech anxiety in the face and tell yourself that it is an unavoidable part of your job as a scientist.
The good news is however, that this doesn’t need to affect the quality of your talk at all. On the contrary, your anxious state also enables you to be very alert and focused, which may actually help you delivering an excellent talk.
Besides your mental state, there are plenty of other issues, which influence the effectiveness of information-flow to your audience. Some of those are behavioral, such as making eye contact with your audience, stress parts of your talk by using your voice dynamically, or simply avoiding some ineffective habits like non-stop lightsabering your slides with a laser pointer.
On the more material side, you may consider structuring your presentation using narratives, making use of effective graphics and trying to eliminate those parts from your presentation which do not contribute to it. Just that someone at Microsoft thought it was a cool idea to offer transition effects like slides disintegrating in blocks and stars, doesn’t mean it was a good idea. Most people, including me, are not entertained by it but respond allergically to such slide transitions, resulting in an instant distraction from what you’re telling. You may also consider avoiding some fonts, which have the potential to cause political uproar. For some, seeing comicsans in your scientific presentation is like saying you like Obamacare in a GOPdebate.
In a nutshell, the single piece of advice for making a good presentation is the old boyscout credo: “Be prepared”. Once you prepared your presentation, you may want to check if it is effective and can be finished within the allocated time slot. Weathered speakers may know this from their experience, but even better is to rehearse your talk in front of group of friendly but critical peers. For a conference talk, the group would ideally consists of scientists from varying research fields, such that the audience better resembles reality. By rehearsing you (1) know whether your talk fits in the 12 minute limit, (2) can check if your main points came accross, and (3) see if your presentation material looks the way it is supposed to do.
The short course is set up as follows. After kicking off with an entertaining talk on presenting, registered participants will give their presentation after which there is time to receive feedback from the organizers and audience. In contrast to your actual conference talk, more time is scheduled for feedback and topics related to presentation skills will be given plenty of attention. Everybody is welcome to attend the short course, but we specifically invite scientists, notably early career scientists, to sign up for a try-out of their EGU talk (PICO or oral).
But even after attending our shortcourse, and being very well prepared, you’re climbing that podium and are still nervous. But now, you got what it takes to deliver a conference highlight.
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english
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var searchData=
[
['edge',['Edge',['../classcompute__modularity_1_1_edge.html',1,'compute_modularity']]]
];
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code
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# kea-pxe-replace-mod
Kea module utilizing hooks api to replace pxe options (siaddr, option 66, 67) via web request.
# Usage
Install the compiled library somewhere sensible (such as /usr/local/lib/kea-pxe-replace4.so)
Configure /etc/kea/kea-pxe-replace4.conf
Please see the example config in the etc/ directory. Each key is mandatory,
however the value can be a value of your choosing. These refer to the fields
in the json response that this module will use for each parameter; this is so
you can configure an existing API service to interact with the module. Nested
fields should be dot (.) separated. IE: 'myserver.params.siaddr'.
For the response, the field does not have to be present. If the field is not
present, it will not attempt to override the value.
These values are only applied to fields already set. If next-server is not defined
somewhere in your kea config, or otherwise not set during packet processing, it
will not be overridden. The same logic applies to the option fields; they are
only updated if they are set in the first place.
Configure /etc/kea/kea-dhcp4.conf
Will result in requesting json from "url" in the form of "url+mac"
Example: http://myurl.com/aa:bb:cc:dd:ee:ff
Note: a '/' is not inserted between the url and the mac address for you, you
must provide that in the url if necessary.
## Build requirements
This software has been developed on Debian Stretch and Ubuntu 16.04. This
software is currently experimental, use at your own risk (as always).
apt-get install g++ libcurl4-gnutls-dev libboost-dev kea-dev
./build.sh
## Testing locally
You can create a veth pair:
ip link add veth0 type veth peer name veth1
ip l set veth0 up && ip l set veth1 up
ip a add dev veth0 192.0.2.1/24
dhclient -d -v veth1
## Testing with libvirt locally
You probably need to kill the dnsmasq agents that are running.
brctl addbr brx && brctl addif brx veth1
virt-install --pxe --network bridge=brx --name pxe1 --memory 128 --disk none
Also checkout dhcpdump!
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code
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صلیبی جنگہٕ (1096-1271) أس اکھ مقدس جنگ کس تتھ عزائمس پیٹھ مبنی یمن خلاف چرچ اوس کافرن ہنٛد پأنٹھ وژان۔
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kashmiri
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List of famous investors, with photos, bios, and other information when available. Who are the top investors in the world? This includes the most prominent investors, living and dead, both in America and abroad. This list of notable investors is ordered by their level of prominence, and can be sorted for various bits of information, such as where these historic investors were born and what their nationality is. The people on this list are from different countries, but what they all have in common is that they're all renowned investors.
The list you're viewing has a variety of people in it, like Michael Jordan and Donald Trump.
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english
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बीजेपी ने क्यों लिया अजित पवार का साथ? जानें पूर्व मुख्यमंत्री देवेंद्र फडणवीस का जवाब | लाटेस्ली हिन्दी
बीजेपी ने क्यों लिया अजित पवार का साथ? जानें पूर्व मुख्यमंत्री देवेंद्र फडणवीस का जवाब
महाराष्ट्र (महाराष्ट्र) के पूर्व मुख्यमंत्री और बीजेपी नेता देवेंद्र फडणवीस (देवेन्द्र फादनाविस) ने राष्ट्रवादी कांग्रेस पार्टी (नप) के नेता अजित पवार (अजित पॉवर) के साथ गठबंधन पर बयान दिया है. दरअसल, बुधवार को देवेंद्र फडणवीस से जब पूछा गया कि बीजेपी का अजित पवार के साथ जाना क्या एक गलत कदम था. इस पर उन्होंने कहा कि चिंता न करें, मैं सही वक्त आने पर बयान दूंगा. दरअसल, पुणे (पुणे) की बारामती सीट से १.६५ लाख वोटों के अंतर से विधानसभा चुनाव जीतने वाले अजित पवार ने अपनी पार्टी एनसीपी और परिवार को पिछले शनिवार को उस समय अचंभे में डाल दिया था जब उन्होंने बीजेपी से हाथ मिला लिया और वह देवेंद्र फडणवीस के नेतृत्व वाली सरकार में उपमुख्यमंत्री बने.
इसके बाद अजित पवार ने मंगलवार को निजी वजहों का हवाला देते हुए उपमुख्यमंत्री पद से इस्तीफा दे दिया. इसके बाद देवेंद्र फडणवीस ने भी मुख्यमंत्री पद से इस्तीफा दे दिया जिससे बीजेपी के नेतृत्व वाली सरकार गिर गई. यह भी पढ़ें- महाराष्ट्र में ८० साल के शरद पवार ने इस तरह पलटी बाजी, देवेंद्र फडणवीस समेत पूरी बीजेपी को किया चारों खाने चित.
वहीं, अजित पवार ने बुधवार को कहा कि वह एनसीपी में बने रहेंगे और इसके बारे में भ्रम पैदा करने की कोई वजह नहीं है. उन्होंने पत्रकारों से कहा, अभी मेरे पास कहने के लिए कुछ नहीं है, मैं सही समय आने पर बोलूंगा. मैंने पहले भी कहा था कि मैं एनसीपी में हूं और मैं एनसीपी में ही रहूंगा. भ्रम पैदा करने की कोई वजह नहीं है.
गौरतलब है कि महाराष्ट्र के राज्यपाल भगत सिंह कोश्यारी (गवर्नर भगत सिंह कोश्यारी) द्वारा बुधवार को बुलाए गए विधानसभा के विशेष सत्र में देवेंद्र फडणवीस, अजित पवार व अन्य नवनियुक्त सदस्यों ने विधायक के तौर पर शपथ ली. वहीं, उद्धव ठाकरे गुरुवार शाम महाराष्ट्र के मुख्यमंत्री पद की शपथ लेंगे.
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hindi
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होम > उत्पादों > चमड़ा एलईडी कुत्ते पट्टा
(चमड़ा एलईडी कुत्ते पट्टा के लिए कुल २४ उत्पादों)
चमड़ा एलईडी कुत्ते पट्टा - निर्माता, कारखाने, आपूर्तिकर्ता चीन से
हम विशेष हैं चमड़ा एलईडी कुत्ते पट्टा निर्माताओं और आपूर्तिकर्ताओं / कारखाने चीन से। कम कीमत / सस्ते के रूप में उच्च गुणवत्ता के साथ थोक चमड़ा एलईडी कुत्ते पट्टा, चीन से अग्रणी ब्रांडों में से एक चमड़ा एलईडी कुत्ते पट्टा में से एक, ऐडी फ्लाशिंग प्रोडक्ट को.लिमिटेड।
थोक चीन से चमड़ा एलईडी कुत्ते पट्टा , लेकिन कम कीमत के अग्रणी निर्माताओं के रूप में सस्ते चमड़ा एलईडी कुत्ते पट्टा खोजने की आवश्यकता है। बस चमड़ा एलईडी कुत्ते पट्टा पर उच्च गुणवत्ता वाले ब्रांडों पा कारखाना उत्पादन, आप आप क्या चाहते हैं, बचत शुरू करते हैं और हमारे चमड़ा एलईडी कुत्ते पट्टा का पता लगाने के बारे में भी राय, आप में सबसे तेजी से उत्तर हम करूँगा कर सकते हैं।
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hindi
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Learn how to be a better soccer player with the soccer drills and soccer tricks demonstrated by Deejae Johnson in these Howcast videos.
Since 2004, when Deejae Johnson took over the UHS soccer program as the head boys' and girls' coach, his teams have won eight of 10 league titles. Prior to joining UHS, Deejae was a member of the Cal coaching staff for both the men's and women's teams. He also coaches the Marin FC U15, Marin United u23 and u21 and The Olympic Club women's team. Deejae holds a USSF A license, NSCAA Advanced National Diploma, English FA prelim badge and a Dutch KNVB diploma. Deejae is a scout for the USSF Development Academy League and has been a member of the Cal North Olympic Development staff.
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english
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"अच्छा और स्तरीय काम"
ये मल्लिका शेरावत कह रही है?
४५ या ३५ जलवे तो वो ही हैं ना २५ वाले और यही मायने रखता है
"लोग मुझसे जुड़ी सभी चीज़ों को इसी तरह बढ़ा-चढ़ा के दिखाते हैं"
कमाल हैं दिखाती तो आप हैं या लोग तो सिर्फ देखते हैं?
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hindi
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میٲنِس سیٚنٛٹرٛل بیٚنٛک آف اِنٛڈیا اکاونٹَس منٛز کوٗتہ روٗد
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kashmiri
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मोदी सरकार के आम बजट में महिला सशक्तीकरण पर जोर | परफॉर्म इंडिया
होम समाचार मोदी सरकार के आम बजट में महिला सशक्तीकरण पर जोर
मोदी सरकार के आम बजट में महिला सशक्तीकरण पर जोर
प्रधानमंत्री नरेंद्र मोदी की हर योजना और कदम के केंद्र में देश की आधी आबादी यानी महिलाएं रहती हैं। मोदी सरकार में महिलाओं के सामाजिक और आर्थिक विकास के लिए जितना काम किया है, उतना पहले किसी भी सरकार ने नहीं किया। गुरुवार को लोकसभा में पेश किए गया आम बजट भी महिला सशक्तीकरण पर केंद्रित रहा। बजट में ग्रामीण, शहरी और नौकरीपेशा महिलाओं के उत्थान के लिए तमाम योजनाओं का ऐलान किया गया। एक नजर डालते हैं-
प्रेवियस आर्टियलमोदी सरकार के बजट में दिखा सबका साथ सबका विकास का मंत्र, मध्यम वर्ग को मिलेगा लाभ
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hindi
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\begin{document}
\title[MA ROM]{Manifold Alignment-Based Multi-Fidelity Reduced-Order Modeling Applied to Structural Analysis}
\author[1]{\fnm{Christian} \sur{Perron}}\email{[email protected]}
\equalcont{These authors contributed equally to this work.}
\author[1]{\fnm{Darshan} \sur{Sarojini}}\email{[email protected]}
\equalcont{These authors contributed equally to this work.}
\author*[1]{\fnm{Dushhyanth} \sur{Rajaram}}\email{[email protected]}
\equalcont{These authors contributed equally to this work.}
\author[1]{\fnm{Jason} \sur{Corman}}\email{[email protected]}
\equalcont{These authors contributed equally to this work.}
\author[1]{\fnm{Dimitri} \sur{Mavris}}\email{[email protected]}
\equalcont{These authors contributed equally to this work.}
\affil*[1]{\orgdiv{Daniel Guggenheim School of Aerospace Engineering},
\orgname{Georgia Institute of Technology}, \orgaddress{\street{North Avenue NW}, \city{Atlanta}, \postcode{30332}, \state{Georgia}, \country{USA}}}
\abstract{
This work presents the application of a recently developed parametric, non-intrusive, and multi-fidelity reduced-order modeling method on high-dimensional displacement and stress fields arising from the structural analysis of geometries that differ in the size of discretization and structural topology.
The proposed approach leverages manifold alignment to fuse inconsistent field outputs from high- and low-fidelity simulations by individually projecting their solution onto a common latent space.
The effectiveness of the method is demonstrated with two multi-fidelity scenarios involving the structural analysis of a benchmark wing geometry.
Results show that outputs from structural simulations using incompatible grids, or related yet different topologies, are easily combined into a single predictive model, thus eliminating the need for additional pre-processing of the data.
The new multi-fidelity reduced-order model achieves a relatively higher predictive accuracy at a lower computational cost when compared to a single-fidelity model.
}
\keywords{
Multi-fidelity,
Reduced Order Model,
Wing Structural Analysis,
Machine Learning
}
\maketitle
\renewcommand*{\arraystretch}{1.3}
\section{Introduction}\label{sec:intro}
The design and certification of aircraft are increasingly relying on numerical simulation instead of physical experiments owing to the former's relatively rapid turnaround time and high accuracy.
Despite the increased reliance on numerical simulations, routinely encountered many-query scenarios such as design space exploration, optimization, uncertainty quantification, etc., require repetitive evaluation of computationally expensive codes.
Consequently, engineering workflows with expensive numerical simulations \emph{in-the-loop} can suffer from impractical compute times.
Surrogate models have provided respite to this computational burden by trading the high accuracy of physics-based numerical models for faster evaluation times.
While conventional surrogate models have mainly focused on the prediction of scalar quantities~\cite{Queipo2005, Forrester2009, Yondo2018, Bhosekar2018}, reduced-order models (ROMs) have enabled the prediction of spatially and temporally distributed field quantities.
This class of methods reduces the high-dimensional nature of the problem by extracting a low-dimensional subspace (or more generally, latent space) that best captures the most dominant physical features of the solution space.
In a sense, by preserving and capturing the underlying spatial and temporal correlation within the high-dimensional solutions, ROMs can offer physically richer predictions.
This trait renders ROMs useful in enabling rapid multidisciplinary analyses where high-dimensional coupling variables must be passed between simulations to achieve consistency between, for instance, the aerodynamics and the structures disciplines.
The construction of surrogate models for both scalar and field quantities suffers from a major drawback when the underlying response is highly nonlinear and consists of many independent parameters.
In such cases, most methods require a large amount of training data to build a model with reasonable predictive accuracy.
Training data generated from high-fidelity simulations usually incur a substantial computational cost, which may arguably defeat the purpose of constructing the surrogate model.
Compared to scalar quantities, the multi-fidelity combination of fields brings more challenges because of the potential inconsistency between results produced by different analyses~\cite{Perron2020}.
For instance, it is well known that a fine mesh FEM model generally produces more accurate results than a coarse mesh FEM model but at a higher computational cost.
Although fine mesh and coarse mesh FEM models represent the same structure, they have inconsistent field outputs.
This inconsistency non-trivializes the process of combining their high-dimensional results into a single predictive ROM.
Recent work by Perron et al.~\cite{Perron2020, Perron2020a, perron2021multi} introduced a novel ROM method based on manifold alignment (MA-ROM) to overcome the issues caused by inconsistent multi-fidelity field outputs.
The proposed method employs manifold alignment to find a common low-dimensional representation of two datasets composed of heterogeneous high-dimensional field outputs.
The method's capabilities and performance in comparison to other multi-fidelity ROM methods were demonstrated on aerodynamic problems, and the results have shown that multi-fidelity fields having distinct sizes and topologies could be readily combined into a single model with superior predictive accuracy.
In an effort to investigate the effectiveness of MA-ROM against the unique challenges presented by structural simulations, this work primarily focuses on empirically assessing the performance of the MA-ROM by constructing surrogates on high-dimensional outputs from related yet distinct 3D structural analyses.
The Common Research Model (CRM) wing geometry serves as a testbed to showcase the method.
This demonstration aims to construct multi-fidelity ROMs with wing planform shape parameters as input and structural field responses as outputs for two challenging multi-fidelity scenarios.
The first scenario captures the differences in the outputs arising from the coarse and fine mesh FEM representation discussed earlier.
Few samples from the more accurate but expensive fine mesh field are combined with many coarse mesh samples.
The second scenario is a novel demonstration that tackles field outputs from models with inconsistent topologies;~one, a primary dataset from a geometry with a high rib count, and two, an auxiliary dataset from a low rib count wing.
This scenario is designed to test whether the MA-ROM can \emph{warm-start} the training of a prediction model using existing data from a related, yet different problem.
Results show that the proposed MA-ROM can effectively combine field outputs having inconsistent dimensions and topologies from both the scenarios described above.
The method produces relatively more accurate results when applied to the displacement rather than the von-Mises stress field.
The remainder of the manuscript is organized as follows.
Section~\ref{sec:lit} briefly discusses multi-fidelity surrogate models and ROMs in the realm of structural analysis.
Section~\ref{sec:formulation} reviews the conventional POD-based ROM and presents a succinct summary of the multi-fidelity ROM method based on manifold alignment.
Section~\ref{sec:application} introduces the details of the application problem and the two multi-fidelity scenarios considered in this work.
Finally, Section~\ref{sec:results} presents the results of the experiments and discusses the performance of the MA-ROM in comparison to a single-fidelity POD and interpolation-based approach.
\section{Literature Review}\label{sec:lit}
Several authors have adopted a surrogate-based optimization strategy to perform an aero-structural optimization of an aircraft wing.
Abdelkader and Stanislaw~\cite{benaouali2019multidisciplinary} created radial basis function (RBF) models for scalar outputs (e.g., lift, drag, weight, and aggregated structural failure) from the high-fidelity analysis.
Bordogna et al.~\cite{bordogna2017surrogate} created surrogate models trained with lift and drag coefficients from rigid RANS computations as constraints in optimization.
Rasmussen et al.~\cite{rasmussen2009optimization} used a polynomial regression to explore a joined-wing aircraft's optimal regions and design trends.
Joined-wing planform parameters and structural component thicknesses were the inputs to the surrogate model, and the sized total weight was the output.
Cipolla et al.~\cite{cipolla2021doe} created regression models for predicting the wing structural mass, estimating the maximum von-Mises stress in each main wing, and evaluating the highest wing-tip displacement between the front and rear wings.
Surrogate models have also been employed in other many-query applications such as reliability-based optimization of aircraft wing structural components~\cite{li2019reliability,neufeld2010aircraft}, evaluation of certification constraints~\cite{sarojini2020certification}, or uncertainty quantification of dynamic loads on the airframe~\cite{duca2018effects}.
In the literature mentioned so far, surrogate models have been used to approximate scalar quantities as a function of wing planform, geometrical parameters, aerodynamic parameters, and structural thicknesses.
While the wing's weight and global failure criteria provide practical metrics for overall optimization purposes, these aggregated quantities hide many details useful to a designer.
One such detail, for instance, is the location of maximum stress or strain and its spatial shifting as design parameters change.
Furthermore, in an aero-structural context, the full displacement of a wing structure under loads must be provided to an aerodynamic analysis code in order to adequately compute the aerodynamic forces for a given flight shape.
ROMs also have other applications, such as real-time estimation of a vehicle’s evolving structural state from sensor data~\cite{mainini2015surrogate}.
As for the prediction of structural field quantities, Lieu et al.~\cite{lieu2006reduced,lieu2007adaptation} proposed a POD-based ROM for CFD-based aeroelastic analysis using the F-16 aircraft as a test case.
Mainini and Willcox~\cite{mainini2015surrogate} used the POD to construct ROMs to predict high-fidelity FEM solutions for the estimation of structural capability from sensor data.
ROMs have also been extensively leveraged for the prediction of loads.
Bekemeyer and Timme~\cite{bekemeyer2019flexible} used a subspace projection model reduction to obtain transonic gust loads.
Ripepi et al.~\cite{ripepi2018reduced} created a ROM of the aerodynamic influence coefficients and used steady CFD solutions to obtain loads preserving non-linearities.
ROMs have also been used to reduce the computational cost of performing flutter and limit cycle oscillations analysis~\cite{zhang2012efficient}.
Note that the literature referenced above pertains to single-fidelity ROMs, i.e., constructed using data from one source.
The idea of leveraging a lower-fidelity analysis to reduce the computational cost of generating surrogate models has been shared by multiple authors in the past.
For example, Forrester et al.~\cite{Forrester2007} used multi-fidelity surrogate models for the aerodynamic optimization of a wing, and Toal~\cite{Toal2015} followed a similar process for the optimization of a high-pressure compressor.
Whereas multi-fidelity methods are relatively mature for scalar surrogate models, the prediction of field quantities brings additional challenges, as pointed out in Section~\ref{sec:intro}.
As described by Perron~\cite{Perron2020a}, these discrepancies can be due to differences in grid sizes, grid topologies, and field features.
A few efforts have attempted to avoid this issue by enforcing the same discretization for both the high- and low-fidelity analysis and have achieved computational savings by using a simplified physics model.
This approach was used by Mifsud et al.~\cite{Mifsud2016, Mifsud2008} for the aerodynamic analysis of a projectile and by Bertram et al.~\cite{Bertram2018} for the aerodynamic analysis of a road vehicle.
Alternatively, one could force the high- and low-fidelity field representations to be consistent by mapping all the results on a common grid, assuming that both fidelities have compatible field topologies.
Malouin et al.~\cite{Malouin2013} demonstrated this strategy for the analysis of a transonic airfoil.
Similarly, Benamara et al.~\cite{Benamara2016,Benamara2017} applied this approach to the analysis of a compressor blade.
More recently, Perron et al.~\cite{Perron2020a} proposed the MA-ROM method that leverages manifold alignment~\cite{Ham2005,Wang2009} to resolve discrepancies between the high- and low-fidelity fields.
The effectiveness of this method was tested with the aerodynamic analysis of an airfoil and a wing.
Furthermore, results have shown that even when fields with consistent representation are used (i.e., by mapping results to a common grid), the MA-ROM method offered an accuracy on par or superior to previous multi-fidelity ROM methods~\cite{Perron2020}.
Although multi-fidelity ROM methods have been successfully applied to aerodynamic problems, the viability of these approaches has not yet been established for structural applications.
Moreover, topological differences, in terms of model representation and structural layout, are much more prevalent in structural than in aerodynamic problems.
Previous data suggest that the MA-ROM is especially well suited for dealing with multi-fidelity structural data with possibly inconsistent representations.
Empirically testing this hypothesis is the primary goal of what follows.
\section{Formulation}\label{sec:formulation}
The MA-ROM method can be interpreted as a combination of three techniques: proper orthogonal decomposition (POD), manifold alignment (MA), and hierarchical Kriging (HK).
The following section briefly presents each of these techniques and describes how these are combined to form the MA-ROM method.
\subsection{Proper Orthogonal Decomposition}
The POD method~\cite{Lumley67} is a linear and unsupervised dimensionality reduction technique that is also commonly referred to as principal component analysis (PCA) in the field of machine learning.
It is arguably the most common dimensionality reduction method used in the context of projection-based ROM.
Let us consider the data matrix $\mathbf{X} \in \mathbb{R}^{d\times n}$ that contains $n$ samples $\mathbf{x}_j$ of dimension $d$ organized column-wise.
Let us further assume that $\mathbf{X}$ has been pre-processed such that its sample mean
\begin{equation}
\overline{\mathbf{x}} = \frac{1}{n}\sum_{j=1}^{n} \mathbf{x}_j
\end{equation}
is zero, meaning that the mean has been subtracted from the original data set beforehand.
The POD method attempts to construct a simpler representation of $\mathbf{X}$ that captures most of the data variance.
This is achieved by finding basis vectors $\boldsymbol{\phi}_j \in \mathbb{R}^d$ (also called POD modes) that maximizes the following \emph{Rayleigh quotient}~\cite{Magnus1997}
\begin{equation}\label{eq:Rayleigh}
\max_{\boldsymbol{\phi}_j} \frac{\boldsymbol{\phi}_j^T \mathbf{S} \, \boldsymbol{\phi}_j}{\boldsymbol{\phi}_j^T \boldsymbol{\phi}_j}
\end{equation}
where $\mathbf{S}= \frac{1}{n} \mathbf{X}\mathbf{X}^T \in \mathbb{R}^{d \times d}$ is the sample covariance matrix.
The solution of to the above objective is given by the eigendecomposition of $\mathbf{S}$~\cite{DeBie2005} such that
\begin{equation}
\mathbf{S} = \boldsymbol{\Phi} \boldsymbol{\Lambda} \boldsymbol{\Phi}^T
\end{equation}
where $\boldsymbol{\Phi} \in \mathbb{R}^{d \times d}$ contains the eigenvectors, and $\boldsymbol{\Lambda} \in \mathbb{R}^{n \times n}$ is a diagonal matrix where the entries $\lambda_j$ are the corresponding eigenvalues.
We define the POD basis $\boldsymbol{\Phi}_k \in \mathbb{R}^{d \times k}$ as a low-rank approximation of $\boldsymbol{\Phi}$ whose columns are composed of the $\boldsymbol{\phi}_j$ associated with the $k$-largest $\lambda_j$, where $k \ll d$.
Using $\boldsymbol{\Phi}_k$, the original data can be projected into the POD latent space such that
\begin{equation}
\mathbf{Z} = \boldsymbol{\Phi}_k^T \mathbf{X}
\end{equation}
where $\mathbf{Z} \in \mathbb{R}^{k \times n}$ contains the latent coordinates $\mathbf{z}_j \in \mathbb{R}^k$ (also known as POD coefficients) corresponding to $\mathbf{x}_j$.
Conversely, the original data can be approximately reconstructed from $\mathbf{Z}$ such that
\begin{equation}
\Xtilde = \boldsymbol{\Phi}_k \mathbf{Z}
\end{equation}
Note that $\Xtilde \in \mathbb{R}^{d \times n}$ is an approximation of $\mathbf{X}$ since only the $k$-first POD modes are used.
Lastly, the dimension $k$ of the POD basis is usually determined using the relative information content (RIC) criterion~\cite{Pinnau2008,Vendl2013} defined as
\begin{equation}\label{eq:ric}
\text{RIC} = \frac{\sum_{j=1}^{k} \lambda_j}{\sum_{j=1}^{d} \lambda_j}
\end{equation}
The RIC criterion essentially represents the ratio of the total variance captured by the POD basis.
A higher RIC implies a more accurate representation of $\mathbf{X}$, but typically results in a larger latent space dimension.
Users typically select $k$ such that the RIC is greater than some threshold, and values of 99.9\% or higher are fairly common in the ROM literature.
\subsection{Manifold Alignment}
Computing the POD of different datasets, say a high- and low-fidelity dataset, will clearly result in distinct POD bases and coefficients.
The purpose of manifold alignment~\cite{Ham2005,Wang2009} is to uncover a common latent space, instead of distinct latent spaces obtained via the POD, where disparate data can be more easily compared.
In the current context, the manifold alignment allows us to combine field results having different representations.
Specifically, the MA-ROM method leverages the Procrustes manifold alignment~\cite{Wang2008} method as it can readily be combined with the established POD-based ROM methodology~\cite{Perron2020}.
Consider $\mathbf{X} \in \mathbb{R}^{d \times n}$ and $\mathbf{Y} \in \mathbb{R}^{q \times m}$ as matrices containing the high- and low-fidelity samples $\mathbf{x}_j \in \mathbb{R}^d$ and $\mathbf{y}_j \in \mathbb{R}^q$ respectively.
Note that in view of a realistic scenario, it is assumed there are more low- than high-fidelity samples, i.e., $ m \gg n$.
However, no assumption is made regarding the dimensionality of both sets of fields;~they can be different, i.e., $d \neq q$.
Furthermore, let the first $n$ samples $\mathbf{x}_j$ and $\mathbf{y}_j$ have a pair-wise correspondence;~meaning they are generated by evaluating the respective simulation at identical design parameter values.
Consequently, the low-fidelity data can be partitioned as $\mathbf{Y} = [\mathbf{Y}_\text{L}, \mathbf{Y}_\text{U}]$, where $\mathbf{Y}_\text{L} \in \mathbb{R}^{q \times n}$ contains the solutions \emph{linked} to $\mathbf{X}$, while $\mathbf{Y}_\text{U} \in \mathbb{R}^{q \times (m - n)}$ holds the \emph{unlinked} data.
The first step of the Procrustes manifold alignment is to find the latent space, i.e., the POD basis, of both the high- and low-fidelity datasets.
Let us begin by defining
\begin{align}\label{eq:mf-pod}
\mathbf{Z} &= \boldsymbol{\Phi}_k^T \mathbf{X}\\
\mathbf{W} &= \boldsymbol{\Psi}_k^T \mathbf{Y}
\end{align}
where $\mathbf{Z} \in \mathbb{R}^{k \times n}$ and $\mathbf{W} \in \mathbb{R}^{k \times m}$ are the low- and high-fidelity POD coefficients respectively, while $\boldsymbol{\Phi}_k \in \mathbb{R}^{d \times k}$ and $\boldsymbol{\Psi}_k \in \mathbb{R}^{q \times k}$ are the associated POD modes.
Note that despite having the same dimension $k$, $\mathbf{Z}$ and $\mathbf{W}$ still reside in different subspaces.
As with $\mathbf{Y}$, we split the low-fidelity data into $\mathbf{W} = [\mathbf{W}_\text{L}, \mathbf{W}_\text{U}]$, where $\mathbf{W}_\text{L} \in \mathbb{R}^{k \times n}$ and $\mathbf{W}_\text{U} \in \mathbb{R}^{k \times (m - n)}$ are the linked and unlinked low-fidelity POD coefficients.
Then, the POD coefficients of both datasets are aligned using Procrustes analysis~\cite{Gower2010}.
The goal of this step is to find an affine transformation of $\mathbf{W}_\text{L}$ that would optimally align it with $\mathbf{Z}$.
This is expressed by the following objective
\begin{equation}
\begin{split}
\min_{s, \mathbf{t}, \mathbf{Q}} \quad& \lVert \mathbf{Z} - s \mathbf{Q} (\mathbf{W}_\text{L} -\mathbf{t}) \lVert_F \\
\text{s.t.} \quad& \mathbf{Q}^T \mathbf{Q} = \mathbf{I}
\end{split}
\end{equation}
where $s$ is an isotropic scaling factor, $\mathbf{t} \in \mathbb{R}^{k}$ is a translation vector, and $\mathbf{Q} \in \mathbb{R}^{k \times k}$ is an orthogonal transformation matrix.
The optimal $\mathbf{t}$ is obtained by shifting $\textbf{W}_\text{L}$ such that its centroid coincides with that of $\textbf{Z}$.
It is assumed that $\mathbf{X}$ has been centered prior to computing the POD such that the mean of $\mathbf{Z}$ is also zero.
Thus, the optimal $\mathbf{t}$ is given by
\begin{equation}
\textbf{t} = \frac{1}{n}\sum_{j=1}^{n} \mathbf{w}_j
\end{equation}
For convenience, $\textbf{W}'_\text{L} = \textbf{W}_\text{L} - \textbf{t}$ is defined as the shifted low-fidelity latent variable.
The remaining optimal parameters of the Procrustes analysis are then obtained by the following relations
\begin{align}
s &= \frac{\text{tr}(\boldsymbol{\Sigma})}{\text{tr}\left(\mathbf{W}'_\text{L} (\mathbf{W}'_\text{L})^T\right)} \\
\textbf{Q} &= \textbf{V} \textbf{U}^T
\end{align}
In the results shown above, the matrices $\mathbf{U}$, $\boldsymbol{\Sigma}$, and $\mathbf{V}$, are the outputs of the following singular value decomposition (SVD)
\begin{equation}
\textbf{U} \boldsymbol{\Sigma} \textbf{V}^T = \textbf{W}'_\text{L} \textbf{Z}^T
\end{equation}
Finally, the optimal transformation obtained using $\mathbf{W}_\text{L}$ is applied to all the low-fidelity data $\mathbf{W}$.
The aligned POD coefficients are thus given by
\begin{equation}\label{chap3-eq:z_align}
\textbf{Z}_\text{lo} = s \, \textbf{Q} \left(\textbf{W} - \textbf{t}\right)
\end{equation}
where $\textbf{Z}_\text{lo} \in \mathbb{R}^{k \times m}$ contains the low-fidelity POD coefficients mapped onto the high-fidelity POD basis.
For the sake of clarity, the high-fidelity data $\textbf{Z}$ are referred to as $\textbf{Z}_\text{hi}$ from here onward.
\subsection{Hierarchical Kriging}
Following the POD and the manifold alignment, the high- and low-fidelity data have been reduced to the latent coordinates $\textbf{Z}_\text{hi}$ and $\textbf{Z}_\text{lo}$.
The multi-fidelity data must then be combined into a regression model in order to predict new field solutions at unseen design parameters.
In practice, any multi-fidelity regression technique could be used for this purpose and the literature contains multiple options such as bridge functions~\cite{Choi2009, Eldred2004, Tang2005} and cokriging~\cite{Myers1982,Kennedy2000,Han2012}.
In this study, the hierarchical Kriging (HK) method~\cite{Han2012a} is used to remain consistent with previous MA-ROM studies~\cite{Perron2020,Perron2020a,Decker2021}.
As defined by Han and G\"{o}rtz~\cite{Han2012a}, the HK formulation is nearly identical to cokriging~\cite{Kennedy2000}, although the former is marginally simpler to implement and provides a smoother posterior variance prediction.
Given a high-fidelity function $g_\text{hi}: \mathbf{p} \mapsto z_\text{hi}$, where $\mathbf{p} \in \mathbb{R}^b$ is a design parameter vector and $z_\text{hi}$ is a high-fidelity response, the HK predictor takes the following form
\begin{equation}\label{eq:hk-predictor}
\gtilde_\text{hi}(\mathbf{p}) = \beta\,\gtilde_\text{lo}(\mathbf{p}) + \mathbf{w}^T \mathbf{r}(\mathbf{p})
\end{equation}
where $\beta$ is a multi-fidelity scaling factor, $\mathbf{w} \in \mathbb{R}^n$ is a weight vector, and $\mathbf{r}(\mathbf{p})$ is a kernel function.
In this example, low-fidelity information is included in the form of $\gtilde_\text{lo}(\mathbf{p})\mathbf{p} \mapsto \tilde{z}_\text{lo}$, which is a Kriging predictor of some low-fidelity response.
Loosely speaking, the HK predictor of equation~\eqref{eq:hk-predictor} can be viewed as a Kriging model where the response trend is represented by $\beta\,\gtilde_\text{lo}(\mathbf{p})$.
Note that for the current study, both $\gtilde_\text{hi}(\mathbf{p})$ and $\gtilde_\text{lo}(\mathbf{p})$ use a Mat\'ern~3/2 kernel and the model parameters are obtained with a maximum likelihood approach.
In the context of MA-ROM, the HK model is used to predict a new latent coordinate vector $\textbf{z}_\text{hi}$ given some design parameters $\mathbf{p}$.
Since each coordinate contains $k$ individual components, an equal number of regression models $\gtilde_\text{hi}(\mathbf{p})$ must be trained in order to predict the entire $\textbf{z}_\text{hi}$ vector.
The prediction of the high-dimensional field output is then given by
\begin{equation}
\xtilde = \sum_{i=1}^{k} \boldsymbol{\phi}_i \, \gtilde_{i,\text{hi}}(\mathbf{p})
\end{equation}
where $\gtilde_{i,\text{hi}}(\mathbf{p})$ is the HK model predicting the $i$-th coordinated in the POD latent space associated with the POD mode $\boldsymbol{\phi}_i$.
\subsection{MA-ROM Formulation}
Consider a high- and low-fidelity models defined as $f_x: \mathbf{p} \mapsto \mathbf{x}$ and $f_y: \mathbf{p} \mapsto \mathbf{y}$ respectively.
The MA-ROM formulation can be divided into an \emph{offline} phase where the model is trained, and an \emph{online} phase where predictions are made using the generated model.
The offline phase is summarized through the following steps
\setlist[enumerate,1]{label={\arabic*.}}
\begin{enumerate}
\item \textbf{Generate Linked Data:} Select a set of $n$ distinct designs $\mathbf{P}_\text{L} = [\mathbf{p}_1, \dots, \mathbf{p}_n] \in \mathbb{R}^{b\times n}$ for the linked data.
Then, sample both $f_x$ and $f_y$ to generate the matrices $\mathbf{X}$ and $\mathbf{Y}_\text{L}$.
\item \textbf{Generate Unlinked Data:} Select a set of $m - n$ additional designs $\mathbf{P}_\text{U} = [\mathbf{p}_{n+1}, \dots, \mathbf{p}_m] \in \mathbb{R}^{b\times (m-n)}$ for the unlinked data.
Ideally, $\mathbf{P}_\text{U}$ should not repeat points in $\mathbf{P}_\text{L}$.
Then, sample only $f_y$ with $\mathbf{P}_\text{U}$ to generate the matrix $\mathbf{Y}_\text{U}$.
\item \textbf{Align Manifolds:} Apply the Procrustes manifold alignment on $\mathbf{X}$, $\mathbf{Y}_\text{L}$, and $\mathbf{Y}_\text{U}$, to obtain $\mathbf{Z}_\text{hi}$ and $\mathbf{Z}_\text{lo}$. This is divided into the following three sub-steps:
\begin{enumerate}
\item Use POD to extract $\boldsymbol{\Phi}_k$ and $\boldsymbol{\Psi}_k$, as well as $\mathbf{Z}_\text{hi}$ and $\mathbf{W}$.
\item Perform Procrustes analysis on $\mathbf{Z}_\text{hi}$ and $\mathbf{W}_\text{L}$ to evaluate the optimal value of $s$, $\mathbf{t}$, and $\mathbf{Q}$.
\item Apply the optimal transformation to $\mathbf{W}$ to compute $\mathbf{Z}_\text{lo}$.
\end{enumerate}
\item \textbf{Fit HK Model:} Train a set of $k$ HK models $\gbtilde(\mathbf{p}) = [\gtilde_1(\mathbf{p}), \dots, \gtilde_k (\mathbf{p})]$.
These models combine the information from both $\mathbf{Z}_\text{hi}$ and $\mathbf{Z}_\text{lo}$, together with their associated design parameters $\mathbf{P} = [\mathbf{P}_\text{L}, \mathbf{P}_\text{U}] \in\mathbb{R}^{b\times m}$.
\end{enumerate}
It should be noted that in this study, the design samples in steps 1 and 2 are obtained using Latin hypercube sampling (LHS)~\cite{Santner2003}.
As for the online phase, the process is identical to most non-intrusive ROMs and is given by
\begin{enumerate}[resume*]
\item \textbf{Predict Latent Variable:} Given an out-of-sample design parameter $\mathbf{p}^*$, predict the corresponding latent variable $\ztilde^*$ using $\gbtilde(\mathbf{p})$.
\item \textbf{Reconstruct Field:} Using $\boldsymbol{\Phi}_k$ and $\ztilde^*$, compute the predicted value of the new high-fidelity field $\xtilde^*$.
\end{enumerate}
\section{Application Problem}\label{sec:application}
The application problem considers the analysis of a wing structure to demonstrate the relevance of MA-ROM for large-scale problems representative of an industrial design process.
The baseline geometry considered for this application problem is the wing of the NASA CRM~\cite{kenway2014aerostructural}.
The CRM is representative of a transonic wide-body transport aircraft.
\subsection{Aero-Structural Analysis}\label{subsec:crm_baseline}
The baseline OML\footnote{An OpenVSP file of the baseline CRM OML is included with this paper} and flight conditions considered are given in Table~\ref{tab:crm_baseline_oml_fc} and shown in Figure~\ref{fig:crm_wing_oml}.
\begin{table}[ht]
\centering
\caption{Baseline OML \& flight condition values}\label{tab:crm_baseline_oml_fc}
\begin{tabular}{lcc}
\hline
\textbf{Parameter} & \textbf{Value} & \textbf{Unit} \\
\hline\hline
Planform area & 4444.38127 & \si{\square\ft} \\
Span & 193.34768 & \si{\ft} \\
Aspect ratio & 8.3325 & - \\
Taper ratio & 0.86031 & - \\
Leading edge sweep & 37 & \si{\deg} \\
TOGW & 650000 & \si{\lb} \\
Altitude & 30000 & \si{\ft} \\
Mach number & 0.8 & - \\
\hline
\end{tabular}
\end{table}
\begin{figure*}
\caption{Top, rear, and side view of the CRM wing. The dimensions provided are normalized by the mean aerodynamic chord whose value is 275.8~in. The top right isometric view shows the wing in context of the full aircraft configuration.}
\label{fig:crm_wing_oml}
\end{figure*}
The aerodynamic loads on the wing are obtained from a panel representation of the aircraft OML.
This panel model has the associated sectional airfoils, local incidence angles, etc. that can be used for creating lower order aerodynamic models for tools like AVL.
The baseline panel geometry for the AVL model
\footnote{The AVL model of the baseline CRM OML is provided as supplementary material.} is shown in Figure~\ref{fig:pegasus_avl}.
\begin{figure}
\caption{Panel geometry and discretization for CRM AVL model}
\label{fig:pegasus_avl}
\end{figure}
Figure~\ref{fig:crm_mesh} shows the structural geometry layout and mesh for the wing, horizontal tail, and vertical tail within a fused representation of the OML. In this work, we consider only the wing whose structural topology values are given in Table~\ref{tab:crm_baseline_strcTop}.
\begin{figure}
\caption{Geometry and mesh of the wing and empennage of the CRM}
\label{fig:crm_mesh}
\end{figure}
\begin{table}
\centering
\caption{Baseline structural topology values (cn: chord normalized and sn: span normalized).}\label{tab:crm_baseline_strcTop}
\begin{tabular}{lcc}
\hline
\textbf{Parameter} & \textbf{Value} & \textbf{Unit} \\
\hline\hline
Front spar location & 0.17 & cn \\
Rear spar location & 0.68 & cn \\
Rib spacing & 32 & \si{\in} \\
Rib roll & 0 & \si{\deg} \\
Rib yaw & -10 & \si{\deg} \\
Root rib location & 0.1 & sn \\
Tip rib location & 0.995 & sn \\
\hline
\end{tabular}
\end{table}
The structure is modeled using CQUAD4 shell elements in Nastran\footnote{The Nastran model (i.e., the BDF file) of the baseline CRM OML is provided as supplementary material.} for the ribs, spars, and skins.
All shell panels are assigned aluminium material with a density of \num[round-mode=places,round-precision=3]{2698.79} \si{\kg\per\meter^3}, a Young's Modulus of \num[round-mode=places,round-precision=3]{71016000118} \si{\newton\per\meter^2}, and a Poisson's ratio of $0.33$.
The following assumptions are made in the structural model:
\begin{enumerate}
\item No stiffeners and stringers
\item The thicknesses of the panels are fixed. The values are obtained by averaging thicknesses along the span for a baseline sizing case where the CRM was sized for 2.5g and -1.0g static maneuvers at flight conditions given in Table~\ref{tab:crm_baseline_oml_fc}.
\begin{itemize}
\item Front spar: \num[round-mode=places,round-precision=3]{0.465988771929824} \si{\in}
\item Rear spar: \num[round-mode=places,round-precision=3]{0.257907054747054} \si{\in}
\item Upper skin: \num[round-mode=places,round-precision=3]{0.923677840755735} \si{\in}
\item Lower skin: \num[round-mode=places,round-precision=3]{0.657111039136302} \si{\in}
\item Rib: \num[round-mode=places,round-precision=3]{0.148520056980057} \si{\in}
\end{itemize}
\end{enumerate}
These assumptions are merely simplifications to make the modeling easier.
The proposed multi-fidelity method's performance is neither affected nor depends on this simplification.
Stiffeners can be included by using the smeared-stiffener calculations as done by Kennedy and Martins~\cite{kennedy2014parallel}.
The transfer of aerodynamic loads from rigid-wing aerodynamics in AVL to the structure is done by integrating pressure differentials for elements on the aerodynamic panels to reference points along the span of a wing-like component.
The centers of mass of the ribs are chosen as reference points along the span of the wing.
The forces and moments associated with each reference point are then distributed to the structure via rigid body elements (RBEs) at key intersecting nodes between the ribs and skin.
The structure is analyzed in Nastran using SOL101. Field outputs such as internal loads, stresses, strains, and displacements are obtained.
The proposed MA-ROM approach is used to obtain a surrogate of these field outputs as a function of the input design variables that will be described in Section~\ref{subsec:dv_parametrization}.
It should be noted that in this work each shell element has an assumed thickness.
The Nastran model can be used to size the structure for the applied loads using Hypersizer~\cite{solano2020structural,solano2021pegasus}.
Other fields can be generated from Hypersizer outputs such as optimal sectional dimensions (skin thickness, stiffener height, etc.) and margins of safety for multiple failure criteria.
\subsection{Parametrization}\label{subsec:dv_parametrization}
In the proposed problem, global wing parameters are considered for the design parametrization.
These include the wing sweep, dihedral, tip twist, and span as shown in Table~\ref{tab:ffd-param}.
Note that the bounds are represented as percentage increments/decrements with respect to the original aircraft OML.
\begin{table}
\centering
\caption{Design parameters and bounds considered for the application problem. Values are increments to the original CRM geometry}\label{tab:ffd-param}
\begin{tabular}{lc}
\hline
\textbf{Parameter} & \textbf{Bounds} \\
\hline\hline
Wing sweep & $\pm 10^\circ$ \\
Wing dihedral & $\pm 10^\circ$ \\
Wing tip twist & $\pm 10^\circ$ \\
Wing span & $\pm 20\%$ \\
\hline
\end{tabular}
\end{table}
While the MA-ROM approach can combine fields of high- and low-fidelity having different representations, it still requires each fidelity level to internally use a consistent representation.
For instance, a similar grid topology, in terms of size and connectivity, should be used for all the high-fidelity results, and likewise for the low-fidelity results.
However, this grid topology does not need to be between both high- and low-fidelity results since any discrepancy between fidelity levels will be addressed with the manifold alignment.
For the aforementioned reason, a grid deformation approach is used to apply the design parameters of Table~\ref{tab:ffd-param} on both the structural and aerodynamic grids.
In the current study, the grid is deformed using a free-form deformation (FFD) approach~\cite{Kenway2010}.
This technique encases a geometry inside a lattice or box as shown in Figure~\ref{fig:ffd-crm}.
The FFD box behaves as a flexible rubber-like material such that displacement of the boundary nodes is propagated to the geometry within.
For instance, given the current problem, the wing sweep is modified by translating backward or forward all the end nodes of the FFD box.
In addition to maintaining the grid topology, the FFD techniques can effectively be applied to any discretized geometry and without the need for CAD software.
Note that this technique is popular in the context of aerodynamic shape optimization, some examples of which can be found in Lyu et al.~\cite{Lyu2014} and Koo and Zing~\cite{Koo2016}.
\begin{figure}
\caption{Structural grid of the CRM wing encased in an FFD box (red).}
\label{fig:ffd-crm}
\end{figure}
\subsection{Multi-Fidelity Scenarios}
The usefulness of the MA-ROM method is demonstrated using two multi-fidelity scenarios based on the CRM test case with an FFD parametrization described earlier.
Each of these scenarios is aimed at combining multi-fidelity fields having different types of inconsistencies.
\subsubsection{Scenario~1: Inconsistent Grid Sizes}\label{sec:scn-1}
In this scenario, structural results computed on a fine grid are enhanced with data obtained from a similar analysis, yet using a coarser grid.
Specifically, the high- and low-fidelity simulation use grids with a resolution of roughly 79,000 and 5,000 elements respectively as shown in Figure~\ref{fig:scn-1:grid}.
Using a coarser grid, the computational cost of the low-fidelity analysis is reduced from 5.5 to 0.6 CPU-sec on a modern desktop, which represents an order of magnitude reduction.
In this scenario, the purpose of the multi-fidelity method is to reduce the overall training cost of the model by leveraging a computationally cheaper tool.
As such, the multi-fidelity method's predictive performance will be contrasted with that of a single-fidelity model constructed using the results from the fine grid exclusively.
This comparison will look at these models' accuracy for similar training costs, which for the multi-fidelity model, includes the cost of generating both the high- and low-fidelity data.
\begin{figure}
\caption{Comparison the (A) high- and (B) low-fidelity grids used for Scenario~1.}
\label{fig:scn-1:grid}
\end{figure}
\subsubsection{Scenario 2: Inconsistent Topologies}\label{sec:scn-2}
For the second scenario, the MA-ROM method is used to combine results from similar structural simulations, yet using different structural layouts as shown in Figure~\ref{fig:scn-2:layout}.
Unlike Scenario~1, both simulations are solved using grids with a similar discretization, and as a result, do not follow the usual high- and low-fidelity breakdown.
However, for this scenario, we assume that the results for one of the two structural layouts are readily available in abundance.
For instance, these can correspond to the results previously generated for a different aircraft, or similarly, generated for an earlier version of the aircraft at a point where a different layout was considered.
Therefore, the purpose of the multi-fidelity approach is to \emph{jump-} or \emph{warm-start} the surrogate model generation of a new structural layout using existing data from a different, yet similar application.
Instead of a high- and low-fidelity dataset, the training data is separated into a main and an auxiliary dataset, the former representing the fields that must be predicted by the MA-ROM.
In this example, the main dataset corresponds to the same structural layout used in Scenario~1, and the auxiliary dataset is a structure for the CRM wing using a lower rib count.
The intent behind this scenario is to demonstrate the capability of the MA-ROM to combine structural results despite having inconsistent field topologies.
\begin{figure}
\caption{Comparison the structural layouts used for the (A) main and (B) auxiliary datasets in Scenario 2.}
\label{fig:scn-2:layout}
\end{figure}
\subsection{Performance Metrics}\label{sec:metrics}
To establish the benefits of the MA-ROM method for structural problems, its predictive performance is compared to that of a conventional single-fidelity ROM.
This performance is measured in terms of prediction error and is contrasted with the computational cost required to train the model.
As with any empirical model, one can expect MA-ROM to be more accurate when provided with more data, but for it to be cost-effective, it should achieve this accuracy with lesser resource consumption than a single-fidelity method.
\subsubsection{Field Prediction Error}
The prediction error of a ROM or MA-ROM model is typically measured as the difference between the predicted and the actual field responses on previously unseen data, i.e., on data that was left out of the training process.
Using $n_t$ test samples held out from the training data, we define the field prediction error of some field $\mathbf{x} \in \mathbb{R}^d$ as
\begin{equation}
E(\mathbf{x}) = \sqrt{\frac{\sum_{j=1}^{n_t} \lVert \mathbf{x}^*_j - \widetilde{\mathbf{x}}_j \rVert_2^2}{n_t}}
\end{equation}
where $\widetilde{\mathbf{x}}_j \in \mathbb{R}^d$ is the model prediction from either a ROM or MA-ROM model, and $\mathbf{x}^*_j \in \mathbb{R}^d$ is the corresponding exact solution.
We use the $L^2$-norm of the error field to provide an integrated measure of the prediction error for a given sample.
The test set sample errors are then aggregated using a root-mean-squared operation to represent the overall prediction error of a given model.
Note that the magnitude of $E(\textbf{x})$ is dependent on the field quantity being considered.
To compare errors between different design problems, it is useful to define a normalized prediction error such as
\begin{equation}\label{eq:norm-pred-error}
\widehat{E}(\textbf{x}) = \frac{E(\textbf{x})}{\sqrt{\sum_{j = 1}^{n_t} \lVert \mathbf{x}_j^* - \overline{\mathbf{x}}\rVert^2_2/n_t}}
\end{equation}
It is worth noting that the normalization term used in equation~\eqref{eq:norm-pred-error} is essentially the total standard deviation of the testing dataset.
As a result, the definition of $\widehat{E}(\textbf{x})$ represents the ratio between the variation of $\textbf{x}$ unexplained by the ROM and the total variation of the data.
\subsubsection{Computational Training Cost}
The computational cost of a ROM can be separated into its evaluation and training costs, which are also referred to as the online and offline costs.
The former is designated as an indicator of the cost of predicting new results and is typically negligible, and as such, we only focus on and report the training cost.
In comparison, the time needed to construct the ROM or MA-ROM model once the data is available is relatively large.
The offline computational cost can be measured in terms of CPU time, usually in CPU-hr or CPU-sec, which is the total amount of computational time used by each processing unit involved.
The CPU time can then be divided by the number of CPUs available to obtain an estimate of the wall-time, neglecting any performance loss from the parallelization.
The cost of each sample is assumed to be a constant in this work, and for Scenario~1 for instance, it is estimated at 5.4402 and 0.5998 CPU-sec for the baseline and coarse grid respectively\footnote{
CPU times are based on an Intel Core i9-10885H CPU.
}.
The total training cost of ROMs in this study is then taken as the number of training samples times the constant cost per sample.
Note that the computational cost can also have a monetary meaning as most high-performance computing facilities provide a price per CPU-hr used.
This price is a result of the operation, maintenance, and acquisition costs of a given supercomputer.
For instance, at the time of writing this manuscript, the Phoenix cluster at the Georgia Institute of Technology offered a preferential rate of \$0.0068/CPU-hr for internal users~\cite{PACE}.
\section{Results}\label{sec:results}
This section presents the predictive performance of the MA-ROM method applied to the CRM wing application problem.
The results for both Scenarios~1 and 2, i.e., combining inconsistent grid sizes and structural topologies respectively, are presented and discussed separately.
For each of the fidelity levels considered, a data set of 1,000 cases
\footnote{The field data generated with each fidelity level are provided as supplementary material.}
is generated using an optimized Latin hypercube sampling method~\cite{Santner2003} and the design bounds defined in Table~\ref{tab:ffd-param}.
From the high-fidelity dataset, a subset of $n$ cases are selected to form the high-fidelity training data.
Afterward, the same $n$ cases and an additional $m - n$ cases are randomly selected from the auxiliary dataset to form the low-fidelity training set.
Once the MA-ROM model is trained, the remainder of the high-fidelity data is used to compute the model prediction error.
The accuracy of the MA-ROM prediction is also compared to that of a conventional single-fidelity ROM, where the latter is only trained with high-fidelity data.
It should be noted that for a given $n$, the above training process is repeated at least 100 times with different subsets of the overall data, and the prediction errors of all models are averaged and reported.
The purpose of these repetitions is to decouple the measured error from the choice of training samples.
In doing so, presented results represent the expected accuracy of the MA-ROM (or single-fidelity ROM) method rather than the performance of one specific model instance.
In the current study, the field outputs of interest are the von-Mises stress distribution $\sigma_\text{vm}$ and the structural displacement $\delta$, although the MA-ROM could be applied to any field responses.
Furthermore, high-fidelity training sample sizes $n$ between 10 and 800 are considered, leaving at least 200 cases for the error calculation.
As for the low-fidelity training set, sample size $m$ is kept proportional to $n$ using the multi-fidelity ratio $\tau = m/n$, and ratio of $\tau = \{2, 4, 8\}$ are considered.
Lastly, all POD bases are computed with an RIC of 99.9999\% or greater for each fidelity level.
\subsection{Scenario~1: Inconsistent Grid Sizes}\label{sec:results:scn-1}
\begin{figure*}
\caption{Prediction errors of the von-Mises stress $\sigma_\text{vm}
\label{fig:scn-1:stress}
\end{figure*}
\begin{figure*}
\caption{Prediction errors of the structural displacement $\delta$ using the MA-ROM method to combine fine and coarse grid results.}
\label{fig:scn-1:displacement}
\end{figure*}
As defined in Section~\ref{sec:scn-1}, the first multi-fidelity scenario combines the structural results of the CRM wing obtained from a fine and coarse grid.
The normalized prediction errors on the von-Mises stress distribution $\sigma_\text{vm}$ are presented in Figure~\ref{fig:scn-1:stress} for this scenario.
The blue, red, and green lines represent the results using a multi-fidelity ratio $\tau$ of 2, 4, and 8, respectively.
For comparison, the results for a single-fidelity ROM trained with only the fine grid results are shown using a black line.
The left side of Figure~\ref{fig:scn-1:stress} presents the errors as a function of $n$, while the right side shows the results scaled with the overall training cost of the various models in CPU-sec.
Note that for the multi-fidelity models, the training cost is comprised of the cost for both the high- and low-fidelity training data.
Following a similar presentation, Figure~\ref{fig:scn-1:displacement} also presents the prediction errors of the MA-ROM and ROM methods, although for the structural displacement $\delta$ instead of $\sigma_\text{vm}$.
For the reader's convenience, some of the data points in Figures~\ref{fig:scn-1:stress} and~\ref{fig:scn-1:displacement} are also tabulated in Table~\ref{tab:scn-1}.
\begin{table}[]
\centering
\caption{Prediction error of the MA-ROM method applied to Scenario~1.}
\label{tab:scn-1}
\begin{tabular}{ccccc}
\hline
$n$ & $m$ & $\widehat{E}(\sigma_\text{vm})$ & $\widehat{E}(\delta)$ & CPU-sec \\
\hline\hline
100$^*$ & - & 0.805\% & 0.587\% & 544.0 \\
100 & 200 & 0.616\% & 0.284\% & 664.0 \\
100 & 400 & 0.568\% & 0.174\% & 783.9 \\
100 & 800 & 0.528\% & 0.119\% & 1,023.9 \\
\hline
183$^*$ & - & 0.609\% & 0.321\% & 1,000.0 \\
150 & 300 & 0.526\% & 0.210\% & 1,000.0 \\
127 & 508 & 0.514\% & 0.145\% & 1,000.0 \\
97 & 779 & 0.536\% & 0.120\% & 1,000.0 \\
\hline
\multicolumn{5}{l}{$^*$\footnotesize{Results for a single-fidelity ROM}}
\end{tabular}
\end{table}
As can be expected, both Figures~\ref{fig:scn-1:stress} and~\ref{fig:scn-1:displacement} show that the accuracy of both the single- and multi-fidelity methods monotonically improves with additional training data.
More importantly, these figures also demonstrate that for all $n$ values, the MA-ROM outperforms the single-fidelity ROM by enhancing the high-fidelity data with additional low-fidelity results.
When compared to a single-fidelity model trained with the same computational budget, say $1,000$~CPU-sec, a relative reduction in the error of up to 15.6\% and 62.6\% is observed for the prediction of $\sigma_\text{vm}$ and $\delta$ respectively as shown in Table~\ref{tab:scn-1}.
However, this improvement appears to depend on the specific field quantity of interest the MA-ROM is applied to.
For the von-Mises stress field, Figure~\ref{fig:scn-1:stress} shows a reasonable reduction in $\widehat{E}(\sigma_\text{vm})$ with $\tau=2$, yet increasing $\tau$ further only provides a marginal improvement.
Specifically, for $n=100$, the normalized prediction error is 0.630\% and 0.574\% for $m=200$ and $800$ respectively, which represents a relative improvement of only 8.9\% for almost double the cost.
When including the additional computational cost of the low-fidelity data into the analysis, the right side of Figure~\ref{fig:scn-1:stress} indicates that for a given prediction error, the MA-ROM can reduce the overall training cost of the model over a single-fidelity approach.
Although, the results also show that predicting $\sigma_\text{vm}$, using $\tau > 4$ is likely not cost-effective.
This suggests that the high- and low-fidelity fields don't share a strong statistical dependence and that the transfer of low-fidelity information within the multi-fidelity method saturates.
On the other hand, for the prediction of the structural displacement, the results of Figure~\ref{fig:scn-1:displacement} exhibit a more pronounced decrease in $\widehat{E}(\delta)$ with additional low-fidelity data.
For $n=100$, the normalized prediction error is 0.275\% and 0.112\% for $m=200$ and $800$ respectively, which represents a relative improvement of 59.3\%.
Even with the additional cost of the low-fidelity samples, the right side of Figure~\ref{fig:scn-1:displacement} presents a clear reduction in $\widehat{E}(\delta)$ for a given computational cost, or equivalently, a reduction in cost for a target prediction error.
While the results show reduced benefits with larger $\tau$, using more low-fidelity data still appears to be cost-effective, especially when compared to the single-fidelity results.
The better performance of the MA-ROM method for the prediction of the displacement field than for the von-Mises stress field can be explained by the higher non-linearity and complexity of the latter.
As one can expect, this suggests that the effectiveness of the multi-fidelity method ultimately depends on the application.
That being said, the results demonstrate that MA-ROM can be a suitable method to predict structural results and has the potential to offer a lower training cost than a single-fidelity method.
Furthermore, these benefits are achieved despite the high- and low-fidelity fields being represented on grids with different discretizations.
\subsection{Scenario~2: Inconsistent Topologies}\label{sec:results:scn-2}
The second scenario of this study combines the results of wing structures with different structural layouts.
Unlike Scenario~1, the fields of the main and auxiliary datasets have inconsistent topologies.
As discussed in Section~\ref{sec:results:scn-1}, the auxiliary analysis is the one using a structural layout with a low-rib count (see Figure~\ref{fig:scn-2:layout}).
It is assumed that this data is readily available from a previous study, and is used to enhance the main data corresponding to a new design study.
The results in Figure~\ref{fig:scn-2} present the prediction error obtained with the MA-ROM method applied to the current scenario.
As in the previous scenario, the normalized error is shown for a range of high-fidelity samples $n$ and multiple multi-fidelity ratios $\tau$.
The error achieved with a single-fidelity approach is also shown for comparison.
The top part of Figure~\ref{fig:scn-2} gives the prediction error for the von-Mises stress field $\sigma_\text{vm}$, while the bottom part shows the error for the structural displacement field.
The numerical values for the cases at $n=100$ are also listed in Table~\ref{tab:scn-2} for an easier comparison.
Unlike in Figure~\ref{fig:scn-1:stress} and~\ref{fig:scn-1:displacement} in Scenario~1, the computational cost is not shown here since the auxiliary data is assumed to already exist.
\begin{figure}
\caption{Prediction errors of $\sigma_\text{vm}
\label{fig:scn-2}
\end{figure}
\begin{table}[]
\centering
\caption{Prediction error of the MA-ROM method applied to Scenario~2.}
\label{tab:scn-2}
\begin{tabular}{cccc}
\hline
$n$ & $m$ & $\widehat{E}(\sigma_\text{vm})$ & $\widehat{E}(\delta)$ \\
\hline\hline
100$^*$ & - & 0.805\% & 0.587\% \\
100 & 200 & 0.630\% & 0.275\% \\
100 & 400 & 0.595\% & 0.167\% \\
100 & 800 & 0.574\% & 0.112\% \\
\hline
\multicolumn{4}{l}{$^*$\footnotesize{Results for a single-fidelity ROM}}
\end{tabular}
\end{table}
When compared to the results reported in Section~\ref{sec:results:scn-1}, the current results exhibit trends in agreement with those of Scenario~1.
Once again, the normalized error decreases monotonically with additional training data and adding auxiliary data produces better results than using the main dataset only.
For $n=100$, the prediction error is up to 28.70\% and 80.92\% smaller for $\sigma_\text{vm}$ and $\delta$ respectively, which is comparable to the results presented in Table~\ref{tab:scn-1}.
Furthermore, the results of Figure~\ref{fig:scn-2} indicate once more that improving the prediction of $\sigma_\text{vm}$ with the MA-ROM method is more challenging than it is for $\delta$.
This reinforces the suspicion that the higher complexity and non-linearity of the von-Mises stress field renders the main and auxiliary fields more difficult to fuse.
The main outcome of the results of Scenario~2 is that combining data with disparate topologies does not appear to alter the performance of the MA-ROM method.
Therefore, the accuracy of a field prediction model can readily be increased using MA-ROM if data from a similar, yet different problem is available even if their field size or topology are inconsistent.
For instance, this could be useful if a ROM was previously trained for an earlier iteration of the design, and is no longer adequate due to changes to the structural layout.
Consequently, the MA-ROM method gives renewed value to old, discarded data by using them to enhance the prediction of models trained on new, expensive data.
\subsection{Field Visualization}
This section presents a few key figures for a visual depiction of the predictive accuracy provided by the MA-ROM.
For reference, Figure~\ref{fig:worst_point} shows the actual solution for the test point with the largest observed prediction error.
Recollect that single- and multi-fidelity ROMs are constructed to predict both the stress and the displacement fields.
It should be noted that the test case presented corresponds to a design that would normally be infeasible due to the high stress and displacement values computed.
However, other than being an extreme design, this does not impact the ROM models as they merely reproduce the analysis results.
Figures~\ref{fig:stress_visuals} and~\ref{fig:displacement_visuals} then present the predicted fields and the corresponding prediction errors for both quantities of interest.
The results for both multi-fidelity scenarios are shown, as well as the results from a single-fidelity approach for comparison sake.
Comparing Figures~\ref{fig:stress_visuals} and~\ref{fig:displacement_visuals} to Figure~\ref{fig:worst_point} shows that all ROM methods can recover the actual field of interest with a reasonable accuracy.
Yet, an inspection of the error fields reveals that for the prediction of both the stress and the displacement fields, the MA-ROM outperforms the single-fidelity POD-based ROM even for the worst prediction in the test set.
These visualizations also show that the error tends to be higher near regions of high stress (e.g., the wing root) or large displacement (e.g., the wing tip)
One should further note that the error levels for both multi-fidelity scenarios have similar magnitudes.
These observations support previous discussions and reinforce MA-ROM's efficacy when applied to high-fidelity structural analyses with grid inconsistencies and differences in topology.
\begin{figure}
\caption{Exact simulation results of the von-Mises stress and the $z$ displacement fields for the test case with the highest prediction error.}
\label{fig:worst_point}
\end{figure}
\begin{figure*}
\caption{Prediction results of the von-Mises stress field for the test case with the highest prediction error. Shown results consider a single-fidelity ROM model~(A), MA-ROM model combining inconsistent grid sizes~(B), and a MA-ROM model combining inconsistent structural topologies~(C).}
\label{fig:stress_visuals}
\end{figure*}
\begin{figure*}
\caption{Prediction results of the $z$ displacement field for the test case with the highest prediction error. Shown results consider a single-fidelity ROM model~(A), MA-ROM model combining inconsistent grid sizes~(B), and a MA-ROM model combining inconsistent structural topologies~(C).}
\label{fig:displacement_visuals}
\end{figure*}
\section{Conclusion}\label{sec:conclusion}
Reduced-order models are a popular solution to create a computationally cheap model useful for many-query applications.
The usefulness and adoption of these \emph{quick-to-evaluate} models have increased further due to multi-fidelity techniques that use sparse high-fidelity and abundant low-fidelity data.
Such models tackle the trade-off between predictive accuracy and the cost of generating a sufficient amount of expensive training data.
The structural design and sizing of aircraft components at the preliminary design phase is a scenario of many-query applications. A feature of this problem is the necessity to accurately predict the high-dimensional field responses. The field responses from physics-based structural simulation models of different fidelity have disparate representations. These inconsistencies are an obstacle to the use of contemporary approaches for multi-fidelity ROMs.
In this work, we proposed the use of a recent multi-fidelity ROM method that is capable of merging fields with inconsistent dimensionalities, topologies, and features. The method projects the low- and high-fidelity data to their respective reduced subspaces. Then, a Procrustes analysis is used to align the two manifolds. The alignment allows for the two datasets to be combined into a multi-fidelity ROM.
The proposed method is demonstrated on two structural scenarios that have multi-fidelity fields of inconsistent representations. The first scenario tackled inconsistent grid sizes by combining data from a fine structural mesh with that from a coarse structural mesh. The second scenario tackled inconsistent topologies by combining data from a high-rib count structure with that from a low-rib count structure.
Results showed that the method could improve the accuracy by up to 62.6\% when compared to a single-fidelity ROM trained with a similar computational budget.
While lower errors were observed for the prediction of both the von-Mises stress and the structural displacement fields, the MA-ROM was noticeably more suited to the evaluation of the latter.
Indeed, for a set of high-fidelity samples, the proposed method would become more accurate with additional low-fidelity results, but when applied to the stress field, the lowest cost for a specified accuracy is found at $\tau = 4$.
The reduced effectiveness of the MA-ROM method for the prediction of the von-Mises is explained by the higher complexity and non-linearity of that field when compared to the structural displacement.
An extension of the present work is to explore creating a multi-fidelity ROM combining results from beam and shell models.
The computationally efficient beam model would provide low-fidelity data owing to its inaccuracy at capturing the torsional response due to the ribs' absence.
On the other hand, the computationally inefficient, and generally more accurate, shell model would provide the high-fidelity data.
The authors also intend to extend the current method to adaptive sampling scenarios.
This would enable even greater cost reduction in the context of surrogate-based optimization.
\backmatter
\section*{Declarations}
\bmhead{Supplementary information}
This article has the following accompanying material that will be made available after receiving peer reviews: 1) Computer program that was developed for constructing the predictive model; 2) Data that was used to construct the models.
\bmhead{Funding}
No funding, grants, or other monetary support was received.
\bmhead{Conflict of Interest/Competing Interests}
The authors declare that there is no conflict of interest.
\bmhead{Availability of Data and Materials}
The data and accompanying material used to generate results presented in this work will be made available as electronic supplementary material.
\bmhead{Code Availability}
The computer program that implements the method presented in this work will be made available through a repository link.
\bmhead{Financial/Non-Financial Interests}
The authors have no relevant financial or non-financial interests to disclose.
\end{document}
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राजस्थान: स्वाइन फ्लू से १२७ लोगों की मौत, ३५०८ लोगों में दिखे बीमारी के लक्षण | क्रिस्टल टेक न्यूज
स्वाइन फ्लू एक संक्रामक बीमारी है, जो काफी तेजी और आसानी से फैलती है लेकिन इसका इलाज संभव है. अगर बीमारी के दौरान और बाद में कुछ बातों
का ध्यान रखा जाए तो इसको गंभीर होने से बचाया जा सकता है. दरअसल ये एक संक्रमण है, जो इंफ्लूएंजा ए वायरस के कारण फैलता है.
प्रेवियस आर्टियलआयरलैंड: ऊंची चट्टान से सेल्फी के चक्कर में भारतीय छात्र ने गंवाई जान
नेक्स्ट आर्टियलमुस्लिमों से की थी अपील, सिद्धू पर ७२ घंटे का बैन
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विशेषज्ञों के अनुसार, हमारे जीनोम का लगभग आधा हिस्सा रेट्रो-वायरस से बना है। ये सभी डीएनए एलिमेंट्स हैं और हमारी कोशिकाओं को नुकसान पहुंचा सकते हैं। हेट्रोक्रोमैटिन उन रेट्रो-वायरस को अंदर बहुत अधिक गहराई में रखता है ताकि उनसे हमारी कोशिकाओं को किसी तरह का नुकसान ना हो।
हाल ही हुई एक रिसर्च में एक हैरान करनेवाली बात सामने आई है। सेलुलर और मोलेक्यूलर मेडिसिन की दुनिया के लिए यह खबर किसी नए चमत्कार की तरह है। हाल ही एक रिसर्च में यह बात साबित हुई है कि कुछ डीएनए स्ट्रक्चर बहुत अधिक अनस्टेबल या अस्थाई होते हैं। इनका स्थायित्व एक्सपर्ट्स के उस अनुमान से भी कम होता है, जितना अभी तक वे इसके अनस्टेबिलिटी टाइम को मानते थे।
विशेषज्ञों के अनुसार, हमारे जीनोम का लगभग आधा हिस्सा रेट्रो-वायरस से बना है। ये सभी डीएनए एलिमेंट्स हैं और हमारी कोशिकाओं को नुकसान पहुंचा सकते हैं। हेट्रोक्रोमैटिन उन रेट्रो-वायरस को अंदर बहुत अधिक गहराई में रखता है ताकि उनसे हमारी कोशिकाओं को किसी तरह का नुकसान ना हो और हमारी बॉडी सेल्स हार्मलेस रहें।
यह भी पढ़ें: पुरुष नहीं महिलाओं की मेंटल हेल्थ ज्यादा खराब करती है यह शिफ्ट
यूएस के सेलुलर और मोलेक्यूलर मेडिसिन में असोसिएट प्रफेसर केथ मागर्ट के अनुसार, हमने हमेशा हेट्रोक्रोमैटिन की परवाह की है और इस पर स्टडी करते हैं क्योंकि हेट्रोक्रोमैटिन के बिना हमारा शरीर सभी प्रकार की बीमारियों के लिए खुद को खोल देगा और हम किसी बीमारी से बच नहीं पाएंगे। केथ पिछले २५ वर्षों से हेट्रोक्रोमैटिन पर रिसर्च कर रहे हैं और अब उनका फोकस ड्रोसोफिला-थे फ्रूट फ्लाय के अध्ययन पर है।
यह भी पढ़ें:खुद तय कीजिए रेड मीट आपको फायदा दे रहा है या नुकसान
हेट्रोक्रोमैटिन का स्थायित्व हमारी बॉडी सेल्स को रेट्रो वायरस से सुरक्षित रखने के लिए बहुत जरूरी है। केथ सहित इस फील्ड से जुड़े विशेषज्ञों ने करीब ५ साल पहले एक रिसर्च में यह अनुभव किया था कि यह उतना स्टेबल नहीं है, जितना उन्हें लगा था। केथ का कहना है कि हमने उस समय मूल रूप से सोचा था कि यह बहुत स्थिर है और एक बार इन विषैले टॉक्सिन्स पर निर्मित होने के बाद उन्हें बंद कर देता है और हम इनके खतरे से मुक्त हो जाते हैं। लेकिन अब यह थिअरी इस रिसर्च के साथ बदल गई है।
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\begin{document}
\title{An Incremental Knowledge Compilation in First Order Logic}
\author{Manoj K. Raut\\ School of
Computing Sciences\\ VIT University, Vellore
\\ Email: {\tt [email protected]}}
\date{}
\maketitle
\begin{abstract}
An algorithm to compute the set of prime implicates of a
quantifier-free clausal formula $X$ in first order logic had been
presented in earlier work. As the knowledge base $X$ is dynamic, new
clauses are added to the old knowledge base. In this paper an
incremental algorithm is presented to compute the prime implicates of $X$
and a clause $C$ from $\pi(X)\cup C$. The
correctness of the algorithm is also proved.
\end{abstract}
\noindent{\it Keywords:} Knowledge Compilation, Prime Implicates
\section{Introduction}
Propositional entailment is a central issue in artificial
intelligence due to its high complexity. Determining the logical
entailment of a given query from a knowledge base is intractable in
general \cite{Cook} as all known algorithms run in time exponential in
the size of the given knowledge base. To overcome such computational
intractability, the propositional entailment problem is split into two
phases such as {\em off-line} and {\em on-line}. In the off-line phase
the original knowledge base $X$ is transformed into another knowledge
base $X^{'}$ and the queries are answered in the on-line phase from
the new knowledge base in polynomial time in the size of $X^{'}$. In
such type of compilation most of the computational overhead shifted
into the off-line phase is amortized over on-line query
answering. The off-line computation is known as {\em knowledge
compilation}.
Several algorithms in knowledge compilation have been suggested so
far, see for example, \cite{Coudert1, Jackson1, Kleer1, Kleer2, Kean1,
Ngair, Reiter, Shiny, Slagle, Strzemecki, Tison, delVal1}. In these
approaches of knowledge compilation, a knowledge base $KB$ is compiled
off-line into another {\it equivalent} knowledge base $\pi(KB)$, i.e,
the set of prime implicates of $KB$, so that queries can be answered
from $\pi(KB)$ in polynomial time. Most of the work in knowledge
compilation have been restricted to propositional knowledge bases in
spite of the greater expressing capacity of a first order knowledge
base. Due to lack of expressive power in propositional logic, first
order logic is required to represent knowledge in many problems. An
algorithm to compute the set of prime implicates of a first order
formula in SCNF had been proposed in \cite{Manoj}.
As a knowledge base is not static, new clauses are added to the
existing knowledge base. It will be inefficient to compute the set of
prime implicates of the new knowledge base from the scratch. On the
other hand properties of the old $\pi(KB)$ can be utilized for
computing the new $\pi(KB)$. In this paper, we suggest an incremental
method to compute the set of prime implicates of the new knowledge
base from the prime implicates of the old knowledge base. The
incremental method based upon the algorithm discussed in \cite{Manoj}.
The paper is organized as follows. We present the definitions in
Section 2. In Section 3, we describe the properties and main results
for computing the prime implicates incrementally. In Section 4, we
present the incremental algorithm and its correctness. Section 5
concludes the paper.
\section{Preliminary Concepts}
\ni The alphabet of first order language contains the symbols
$x,y,z,\ldots$ as variables, $f,g,h,\ldots$ as function symbols,
$P,Q,R,\ldots$ as predicates, $\neg,\vee,\wed$ as connectives, $(,)$
and `,' as punctuation marks and, $\forall$ as the universal
quantifier. We assume the syntax and semantics of first order
logic. For an interpretation or a first order structure $i$ and a
formula $X$, we write $i\md X$ if $i$ is a model of $X$. For a set of
formulas $\Sg$ (or a formula) and any formula $Y$ we write $\Sg\md Y$
to denote that for every interpretation $i$ if $i$ is a model of every
formula in $\Sg$ then $i$ is a model of $Y$. In this case, we call $Y$
a logical consequence of $\Sg$. When $\Sg=\{X\}$, we write $X\md Y$
instead of $\{X\}\md Y$. If $X\md Y$ and $Y\md X$ then $X$ and $Y$
are said to be equivalent which is denoted by $X\equiv Y$.
A literal is an atomic formula or negation of an atomic formula. A
disjunctive clause is a finite disjunction of literals which is also
represented as a set of literals. A quantifier-free formula is in
conjunctive normal form (CNF, infact, SCNF) if it is a finite conjunction of
disjunctive clauses. For convenience, a formula is also represented as
a set of clauses. In this paper, we consider formulas only in clausal
form. In this representation, all variables are considered universally
quantified.
Two literals $r$ and $s$ are said to be {\it complementary} to
each other iff the set $\{r,\neg s\}$ is unifiable with respect to
a most general unifier $\xi$. We call $\xi$ a complementary
substitution of the set $\{r,\neg s\}$. For example, $Pxf(a)$ and
$\neg Pby$ are complementary to each other with respect to the
complementary substitution (most general unifier or mgu, for
short) $[x/b,y/f(a)]$. So a most general unifier bundles upon
infinite number of substitutions to a finite number.
A clause which does not contain a literal and its negation is said
to be {\it fundamental}. Thus a non-fundamental clause is valid.
We avoid taking non-fundamental clauses in clausal form because the
universal quantifiers appearing in the beginning of the formula
can appear before each conjunct of the CNF since $\forall$
distributes over $\wed$. So each clause in a formula of the
knowledge base is assumed to be non-fundamental.
Let $C_{1}$ and $C_{2}$ be two disjunctive clauses. Then $C_{1}$ {\em
subsumes} $C_{2}$ iff there is a substitution $\sigma$ such that
${C_{1}}{\sigma}\subseteq C_{2}$, i.e, ${C_{1}}{\sigma}\models
C_2$. For example, $\{\neg{R}{x}{f}(a), \neg{P}{y}\}$ subsumes
$\{\neg{R}{g}(a){f}(a), \neg{P}{y},$ ${Q}{z}\}$ with respect to a
$\sigma=[x/{g}(a)]$. A disjunctive clause $C$ is an {\it implicate}
of a finite set of formulas $X$ (assumed to be in CNF) if $X\sg\md C$
for a substitution $\sg$. We write $I(X)$ as the set of all
implicates of $X$. A clause $C$ is a {\it prime implicate} of $X$ if
$C$ is an implicate of $X$ and there is no other implicate $C^{'}$ of
$X$ such that $C^{'}\tau\md C$ for a substitution $\tau$ (i.e., if no
other implicate $C^{'}$ subsumes $C$). $\Pi(X)$ denotes the set of all
prime implicates of $X$. It may be observed that if $C$ is not prime
then there exists a prime implicate $D$ of $X$ such that $D\tau\md
C$. The set of all implicates of $X$ is denoted by $\Psi(X)$
Note that the notion of prime implicate is well
defined as the knowledge base contains clauses unique up to
subsumption. Let $Y$ be a set of fundamental clauses. The residue of
subsumption of $Y$, denoted by ${Res}(Y)$ is a subset of $Y$ such
that for every clause $C\in Y$, there is a clause $D\in {Res}(Y)$
where $D$ subsumes $C$; and no clause in ${Res}(Y)$ subsumes any
other clause in ${Res}(Y)$.
Let $C_1$, $C_2$ be two clauses in $X$ and $r\in C_1$, $s\in C_2$ be a
pair of complementary literals with respect to a most general unifier
$\sg$. The resolution of $C_1$ and $C_2$ is
$C=[(C_1-\{r\})\cup(C_2-\{s\})]\sg$. If $C$ is fundamental, it is
called {\em consensus} of $C_1$ and $C_2$ denoted by $CON(C_1,C_2)$.
$C$ can also be written as $[(C_1\sg-\{t\})\cup(C_2\sg-\{\neg t\})]$
for a literal $t$, provided $r\sg=t$ and $s\sg=\neg t$. We can also
say that $C$ is the propositional consensus of $C_1\sg$ and $C_2\sg$.
For example, if $C_{1}=\{{R}{b}{x}, \neg {Q}{g}(a)\}$ and
$C_{2}=\{{R}{a}{b}, {Q}{z}\}$ then $CON(C_{1}, C_{2})=\{{R}{b}{x},
{R}{a}{b}\}=$ propositional consensus of $C_1[z/g(a)]$ and
$C_2[z/g(a)]$. If $C$ is the consensus of $C_1$ and $C_2$ with respect
to a most general unifier $\sg$ then $C$ is said to be associated with
$\sg$. By default, each clause $C$ of a set of formulas $X$ is
associated with the empty substitution $\ep$. Let $C_1$ and $C_2$ be
two resolvent clauses associated with substitutions $\sg_1$ and
$\sg_2$, respectively. Then their consensus with respect to $\sg$ is
defined provided $\sg_1\sg=\sg_2\sg$. In that case the consensus is
the propositional consensus of $C_1\sigma$ and $C_2\sigma$ and the
consensus is associated with the substitution
$\sigma_{3}=\sigma_{1}{\sigma}=\sigma_{2}{\sigma}$.
\section{Computation of Prime Implicates}
\ni Besides presenting some main results from \cite{Manoj}, we
describe the computation of prime implicates incrementally of
quantifier free first order formulas in clausal form. Let $X=\{C_{1},
\ldots, C_{n}\}$ be a formula where each disjunctive clause $C_{i}$
is fundamental. Each $C_{i}$ is an implicate of $X$ with respect to
the empty substitution, but each may not be a prime implicate. The
key is the subsumption of implicates of $X$. As the clauses we deal
with are disjunctive, if $C_{1}$ subsumes $C_{2}$ then there is a
substitution $\sigma$ such that ${C_{1}}{\sigma}\models C_2$. We will
see that computation of prime implicates is the result of deletion of
subsumed clauses from the consensus closure. We also explore the
relationship between consensus closure and prime implicates of a
formula. We can derive the following two results from \cite{Manoj}.
\begin{lemma}
\label{prime11} A clause $D$ is an implicate of $X\cup C$ if and
only if there is a prime implicate $D^{'}$ of $X\cup C$ such that
$D^{'}$ subsumes $D$.
\end{lemma}
\begin{lemma}
\label{prime12} $X\cup C\equiv \Psi(X\cup C)\equiv\Pi(X\cup C)$
\end{lemma}
The computational aspects of prime implicates is given below. For a
set of clauses $X$, let ${L}(X)$ be the set of all consensus among
clauses in $X$ along with the clauses of $X$, {\it i.e.},
${L}(X)=X\cup \{ S : S$ is a consensus of each possible pair of
clauses in $X \}.$ We construct the sequence $X, {L}(X), {L}({L}(X)),
\ldots$, i.e, ${L^{0}}(X)=X$, ${L^{n+1}}(X)={L}({L^{n}}(X))$ for
$n\geq 0$. Define the {\it consensus closure} of $X$ as
$\overline{L}(X)= \cup\{{L^{i}}(X):i\in \mathbb N\}$.
\begin{example}
Let $X=(Pxa\vee\neg Qaf(x))\wed(\neg Pba\vee Rbz)\wed(\neg Rxf(a)\vee
Qzf(a))=C_1\wed C_2\wed C_3$.
\end{example}
The consensus of $C_1$ and $C_2$ with respect to the substititon
$[x/b]$ is $C_4=(\neg Qaf(b)\vee Rbz)$, of $C_1$ and $C_3$ with
respect to the substitution $[z/a, x/a]$ is $C_5=(Paa\vee \neg
Raf(a))$, of $C_2$ and $C_3$ with respect to the substitution
$[x/b,z/f(a)]$ is $C_6=(\neg Pba\vee Qf(a)f(a))$. So $C_4$, $C_5$,
$C_6$ are associated with substitutions $[x/b]$, $[z/a,x/a]$, $[x/b,
z/f(a)]$ and each of $C_1$, $C_2$, $C_3$ is associated with the empty
substitution $\epsilon$. Then \\
\noindent $L^{1}(X)= (Pxa\vee\neg Qaf(x))\wed(\neg Pba\vee
Rbz)\wed(\neg Rxf(a)\vee Qzf(a))\wed $ \\
$~~~~~~~~~~~~~$ $(\neg
Qaf(b)\vee Rbz)\wed (Paa\vee \neg Raf(a))\wed (\neg Pba\vee
Qf(a)f(a))$\\
$~~~~~~~~~~~~$ $=C_1\wed C_2\wed C_3\wed C_4\wed C_5\wed C_6$.\\
The consensus of $C_3$ and $C_4$ with respect to the
substitution $[x/b, z/f(a)]$ is $C_7=(Qf(a)f(a)\vee \neg Qaf(b))$. So
$C_7$ is associated with the substitution $[x/b,z/f(a)]$. Note that
the consensus of $C_1$ and $C_6$ with respect to the substitution
$[x/b]$ is not possible as the composition of substitution is not well
defined, i.e, $[\epsilon][x/b]\neq [x/b,z/f(a)][x/b]$. Similarly the
consensus between $C_3$ and $C_5$ is not possible as the composition
of substitution is not well defined. Hence,\\
\noindent $L^{2}(X)= ((Pxa\vee\neg
Qaf(x))\wed(\neg Pba\vee Rbz)\wed(\neg Rxf(a)\vee Qzf(a))\wed (\neg
Qaf(b)\vee $\\
$~~~~~$ $ Rbz)\wed (Paa\vee \neg Raf(a))\wed (\neg Pba\vee
Qf(a)f(a))\wed (Qf(a)f(a)\vee \neg Qaf(b)))$.\\
\noindent Since
$L^{2}(X)=L^{3}(X)$, we have, $\overline{L}(X)= L^{2}(X)$.
{$\Box$}
As each clause of a formula $X$ is itself an implicate, the following
result shows that other implicates can be computed by taking
consensus among the clauses of a formula $X$.
\begin{theorem}
\label{prime13} Consensus of two implicates of $\pi(X_1)\cup X_2$
is an implicate of the formula $X_1\cup X_2$ where $X_1$ and $X_2$ are
sets of clauses.
\end{theorem}
\begin{proof}
let $C_1$ and $C_2$ be two implicates of $\pi(X_1)\cup X_2$ associated
with $\sg_1$ and $\sg_2$ respectively. $(\pi(X_1)\cup X_2)\sg_1\md
C_1$, $(\pi(X_1)\cup X_2)\sg_2\md C_2$. $CON(C_{1},
C_{2})=PCON(C_{1}\sigma, C_{2}\sigma)$ provided
${\sigma_{1}}{\sigma}={\sigma_{2}}{\sigma}$ for some substitution
$\sigma$. So $C=CON(C_{1},C_{2}) =
((C_{1}\sigma-\{t\})\,\cup\,(C_{2}\sigma-\{\neg t\}))$. Let
$i\md(X_1\cup X_2)$. $i\md (X_1\cup X_2)\sg = X_1\sg\cup X_2\sg$. So
$i\md X_1\sg$ or $i\md X_2\sg$. Let $i\md X_1\sg$. Then,
$i\md\pi(X_1)\sg$ (by Lemma \ref{prime12}. Moreover, $i\md \pi(X_1)\sg\cup
X_2\sg$. $i\md (\pi(X_1)\cup X_2)\sg$. $i\md C_1$ and $i\md
C_1\sg$. Similarly, if $i\md X_2\sg$ then $i\md C_2\sg$. Suppose $i\md
t$. Then $i\md C_2\sg-\{\neg t\}$, i.e., $i\md
C$. Similarly, if $i\md\neg t$, then $i\md
C_1\sg-\{t\}$, i.e, $i\md c$. Hence $C$ is an implicate of $X_1\cup X_2$
{$\Box$}
\end{proof}
The following result tells that we can add the prime implicates one by
one to $X$ as it preserves the truth value.
\begin{theorem}
\label{prime13_1}
Let $X=\{C_1,C_2,\ldots,C_n\}$ be a formula and $C$ be the consensus
of a pair of clauses from $X$ then $X\equiv X\wed C$.
\end{theorem}
\begin{proof}
As $X$ is in SCNF and $C$ is a disjunctive clause, $X\wed C\md
X$. Conversely, let $i\md X$. Then $i\md C_k$ for $1\leq k\leq n$, and $i\md
C_i\wed C_j$ for $1\leq \{i, j\}\leq n$. Let $C= CON(C_i, C_j)$, where
$r$ and $s$ be a pair of complementary literals and $r\in C_i$,
$s\in C_j$, $r\sg=t$ and $s\sg=\neg t$. Let $i\md C_i\wed
C_j=(D_1\vee r)\wed(D_2\vee s)$, where $D_1$ is a disjunctive clause
in $C_i$ leaving $r$ and $D_2$ is a disjunctive clause from $C_j$
leaving $s$. $C= CON(C_i, C_j)=CON(D_1\vee r, D_2\vee s)=D_1\sg\vee
D_2\sg$. Further, $i\md (D_1\vee r)$ and $i\md(D_2\vee s)$. If $i\md D_1\vee r$
then $i\md D_1$. $i\md D_1\sg$. $i\md D_1\sg\vee D_2\sg$ and $i\md
C$. Similarly if $i\md D_2\vee s$ then $i\md C$. Let $i\md r$. Then
$i\not\md s$, so $i\md D_2$ and $i\md D_2\sg$. This implies, $i\md C$. If
$i\md s$ then $i\not\md r$, $i\md D_1$. $i\md D_1\sg$, and $i\md
C$. This implies $i\md X\wed C$.
{$\Box$}
\end{proof}
\begin{theorem}
\label{prime13_2}
$Res(\overline{L}(\pi(X)\cup C))= Res(\overline{L}(X\cup C))$
\end{theorem}
\begin{proof}
Obviously, $\overline{L}(\pi(X)\cup C)\se \overline{L}(X\cup C)$. This
implies $Res(\overline{L}(\pi(X)\cup C))\se Res(\overline{L}(X\cup C))$
Conversely, let $C_1\in Res(\overline{L}(X\cup C))$. So $C_1\in
\overline{L}(X\cup C)$ and there exists no $D\in \overline{L}(X\cup
C)$ such that $D$ subsumes $C_1$. If $C_1\not\in
Res(\overline{L}(\pi(X)\cup C))$ then $D_1\in \overline{L}(\pi(X)\cup
C)$ such that $D_1$ subsumes $C_1$. As $\overline{L}(\pi(X)\cup C)\se
\overline{L}(X\cup C)$, there exists $D_1\in \overline{L}(X\cup C)$
such that $D_1$ subsumes $C_1$. This gives a contradiction. So $C_1\in
Res(\overline{L}(\pi(X)\cup C))$. This implies $\overline{L}(X\cup
C)\se Res(\overline{L}(\pi(X)\cup C))$.
{$\Box$}
\end{proof}
The following results are consequences of Lemma \ref{prime11} and
\ref{prime12} and Theorem \ref{prime13_2}.
\begin{theorem}
\label{prime14} A clause $D_1$ is an implicate of $X\cup C$ iff
there is $D_2\in\overline {L}(\pi(X)\cup C)$ such that $D_2$ subsumes
$D_1$.
\end{theorem}
\begin {theorem}
\label{prime15} The set of all prime implicates of $X\cup C$ is a subset of
the consensus closure of
$\pi(X)\cup C$, i.e, $\pi(X\cup C)\se \overline
{L}(\pi(X)\cup C)$. Moreover, $\pi(X\cup C)= Res(\overline
{L}(\pi(X)\cup C))$
\end{theorem}
Moreover, the sets $\pi(X_1\cup X_2)$ and $\pi(\pi(X_1)\cup X_2)$ are
not only equivalent but also identical, as the next result shows.
theorem.
\begin{theorem}
\label{prime16} $\pi(X_1\cup X_2)=\pi(\pi(X_1)\cup X_2)$
\end{theorem}
\begin{proof}
let $C\in \pi(X_1\cup X_2)$. $(X_1\cup X_2)\sg_1\md C$ and there does
not exist any implicate $D$ of $X_1\cup X_2$ such that $D$ subsumes
$C$. $X_1\sg\cup X_2\sg\md C$ and there doesnot exist any implicate
$D$ of $X_1\cup X_2$ such that $D$ subsumes $C$. As $X_1\equiv
\pi(X_1)$, $\pi(X_1)\sg\cup X_2\sg\md C$ and there does not exist
any implicate $D$ of $\pi(X_1)\cup X_2$ such that
$D$ subsumes $C$. $(\pi(X_1)\cup X_2)\sg\md C$ and there does not exist
any implicate $D$ of $\pi(X_1)\cup X_2$ such that
$D$ subsumes $C$. $C\in\pi(\pi(X_1)\cup X_2)$. So $\pi(X_1\cup X_2)\se
\pi(\pi(X_1)\cup X_2)$. Similarly the inclusion $\pi(\pi(X_1)\cup
X_2)\se \pi(X_1\cup X_2)$ is proved.
{$\Box$}
\end{proof}
\section{Incremental Algorithm}
The following algorithm computes the set of prime implicates of
$\pi(X_1)\wed \Sg$ (i.e, of $(\pi(X_1)\cup \Sg)$ by consensus
subsumption method in first order logic. Recall that for a set of
clauses $A$, $L(A)$ denotes the set of clauses of $A$ along with the
consensus of each possible pair of clauses of $A$. The algorithm
computes the consensus $L(\pi(X_1)\cup \Sg)$ by taking clauses from
$\pi(X_1)$ and clauses from $\Sg$. It does not compute the consensus
between two clauses of $\pi(X_1)$ as it is wasteful. $L(\pi(X_1)\wed
\Sg)=\{CON(D_1, D_2)$ : $D_1\in \pi(X_1)\cup \Sg$ and $D_2\in \Sg\}$. The
algorithm applies subsumption on $L(\pi(X_1)\wed \Sg)$ and keeps the
residue $Res(L(\pi(X_1)\wed \Sg))$ and repeat the steps till two
iteration steps produce the same result.
\ni{\bf Algorithm} \hs{.2cm} {\em INCRPI}
\vs{.2cm}
\ni Input: The set of prime implicates $\pi(X_1)$ and a clause $C$\\
Output: The set of prime implicates of $X_1\cup C$\\
{\tt begin\\
$~~~~$if $C$ is a non-fundamental clause, then\\
$~~~~~~~~$$\pi(X_1\cup C)=\pi(X_1)$\\
$~~~~$else\\
$~~~~~~~~$$\Sg=\{C\}$;\\
$~~~~~~~~$$\eta_0=\emptyset$;\\
$~~~~~~~~$$i=1$;\\
$~~~~~~~~$$\gamma=\pi(X_1)\cup \Sg$;\\
$~~~~~~~~$$\eta_i=Res(\pi(X_1)\cup\Sg)$;\\
$~~~~~~~~$if $\Sg$ is deleted\\
$~~~~~~~~~~$then stop\\
$~~~~~~~~$else\\
$~~~~~~~~~~$while $\eta_i\neq \eta_{i-1}$\\
$~~~~~~~~~~~~~$compute $R=CON(D_1,D_2)$ s.t.
$D_1\in \eta_i$ and $D_2\in \Sg$;\\
$~~~~~~~~~~~~~$$\Sg=\Sg\cup R$;\\
$~~~~~~~~~~~~~$$L(\eta_i)=\eta_i\cup \Sg$;\\
$~~~~~~~~~~~~~$$\eta_{i+1}=Res(L(\eta_i))$;\\
$~~~~~~~~~~~~~$if any clause of $\Sg$ is deleted\\
$~~~~~~~~~~~~~~~~~~~~$then update $\Sg$;\\
$~~~~~~~~~~~~~$$i=i+1$;\\
$~~~~~~~~~~$$\pi(\gamma)=\eta_{i+1}$;\\
end}
\begin{theorem}
\label{correctness} Let $\pi(X_1)$ be a set of prime implicates of a
formula $X_1$ and $C$
be any clause. The incremental algorithm generates the set of prime
implicates of $X_1\cup \{C\}.$
\end{theorem}
\begin{proof}
Let $\gamma=\pi(X_1)\cup \{C\}$. $\eta_1$ is computed
by taking residue of subsumption on $\gamma$. If a clause
$C\in \gamma$ subsumes a clause $D \in \gamma$,
any clause entailed by $D$ is entailed by $C$. So $D$ can be
discarded from $\gamma$ without changing its deduction closure.
Let $\eta_1= Res(\gamma)$.
$\gamma =\pi(\eta_1)$, by Theorem \ref{prime15}.
If the clause $C$ is deleted while taking residue then
$\pi(\gamma)=\pi(X_1\cup \{C\})=\pi(X_1)$. If $C$ is not deleted from
$\Sg$, then the algorithm computes the consensus $R$ between a pair of
clauses from $\eta_1$ (i.e,$Res(\gamma)$) and $C$. It does not take
consensus between two clauses in $\pi(X_1)$ as they are prime
implicates. Any attempt to take consensus between them will increase
the number of uncessary operations. As every clause of $\eta_1$ is an
implicate of $\eta_1$, by Theorem \ref{prime13}, $R$ is an implicate
of $\eta_1$, i.e, of $\gamma$. $R$ can be added to $\eta_1$ without
changing its deduction closure by Theorem \ref{prime13_1}. We add $R$
to $\Sg=\{C\}$ as the new clauses formed can also take part in further
consensus. We maintain $\pi(X_1)$ and $\Sg$ separately so that while
taking consensus next time at least one clause will be from $\Sg$,
i.e, $L(\eta_1)=\eta_1\cup \Sg$. By Theorem \ref{prime15},
$\pi(\gamma)=\pi(\eta_1)\se\overline{L}(\eta_1)$. If any clause $C$
subsumes a clause $D$ in $L(\eta_1)$ then $D$ can be discarded without
changing the deduction closure as $C\md D\sg$. Note $\eta_2=
Res(L(\eta_1))$. By Theorem \ref{prime15}, $\pi(\gamma)=\pi(\eta_1)\se
\overline{L}(\eta_2)$. In general, $\pi(\gamma)=\pi(\eta_1)\se
\eta_i=Res(L(\eta_{i-1}))\se\overline{L}(\eta_{i-1})$. If the
algorithm terminates, at some stage then
$\overline{L}(\eta_i)=\overline{L}(\eta_{i-1})$ and
$Res(\overline{L}(\eta_i))= Res(\overline{L}(\eta_{i-1}))$. By Theorem
\ref{prime15}, $\pi(\gamma)= Res(\overline{L}(\eta_i))$. Since the
algorithm computes $\eta_{i+1}=Res(L(\eta_i))$,
$\pi(\gamma)=\eta_{i+1}$. By Theorem \ref{prime16} we obtain the set
of prime implicates of $X_1\cup C$ as $\pi(X_1\cup C)=\pi(\pi(X_1)\cup
C)=\pi(\gamma)=\eta_{i+1}$.
{$\Box$}
\end{proof}
\begin{example}\label{exp}
Let $X=\{\{Qy\},\{\neg Rf(x)b\},\{Px\vee Ryb\vee\neg
Qz\}\}$$=\eta_1(say)$ and $C$ be the clause $C=\{\neg Pa\vee\neg
Qz\}$$=\Sg$, another clause. Take $\Sg=\{C\}$
\end{example}
As computed in \cite{Manoj}, the set of prime implicates of the formula $X$
is $\pi(X)=\{\{Qy\},\{\neg Rf(x)b\}, \{Px\vee Rzb\},
\{Px\vee\neg Qz\}\}$, where the clause $Qy$ is associated with $\epsilon$,
$\neg Rf(x)b$ is associated with $\epsilon$, $Px\vee Rzb$ is associated with
$[y/z]$ and $Px\vee\neg Qz$ is associated with $[y/f(x)]$. \\
\noindent Let $\eta_1=\pi(X)\cup \Sg=\{\{Qy\},\{\neg Rf(x)b\},
\{Px\vee Rzb\}, \{Px\vee\neg Qz\},\{\neg Pa\vee\neg Qz\}\}$.\\
\noindent As the literal $Qy$ in $\{Qy\}$ and $\neg Qz$ in $\{\neg
Pa\vee\neg Qz\}$ are a pair of complementary literals with respect to
the substitution $[y/z]$, the consensus between the clauses
$\{Qy\}$ and $\{\neg Pa\vee\neg Qz\}$ is $\{\neg Pa\}$. Here, the
substitution $[y/z]$ is a most general unifier. Note that we can not
take consensus between $\{Px\vee Rzb\}$ and $\{\neg Pa\vee\neg Qz\}$,
between $\{Px\vee\neg Qz\}$ and $\{\neg Pa\vee\neg Qz\}$ as
composition of substitution are not well defined. Now updating $\Sg$ we
get, $\Sg= \{\{\neg Pa\vee\neg Qz\},\{\neg Pa\}\}$. The new clause
formed is added to $\eta_1$ to get $L(\eta_1)$.\\
\noindent $L(\eta_1)=\{\{Qy\},\{\neg Rf(x)b\}, \{Px\vee Rzb\},
\{Px\vee\neg Qz\},\{\neg Pa\vee\neg Qz\}, \{\neg Pa\}\}$.\\
\noindent As $\{\neg Pa\}$ subsumes
$\{\neg Pa\vee\neg Qz\}$ in $L(\eta_1)$, we get the residue as\\
\noindent $\eta_2=Res(L(\eta_1))=\{\{Qy\},\{\neg Rf(x)b\}, \{Px\vee
Rzb\}, \{Px\vee\neg Qz\}, \{\neg Pa\}\}$.\\
\noindent Now the clause $\{Px\vee Rzb\}$ associated with $[y/z]$ and
$\{\neg Pa\}$ associated with $[y/z]$ contain a pair of complementary
literals. The consensus between $\{Px\vee Rzb\}$ and $\{\neg Pa\}$
with respect to the substitution (mgu) $[y/z,x/a]$ is $Rzb$. Again we
can not take consensus between $\{Px\vee\neg Qz\}$ associated with
$[y/f(x)]$ and $\{\neg Pa\}$ associated with $[y/z]$ as composition of
substitution is not well defined. Now $\Sg=\{\{\neg Pa\}, \{Rzb\}\}$.
We now add the new clause to $\eta_2$ to get $L(\eta_2)$.\\
\noindent $L(\eta_2)=\{\{Qy\},\{\neg Rf(x)b\}, \{Px\vee Rzb\},
\{Px\vee\neg Qz\}, \{\neg Pa, Rzb\}\}$.\\
\noindent As $\{Rzb\}$ subsumes $\{Px\vee Rzb\}$, the residue becomes\\
\noindent $\eta_3=Res(L(\eta_2))=\{\{Qy\},\{\neg Rf(x)b\},
\{Px\vee\neg Qz\}, \{\neg Pa\}, \{Rzb\}\}$.\\
\noindent We cannot take any more consensus among the clauses of
$\eta_3$ as the composition of substitution is not well defined
between clauses. So $\eta_3 = L(\eta_3)=\eta_4$.That is,\\
\noindent $\pi(X\cup C)=\pi(\pi(X)\cup C)= \{\{Qy\},\{\neg Rf(x)b\},
\{Px\vee \neg Qz\}, \{\neg Pa\}, \{Rzb\}\}$.
\section{Conclusion}
In this paper, we have suggested an incremental algorithm to compute
the set of prime implicates of a knowledge base $X$ and a clause
$C$. We have also proved the correctness of the algorithm. In
Example \ref{exp}, when new clauses or clause sets are added,
computation of the prime implicates uses the primeness of the
already computed clauses. The algorithm adds one clause at a time
and compiles the enhanced knowledge base. A simple modification of
the algorithm can be made to accomodate a set of clauses instead of
one by putting {\it INCRPI} inside an iterative loop.
If we compute the prime implicates of $X\cup C$ directly by using the
algorithm from \cite{Manoj}, we obtain the same prime implicates,
though it involves wasteful computations. Efficiency of the proposed
algorithm {\it INCRPI} results in exploiting the properties of $\pi(X)$
instead of using $X$ directly.
\end{document}
|
math
|
// __BEGIN_LICENSE__
// Copyright (c) 2009-2013, United States Government as represented by the
// Administrator of the National Aeronautics and Space Administration. All
// rights reserved.
//
// The NGT platform is licensed under the Apache License, Version 2.0 (the
// "License"); you may not use this file except in compliance with the
// License. You may obtain a copy of the License at
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
// __END_LICENSE__
/// \file stereo.h
///
#ifndef __ASP_STEREO_H__
#define __ASP_STEREO_H__
#include <boost/algorithm/string.hpp>
#include <asp/Core/StereoSettings.h>
#include <asp/Core/MedianFilter.h>
#include <asp/Core/Macros.h>
#include <asp/Core/Common.h>
// Support for ISIS image files
#if defined(ASP_HAVE_PKG_ISISIO) && ASP_HAVE_PKG_ISISIO == 1
#include <asp/IsisIO/DiskImageResourceIsis.h>
#endif
// Boost headers
#include <boost/thread/xtime.hpp>
// Posix time is not fully supported in the version of Boost for RHEL Workstation 4
#ifdef __APPLE__
#include <boost/date_time/posix_time/posix_time.hpp>
#else
#include <ctime>
#endif
namespace po = boost::program_options;
namespace fs = boost::filesystem;
/// The stereo pipeline has several stages, which are enumerated below.
enum { PREPROCESSING = 0,
CORRELATION,
REFINEMENT,
FILTERING,
POINT_CLOUD,
WIRE_MESH,
NUM_STAGES};
// Allows FileIO to correctly read/write these pixel types
namespace asp {
/// Load the D_sub file in a consistent format.
/// - Returns false if the file does not exist.
bool load_sub_disp_image(std::string const& sub_disp_path,
vw::ImageViewRef<vw::PixelMask<vw::Vector2f> > &sub_disp);
/// Transform the crop window to be in reference to L.tif
vw::BBox2i transformed_crop_win(ASPGlobalOptions const& opt);
/// Parse the command line options for multi-view stereo
void parse_multiview(int argc, char* argv[],
boost::program_options::options_description const&
additional_options,
bool verbose,
std::string & output_prefix,
std::vector<ASPGlobalOptions> & opt_vec,
bool exit_early = false);
/// Parse input command line arguments
void handle_arguments( int argc, char *argv[], ASPGlobalOptions& opt,
boost::program_options::options_description const&
additional_options,
bool allow_unregistered,
std::vector<std::string> & unregistered,
std::string & usage, bool exit_early = false);
/// Register DiskImageResource types that are not included in Vision Workbench.
void stereo_register_sessions();
/// Checks for obvious user mistakes
/// - Throws if any incompatible settings are found.
void user_safety_checks(ASPGlobalOptions const& opt);
bool skip_image_normalization(ASPGlobalOptions const& opt);
} // end namespace vw
#endif//__ASP_STEREO_H__
|
code
|
\begin{document}
\title{Bypassing state initialisation in perfect state transfer protocols on spin-chains}
\author{C. Di Franco\inst{1} \and M. Paternostro\inst{2} \and M. S. Kim\inst{3}}
\institute{Department of Physics, University College Cork, Cork, Republic of Ireland \and School of Mathematics and Physics, Queen's University, Belfast BT7 1NN, United Kingdom \and Institute for Mathematical Sciences, Imperial College London, SW7 2PG, United Kingdom \\ and \\ QOLS, The Blackett Laboratory, Imperial College London, Prince Consort Road, SW7 2BW, United Kingdom}
\maketitle
\begin{abstract}
Although a complete picture of the full evolution of complex quantum systems would certainly be the most desirable goal, for particular Quantum Information Processing schemes such an analysis is not necessary. When quantum correlations between only specific elements of a many-body system are required for the performance of a protocol, a more distinguished and specialised investigation is helpful. Here, we provide a striking example with the achievement of perfect state transfer in a spin chain without state initialisation, whose realisation has been shown to be possible in virtue of the correlations set between the first and last spin of the transmission-chain.
\end{abstract}
\section{Introduction}
Quantum Information Theory (QIT) is having a remarkable impact from a fundamental viewpoint by providing alternative perspectives to physical problems using new conceptual instruments. The study of quantum correlations shared by many distinctive objects is helping us in understanding their behaviour at critical points~\cite{criticalpoints} and quantifying the resources required in order to efficiently simulate such situations~\cite{simulation}. The simulation of complex quantum systems is usually a prohibitive task for even the most powerful classical machine due to the exponential growth of its Hilbert space with respect to the number of elements. In this context, several advances have recently been made in the study of the ground state of particular many-particle systems~\cite{groundstate}. While all the proposed methods have found use in simulating static properties of ground states, their application to the investigation of time evolution is, in general, problematic. However, although the analysis of the complete behaviour of quantum many-particle systems will be a fundamental task in QIT, for the study of some particular Quantum Information Processing (QIP) schemes this is not necessary. When quantum correlations between only specific elements of a many-body system are required for the performance of a protocol, a more distinguished and specialised approach is helpful. The problems related to simulating the dynamics of many-particle systems can be solved, in this context, by considering not their whole evolution, but only the behaviour of a few characteristic features. Here we provide a striking example in the achievement of perfect Quantum State Transfer (QST) in a spin chain without state initialisation, whose realisation has been shown to be possible in virtue of the correlations set between the first and last spin of the transmission-chain~\cite{roastedchicken}.
This result is also important on a more pragmatic ground. Recently, it has been shown that the control over multipartite registers for the purposes of QIP can be sensibly reduced in a way so as to avoid the generally demanding fast and accurate inter-qubit switching and gating. In this case, the price to pay for the performance of efficient operations is the pre-engineering of appropriate patterns of couplings~\cite{preengineering}. The preparation of a fiducial state for the initialisation of a QIP device can be however experimentally demanding. This is mainly due to the difficulty of preparing pure states of multipartite systems, which is one of DiVincenzo's criteria~\cite{divincenzo}, a set of requirements that any QIP system should meet. Remarkably, our proposal is able to bypass the initialisation of the spin-medium in a known pure state. The scheme requires only end-chain single-qubit operations and a single application of a global unitary evolution and is thus fully within a scenario where the control over the core part of the spin medium is relaxed in favour of controllability of the first and last element of the chain. The relaxation of the conditions necessary for manipulating information is a necessary step in order to shorten the time for the achievement of realistic QIP. This allows us to loose the requirements for information protection from environmental effects. Instead of utilising demanding always-on schemes for the shielding of the information content of a system, this could be done only during the running-time of the protocol.
\section{Perfect state transfer without state initialisation for the $X\!X$ model}
Spin chains have recently emerged as remarkable candidates for the realisation of faithful short-distance transmission of quantum information~\cite{unmodulated}. Here, the system under investigation is a nearest-neighbour $X\!X$ coupling involving $N$ spin-$1/2$ particles. Its Hamiltonian reads
\begin{equation}
\hat{{\cal H}}=\sum^{N-1}_{i=1}J_{i}(\hat{X}_{i}\hat{X}_{i+1}+\hat{Y}_i\hat{Y}_{i+1}),
\end{equation}
where $J_i$ is the interaction strength between spin $i$ and $i+1$ and $\hat{X},\,\hat{Y}$ and $\hat{Z}$ denote the $x,\,y$ and $z$ Pauli matrix, respectively. Let us start considering
\begin{equation}
J_{i}=J\sqrt{i(N-i)}
\end{equation}
with $J$ being a characteristic energy scale that depends on the specific physical implementation of the model (we choose units such that $\hbar=1$ throughout the paper). This model has been extensively analysed~\cite{cambridge}: $1\rightarrow{N}$ perfect QST is achieved, through this coupling, when the initial state of all the spins but the first one is $\ket{0}$.
In our investigation, however, we drop the condition on the state of the central qubits, and we just assume control over the external ones. For the understanding of what follows, it is useful to analyse the time-evolution, in the Heisenberg picture, of the two-site operators $\hat{\;\mbox{$\rm{I} \hspace{-1.0mm} {\bf I}$}\,}_i\hat{Z}_{N-i+1}$, $\hat{X}_i\hat{X}_{N-i+1}$, and $\hat{X}_i\hat{Y}_{N-i+1}$. We define $\hat{{\cal O}}(t)$ as the time-evolved form of a given operator $\hat{O}$. By solving a set of Heisenberg equations, we have that, at time $t^*=\pi/4J$ and for any $N$,
\begin{equation}
\hat{\;\mbox{$\rm{I} \hspace{-1.0mm} {\bf I}$}\,}_i(t^*)\hat{\cal Z}_{N-i+1}(t^*)=\hat Z_i\hat{\;\mbox{$\rm{I} \hspace{-1.0mm} {\bf I}$}\,}_{N-i+1}.
\label{eq:evolutioncambridgez}
\end{equation}
On the other hand, for an even $N$ we find
\begin{eqnarray}
\nonumber&\hat{\cal X}_i(t^*)\hat{\cal X}_{N-i+1}(t^*)=\hat X_i\hat X_{N-i+1},\\
&\hat{\cal X}_i(t^*)\hat{\cal Y}_{N-i+1}(t^*)=\hat Y_i\hat X_{N-i+1},
\label{eq:evolutioncambridge}
\end{eqnarray}
while for an odd $N$
\begin{eqnarray}
\nonumber&\hat{\cal X}_i(t^*)\hat{\cal X}_{N-i+1}(t^*)=\hat Y_i\hat Y_{N-i+1},\\
&\hat{\cal X}_i(t^*)\hat{\cal Y}_{N-i+1}(t^*)=-\hat X_i\hat Y_{N-i+1}.
\label{eq:evolutioncambridge2}
\end{eqnarray}
These results can also be easily obtained from the analysis presented in Ref.~\cite{ijqi}, where it is shown that the evolution of single-qubit operators can be evaluated by means of a method based on oriented graphs. For instance, in the case $N=5$, $\hat{\cal X}_1(t)$ can be decomposed as
\begin{eqnarray}
\nonumber&\hat{\cal X}_1(t)=\alpha_1(t)\hat{X}_1+\alpha_2(t)\hat{Z}_1\hat{Y}_2+\alpha_3(t)\hat{Z}_1\hat{Z}_2\hat{X}_3+\\
&\alpha_4(t)\hat{Z}_1\hat{Z}_2\hat{Z}_3\hat{Y}_4+\alpha_5(t)\hat{Z}_1\hat{Z}_2\hat{Z}_3\hat{Z}_4\hat{X}_5.
\end{eqnarray}
Similarly,
\begin{eqnarray}
\nonumber&\hat{\cal X}_5(t)=\beta_1(t)\hat{X}_5+\beta_2(t)\hat{Z}_5\hat{Y}_4+\beta_3(t)\hat{Z}_5\hat{Z}_4\hat{X}_3+\\
&\beta_4(t)\hat{Z}_5\hat{Z}_4\hat{Z}_3\hat{Y}_2+\beta_5(t)\hat{Z}_5\hat{Z}_4\hat{Z}_3\hat{Z}_2\hat{X}_1.
\end{eqnarray}
The time-evolution of the two-site operator $\hat{X}_1\hat{X}_5$ can be therefore obtained by considering the sum of all the possible products of elements of these two sets of operator. For $J_{i}=J\sqrt{i(N-i)}$, the time-dependent coefficients $\alpha_i(t)$ have the behaviour shown in Fig.~\ref{fig2}.
\begin{figure}
\caption{Coefficients $\alpha_1$ (red dashed line), $\alpha_2$ (green dashed line), $\alpha_3$ (blue dashed line), $\alpha_4$ (purple dashed line) and $\alpha_5$ (black line) against dimensionless time $Jt$, for $N=5$ and $J_{i}
\label{fig2}
\end{figure}
By symmetry, $\beta_i(t)=\alpha_i(t)$ for all values of $i$ and $t$. It is easy to notice, in Fig.~\ref{fig2}, that at the time $t^*=\pi/4J$, $\alpha_5(t^*)=\beta_5(t^*)=1$, while all the other coefficients are equal to $0$. For that particular time, therefore, the evolved operator $\hat{\cal X}_1(t^*)\hat{\cal X}_5(t^*)$ is just the product of $\hat{Z}_1\hat{Z}_2\hat{Z}_3\hat{Z}_4\hat{X}_5$ times $\hat{Z}_5\hat{Z}_4\hat{Z}_3\hat{Z}_2\hat{X}_1$. We have $\hat{\cal X}_1(t^*)\hat{\cal X}_5(t^*) = \hat Y_i\hat Y_{N-i+1}$. In the same way, all the other evolved operators in Eqs.~(\ref{eq:evolutioncambridgez}) and (\ref{eq:evolutioncambridge2}) can be obtained.
Each of the two-site operators in Eqs.~(\ref{eq:evolutioncambridgez})-(\ref{eq:evolutioncambridge2}) evolves into operators acting on the same qubits, without any dependence on other operators of the chain. This paves the way to the core of our protocol, which we now describe qualitatively. Qubit $1$ is initialised in the input state $\rho^{in}$ (either a pure or mixed state) we want to transfer and qubit $N$ is projected onto
\begin{equation}
\ket{\pm_N}=\frac{1}{\sqrt{2}}(\ket{0}\pm e^{iN\frac{\pi}{2}}\ket{1}).
\end{equation}
In what follows, we say that outcome $+1$ ($-1$) is found if a projection onto $\ket{+_N}$ ($\ket{-_N}$) is performed. Then the interaction encompassed by $\hat{\cal H}$ is switched on for a time $t^*=\pi/4J$, after which we end up with an entangled state of the chain. The amount of entanglement shared by the elements of the chain depends critically on their initial state. Regardless of the amount of entanglement being set, an $\hat{X}$-measurement over the first spin projects the $N$-th one onto a state that is locally-equivalent to $\rho^{in}$. More specifically, if the product of the measurement outcomes at $1$ (after the evolution) and $N$ (before the evolution) is $+1$ ($-1$), the last spin will be in $(\hat{T}^N)^\dag\rho^{in}(\hat{T}^N)$ [$\hat{Z}(\hat{T}^N)^\dag\rho^{in}(\hat{T}^N)\hat{Z}$], where $\hat{T}=\ket{0}\bra{0}+e^{i\frac{\pi}{2}}\ket{1}\bra{1}$ (therefore, $\hat{T}^2=\hat{Z}$)~\cite{comment}. In any case, apart from a simple single-spin transformation, perfect QST is achieved. A sketch of the scheme is presented in Fig.~\ref{fig3}.
\begin{figure}
\caption{Sketch of the scheme for perfect QST. $M_1$ and $M_2$ are measurements performed over a fixed basis and $\Sigma$ is a conditional operation. $\hat{\cal H}
\label{fig3}
\end{figure}
The crucial point here is that, regardless of the amount of entanglement established between the spin-medium and the extremal elements of the chain ({\it i.e.} spins $1$ and $N$), upon $\hat{X}$-measurement of $1$, the last spin is disconnected from the rest of the system, {\it whose initial state is inessential to the performance of the protocol} and could well be, for instance, a thermal state of the chain in equilibrium at finite temperature. In fact, the key requirements for our scheme are the arrangement of the proper time-evolution (to be accomplished within the coherence times of the system) and the performance of clean projective measurements on spin $1$ and, preventively, on $N$.
\section{General conditions}
In general, the protocol can be adapted to any Hamiltonian for which we can find a triplet of single-spin operators $\hat{\cal B},\hat{C},\hat{D}$ such that, for symmetric spin pairs, we have
\begin{equation}
\hat{\cal B}^{j_{O}}_i(t^*)\hat{C}_{N-i+1}\hat{\cal O}_{N-i+1}(t^*)=\hat{O}_{i}\hat{D}^{k_O}_{N-i+1}.
\end{equation}
Here, $\hat{\cal B}_{i}$ ($\hat{D}_{N-i+1}$) provides the eigenbasis for the measurement over spin $i$ ($N-i+1$) of the chain after (before) the evolution, $\hat{C}_{N-i+1}$ is a decoding operation, $\hat{O}_i=\hat{\cal O}_i(0)=\hat{X},\hat{Y},\hat{Z}$ and $j_O,k_O=0,1$, depending on the coupling model. We point out that, when these conditions are not fulfilled, our protocol can still be rather successful. In these cases, through an information flux approach, we can still estimate the average transfer fidelity~\cite{informationflux}. For instance, we can consider the case in which we are able to engineer the strength of the coupling rates of just the extremal qubits ($J_1$ and $J_{N-1}$). Therefore, we take $J_i = J$ (for ${i}=2,...,N-2$), $J_1 = J_{N-1} = \eta{J}$. The behaviour of this system against the dimensionless interaction time $J t$ and the inhomogeneity parameter $\eta$ has already been studied in Ref.~\cite{informationflux}. For simplicity, here we consider the time-dependent coefficients $\alpha_i(t)$ in the case $N=5$, for the value of $\eta$ which maximises QST fidelity ($\eta\sim0.815$). Also this system is centro-symmetric, therefore we have $\beta_i(t)=\alpha_i(t)$ for all values of $i$ and $t$. In this case, however, there is no time for which $\alpha_5(t)=\beta_5(t)=1$, while all the other coefficients are equal to $0$. Nevertheless, for a dimensionless time $J t'\sim1.9$, the value of $\alpha_5$ and $\beta_5$ are close to $1$, while all the other ones are close to $0$. Our estimate gives an average transfer fidelity via our protocol of $F\sim\alpha_5^2(t')>99.9\%$.
\section{Remarks}
We have shown the existence of a simple control-limited scheme for the achievement of perfect QST in a system of interacting spins without the necessity of demanding state initialisation. Our protocol requires just {\it one-shot} unitary evolution and end-chain local operations. Its efficiency arises from the establishment of {\it correlations} between the first and last spin of the transmission-chain. With the exception of limiting cases where the transfer is automatically achieved (as for the transfer of eigenstates of $\hat{X}_1$ when model $\hat{\cal H}$ is used), these are set regardless of the state of the spin medium, their amount being a case-dependent issue. The end-chain measurements, which are key to our scheme, ``adjust'' such correlations in a way so as to achieve perfect QST.
Due to the dependence of this protocol only on the correlations established between the first and last spin, the analysis of this scheme just requires the investigation of the time evolution of two-site operators. The exponential growth of the Hilbert space of the total state of the system does not affect our analysis, that can thus be done by means of only ``slowly-growing'' computational effort. In fact, the number of elements in the decomposition of the relevant two-site operators grows as $N^2$. Moreover, in this way, we were able to obtain our results removing any dependance on the state of all the central qubits.
We would like to conclude this contribution by remarking that our protocol for state transfer without initialisation is already encountering the attention of the community interested in quantum spin-chain dynamics. In fact, a recent proposal by Markiewicz and Wiesniak~\cite{MW} has addressed a scheme for perfect state transfer without initialisation where the necessity for ``remote coordination'' between sending and receiving agents is bypassed.
\section{Acknowledgments}
We acknowledge support from the UK EPSRC. C.D.F. is supported by the Irish Research Council for Science, Engineering and Technology. M.P. thanks the UK EPSRC (EP/G004579/1) for financial support.
\end{document}
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math
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जली हई कार को देखते ग्राणीण
कहा जा रहा है कि सभी लोग शादी समारोह में हिस्सा लेने के लिए एटा जा रहे थे. तभी कार की एटा सड़क पर ट्रक से टक्कर हो गई. इसके बाद कार में आग लग गई.
एटा. उत्तर प्रदेश (उत्तर प्रदेश) के एटा में भीषण सड़क हादसा (रोड एक्सीडेंट) हुआ है, जहां तेज रफ्तार कार की सड़क किनारे खड़े ट्रक (ट्रक) से टक्कर हो गई. इसके बाद कार में भीषण आग लग गई. इससे कार सवार ६ लोगों में से पांच की जिंदा जलकर (बर्न्ट) मौत हो गई है. वहीं, एक युवक को गंभीर हालत में आगरा रेफर कर दिया गया है. मौके पर पहुंचे ग्रामीणों ने पुलिस की सहायता से आग पर काबू पा लिया है.
घटना जिले के बागवाला थाना क्षेत्र की है. जानकारी के मुताबिक, शादी समारोह में हिस्सा लेने के लिए छह लोग कार पर सवार होकर नोएडा से एटा के रामपुर जा रहे थे. तभी कार की एटा रोड पर खड़े ट्रक से टक्कर हो गई. टक्कर इतनी भयंकर थी कि कार में आग लग गई.
सूचना के बाद स्थानीय ग्रामीण और पुलिस मौके पर पहुंच गए और बचाव कार्य शुरू कर दिया गया. लेकिन तब तक काफी देर हो चुकी थी और कार में सवार ६ लोगों में से ५ की जिंदा जलकर मौत हो गई. बताया जा रहा है कि हादसे के दौरान कार सवार एक युवक गाड़ी से बाहर गिर गया था, जिससे वह गंभीर रूप से घायल हो गया. उसका आगरा स्थित एक अस्पताल में इलाज चल रहा है. वहीं, पुलिस ने पांचों मृतकों के शवों को कब्जे में लेकर पोस्टमार्टम को भेज दिया है. साथ ही पुलिस घटना की जांच में जुटी हुई है.
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hindi
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\begin{enumerate}gin{document}
\title{
\bf The Weak Bruhat Order and Separable Permutations
\footnote{This research was carried out under the
direction of R. Stanley when the author was an undergraduate at
M.I.T.}
}
\author{Fan Wei\footnote{fan\[email protected], MIT}}
\date{September 21, 2010}
\maketitle{\small \textsc{\bf Abstract}
In this paper we consider the rank generating function of a
separable permutation $\pi$ in the weak Bruhat order on the two
intervals $[\text{id}, \pi]$ and $[\pi, w_0]$, where $w_0 =
n,(n-1),\dots, 1$. We
show a surprising result that the product of these two
generating functions is the generating function for the
symmetric group with the weak order. We then
obtain explicit formulas for the rank generating functions
on $[\text{id}, \pi]$ and $[\pi, w_0]$, which leads to
the rank-symmetry and unimodality of the two graded posets. }
\vskip 5mm
\section{Introduction and Definitions} \label{sec1}
Let $\mathfrak{S}_n$ denote the symmetric group of all permutations of
$1,2,\dots,n$. Define the \emph{length} of the permutation
$\pi=a_1a_2\cdots a_n\in\mathfrak{S}_n$ by $$ \ell(\pi)=\#\{ 1\leq i<j\leq
n\colon a_i>a_j\}, $$ which is the number of inversions of $\pi$. One
of the fundamental partial orderings of $\mathfrak{S}_n$ is the \emph{weak
(Bruhat) order}. A cover relation $\pi\lessdot \sigma$ in weak order,
i.e., $\pi<\sigma$ and nothing is in between, is defined by
$\sigma=\pi s_i$ for some adjacent transposition $s_i=(i,i+1)$,
provided that $\ell(\sigma)> \ell(\pi)$. We are multiplying
permutations right-to-left, so for instance $2413s_2=2143$. The weak
order makes $\mathfrak{S}_n$ into a graded poset of rank $\binom n2$. If
$\pi=a_1a_2\cdots a_n\in\mathfrak{S}_n$, then the rank function of $\mathfrak{S}_n$ (which
will have the weak order unless stated otherwise) is the function
$\ell$. The rank generating function is then given by
$$ F(\mathfrak{S}_n,q) =\sum_{\pi\in\mathfrak{S}_n}q^{\ell(\pi)} = [n]!, $$
where $[n]!=[1][2]\cdots[n]$ and $[i]=1+q+q^2+\cdots+q^{i-1}$.
A permutation $\pi=a_1a_2\cdots a_n\in\mathfrak{S}_n$ is \emph{3142-avoiding} and
\emph{2413-avoiding} if there do not exist $i<j<k<h$ with
$a_{j}<a_{h}<a_{i}<a_{k}$ or $a_{k}<a_{i}<a_{h}<a_{j}$. Such
permutations are also called \emph{separable}. For a general
introduction to pattern avoidance, see \cite{bona}. Separable
permutations first arose in the work of Avis and Newborn
\cite{av-ne} and have subsequently received a lot of attention. A
survey of some of their properties appears in \cite{albert}. In
particular, the number of separable permutations in $\mathfrak{S}_n$ is the
(large) Schr\"oder number $r_{n -1}$. Let id denote the identity element
of $\mathfrak{S}_n$ (the unique minimal element in weak order), and let
$w_0=n,n-1,\dots,1$, the unique maximal element. For $\pi\in\mathfrak{S}_n$,
let $\Lambda_\pi$ denote the interval $[\mathrm{id},\pi]$ (in weak
order), and let $V_\pi=[\pi,w_0]$. Thus $\Lambda_\pi$ and $V_\pi$ are
themselves graded posets (with rank$(\pi)=0$ in $V_\pi$). The main
result of this paper is the surprising formula
\begin{enumerate}q F(\Lambda_\pi,q)F(V_\pi,q) = F(\mathfrak{S}_n,q) =[n]!. \label{eqmain} \end{enumerate}q
Equation~\eqref{eqmain} was conjectured by R. Stanley. It was inspired
by an observation of Steven Sam that if $\pi$ is 231-avoiding, then
$\Lambda_\pi$ appears to be rank-symmetric and rank-unimodal. These
two properties are simple consequences of Theorem \ref{formula}. (See
Corollary \ref{sym and unim}.) Figure~\ref{fig1} shows the Hasse
diagram of $\mathfrak{S}_4$. If for instance $\pi=4132$ (which is
separable), then
$F(\Lambda_\pi,q)=1+2q+2q^2+2q^3+q^4$ and $F(V_\pi,q)=1+q+q^2$. Then
multiplying $F(\Lambda_\pi,q)$ and $F(V_\pi,q)$ gives us $[4]!$.
We also give a convenient method to find an explicit formula for
$F(\Lambda_\pi,q)$ and $F(V_\pi,q)$. In fact, when $\pi=a_1a_2\cdots
a_n\in\mathfrak{S}_n$ is 231-avoiding, meaning that there do not exist $i<j<k$
with $a_{k}<a_{i}<a_{j}$, the explicit formula for
$F(\Lambda_\pi,q)$ is given by
\begin{enumerate}q F(\Lambda_\pi,q) = \prod_{i=1}^n [c_i], \end{enumerate}q
where $a_{c_i+i}$ is
the first element to the right of $a_i$ in $\pi$ satisfying $a_{c_i+i}
> a_i$, setting $a_{n+1}=\infty$.
$$\xymatrix@R=18pt@C=-5pt@M=0.5pt@H=1pt@W=1pt{
& & & & & 4321 \ar@{-}[dll] \ar@{-}[d] \ar@{-}[drr] & & & & &\\
& & & 4312 \ar@{-}[dll] \ar@{-}[drr] & & 4231 \ar@{-}[dll] \ar@{-}[drr] & & 3421 \ar@{-}[dll] \ar@{-}[drr] & & & \\
& 4132 \ar@{-}[dl] \ar@{-}[dr] & & 4213 \ar@{-}[dl] \ar@{-}[dr] & & 3412 \ar@{-}[dr] & & 2431 \ar@{-}[dlll] \ar@{-}[dr] & & 3241\ar@{-}[dl] \ar@{-}[dr] & \\
1432 \ar@{-}[dr] \ar@{-}[drrr] & & 4123 \ar@{-}[dl] & & 2413 \ar@{-}[dr] & & 3142 \ar@{-}[dlll] \ar@{-}[dr] & & 3214 \ar@{-}[dl] \ar@{-}[dr] & & 2341 \ar@{-}[dl] \\
& 1423 \ar@{-}[drr] & & 1342 \ar@{-}[drr] & & 2143 \ar@{-}[dll] \ar@{-}[drr] & & 3124 \ar@{-}[dll] & & 2314\ar@{-}[dll] & \\
& & & 1243 \ar@{-}[drr] & & 1324 \ar@{-}[d] & & 2134 \ar@{-}[dll] & & & \\
& & & & & 1234 & & & & &\\ }\label{fig1}$$
\begin{enumerate}gin{center}{Figure 1. The graded poset $\mathfrak{S}_4$ under
weak order} \end{center}
The \emph{inversion poset} $P_\pi$ of $\pi = a_1a_2\cdots a_n\in \mathfrak{S}_n$
has the relations $a_i<a_j$ in $P$ if $i<j$ and $a_i<a_j$ in
$\mathbb{Z}$. Figure 2 is the diagram of the inversion posets of
the permutations $34125$ and $31425$.
$$\xymatrix@R=18pt@C=8pt@M=3pt@H=1pt@W=1pt{
& & & 5 \ar@{-}[dl]\ar@{-}[dll] & & & & & 5 \ar@{-}[dl]\ar@{-}[dll]\\
& 4 \ar@{-}[dl] & 2 \ar@{-}[dl] & & & & 4 \ar@{-}[d] \ar@{-}[dl] & 2 \ar@{-}[dl] & \\
3 & 1 & & & & 3 & 1 & & }$$
\label{fig2}
\begin{enumerate}gin{center}
{Figure 2. The inversion posets of 34125 (left) and 31425 (right)}
\end{center}
Let $P$ and $Q$ be posets on disjoint sets. The \emph{disjoint union}
$P+Q$ is the poset on the union $P\cup Q$ such that $s\leq t$ in $P+Q$
if either $s,t\in P$ and $s\leq t$ in $P$, or $s,t\in Q$ and $s\leq t$
in $Q$. The \emph{ordinal sum} $P\oplus Q$ is the poset on the union
$P\cup Q$ such that $s\leq t$ in $P\oplus Q$ if either $s,t\in P$ and
$s\leq t$ in $P$, or $s,t\in Q$ and $s\leq t$ in $Q$, or $s\in P$ and
$t\in Q$.
The following lemma is easy to prove, so we omit the proof here.
\begin{enumerate}gin{lemma} \label{sum}
Let $\pi \in \mathfrak{S}_n$ with $\pi = \pi_A \pi_B$, where $\pi_A$ is a
permutation of size $m$ and $\pi_B$ is a permutation of size $n-m$ for
some $m < n$. Then
\begin{enumerate}gin{itemize}
\item $P_\pi = P_{\pi_A} + P_{\pi_B}$ if and only if $\pi_B$ is a
permutation of the letters $\{1,2,\dots, m\}$ and $\pi_A$ is a
permutation of the letters $\{m+1, m+2, \dots, n\}$.
\item $P_\pi = P_{\pi_A} \oplus P_{\pi_B}$ if and only if $\pi_A$ is a
permutation of the letters $\{1,2,\dots, m\}$ and $\pi_B$ is a
permutation of the letters $\{m+1, m+2, \dots, n\}$.
\end{itemize}
\end{lemma}
A \emph{linear extension} of a poset $P$ on the set $\{1,2,\dots,n\}$
is a permutation $\pi=a_1\cdots a_n\in\mathfrak{S}_n$ such that if $i<j$ in $P$,
then $i$ precedes $j$ in $\pi$.
We use $\mathcal{L}(P)$ to denote the set of linear extensions of
$P$. Since a linear extension $\pi$ of a poset $P$ on $\{1,\dots,n\}$
has been defined as a permutation of $\{1,\dots,n\}$, it has length
$\ell(\pi)$ as defined above. We define
$$ F(\mathcal{L}(P), q) = \sum_{\pi \in
\mathcal{L}(P)}q^{\ell(\pi)}. $$
We have the following rules for the operation on $F(\mathcal{L}(P), q).$
\begin{enumerate}gin{lemma}\label{op}
Let $P$ and $Q$ be two posets, where $P$ is on $\{1,2,\dots,
m\}$ and $Q$ is on $\{m+1, \dots, m+n\}$. Then
\begin{enumerate}q F(\mathcal{L}(P \oplus Q), q) = F(\mathcal{L}(P,
q))F(\mathcal{L}(Q, q) ) \label{opsum},\end{enumerate}q
\begin{enumerate}q F(\mathcal{L}(P + Q), q) = F(\mathcal{L}(P, q))F(\mathcal{L}(Q,
q)) \begin{enumerate}gin{bmatrix}m+n\\ m \end{bmatrix}\label{opunion}, \end{enumerate}q
where $\begin{enumerate}gin{bmatrix}m+n\\ m \end{bmatrix} = \dfrac{[m+n]!}{[m]![n]!}.$
\end{lemma}
The proof of (\ref{opsum}) is immediate by considering the definition
of ordinal sum and counting the number of inversions. The proof of
(\ref{opunion}) follows from the theory of $P$-partitions,
a straightforward extension of the second proof of Proposition 1.3.17
of \cite{EC1}.
A \emph{reduced decomposition} of a permutation $\pi\in\mathfrak{S}_n$ is a
sequence $(i_1,i_2,\dots,i_\ell)$ such that $\pi =s_{i_1}s_{i_2}\cdots
s_{i_\ell}$ and $\ell$ is minimal, viz., $\ell=\ell(\pi)$. If
$\pi=\pi_0\lessdot \pi_1\lessdot\cdots\lessdot \pi_m= \sigma$ is a saturated
chain $C$ from $\pi$ to $\sigma$, where $\pi_j=\pi_{j-1}s_{i_j}$, then
$r(C):=(i_1,\dots,i_\ell)$ is a reduced decomposition of
$\pi^{-1}\sigma$.
Write $R(\pi)$ for the set of reduced decompositions of $\pi$. Thus
the map $C\mapsto r(C)$ is a bijection between saturated chains from
$\mathrm{id}$ to $\pi$ and reduced decompositions of $\pi$.
With the definitions above, we proceed to the proofs of the main
theorem and the explicit formula for $F(\Lambda_\pi,q)$.
\section{Preliminary Results}
The following lemma states a property of separable permutations
which is of great importance to our proof of the main theorem.
\begin{enumerate}gin{lemma}\label{separable}
If $n>1$ and $\pi=a_1a_2\cdots a_n\in \mathfrak{S}_n$ is a separable permutation,
then we can write $\pi=\pi_A\pi_B$ (concatenation of words), where
$\pi_A$ and $\pi_B$ are both separable permutations satisfying one of
the two following properties:
\begin{enumerate}gin{itemize}
\item$\pi_A$ is a permutation of $1,2,\dots, m$ and $\pi_B$ is of
$m+1,\dots, n$ for some $m$ with $1 \leq m < n$;
\item$\pi_A$ is a permutation of $m+1,\dots, n$ and $\pi_B$ is of
$1,2,\dots, m$ for some $m$ with $1 \leq m < n$.
\end{itemize}
\end{lemma}
Lemma \ref{separable} is well-known and easy to prove; thus we omit
the proof here.
The following lemma is an immediate consequence of Lemma \ref{separable}
\begin{enumerate}gin{corollary} \label{cor:ppi}
If $n>1$ and $\pi=a_1a_2\cdots a_n\in \mathfrak{S}_n$ is a separable permutation,
then there exist two disjoint nonempty posets $P_{\pi_A}, P_{\pi_B}$
such that $P_\pi = P_{\pi_A} + P_{\pi_B}$ or $P_\pi = P_{\pi_A} \oplus
P_{\pi_B}$.
\end{corollary}
The following lemma is a special case of a result of Bj\"orner and
Wachs \cite[Thm.~6.8]{bj-wa}.
\begin{enumerate}gin{lemma} \label{FF}
Let $\pi$ be any permutation in $\mathfrak{S}_n$, then $F(\mathcal{L}(P_\pi), q) = F(\Lambda_\pi, q)$.
\end{lemma}
\vskip 0.2cm
Now we arrive at one of the main preliminary results of this section.
\begin{enumerate}gin{proposition} \label{below}
If $\pi =a_1a_2\cdots a_n \in \mathfrak{S}_n$ is a separable
permutation, then the following hold:
\paragraph{(i)} When $a_1 < a_n$, we can write $\pi = \pi_A\pi_B$
where $\pi_A$ is a permutation of size $m$ for some $m$ with $1 \leq m
< n$, and
\begin{enumerate}q F(\Lambda_\pi, q) = F(\Lambda_{\pi_A}, q) \cdot F(\Lambda_{\pi_B}, q) \label{belowsb}. \end{enumerate}q
\paragraph{(ii)} When $a_1 > a_n$, we can write $\pi = \pi_A
\pi_B$, where $\pi_A$ is a permutation of size $m$ for some $m$ with
$1 \leq m < n$, and \begin{enumerate}q F(\Lambda_\pi, q) = \begin{enumerate}gin{bmatrix}n\\ m
\end{bmatrix} F(\Lambda_{\pi_A}, q) \cdot F(\Lambda_{\pi_B},
q). \label{belowbs} \end{enumerate}q
\end{proposition}
\begin{enumerate}gin{proof}
Let $P$ be the inversion poset of $\pi$, $P_{A}$ be the inversion
poset of $\pi_A$, and $P_{B}$ be the inversion poset of $\pi_B$.
When $a_1 < a_n$, it follows from Lemma \ref{separable} that we can
write $\pi = \pi_A \pi_B$, where $\pi_A$ is a permutation of $\{1,2,
\dots, m\}$ and $\pi_B$ is a permutation of $\{m+1, m+2, \dots,
n\}$. By Lemma \ref{sum}, we have $P = P_{A} \oplus P_{B}$. It
follows from Lemma \ref{op} that $$F(\mathcal{L}(P_{A} \oplus
P_{B}), q) = F(\mathcal{L}(P_{A}), q)F(\mathcal{L}(P_{B}),
q). $$
Since $\pi, \pi_A,\pi_B$ are all separable permutations, by Lemma
\ref{FF}, we have \begin{enumerate}gin{eqnarray*}
F(\Lambda_\pi, q) = F(\mathcal{L}(P, q)) &=& F(\mathcal{L}(P_{A}
\oplus P_{B}), q) \\ &= &F(\mathcal{L}(P_{A}),
q)F(\mathcal{L}(P_{B}), q)\\ &= &F(\Lambda_{\pi_A}, q)
F(\Lambda_{\pi_B}, q). \end{eqnarray*}
The proof of (ii) is similar.
\end{proof}
\begin{enumerate}gin{proposition}\label{above}
If $\pi =a_1a_2\cdots a_n \in \mathfrak{S}_n$ is a separable
permutation, then the following hold:
\paragraph{(i)} If $a_1 < a_n $, then we can write $\pi = \pi_A\pi_B$
where $\pi_A$ is a permutation of size $m$ for some $m$ with $1 \leq m
< n$, and
\begin{enumerate}q F(V_{\pi}, q) =\begin{enumerate}gin{bmatrix}n\\ m \end{bmatrix} F(V_{\pi_A},
q) \cdot F(V_{\pi_B}, q) \label{abovesb}.\end{enumerate}q
\paragraph{(ii)} If $a_1 > a_n$, then we can write $\pi = \pi_A
\pi_B$, where $\pi_A$ is a permutation of size $m$ for some $m$ with
$1 \leq m < n$, and \begin{enumerate}q F(V_\pi, q) = F(V_{\pi_A}, q) \cdot
F(V_{\pi_B}, q) \label{abovebs}.\end{enumerate}q
\end{proposition}
The proof is similar to that of Proposition \ref{below} by using the
\emph{complement} $\pi^c$ of a permutation $\pi = a_1a_2\cdots a_n \in
\mathfrak{S}_n$ defined by $\pi^c = a_1'a_2'\cdots a_n'$ where $a_i' = n+1-a_i$
for all $1 \leq i \leq n$.
A standard property of $\pi^c$ and weak order is stated in the following lemma, and we omit the easy proof here.
\begin{enumerate}gin{lemma} \label{up bij}
The rank relation between a permutation and its complement is given by
$$\ell(\pi^c) = \binom n2 - \ell(\pi).$$
In fact, there exists a bijection $\mu: [\pi, w_0] \to [\mathrm{id},
\pi^c]$ defined by $\mu(w) = w^c$ for all $w \in [\pi, w_0]$.
\end{lemma}
\begin{enumerate}gin{proof}[Proof of Proposition \ref{above}]
For any $\omega \in [\pi, w_0]$, by Lemma \ref{up bij} and the fact
that
$$\ell(w^c) = \binom n2 - \ell(w) = \ell(\pi^{-1}w_0) -
\ell(\pi^{-1}w),$$
any $q^{\ell(\pi^{-1}\omega)}$ in $F(V_\pi, q)$
corresponds uniquely to a term
$q^{\ell(\pi^{-1}w_0)-{\ell(\pi^{-1}\omega)}}$ in $F(\Lambda_{\pi^c},
q)$. Thus
\begin{enumerate}q q^{\ell(\pi^{-1}w_0)} F(V_\pi, q^{-1}) = F(\Lambda_{\pi^c}, q).
\label{UD} \end{enumerate}q
We now consider $\pi^c$ in the two cases in Proposition \ref{above}.
\textbf{(i)} When $a_1 < a_n $, by equation (\ref{belowbs}) we have
\begin{enumerate}q F(\Lambda_{\pi^c}, q) = \begin{enumerate}gin{bmatrix}n\\ m \end{bmatrix}
F(\Lambda_{{\pi_A}^c}, q) \cdot F(\Lambda_{{\pi_B}^c},
q)\label{1}.\end{enumerate}q
Combining (\ref{UD}) and (\ref{1}) gives us
\begin{enumerate}q q^{\ell(\pi^{-1}w_0)} F(V_\pi, q^{-1}) = \begin{enumerate}gin{bmatrix}n\\ m
\end{bmatrix}F(V_{{\pi_A}^c}, q^{-1}) F(V_{{\pi_B}^c},
q^{-1}) \cdot q^{\binom m2 - \ell(\pi_A)}\cdot q^{\binom{n-m}{2}
- \ell(\pi_B)}\label{2}. \end{enumerate}q
Since the letters in $\pi_A$ are all smaller than the letters in
$\pi_B$, we have
$\ell(\pi) = \ell(\pi_A) + \ell(\pi_B)$. Substituting $q^{-1}$ for
$q$ in (\ref{2}), which converts $\begin{enumerate}gin{bmatrix}n\\ m \end{bmatrix}$
into $q^{\binom n2-\binom m2 -\binom{n-m}{2}}\begin{enumerate}gin{bmatrix}n\\ m
\end{bmatrix}$, completes the proof of
(\ref{abovesb}).
\textbf{(ii)} Since all the letters in $\pi_A$ are greater than the
letters in $\pi_B$, we have $$\ell(\pi) = (n-m)m + \ell(\pi_A) +
\ell(\pi_B).$$ The rest of (\ref{abovebs}) can be proved analogously.
\end{proof}
\section{Main Results}
\subsection{Main Theorem}
\begin{enumerate}gin{theorem} \label{main theorem}
Let $\pi\in \mathfrak{S}_n$, $\Lambda_\pi = [\mathrm{id}, \pi]$, and
$V_\pi = [\pi, w_0]$. The following equation holds for any separable
permutation $\pi$:
\begin{enumerate}q F(\Lambda_\pi, q)F(V_\pi. q) = F(\mathfrak{S}_n, q) = [n]!.\end{enumerate}q
\end{theorem}
\begin{enumerate}gin{proof}
When $n=2$, it is easy to verify that the expression holds. Suppose
the statement holds when $k < n$ for some $n \geq 3$; we want to show
that when $k = n$, the statement still holds.
Let $\pi_A$ and $\pi_B$ be the same as before.
When $a_1> a_n$ we have by (\ref{belowbs}) and (\ref{abovebs}) that
$$
F(\Lambda_\pi, q)F(V_\pi. q) = \begin{enumerate}gin{bmatrix}n\\ m
\end{bmatrix}F(\Lambda_{\pi_A}, q)F(\Lambda_{\pi_A}, q)\cdot
F(V_{\pi_A}, q)F(V_{\pi_B}, q).$$ Thus by the inductive hypothesis, we
have $$ F(\Lambda_\pi, q)F(V_\pi, q) = \begin{enumerate}gin{bmatrix}n\\ m
\end{bmatrix}[m]! [n-m]! = [n]!.$$
The proof for $a_1< a_n$ is similar.
\end{proof}
\subsection{A Bijection $\varphi\colon \Lambda_w\times V_w\to \mathfrak{S}_n$}
We can also give a bijective proof of Theorem \ref{main
theorem}.
\begin{enumerate}gin{theorem} \label{bijj}
Let $\pi = a_1a_2 \cdots a_n \in \mathfrak{S}_n$ be a separable
permutation. The map
$$ \phi: \Lambda_\pi \times V_\pi \rightarrow
\mathfrak{S}_n $$ defined by $\phi (u,v)= u^{-1}v$, where $u\leq \pi$ and $v\geq
\pi$, is a bijection.
\end{theorem}
Since $(u^{-1}v)^{-1} = v^{-1}u$, it is a direct consequence of
Theorem \ref{bijj} that the map $$ \phi': \Lambda_\pi \times V_\pi
\rightarrow \mathfrak{S}_n $$ defined by $\phi'(u,v) = v^{-1}u$ for $u\leq \pi$
and $v\geq \pi$ is also a bijection.
We use the following lemma to prove this theorem.
\begin{enumerate}gin{lemma} \label{no interaction}
{If $\pi = a_1a_2\cdots a_n \in\mathfrak{S}_n$ is a separable permutation with
$a_1<a_n$, and $(i_1,\dots,i_\ell)\in R(\pi)$, then there exists an
integer $m$ with $1 \leq m < n$ such
that none of the simple transpositions $s_{i_j}$ transposes an element
in $A_\pi = \{1,2,\dots, m\}$ with an element in $B_\pi = \{m+1,
\dots, n\}$. In other words, there is no interaction between the sets
$A_\pi$ and $B_\pi$.}
\end{lemma}
The proof of Lemma \ref{no interaction} can be achieved easily from
the definition of weak order.
\begin{enumerate}gin{proof}
If the lemma does not hold, then in the sequence of all simple
transpositions there exists a nonempty subsequence consisting of
simple transpositions between the letters in $A_\pi$ and the letters
in $B_\pi$. Suppose the last transposition in this subsequence is
between $a \in A_\pi$ and $b \in B_\pi$.
From Proposition \ref{separable} we know that $a$ is to the left of
$b$. Since
$a < b$, by the definition of weak order the permutation after the
transposition is covered by the permutation before swapping $a$ and $b$, which leads to a contradiction.
\end{proof}
\begin{enumerate}gin{proof}[Proof of Theorem \ref{bijj}]
When $a_1 < a_n$, by Lemma \ref{separable} we can write $\pi = \pi_A
\pi_B$ where $\pi_A$ is a separable permutation of $\{1,2,\dots, m\}$
for some $m > 0$.
For the injectivity part, we want to show that there do not exist two
different pairs $(u_1, v_1), (u_2, v_2) \in \Lambda_\pi \times V_\pi$
such that $u_1^{-1}v_1 =
u_2^{-1}v_2$. It is sufficient to show that $u^{-1}\pi \neq \pi^{-1}v$
for all $(u, v) \in \Lambda_\pi \times V_\pi$, and $u, v \neq \pi$.
Let $r_1(C_\Lambda) = (i_1, i_2, \dots, i_{k_1})$ be the reduced
decomposition of $u^{-1}\pi$ and $r_2(C_V) = (j_1, j_2, \dots,
j_{k_2})$ be the reduced decomposition of $\pi^{-1}v$. We need only
consider the situation when $k_1 = k_2$.
Since $\pi_A$ is a permutation
of $\{1,2,\dots, m\}$ and $\pi_B$ is a permutation of $\{m+1, \dots,
n\}$, by Lemma \ref{no interaction} we can write $u = u_Au_B$ where
$u_A$ is a permutation of $\{1,2,\dots, m\}$ and $u_B$ is a
permutation of $\{m+1, \dots, n\}$. Furthermore, we can also write the reduced
decomposition of $u^{-1}\pi$ as a concatenation of the reduced
decompositions of
$u_A^{-1}\pi_A$ and $u_B^{-1}\pi_B$. Accordingly, if there exists $v
\geq \pi$ such that $\pi v = u^{-1}\pi $, then we can write $v$ as a concatenation of two subpermutations $
v_A, v_B$, and the reduced decomposition for $\pi^{-1}u$ is a concatenation of
$\pi_A^{-1}v_A$ and $\pi_B^{-1}v_B$. Hence in order to
have $u^{-1}\pi = \pi v$, we must have
$$u_A^{-1}\pi_A
=\pi_A^{-1}v_A \text{
and } u_B^{-1}\pi_B=\pi_B^{-1}v_B.$$
Thus we need only consider the case in which the size of the
permutation is less than $n$.
For the surjectivity part, we want to show that, for each permutation
$w \in \mathfrak{S}_n$, there exists $(u,v)\in \Lambda_\pi \times V_\pi$ such
that $u^{-1}v = w$.
Let $w\in \mathfrak{S}_n$ be as in Proposition~\ref{above}(ii).
Let $w_1$ be the sub-permutation of $w$ which consists of the letters
$\{1,2,\dots, m\}$, and let $w_2$ be the sub-permutation of $w$ which
consists of $\{m+1,m+2,\dots, n\}$.
By the inductive hypothesis, there exist $(u_1, v_1) \in
\Lambda_{\pi_A} \times V_{\pi_A}$ and $(u_2, v_2) \in
\Lambda_{\pi_B} \times V_{\pi_B}$ such that $$u_1^{-1}v_1 = w_1 \text{
and } u_2^{-1}v_2= w_2.$$ It follows that $$(u_1u_2)^{-1}(v_1v_2) = w_1w_2,$$
and $(u_1u_2, v_1v_2) \in \Lambda_{\pi} \times V_{\pi}.$
We now show that we can find $v' \geq v_1v_2$ such that
$$(v_1v_2)^{-1}v' = (w_1w_2)^{-1}w.$$
Then it follows that for any arbitrary $w$, there exists a $(u_1u_2,
v') \in \Lambda_{\pi} \times V_{\pi}$ and $$(u_1u_2)^{-1}v'
=(u_1u_2)^{-1}(v_1v_2)(v_1v_2)^{-1}v' = (w_1w_2)(w_1w_2)^{-1}w = w.$$
We will show an explicit way to find $v'$.
Let $A_1 < A_2 < \cdots <A_m$ be the positions in $\pi$ that are
occupied by the letters $\{1,2,\dots, m\}$. We start by shifting the letters
$\{1,2,\dots, m\}$ in both $v_1v_2$ and $w_1w_2$ to the positions
indexed by $A_1, A_2, \dots, A_m$. That is, we move the letters at the
$m$th position in $v_1v_2$ and $w_1w_2$ to the position indexed by
$A_m$, and then move the letter at the $(m-1)$-st position to the
position indexed by $A_{m-1}$, and so on. Finally, we move the letter
at the first position to the position indexed by $A_1$.
Recall that $v_1$ and $w_1$ are permutations of $\{1,2,\dots, m\}$
and $v_2$ and $w_2$ are permutations of $\{m+1,\dots, n\}$.
Since $A_1 < A_2 < \cdots < A_m,$ it is easy to show that during the
shifting process, all the transpositions are between a letter in
$\{1, 2, \dots, m\}$ and a letter in $\{m+1, m+2, \dots, n\}$, and
that after each transposition, the length of the permutation increases
by $1$. This process thus turns $w_1w_2$ into $w$ and $v_1v_2$ into
another permutation, which we set to be $v'$. Accordingly, by the inductive
hypothesis and this shifting process, we have an explicit way to find
$v'$ such that $$(u_1u_2)^{-1}v' = w.$$
When $a_1 > a_n$, we use the complement of the permutation, and the
rest of the proof is similar.
\end{proof}
\subsection{Explicit Formulas for $F(\Lambda_\pi, q)$ and\ $F(V_\pi,q)$}
Based on Proposition \ref{below} and Proposition \ref{above}, we
introduce a convenient method to find the explicit formulas for
$F(\Lambda_\pi, q)$ and $F(V_\pi,q)$.
The most convenient way is to use a \emph{separating tree}. We
define it recursively as follows.
Let $\pi=a_1a_2\cdots a_n$ be a separable permutation.
When $n=2$, its separating tree $T_\pi$ is an ordered binary tree with
the left leaf $a_1$ and right leaf $a_2$.
When $n > 2$, by Lemma \ref{separable} we can write $\pi = \pi_A
\pi_B$ where $\pi_A$ and $\pi_B$ are separable permutations with size
strictly smaller than $n$. Then $T_\pi$ is an ordered binary tree,
with the subtree rooted at the left child of the root, being
$T_{\pi_A}$, and the subtree rooted at the right child of the root,
being $T_{\pi_B}$.
Since there might be more than one way to write $\pi = \pi_A\pi_B$, a
separable permutation can have more than one separating tree. Also,
only the separable permutations have separating trees.
The definition of the separating tree $T_\pi$ gives the following
lemma, which is easy to prove.
\begin{enumerate}gin{lemma}
For any node in $T_\pi$, the leaves of the subtree rooted at that node
form a subrange, a set of consecutive integers.
\end{lemma}
This lemma allows us to classify the nodes in $T_\pi$ into two
categories. A node is \emph{negative} if the subrange of
its left child is greater than that of its right child. \emph{Positive}
node is defined analogously.
Figure 3 shows a separating tree for $4231$, which has two negative nodes and one positive node, as labeled in the figure.
\begin{enumerate}gin{center}
\scalebox{0.75}{
$$\xymatrix@!C@!R@R=12pt@C=-26pt@M=0pt@H=0pt@W=0pt{
& & & \txt<4pc>{Negative\\Node} \ar@{-}[dddlll] \ar@{-}[dr] & & & \\
& && & \txt<4pc>{Negative\\Node} \ar@{-}[dl] \ar@{-}[ddrr] & &\\
& & & \txt<4pc>{Positive\\Node} \ar@{-}[dl] \ar@{-}[dr] & & & \\
4 & & 2 & & 3 & & 1 \\
}\label{treeS}$$}
\end{center}
\begin{enumerate}gin{center}{ Figure 3. The separating tree for 4231}\end{center}
\begin{enumerate}gin{theorem}\label{formula}
Let $S^{-}(\pi) = \{$all negative nodes $V_i$ in $T_\pi$ whose
parents are not negative$\}$ and $S^{+}(\pi)=\{$all positive
nodes $V_j$ in $T_\pi$ whose parents are not positive$\}$. Let
$V_0$ be the root of the tree, and $V_0$ is not in either $S^{-}(\pi)$ nor $S^{+}(\pi)$.
Let $N(V_k)$ denote the number of leaves in the subtree rooted at
$V_k$. In particular, we define $\prod_{V_i \in \varnothing}{
[N(V_i)]!} = 1$. Then
\begin{enumerate}gin{eqnarray} \label{formulabelow}
F(\Lambda_w,q) =\left\{\begin{enumerate}gin{array}{lr}
\dfrac{\prod_{V_i \in S^{-}(\pi)}{[N(V_i)]! }} { \prod_{V_j \in
S^{+}(\pi)}{ [N(V_j)]!}}, &V_0\text{ is a positive node;} \\[2em]
\dfrac{\prod_{V_i \in S^{-}(\pi)}{[N(V_i)]! }} { \prod_{V_j \in
S^{+}(\pi)}{ [N(V_j)]!}}[N(V_0)]!, &V_0\text{ is a negative
node}.
\end{array}\right.
\end{eqnarray}
\begin{enumerate}gin{eqnarray} \label{formulaabove}
F(V_w,q) =\left\{\begin{enumerate}gin{array}{lr}
\dfrac{\prod_{V_i \in S^{+}(\pi)}{[N(V_i)]! }} { \prod_{V_j \in
S^{-}(\pi)}{ [N(V_j)]!}}[N(V_0)]!, &V_0\text{ is a positive
node;}\\[2em]
\dfrac{\prod_{V_i \in S^{+}(\pi)}{[N(V_i)]! }} { \prod_{V_j \in
S^{-}(\pi)}{ [N(V_j)]!}}, &V_0\text{ is a negative node}.
\end{array}\right.
\end{eqnarray}
\end{theorem}
\begin{enumerate}gin{example} Let $w = 4231$. Its separating tree is shown in Figure
3. It has one negative node with no parent, one negative node with
a negative parent node, and one positive node with a negative parent
node. Thus $F(\Lambda_w, q) = [4]! / [2]!$, and $F(V_w, q)
= [2]!/[1]!$.
\end{example}
\begin{enumerate}gin{proof}[Proof of Theorem \ref{formula}]
Let $\pi = a_1a_2\cdots a_n$ be a separable permutation. We can use
induction to prove Theorem \ref{formula}.
By the definition of $N(V)$, we have $N(V_0) = n$.
When $a_1 < a_n$, we write $\pi =\pi_A\pi_B$ where $\pi_A$ is a
permutation of $\{1, 2,\dots, m\}$. The root $V$ of $T_\pi$ has two
children with the left child $V_L$ having leaves $\{1,2,\dots, m\}$
and the right child $V_R$ having leaves $\{m+1, m+2, \dots, n\}$. Thus
$V$ is a positive node. Let $T_L$ be the subtree rooted at $V_L$ and
$T_R$ be the subtree rooted at $V_R$. Applying formula
(\ref{formulabelow}) to $\pi_A$ and $\pi_B$, together with
(\ref{belowsb}) and (\ref{abovesb}), we can prove (\ref{formulabelow})
and (\ref{formulaabove}) by induction.
When $a_1 > a_n$, the root of $T_\pi$ is a negative node. The rest of
the proof is similar to the case above when $a_1 < a_n$.
\end{proof}
More specifically, when the permutation $\pi = a_1a_2\cdots a_n$ is
$231$-avoiding (a 231-avoiding permutation requires more restrictions
than a general separable permutation), a more direct formula for
$F(\Lambda_\pi,q)$ can be given.
\begin{enumerate}gin{corollary}[explicit formula for $F(\Lambda_\pi,q)$ for a
231-avoiding permutation]
\label{explicit 231} Let $\pi = a_1 a_2 \cdots a_n$ be $231$-avoiding,
and $a_{c_i+i}$ be the first element to the right of $a_i$ in $\pi$
satisfying $a_{c_i+i}>a_i$, setting $a_{n+1}=\infty$. Then
\[ F(\Lambda_\pi,q) = \prod_{i=1}^n [c_i]. \label{eq:rgf}
\]\end{corollary}
Before proving this proposition, we give an example to explain the
notation in the formula.
\begin{enumerate}gin{example}
Let $\pi = a_1a_2\cdots a_6=142365$. We set $a_7 = \infty$. For $a_1
=1$, letter $4$ is the first one greater than $1$ and to the right of $a_1$;
the distance between these two integers, $c_1$, is thus
$2-1=1$. Similarly, $c_2 = 5-2=3, c_3 = 4-3=1, c_4 = 5-4=1, c_5=7-5=2,
c_6=7-6=1$. Thus the generating function is $F(\Lambda_{142365}, q) =
\prod_{i=1}^{6}{[c_{i}] } = [1][3][1][1][2][1] =(q^2+q+1)(q+1)$.
\end{example}
\begin{enumerate}gin{proof}
By Lemma \ref{no interaction} we know that when $\pi$ is
$231$-avoiding, either $\pi$ has the greatest letter $n$ at its first
position, or $n$ is at the $(m+1)$-st position with $m>0$. Thus we
can write $\pi = \pi_A\pi_B$ where $\pi_A$ is a 231-avoiding
permutation of $\{1,\dots, m\}$ and $\pi_B$ is a 231-avoiding
permutation of $\{m+1,\dots, n\}$. Then we can construct the
separating tree by repeatedly applying the following steps.
For a separating tree with root $V_0$, we first decide its left child
$V_L$ and right child $V_R$ by identifying the position of the
greatest letter in $\pi$, i.e., finding $m$ such that $a_{m+1} =
n$.
When $m=0$, the subtree rooted at $V_L$ has only one leaf $a_1 = n$,
while the subtree rooted at $V_R$ is the separating tree of the
permutation $a_2a_3\cdots a_n$, which we will construct similarly.
When $m > 0$, the subtree rooted at $V_L$ is the separating tree for
the permutation $a_1a_2\cdots a_m$, while the subtree rooted at $V_R$
is the separating tree for the permutation $a_{m+1}\cdots a_n$. We
then construct these two separating tree similarly.
In the first case, $V_0$ is a negative node. We already know that $c_1=n$ and
$$F(\Lambda_{\pi}, q) = [n]\cdot F(\Lambda_{a_2a_3\cdots a_n},q),$$ and
$c_1 = n.$
In the second case, $V_0$ is a positive node. We have $c_1 = n$
and $$F(\Lambda_{\pi}, q) = F(\Lambda_{a_1a_2\cdots a_m},q)
F(\Lambda_{a_{m+1}a_{m+2}\cdots a_n},q).$$
We also know that, for a letter $a$ in $\{1,2,\dots, m\}$, the
distance between $a$ and the first letter greater than $a$ and to its
right is the same in both $\pi$ and $a_1 a_2\cdots a_m$.
The rest of the proof can be completed by induction.
\end{proof}
\vskip 0.2cm
Since we know that $F(\Lambda_{\pi}, q)F(V_{\pi}, q) = [n]!$, as well
as the explicit formula for $F(\Lambda_{\pi}, q)$, we can also obtain
an explicit formula for $F(V_{\pi}, q)$.
By symmetry, we can obtain analogous explicit formulas when the
permutation avoids any of the patterns 132, 231, 312, or 213.
The following two lemmas are standard results about unimodality; see
for instance \cite{rs:unim}.
\begin{enumerate}gin{lemma} \label{young}
The $q$-binomial coefficient $\begin{enumerate}gin{bmatrix} n\\ m \end{bmatrix}$ is
rank-unimodal and rank-symmetric.
\end{lemma}
\begin{enumerate}gin{lemma} \label{product of unimodal}
Let $F(q)$ and $G(q)$ be symmetric unimodal polynomials with
nonnegative real coefficients. Then $F(q)G(q)$ is also symmetric and
unimodal. \end{lemma}
Lemma \ref{young} and Lemma \ref{product of unimodal} imply the following corollary.
\begin{enumerate}gin{corollary} \label{sym and unim}
$F(\Lambda_\pi, q)$ and $F(V_\pi, q)$ are rank symmetric and
unimodal.
\end{corollary}
Theorem~\ref{formula} determines the number of elements of each rank
$k$ of the poset $\Lambda_\pi$ when $\pi$ is separable. We can
also determine the number of elements that cover $k$ elements.
A \emph{descent} of a permutation $\pi = a_1a_2\cdots a_n \in \mathfrak{S}_n$ is a
position $i$ with $1\leq i <n$, such that $a_i > a_{i+1}$. Let
$\mathrm{des}(\pi)$ be the number of descents of $\pi$.
It is easy to see that $\mathrm{des}(\pi)$ is equal to the number of
elements that $\pi$ covers in the weak order on $\mathfrak{S}_n$. If $P_\pi$ is the
inversion poset of $\pi$, then the enumeration of linear extensions of
$P_\pi$ by number of descents is the same as the enumeration of elements
of $\Lambda_\pi$ in weak order by number of covers. Let
$\Omega_P(m)$ denote the number of order-preserving maps $f\colon P\to
\{1,\dots,m\}$. Then we have the following theorem which relates
$\Omega_P(m)$ with the descent number. The proof can be found in
\cite[Thm.~4.5.14]{EC1}.
\begin{enumerate}gin{theorem} \label{des}
For any poset $P$ on $\{1,2,\dots,n\}$, we have
$$ \sum_{m\geq 1}\Omega_P(m)x^m
=\frac{\sum_{\pi\in\mathcal{L}(P)}
x^{\mathrm{des}(\pi)+1}}{(1-x)^{n+1}}. $$
\end{theorem}
Using the recursive structure of $P_\pi$ when $\pi$ is separable
(Corollary~\ref{cor:ppi}) we can give a recursive description of
$\Omega_{P_\pi}(m)$ and thus of the number of elements in
$\Lambda_\pi$ that cover $k$ elements. We do not enter into the
details here.
Our results suggest several open problems. For what permutations
$\pi\in\mathfrak{S}_n$ is the poset $\Lambda_\pi$ rank-symmetric? When is
$[n]!$ divisible by the rank generating function
$F(\Lambda_\pi,q)$? When is $F(\Lambda_\pi,q)$ a product of cyclotomic
polynomials? R. Stanley has verified that for $n\leq 8$, if
$\Lambda_\pi$ is rank-symmetric then $F(\Lambda_\pi,q)$ is a product
of cyclotomic polynomials, but $F(\Lambda_\pi,q)$ need not divide
$[n]!$. For instance, when $n=8$ there are 8558 separable
permutations, 10728 permutations $\pi$ for which $\Lambda_\pi$ is
rank-symmetric (and hence a product of cyclotomic polynomials), and
961 permutations $\pi$ for which $\Lambda_\pi$ is rank-symmetric but
$F(\Lambda_\pi,q)$ does not divide $[8]!$. A further problem is to
extend our work to the weak order of other Coxeter groups.
\begin{enumerate}gin{thebibliography}{9}
\bibitem{albert} M. H. Albert, Aspects of separability,
$\langle$\texttt{http://www.cs.otago.ac.nz/staffpriv/malbert/Talks/Sep.pdf}$\rangle$.
\bibitem{av-ne} D. Avis and M. Newborn, On pop-stacks in series,
\emph{Utilitas Math.}\ \textbf{19} (1981), 129--140.
\bibitem{bj-wa} A. Bj\"orner and M. Wachs, Permutation statistics and
linear extensions of posets, \emph{J. Combinatorial Theory, Ser.~A}\
\textbf{58} (1991), 85--114.
\bibitem{bona} M. Bona, \emph{Combinatorics of Permutations}, Chapman
Hall-CRC, Boca Raton, FL, 2004.
\bibitem{tree} P. Bose, J. Buss, A. Lubiw, Pattern matching for
permutations, \emph{Information Processing Letters},
\textbf{65} (1998), 277--283.
\bibitem{rs:unim} R. Stanley, Unimodal and log-concave sequences in
algebra, combinatorics, and eometry, in \emph{Graph Theory and Its
Applications: East and West}, Ann.\ New York Acad.\ Sci., vol.\
576, 1989, pp.\ 500--535.
\bibitem{EC1} R.Stanley, \emph{Enumerative Combinatorics},
vol.~I, Cambridge University Press, Cambridge/New York, 1997.
\end{thebibliography}
\end{document}
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भ्रष्टाचार, आतंकवाद, #ब्लैकमने, जाली नोटों के गोरखधंधे के खिलाफ निर्णायक लड़ाई का ऐलान करते हुए प्रधानमंत्री नरेन्द्र मोदी ने देश को आश्चर्य में डालते हुए
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किसानों को राहत देने में धन की कमी नहीं होगी- मुख्यमंत्री श्री शिवराज सिंह चौहान
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hindi
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ہٹس منز (Lyrinx) چُھ آوازٕہ ہُندLigments ننہ گژھنہ سيتی۔ يہٕ اشٲرہ ميلانٕز آواز ہُندCord چُھ بڑان۔
|
kashmiri
|
GAITHERSBURG, MD: Patton have announced general availability for two new members of the SmartNode family: the SmartNode 4120 Single/Dual-Port BRI VoIP gateway series and the SmartNode DTA Singe/Dual-Port BRI VoIP Digital Terminal Adapter series for ISDN BRI VoIP users and service providers.
Announced June 2011, the SN-DTA and SN4120 product lines are now in full production, as Patton ships hundreds of units on a monthly basis. The new SmartNode SKUs expand the range of NT/TE port combinations and add the following enhanced features: PSTN fallback, phantom power, and a high-precision clock that eliminates common interoperability problems with DECT, PBX, and FAX in VoIP implementations.
The new single-port TE gateway (SN4120/1BIS2V) and dual-port NT gateway (SN-DTA/2BIS4V) provide cost-effective solutions for specialized customer applications.
The dual-port SN-DTA/2BIS4VHP offers Patton’s high-precision clock, enabling first-line IP-telephony performance with reliable fax and modem services in often-troublesome ISDN and DECT environments.
The SN-DTA/2BIS2V offers a fallback port with built-in phantom power that delivers power to connected ISDN devices whenever the fallback port is not in use or if an existing fallback line is removed.
ISDN users are notoriously picky,” observed Bernhard Fluehmann, Patton’s Converged-IP Product Manager. “Yet SmartNode quality and reliability have earned great respect within the ISDN community. The new BRI solutions are very well received. We expect similar enthusiasm for the new digital-analog hybrid due out soon.
Patton is currently developing an 8-to-24 call VoIP router offering up to 8 BRIs, up to 8 analog FXS/FXO interfaces, and 4 Ethernet ports. The enterprise-class SmartNode is designed to IP-enable legacy ISDN-PBXs, analog phone/fax/answering machines, and/or analog audio-intercom systems with a single device.
Two weeks ago Patton unveiled the SmartNode 10200 carrier-grade VoIP media gateway series, supporting up to 2048 voice or fax calls over T1/E1, DS3 or STM-1/OC3 trunks with SS7 signaling. Last year, Patton expanded the SmartNode PRI gateway line by adding three new model series.
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english
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code
|
\begin{document}
\title{Middle-Four Maps and Net Categories}
\begin{abstract}
We briefly relate the existence of a middle-four interchange map in a category with two monoidal structures, to the standard Cockett and Seely
notion of a weakly distributive category.
I_\tensnd{abstract}
\section{Introduction}
\renewcommand{I_\parr}{I_\partialarr}
\renewcommand{I_\tens}{I_\tens}
From a middle-four map we shall obtain an example of a structure (on the category of $I_\parr$-bimodules) with two weak distributivity maps:
\begin{align*}
d^l &: A \tens (B \partialarr C) \longrightarrow (A \tens B) \partialarr C \\
d^r &: (B \partialarr C) \tens A \longrightarrow B \partialarr (C \tens A)
I_\tensnd{align*}
\noindent satisfying the eleven (planar) conditions (2)-(4) and (7)-(14) of \cite{CS}, but not necessarily \cite{CS} (1), (5), and (6). We call such a
structure a ``net'' category (see \cite{DP}) with the view that it is a proof-net category (\cite{DP}) without negation, or, equivalently,
a (planar) weakly distributive category \cite{CS} without the unit objects.
Moreover, we can talk about a middle-four category with negation, which satisfies four more suitable axioms (c.f. \cite{CS} \S4 (15)-(18)), so that,
on passing to the category of $I_\parr$-bimodules, we get a proof-net category with a middle-four map.
\section{Middle-Four Categories}
First, we consider a category $V$ with two monoidal structures $(\tens, I_\tens, a, l, r)$ and $(\partialarr, I_\parr)$, where we have assumed that
the second structure is strictly monoidal. Suppose this category is equipped with a natural middle-four transformation:
\[ m : (A \partialarr B) \tens (C \partialarr D) \longrightarrow (A \tens C) \partialarr (B \tens D) \]
and maps: \begin{align*}
\mu &: I_\parr \tens I_\parr \longrightarrow I_\parr \\
I_\tensta &: I_\tens \longrightarrow I_\parr
I_\tensnd{align*}
satisfying the following four axioms:
\subsection*{M1}
\[ (a \partialarr a)m(m \tens 1) = m(1 \tens m)a : ((U \partialarr V) \tens (W \partialarr X)) \tens (Y \partialarr Z) \longrightarrow (U \tens (W \tens Y)) \partialarr
(V \tens (X \tens Z)) \]
\subsection*{M2}
\[ (m \partialarr 1)m = (1 \partialarr m)m \] from \[ (U \partialarr V \partialarr W) \tens (X \partialarr Y \partialarr Z) \] to \[ (U \tens X) \partialarr (V \tens Y) \partialarr (W \partialarr Z) \]
and, if we put $I = I_\tens$ and $R = I_\parr$, we want $(R,\mu,I_\tensta)$ to be a $\tens$-monoid in $V$ such that both
\subsection*{M3}
\[ (1 \partialarr (\mu \tens 1))m(m \tens 1) = m \] from \[ ((A \partialarr R) \tens (B \partialarr R)) \tens (C \partialarr D) \] to \[ (( A \tens B) \tens C) \partialarr (R \tens D) \]
\subsection*{M4}
\[ ((1 \tens \mu) \partialarr 1)m(1 \tens m) = m \] from \[ (B \partialarr C) \tens ((R \partialarr A) \tens (R \partialarr B)) \] to
\[ (B \tens R) \partialarr ( C \tens (A \tens D)) \]
Moreover, a middle-four category with negation has maps $\gamma$ and $\tau$, where
\begin{align*}
\gamma &: A \tens {A^\bot} \longrightarrow R \\
\tau &: I \longrightarrow A^\bot \partialarr A
I_\tensnd{align*}
satisfying four axioms similar to:
\[ \bfig
\node a(-500,+500)[(A \partialarr R) \tens I]
\node b(+500,+500)[(A \partialarr R) \tens (A^\bot \partialarr A)]
\node c(-500,0)[A]
\node d(+500,0)[(A \tens A^\bot) \partialarr (R \tens A)]
\node e(-500,-500)[R \tens A]
\node f(+500,-500)[R \partialarr (R \tens A)]
\arrow|a|[a`b;1 \tens \tau]
\arrow|l|[a`c;r]
\arrow|l|[c`e;I_\tensta \tens 1]
\arrow|r|[b`d;m]
\arrow|r|[d`f;\gamma \partialarr 1]
\arrow|b|[f`e;=]
I_\tensfig \]
commutes (cf. \cite{CS} \S4).
\section{The Category of $R$-bimodules as a net category}
The category $V(R)$ of $R$-bimodules in $V$ has two induced (coherent) tensor products. These are:
\begin{itemize}
\item[(1)] $A \tens B$ with action: \begin{align*}
R \tens (A \tens B) \tens R &\oto{\sim} (R \tens A) \tens (B \tens R) \\
&\oto{\sim} (R \tens A \tens I) \tens (I \tens B \tens R) \\
&\oto{(1 \tens I_\tensta) \tens (I_\tensta \tens 1)} (R \tens A \tens R) \tens (R \tens B \tens R) \\
&\oto{\alpha \tens \beta} A \tens B
I_\tensnd{align*}
which is unitless, and
\item[(2)] $A \partialarr B$ with action
\[ R \tens (A \partialarr B) \tens R \oto{\mbox{\tiny``$m$''}} (R \tens A \tens R) \partialarr (R \tens B \tens R) \oto{\alpha \partialarr \beta} A \partialarr B \]
which has the unit $R$.
I_\tensnd{itemize}
Furthermore, it is easily checked that our basic transformation \[ m : (A \partialarr B) \tens (C \partialarr D) \longrightarrow (A \tens C) \partialarr (B \tens D) \]
becomes an $R$-bimodule homomorphism. Thus, on putting $B = R$ and $C = R$ respectively, and using the bimodule actions, we get
two natural transformations:
\begin{align*}
d^l &: A \tens (B \partialarr C) \longrightarrow (A \tens B) \partialarr C \\
d^r &: (B \partialarr C) \tens A \longrightarrow B \partialarr (C \tens A)
I_\tensnd{align*}
of $R$-bimodules.
We now check some \cite{CS} axioms on $d^l$ and $d^r$.
Axioms \cite{CS} (7) and (8) are trivial.
Axiom \cite{CS} (9):
\[ \bfig
\node a(0,+1500)[(A \tens B) \tens (C \partialarr D)]
\node b(+1500,-1000)[A \tens (B \tens (C \partialarr D))]
\node c(-1000,+1500)[((A \tens B) \tens C) \partialarr D]
\node d(-500,-1000)[(A \tens (B \tens C)) \partialarr D]
\node e(+500,-1000)[A \tens ((B \tens C) \partialarr D)]
\node f(+500,+1000)[\cdot]
\node g(0,+500)[\cdot]
\node h(-500,+1000)[((A \tens B) \tens C) \partialarr (R \tens D)]
\node i(-500,0)[\cdot]
\node j(+500,0)[\cdot]
\node k(0,-500)[\cdot]
\node l(+1000,-500)[\cdot]
\partiallace(+1000,+250)[M1]
\partiallace(+200,+1000)[M3]
\arrow|m|[g`i;a \partialarr 1]
\arrow|m|[g`j;a \partialarr a]
\arrow|m|[i`j;1 \partialarr a]
\arrow|m|[i`k;1 \partialarr (\mu \tens 1)]
\arrow|m|[j`k;1 \partialarr (1 \tens \alpha)]
\arrow|m|[a`h;m]
\arrow|m|[a`f;m \tens 1]
\arrow|m|[f`g;m]
\arrow|m|[g`h;1 \partialarr (\mu \tens 1)]
\arrow|a|[a`c;d^l]
\arrow|m|[h`c;1 \partialarr \alpha]
\arrow|m|[l`e;1 \tens (1 \partialarr \alpha)]
\arrow|m|[e`k;m]
\arrow|m|[l`j;m]
\arrow|m|[b`l;1 \tens m]
\arrow|b|[b`e;1 \tens d^l]
\arrow|m|[k`d;1 \partialarr \alpha]
\arrow|b|[e`d;d^l]
\arrow|l|/{@{>}@/_5em/}/[c`d;a \partialarr 1]
\arrow|r|/{@{>}@/^5em/}/[a`b;a]
I_\tensfig \]
where $\alpha(1 \tens \alpha)a = \alpha(\mu \tens 1)$ sine $D$ is an $R$-bimodule.
Axiom \cite{CS} (10) is similar, but uses M1 and M4.
Axiom \cite{CS} (11) uses M2 and the action (*) of $R$ on $A \partialarr B$.
\[ \bfig
\node a(-1000,+750)[( A \partialarr B \partialarr C) \tens D]
\node b(+1000,+750)[A \partialarr (( B \partialarr C) \tens D)]
\node c(0,-1000)[A \partialarr B \partialarr (C \tens D)]
\node d(0,+500)[\cdot]
\node e(-500,-500)[\cdot]
\node f(0,0)[\cdot]
\node g(+500,-500)[\cdot]
\partiallace(-500,+300)[M2]
\partiallace(-300,-500)[(*)]
\arrow|a|[a`b;d^r]
\arrow|m|[a`d;m]
\arrow|m|[a`e;m]
\arrow|l|/{@{>}@/_5em/}/[a`c;d^r]
\arrow|m|[d`b;\alpha \partialarr 1]
\arrow|m|[b`g;1 \partialarr m]
\arrow|r|/{@{>}@/^5em/}/[b`c;1 \partialarr d^r]
\arrow|m|[g`c;1 \partialarr \alpha \partialarr 1]
\arrow|m|[f`c;\alpha \partialarr \alpha \partialarr 1]
\arrow|m|[e`c;\alpha \partialarr 1]
\arrow|m|[d`f;1 \partialarr m]
\arrow|m|[e`f;m \partialarr 1]
\arrow|m|[f`g;\alpha \partialarr 1]
I_\tensfig \]
Axiom \cite{CS} (12) is similar and still uses M2, while Axiom \cite{CS}(13) uses M2, and \cite{CS} (14) uses M1.
Thus, from any category with two monoidal structures related by a middle-four interchange transformation satisfying the standard axioms for such a
map, we obtain an example of a net-category (that is, a proof-net category without negation) which is also equipped with a middle-four interchange map.
Communication on this article can be made through Micah McCurdy (Macquarie University, \[email protected]$)
who kindly typed this manuscript. I would also like to thank
Ross Street for a helpful comment on the middle-four map.
\begin{thebibliography}{9}
\bibitem{CS} Cockett, J.R.B. and R.A.G. Seely, I_\tensmph{Weakly Distributive Categories}, Journal of Pure and Applied Algebra v114 n2 (1997).
\bibitem{DP} Dosen, K., and Z. Petric, ``Proof-net categories'', (Preprint) Math. Inst. Serbian Academy of Sciences and Arts (2005),
arXiv:math/0503301v3
I_\tensnd{thebibliography}
I_\tensnd{document}
|
math
|
\begin{document}
\newtheorem{theorem}{{\bf Theorem}}[section]
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\newtheorem{sublemma}[theorem]{{\bf Sublemma}}
\newtheorem{proposition}[theorem]{{\bf Proposition}}
\newtheorem{corollary}[theorem]{{\bf Corollary}}
\newtheorem{definition}[theorem]{{\bf Definition}}
\newtheorem{notation}[theorem]{{\bf Notation}}
\newtheorem{convention}{{\bf Convention}}
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\newtheorem{subcase}{{\bf Case 1.\hspace{-0.2em}}}
\newtheorem{subcase2}{{\bf Case 2.\hspace{-0.2em}}}
\newtheorem{problem}{{\bf Problem}}
\newenvironment{proof}{{\bf Proof }}{
$\square$}
\newtheorem{remark}[theorem]{{\bf Remark}}
\newtheorem{claim}{{\bf Claim}}
\newcommand\Z{{\mathbb Z}}
\newcommand\N{{\mathbb N}}
\newcommand\Q{{\mathbb Q}}
\newcommand\R{{\mathbb R}}
\newcommand\diff{{\rm Diff}}
\newcommand\sym{{\rm Sym}}
\newcommand\fix{{\rm Fix}}
\newcommand\out{{\rm Out}}
\newcommand\aut{{\rm Aut}}
\newcommand\inn{{\rm Inn}}
\newcommand\id{{\rm id}}
\newcommand\K{{\mathcal K}}
\newcommand\T{{\mathcal T}}
\newcommand\LL{{\mathcal L}}
\newcommand\D{{\mathcal D}}
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\newcommand\pmcg{{\mathcal PM}}
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\relpenalty=10000
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\uchyph=-1
\makeatletter
\def\def\@captype{table}\caption{\def\@captype{table}\caption}
\def\def\@captype{figure}\caption{\def\@captype{figure}\caption}
\makeatother
\title{Characterization of 3-bridge links with infinitely many 3-bridge spheres}
\author{Yeonhee Jang
\footnote{This research is partially supported by Grant-in-Aid for JSPS Research Fellowships for Young Scientists.}\\
Department of Mathematics, Hiroshima University, \\
Hiroshima 739-8526, Japan\\
[email protected]}
\date{\empty}
\maketitle
\begin{abstract}
In \cite{Jan}, the author constructed an infinite family of 3-bridge links
each of which admits infinitely many 3-bridge spheres up to isotopy.
In this paper, we prove that if a prime, unsplittable link $L$ in $S^3$ admits infinitely many 3-bridge spheres up to isotopy
then $L$ belongs to the family.
\end{abstract}
\section{Introduction}\label{intro}
An {\it $n$-bridge sphere} of a link $L$ in $S^3$
is a 2-sphere which meets $L$ in $2n$ points
and cuts $(S^3, L)$ into $n$-string trivial tangles
$(B_1, t_1)$ and $(B_2, t_2)$.
Here, an {\it $n$-string trivial tangle} is
a pair $(B^3, t)$ of the $3$-ball $B^3$ and
$n$ arcs properly embedded in $B^3$ parallel to the boundary of $B^3$.
We call a link $L$ an {\it $n$-bridge link}
if $L$ admits an $n$-bridge sphere and does not admit an ($n-1$)-bridge sphere.
Two $n$-bridge spheres $S_1$ and $S_2$ of $L$ are said to be {\it pairwise isotopic}
({\it isotopic}, in brief)
if there exists a homeomorphism $f:(S^3, L)\rightarrow (S^3, L)$
such that $f(S_1)=S_2$ and
$f$ is {\it pairwise isotopic} to the identity,
i.e., there is a continuous family of homeomorphisms
$f_t:(S^3, L)\rightarrow (S^3, L)$ $(0\leq t\leq 1)$
such that $f_0=f$ and $f_1=\id$.
It is known
by Otal (\cite{Ota1} and \cite{Ota2})
that the unknot (resp. any 2-bridge link)
admits a unique $n$-bridge sphere up to isotopy
for $n\geq 1$ (resp. $n\geq 2$).
These results were recently refined
by Scharlemann and Tomova \cite{Sch2}.
The author constructed an infinite family of links
each of which admits infinitely many 3-bridge spheres up to isotopy in \cite{Jan},
and gave a classification of 3-bridge spheres of 3-bridge arborescent links in \cite{Jan3}.
In this paper, we prove the following theorem.
\begin{theorem}\label{thm-main}
Let $L$ be a prime, unsplittable link in $S^3$.
Then $L$ admits infinitely many 3-bridge spheres up to isotopy
if and only if
$L$ is equivalent to a link $L(q/2p;\beta_1/\alpha_1,\beta_2/\alpha_2)$ (see Figure \ref{fig-3b-inf}) with $q\not\equiv 1\pmod{p}$ and $|\alpha_1|>1$ (or $|\alpha_2|>1$).
\end{theorem}
\begin{figure}\label{fig-3b-inf}
\end{figure}
Here, two links are said to be {\it equivalent}
if there exists an orientation-preserving homeomorphism of $S^3$ which sends one to the other.
The link $L(q/2p;\beta_1/\alpha_1,\beta_2/\alpha_2)$ in Figure \ref{fig-3b-inf}
is obtained as follows.
Let $V_0$ be a solid torus standardly embedded in $S^3$ and $L_0$ a link in $V_0$ obtained by connecting two rational tangles
of slopes $\beta_1/\alpha_1$ and $\beta_2/\alpha_2$ by \lq\lq trivial arcs\rq\rq\ as in the figure.
Let $K_1\cup K_2$ be a 2-bridge link in $S^3$ of type $(2p,q)$ and $V_1$ the regular neighborhood of $K_1$.
For $i=0,1$, let $l_i$ be the preferred longitude of $V_i$, that is, $l_i$ is an essential loop on $\partial V_i$ which is null-homologous in $S^3\setminus \interior(V_i)$.
Let $h:V_0\rightarrow V_1$ be a homeomorphism which carries $l_0$ to $l_1$.
We denote by $L(q/2p;\beta_1/\alpha_1,\beta_2/\alpha_2)$ the union of $h(L_0)$ and $K_2$.
\begin{remark}\label{rmk-main}
{\rm
We can also see that any 3-bridge sphere of $L(q/2p;\beta_1/\alpha_1,\beta_2/\alpha_2)$
is isotopic to $P^i$ for some integer $i$,
where $P^i$ is obtained from $P^0$ by applying the $i$-th power of the \lq\lq half Dehn twist\rq\rq\ along the torus $T$ as illustrated in Figure \ref{fig-3b-inf} (see \cite{Jan} for detailed description of $P^i$).
This implies that any prime, unsplittable link admits only finitely many 3-bridge spheres up to homeomorphism.
}
\end{remark}
Theorem \ref{thm-main} gives a partial answer to an analogy of the Waldhausen conjecture
in terms of knot theory,
namely, a prime, unsplittable link with atoroidal complement admits only finitely many $n$-bridge spheres up to isotopy for a given $n(\in\N)$.
The (original) Waldhausen conjecture asserts that
a closed orientable atoroidal 3-manifold admits only finitely many Heegaard splittings of given genus $g(\in \N)$ up to isotopy
and was proved to be true by Johannson \cite{Joh2} and Li \cite{Li}.
\section{Heegaard splittings of 3-manifolds}\label{sec-hs}
Let $M$ be a closed orientable 3-manifold.
A genus-$g$ {\it Heegaard splitting} of $M$ is a tuple $(V_1, V_2; F)$,
where $V_1$ and $V_2$ are genus-$g$ handlebodies in $M$
such that $M=V_1\cup V_2$ and $F=\partial V_1=\partial V_2=V_1\cap V_2$.
Two Heegaard splittings $(V_1, V_2; F)$ and $(W_1, W_2; G)$ of a 3-manifold $M$
are said to be {\it isotopic}
if there exists a self-homeomorphism $f$ of $M$
such that $f(F)=G$ and $f$ is isotopic to the identity map $\id_M$ on $M$.
For a genus-$2$ Heegaard splitting $(V_1, V_2; F)$ of $M$,
it is known that
there is an involution $\tau_F$ on $M$
satisfying the following condition.
\begin{itemize}
\item[($\ast$)] $\tau_F(V_i)=V_i$ $(i=1,2)$
and $\tau_F|_{V_i}$ is {\it equivalent} to the standard involution $\T$
on a standard genus-2 handlebody $V$ as illustrated in Figure \ref{fig-hs-3b}.
To be precise, there is a homeomorphism $\psi_i:V_i\rightarrow V$
such that $\T=\psi_i(\tau_F|_{V_i})\psi_i^{-1}$ $(i=1,2)$.
\end{itemize}
\begin{figure}\label{fig-hs-3b}
\end{figure}
The strong equivalence class of $\tau_F$
is uniquely determined by the isotopy class of $(V_1, V_2; F)$ (cf. \cite[Proposition 5]{Jan2})
and we call $\tau_F$ the {\it hyper-elliptic involution}
associated with $(V_1, V_2; F)$
(or associated with $F$, in brief).
Here,
two involutions $\tau$ and $\tau'$ are said to be {\it strongly equivalent}
if there exists a homeomorphism $h$ on $M$
such that $h\tau h^{-1}=\tau'$ and that
$h$ is isotopic to the identity map $\id_M$.
Let $L$ be a prime, unsplittable 3-bridge link.
Let $M$ be the double branched covering of $S^3$ branched along $L$
and $\tau_L$ the covering transformation.
Let $\Phi_L$ be the natural map
from the set of isotopy classes of 3-bridge spheres of $L$
to the set of isotopy classes of genus-2 Heegaard surfaces of $M$ whose hyper-elliptic involution is $\tau_L$.
The following proposition is proved in \cite{Jan3}.
\begin{proposition}\label{prop-hs-3b}
$\Phi_L$ is at most $2$-$1$.
Moreover, $\Phi_L$ is injective if $L$ is not a non-elliptic Montesinos link.
\end{proposition}
In the rest of this section,
we recall a characterization of genus-$2$ $3$-manifolds admitting nontrivial torus decompositions due to Kobayashi \cite{Kob} (and \cite{Kob2}).
We use the following notation.
\begin{itemize}
\item[{$D[r]$}] (resp. $M\ddot{o}[r], A[r]$) :
the set of all orientable Seifert fibered spaces
over a disk $D$ (resp. a M\"{o}bius band $M\ddot{o}$, an annulus $A$)
with $r$ exceptional fibers.
\item[$M_K$] :
the set of the exteriors of the nontrivial 2-bridge knots.
\item[$M_L$] :
the set of the exteriors of the nontrivial 2-component 2-bridge links.
\item[$L_K$]: the set of the exteriors of the 1-bridge knots in lens spaces
each of which admits a complete hyperbolic structure or admits a Seifert fibration
whose regular fiber is not a meridian loop.
\item[$KI$]: the twisted $I$-bundle on the Klein bottle.
\end{itemize}
If $K$ is a 2-bridge knot (resp. a 2-bridge link, a 1-bridge knot in a lens space),
$E(K)$ denotes a manifold in $M_K$ (resp. $M_L$, $L_K$)
obtained as the exterior of $K$.
\begin{theorem}\label{thm-kob}
Let $M$ be a closed, connected, orientable Haken $3$-manifold of Heegaard genus $2$
which admits a nontrivial torus decomposition.
Let $(V_1, V_2; F)$ be a genus-$2$ Heegaard splitting of $M$.
Then $M$ satisfies one of the following four conditions (1), (2), (3) and (4), and
$F$ is isotopic to a Heegaard surface, denoted by the same symbol $F$,
as follows
(see Figure \ref{fig-hs}).
\begin{figure}\label{fig-hs}
\end{figure}
\begin{enumerate}
\item[{\rm (1)}]
$M$ is obtained from $M_1\in D[2]$ and $M_2=E(K)\in L_K$
by identifying their boundaries so that
the regular fiber of $M_1$ is identified with the meridian loop of $K$.
Moreover,
\begin{itemize}
\item $M_1\cap F$ is an essential annulus saturated in the Seifert fibration of $M_1$, and
\item $M_2\cap F$ is a 2-holed torus which gives a 1-bridge decomposition of the 1-bridge knot $K$.
\end{itemize}
Moreover, $V_i\cap T$ $(i=1,2)$ consists of a single separating essential annulus,
where $T=\partial M_1=\partial M_2$.
\item[{\rm (2)}]
$M$ is obtained from $M_1\in D[2]\cup D[3]$ and $M_2=E(K)\in M_K$
by identifying their boundaries so that
the regular fiber of $M_1$ is identified with the meridian loop of $K$.
Moreover,
\begin{itemize}
\item $M_1\cap F$ consists of two disjoint essential saturated annuli in $M_1$
which divide $M_1$ into three solid tori, and
\item $M_2\cap F$ is a 2-bridge sphere of the nontrivial 2-bridge knot $K$.
\end{itemize}
Moreover, by exchanging $V_1$ and $V_2$ if necessary,
\begin{itemize}
\item[{\rm (i)}]
$V_1\cap T$ consists of
two disjoint non-separating essential annuli
satisfying the following condition:
there exists a complete meridian disk system $(D_1, D_2)$ of $V_1$
such that $D_1\cap (V_1\cap T)=\emptyset$
and $D_2\cap (V_1\cap T)$
consists of essential arcs properly embedded in each annulus of $V_1\cap T$, and
\item[{\rm (ii)}]
$V_2\cap T$ consists of
disjoint non-parallel separating essential annuli,
\end{itemize}
where $T=\partial M_1=\partial M_2$.
\item[{\rm (3)}]
$M$ is obtained from
\begin{itemize}
\item[{\rm (3-1)}] $M_1\in M\ddot{o}[r]$ $(r=0,1,2)$ and $M_2=E(K)\in M_K$, or
\item[{\rm (3-2)}] $M_1\in A[r]$ $(r=0,1,2)$ and $M_2=E(K)\in M_L$
\end{itemize}
by identifying their boundaries so that
the regular fiber of $M_1$ is identified with the meridian loop of $K$.
Moreover,
\begin{itemize}
\item $M_1\cap F$ consists of two disjoint essential saturated annuli in $M_1$
which divide $M_1$ into two solid tori, and
\item $M_2\cap F$ is a 2-bridge sphere of the 2-bridge link $K$.
\end{itemize}
Moreover, $V_i\cap T$ $(i=1,2)$ consists of two disjoint non-separating essential annuli
satisfying the condition {\rm (i)} of (2),
where $T=\partial M_1=\partial M_2$.
\item[{\rm (4)}]
$M$ is obtained from $M_1, M_2\in D[2]$ and
$M_3=E(K_1\cup K_2)\in M_L$
by identifying their boundaries so that
the regular fiber of $M_i$ is identified with the meridian loop of $K_i$ ($i=1,2$).
Moreover,
\begin{itemize}
\item $M_i\cap F$ is an essential saturated annulus in $M_i$ $(i=1,2)$, and
\item $M_3\cap F$ is a 2-bridge sphere of the 2-bridge link $K_1\cup K_2$.
\end{itemize}
Moreover, $V_i\cap T$ $(i=1,2)$ consists of
two disjoint non-parallel separating essential annuli
satisfying the condition {\rm (ii)} of (2),
where $T=\partial M_1\cup \partial M_2=\partial M_3$.
\end{enumerate}
\end{theorem}
\begin{proof}
Let $\Gamma$ be the union of tori which gives the torus decomposition of $M$.
If each component of $\Gamma$ is separating,
then we see from the proof of the main theorem of \cite{Kob}
that $M$ and $F$ satisfies one of the conditions (1), (2), (3-1) and (4), where $\Gamma=T$.
In the rest of this proof, we assume that $\Gamma$ has a non-separating component
and show that $M$ and $F$ satisfy the condition (3-2).
By the proof of the main theorem of \cite{Kob},
$M$ satisfies the condition (3-2).
In particular, $M$ is obtained by gluing $M_1\in A[r]$ $(r=0, 1, 2)$ and
$M_2=E(K_1\cup K_2)$, where $K_1\cup K_2$ is a nontrivial 2-bridge link.
In the rest of this proof, we see that $F$ satisfies the condition (3-2) in this case.
First assume that neither $M_1$ nor $M_2$ is homeomorphic to $S^1\times S^1\times I$.
Then $\Gamma$ consists of two components.
By an argument similar to that for the main theorem of \cite{Kob},
together with Lemmas 3.1, 3.2 and 3.3 of \cite{Kob2},
we can see that $F$ satisfies the condition (3-2).
In the remainder of this proof,
assume that either $M_1$ or $M_2$ is homeomorphic to $S^1\times S^1\times I$.
Then we may assume that $\Gamma$ is a component of $T$.
If $M_1$ is homeomorphic to $S^1\times S^1\times I$,
then $M\setminus \Gamma$ is homeomorphic to the interior of $M_2$.
By an argument similar to that for the main theorem of \cite{Kob},
together with Lemmas 3.1, 3.2 and 3.3 of \cite{Kob2},
we can see that $\Gamma\cap V_i$ is a non-separating annulus as illustrated in Figure \ref{fig-hs-r0}.
\begin{figure}\label{fig-hs-r0}
\end{figure}
Let $\tau_F$ be the hyper-elliptic involution associated with $F$.
Then $\tau_F(\Gamma)$ is an essential torus in $M\setminus \Gamma$ (see Figure \ref{fig-hs-r0}).
Note that $M_2$ is homeomorphic to $A(1/n)$ for some integer $n$ or is hyperbolic
(see \cite[Lemma 4.4]{Kob} and see \cite[Section 2]{Jan2} for notation).
Thus any essential torus in $M_2$ is $\partial$-parallel,
and hence, $\tau_F(\Gamma)$ is isotopic to $\Gamma$ in $M$.
This implies that $T$ is isotopic to $\Gamma\cup\tau_F(\Gamma)$
since $M_1$ is homeomorphic to $S^1\times S^1\times I$,
and we see that $F$ satisfies the condition (3).
If $M_2=E(K_1\cup K_2)$ is homeomorphic to $S^1\times S^1\times I$,
then $K_1\cup K_2$ is a Hopf link and
$M\setminus \Gamma$ is homeomorphic to the interior of $M_1$.
By an argument similar to that in the previous case,
we see that $\Gamma\cap V_i$ is a non-separating annulus as illustrated in Figure \ref{fig-hs-r0}
and that $\tau_F(\Gamma)$ is also a non-separating essential torus in $M\setminus \Gamma$.
If $M\setminus \Gamma(\cong M_1)\in A[r]$ for $r\leq 1$, then any essential torus in $M_1$ is $\partial$-parallel,
and hence, we see that $F$ satisfies the condition (3) by an argument similar to that in the previous case.
(We use the same symbol $M\setminus \Gamma$ to denote the manifold
obtained by closing-up $M\setminus \Gamma$ with two tori.)
If $M\setminus \Gamma(\cong M_1)\in A[2]$,
then any essential torus in $M_1$ is either $\partial$-parallel
or an essential torus which divides $M_1$ into two Seifert fibered spaces belonging to $A[1]$.
Hence,
$\Gamma\cup\tau_F(\Gamma)$ cuts $M$ into two Seifert fibered spaces which belong to $A[1]$
whose fibrations are identical on $\tau_F(\Gamma)$.
On the other hand, by an argument in \cite[Section 6]{Kob} (see also Figure \ref{fig-hs-r0}),
one of the two Seifert fibered spaces must be the exterior of a 2-bridge link, say $L'$, and
the meridians of $L'$ must be identified with regular fibers of the other Seifert fibered space.
Since the fibrations of the two Seifert fibered space are identical on $\tau_F(\Gamma)$,
this implies that the meridian of a component of $L'$ is a regular fiber of $E(L')(\in A[1])$.
However, this is impossible (cf. \cite[Lemma 1]{Jan2} or \cite[Lemma 4.4]{Kob}).
Hence, any essential torus in $M_1$ is $\partial$-parallel,
and we see that $F$ satisfies the condition (3)
\end{proof}
We need to study the manifolds satisfying one of the following conditions
in the proof of Theorem \ref{thm-main} (cf. Section \ref{sec-proof}).
\begin{itemize}
\item[(M1)] $M$ is obtained by gluing $M_1\in D[2]$ and $M_2=L(p,q)\setminus N(K)$ as in Theorem \ref{thm-kob} (1), where $M_2$ is hyperbolic,
\item[(M2)] $M$ is obtained by gluing $M_1\in D[2]\cup D[3]$ and $M_2=E(K)\in M_K$ as in Theorem \ref{thm-kob} (2), where $M_2$ is hyperbolic,
\item[(M3-1-1)] $M$ is obtained by gluing $M_1\in M\ddot{o}[r]$ $(r=1,2)$ and $M_2=E(K)\in M_K$ as in Theorem \ref{thm-kob} (3),
where $K$ is a torus knot of type $(2,n)$,
\item[(M3-1-2)] $M$ is obtained by gluing $M_1\in M\ddot{o}[r]$ $(r=0,1,2)$ and $M_2=E(K)\in M_K$ as in Theorem \ref{thm-kob}(3), where $M_2$ is hyperbolic,
\item[(M3-2-1)] $M$ is obtained by gluing $M_1\in A[r]$ $(r=0,1,2)$ and $M_2=E(K)\in M_L$ as in Theorem \ref{thm-kob} (3),
where $K$ is a torus link of type $(2,n)$,
\item[(M3-2-2)] $M$ is obtained by gluing $M_1\in A[r]$ $(r=0,1,2)$ and $M_2=E(K)\in M_L$ as in Theorem \ref{thm-kob}(3), where $M_2$ is hyperbolic,
\item[(M4)] $M$ is obtained by gluing $M_1, M_2\in D[2]$ and $M_3=E(K)\in M_K$ as in Theorem \ref{thm-kob} (4), where $M_3$ is hyperbolic.
\end{itemize}
\begin{remark}\label{rmk-dbc}
{\rm
The double branched covering of $S^3$ branched over $L(q/2p;\beta_1/\alpha_1, \beta_2/\alpha_2)$
satisfies the condition (M3-1-2) or (M3-2-2), where $r\geq 1$ (cf. \cite{Jan}).
}
\end{remark}
\section{Mapping class groups}\label{sec-mcg}
In this section,
we calculate certain subgroups of the mapping class groups of
the Seifert fibered spaces and the manifolds which arose in Theorem \ref{thm-kob}.
This enables us to compare the hyper-elliptic involutions of genus-2 Heegaard surfaces of 3-manifolds.
For a hyperbolic 3-manifold $N$,
let $\mcg(N)$ be the (orientation-preserving) mapping class group of $N$.
For a Seifert fibered space $N$,
let $\mcg(N)$ be the subgroup of the (orientation-preserving) mapping class group of $N$
which consists of elements preserving each singular fiber of $N$.
(See \cite{Jan2} for more details.)
When $N$ is a Seifert fibered space over a surface $F$,
let $\mcg^0(N)$ be the subgroup of $\mcg(N)$ which consists of the elements inducing the identity map on $F$,
$\mcg^{\ast}(F)$ the mapping class group of $(F, {\rm exceptional\ points})$.
For a 3-manifold $M$ in Theorem \ref{thm-kob},
let $\mcg(M)$ be the subgroup of the (orientation-preserving) mapping class group of $M$
which consists of the elements preserving each piece of the torus decomposition of $M$ and each singular fiber of the Seifert pieces.
We describe some elements of the mapping class groups of certain Seifert fibered spaces.
Let $N_1$ and $N_2$ be the Seifert fibered spaces
$M\ddot{o}(\beta_1/\alpha_1,\beta_2/\alpha_2)\in M\ddot{o}[2]$ and $A(\beta_1/\alpha_1,\beta_2/\alpha_2)$\\$\in A[2]$, respectively
(see \cite[Section 2]{Jan2} for notation).
We define $g_i, b, D_j\in \mcg(N_1)$ and $h_i, a, D_j'\in \mcg(N_2)$ $(i,j\in\{1,2\})$ as follows.
We denote by $g_i$ and $h_i$ the involutions as illustrated in Figure \ref{fig-mcg}.
The symbols $a$ and $b$ denote the Dehn twist along saturated annuli $A_a$ and $A_b$, respectively, in the direction of a fiber, and
$D_j$ and $D_j'$ are the Dehn twists along saturated tori $T_{D_j}$ and $T_{D_j'}$, respectively, in the direction of loops intersecting regular fibers in one point,
as illustrated in Figure \ref{fig-mcg}.
\begin{figure}\label{fig-mcg}
\end{figure}
For more precise description of the above elements, see \cite[Section 5 and Remark 2]{Jan2}.
\begin{lemma}\label{lem-mcg-sfs2}
{\rm (1)} If $N$ is a Seifert fibered space $M\ddot{o}(\beta_1/\alpha_1,\beta_2/\alpha_2)\in M\ddot{o}[2]$,
then $\mcg(N)=\langle b\rangle$\\$\times(\langle D_1,D_2\rangle\rtimes\langle g_1,g_2\rangle)$
and has a group presentation
\begin{align*}
\begin{array}{rll}
\mcg(N)=\langle D_1,D_2,g_1,g_2,b&|&g_i^2, [g_1,g_2], g_1D_jg_1=D_j^{-1}, g_2D_1g_2=D_2^{-1},\\
&&b^2, [g_i,b], [D_j,b]\ (i,j\in\{1,2\})\rangle.
\end{array}
\end{align*}
In particular, the subgroup $\langle D_1,D_2\rangle$ of $\mcg(N)$
is a free group of rank 2.
{\rm (2)} If $N$ is a Seifert fibered space $A(\beta_1/\alpha_1,\beta_2/\alpha_2)\in A[2]$,
then $\mcg(N)=\langle a\rangle\rtimes(\langle D_1',D_2'\rangle\rtimes$\\$\langle h_1,h_2\rangle)$
and has a group presentation
\begin{align*}
\begin{array}{rll}
\mcg(N)=\langle D_1',D_2',h_1,h_2,a&|&h_i^2, [h_1,h_2], h_1D_j'h_1=D_j'^{-1}, h_2D_1'h_2=D_2'^{-1},\\
&&h_1ah_1=a, h_2ah_2=a^{-1}, D_j'aD_j'^{-1}=a\ (i,j\in\{1,2\})\rangle.
\end{array}
\end{align*}
In particular, the subgroup $\langle D_1',D_2'\rangle$ of $\mcg(N)$
is a free group of rank 2.
\end{lemma}
\begin{proof}
By \cite[Proposition 25.3]{Joh}, we have a split exact sequence
$$
1\rightarrow \mcg^0(N)\rightarrow \mcg(N)\rightarrow \mcg^{\ast}(F)\rightarrow 1.
$$
(1) By \cite[Lemma 25.2]{Joh}, $\mcg^0(N)$ is an order-2 group generated by $b$.
On the other hand, by \cite[Section 4.1]{Bir3},
we have the following exact sequence, called the \lq\lq Birman exact sequence\rq\rq.
$$
1\rightarrow \pi_1(F',x_0)\rightarrow \mcg^{\ast}(F)(\cong \mcg_2)\rightarrow \mcg_1\rightarrow 1,
$$
where
$\pi_1(F',x_0)$ denotes the fundamental group of a once-punctured M\"{o}bius band and
$\mcg_{n}$ denotes the mapping class group of a M\"{o}bius band fixing $n$ specified points.
Recall that $\pi_1(F',x_0)$ is a free group of rank 2
and that $\mcg_1=\langle g_1,g_2 \rangle\cong\Z_2\oplus\Z_2$ (cf. \cite[Lemma 4]{Jan2}).
Moreover, we may take the images of $T_{D_1}$ and $T_{D_2}$ by the projection map as the generators of $\pi_1(F',x_0)$.
Then their images in $\mcg^{\ast}(F)$, by the second map in the above exact sequence, are $D_1$ and $D_2$.
Moreover, the conjugation of $D_j$ by $g_i$ $(i,j\in\{1,2\})$ is as follows:
$$
g_1D_jg_1=D_j^{-1},\ \ g_2D_1g_2=D_2.
$$
Hence, by using an argument in \cite[p.136--139]{Joh3},
we obtain the following group presentation of $\mcg^{\ast}(F)$.
$$
\mcg^{\ast}(F)=\langle D_1,D_2,g_1,g_2\mid g_i^2, [g_1,g_2],
g_1D_jg_1=D_j^{-1},g_2D_1g_2=D_2\ (i,j\in\{1,2\})\rangle.
$$
Since the conjugation of $b$ by $D_j$ or $g_1$ is $b$,
we obtain the desired result by using an argument in \cite[p.136--139]{Joh3} again.
(2) can be proved similarly.
\end{proof}
Let $M$ be a manifold in Theorem \ref{thm-kob},
and $T$ the union of tori
in the theorem.
Let $\D$ be the subgroup of $\mcg(M)$ generated by the all possible Dehn twists along $T$.
Then we obtain the following,
which can be proved by an argument similar to that for \cite[Proposition 15.2]{Bon} or \cite[Lemma 3]{Jan2}.
\begin{lemma}\label{lem-d}
Let $M$ be a manifold in Theorem \ref{thm-kob}.
{\rm (1)} If $M$ satisfies the condition (M1) or (M2),
then $\D$ is an infinite cyclic group generated by $D_l$,
where $D_l$ is the Dehn twist along (a component of) $T$ in the direction of a longitude of $K$.
{\rm (2)} If $M$ satisfies the condition (M3-1-1),
then $\D\cong \langle D_l\rangle\cong\Z$,
where $D_l$ is the Dehn twist along (a component of) $T$ in the direction of a longitude of $K$.
{\rm (3)} If $M$ satisfies the condition (M3-1-2),
then $\D$ is generated by $D_m$ and $D_l$,
where $D_m$ and $D_l$ are the Dehn twists along $T$ in the direction of a meridian and a longitude of $K$, respectively.
Moreover, $\D\cong\langle D_m\rangle\cong\Z$ if $M_1\in M\ddot{o}[0]$, and
$\D\cong \langle D_m, D_l\rangle\cong\Z\oplus\Z$ otherwise.
{\rm (4)} If $M$ satisfies the condition (M3-2-1), namely, $M$ is obtained by gluing $M_1\in A[r]$ ($r=0,1,2$) and $M_2=E(K)=A(1/n)$ so that the regular fibers of $M_1$ are identified with the meridians of $K$,
then $\D$ is an abelian group generated by $D_m$ and $D_l$,
where $D_m$ and $D_l$ are the Dehn twists along (a component of) $T$ in the direction of a meridian and a longitude of $K$, respectively.
{\rm (5)} If $M$ satisfies the condition (M3-2-2),
then $\D$ is generated by $D_{m_1}, D_{l_1}$ and $D_{l_2}$,
where $D_{m_i}$ and $D_{l_i}$ are the Dehn twists along a component $T_i$ ($i=1,2$) of $T$ in the direction of a meridian and a longitude of $K$, respectively.
Moreover, $\D\cong \langle D_{m_1}, D_{l_1}\rangle\cong\Z\oplus\Z$ if $M_1\in A[0]$, and
$\D\cong \langle D_{m_1}, D_{l_1}, D_{l_2}\rangle\cong\Z\oplus\Z\oplus\Z$ otherwise.
{\rm (6)} If $M$ satisfies the condition (M4),
then $\D\cong \langle D_{l_1}, D_{l_2}\rangle\cong\Z\oplus\Z$,
where $D_{l_i}$ is the Dehn twist along a component $T_i$ ($i=1,2$) of $T$ in the direction of a longitude of $K$.
\end{lemma}
We define some self-homeomorphisms of $M$ when $M$ satisfies the condition (M3-1-1) or (M3-2-1) as follows.
\begin{definition}\label{def-homeo}
{\rm
Let $M$ be a manifold which satisfies the condition (M3-1-1) or (M3-2-1) with $r=2$.
(1) When $M$ satisfies the condition (M3-1-1),
we define self-homeomorphisms $G_1$, $G_2$ and $B$ of $M$ as follows.
\begin{eqnarray*}
\begin{array}{lll}
G_1|_{M_1}=g_1, & G_1|_{M_2}=f, & G_1|_{T\times[1,2]}=R,\\
G_2|_{M_1}=g_2, & G_2|_{M_2}=id, & G_2|_{T\times[1,2]}=R_lD^{1/2}_{l},\\
B|_{M_1}=b, & B|_{M_2}=id, & B|_{T\times[1,2]}=R_mD^{1/2}_{m},\\
D_j|_{M_1}=D_j, & D_j|_{M_2}=id, & D_j|_{T\times[1,2]}=id\hspace{5mm}(j=1,2).
\end{array}
\end{eqnarray*}
Here, $g_1$, $g_2$ and $b$ are involutions of $M_1$
as described in Lemma \ref{lem-mcg-sfs2},
$f$ is an involution of $M_2=E(K)$ which gives a strong inversion of the torus knot $K$ (see \cite[Remark 7]{Jan2}),
and $R$ and $R_{\alpha}$ ($\alpha=m$ or $l$) are the self-homeomorphisms of $T\times [1,2]$ defined by
$R([\vec{x}],t)=([-\vec{x}],t)$ and $R_{\alpha}([\vec{x}],t)=([\vec{x}+\frac{1}{2}\vec{\alpha}],t)$, respectively.
Here, we identify $T$ with $\R^2/\Z^2$ and
$[\vec{x}]$ denotes the point of $\R^2/\Z^2$ determined by $\vec{x}\in \R^2$.
In the identity $D_j|_{M_1}=D_j$,
the right-hand side represents the homeomorphisms in Lemma \ref{lem-mcg-sfs2} (1).
The symbols $D_l$ and $D_m$ denote the Dehn twists given in Lemma \ref{lem-d} (2) and (4),
and $D^{1/2}_l$ and $D^{1/2}_m$ denote the half Dehn twists in the direction of $l$ and $m$, respectively
(see \cite[Section 5]{Jan2} for definition of half Dehn twists).
(2) When $M$ satisfies the condition (M3-2-1),
we define self-homeomorphisms $H_1$ and $H_2$ of $M$ as follows.
\begin{eqnarray*}
\begin{array}{lll}
H_1|_{M_1}=h_1, & H_1|_{M_2}=h_1, & H_1|_{T\times[1,2]}=R,\\
H_2|_{M_1}=h_2, & H_2|_{M_2}=h_2, & H_2|_{T\times[1,2]}=h_2|_T\times[1,2],\\
D_j'|_{M_1}=D_j', & D_j'|_{M_2}=id, & D_j'|_{T\times[1,2]}=id.
\end{array}
\end{eqnarray*}
Here, $h_1$ is an involution of $M_1$ as described in Lemma \ref{lem-mcg-sfs2},
$h_2$ is an involution of $M_2=E(K)$ which gives a strong inversion of the torus link $K$ (see \cite[Lemma 4 (3)]{Jan2}),
and $R$ is the self-homeomorphism of $T\times [1,2]$ defined in (1).
In the identity $D_j'|_{M_1}=D_j'$,
the right-hand side represents the homeomorphisms in Lemma \ref{lem-mcg-sfs2} (2).
}
\end{definition}
In \cite[Proposition 6]{Jan2},
the author calculated $\mcg(M)$ for certain manifolds in Theorem \ref{thm-kob}
by using \cite[Theorem 15.1]{Bon} and \cite{Sak3}.
The following theorem can be obtained by a similar argument
together with Lemmas \ref{lem-mcg-sfs2} and \ref{lem-d}.
\begin{theorem}\label{thm-mcg}
Let $M$ be a manifold which satisfies the condition (M3-1-1) or (M3-2-1) with $r=2$.
(1) If $M$ satisfies the condition (M3-1-1),
then the subgroup $\langle D_1,D_2\rangle$ of $\mcg(M)$ generated by $D_1$ and $D_2$
is a free group of rank 2.
(2) If $M$ satisfies the condition (M3-2-1),
then the subgroup $\langle D_1',D_2',D_l\rangle$ of $\mcg(M)$
is the direct product of the infinite cyclic group generated by $D_l$ and the free group of rank 2 generated by $D_1'$ and $D_2'$.
\end{theorem}
\begin{proof}
Recall from \cite[Theorem 15.1]{Bon} and \cite{Sak3} (cf. \cite[Section 5]{Jan2})
that there is an exact sequence
\begin{equation}\label{exact}
1\rightarrow \D\rightarrow \mcg(M)\rightarrow \Delta\rightarrow 1,
\end{equation}
where $\Delta$ is the subgroup of $\mcg(M_1)\times \mcg(M_2)$ consisting of all elements $(f_1,f_2)$
such that $f_1|_T$ is isotopic to $f_2|_T$.
(1) Let $M$ be a manifold which satisfies the condition (M3-1-1).
Then we have $\mcg(M_1)=$\\$\langle b\rangle\times(\langle D_1,D_2\rangle\rtimes\langle g_1,g_2\rangle)$ by Lemma \ref{lem-mcg-sfs2} (1), and $\mcg(M_2)=\langle f\rangle\cong \Z_2$ by \cite[Lemma 4 (1)]{Jan2}.
Note that the subgroup of $\mcg(M_1)$ generated by $D_1$ and $D_2$ is a free group of rank 2.
Hence, we see that
$$\Delta=\langle (b,id)\rangle\times(\langle (D_1,id),(D_2,id)\rangle\rtimes\langle (g_1,f),(g_2,id)\rangle)$$
and the subgroup of $\Delta$ generated by $(D_1,id)$ and $(D_2,id)$ is also a free group of rank 2.
Hence, we see from the exact sequence (\ref{exact}) that the subgroup of $\mcg(M)$ generated by $D_1$ and $D_2$ is a free group of rank 2.
(2) Let $M$ be a manifold which satisfies the condition (M3-2-1).
Then we have $\mcg(M_1)=$\\$\langle a\rangle\rtimes(\langle D_1',D_2'\rangle\rtimes\langle h_1,h_2\rangle)$ by Lemma \ref{lem-mcg-sfs2} (2).
On the other hand, we have $\mcg(M_2)=\langle a\rangle\rtimes\langle h_1,$\\$h_2\rangle$ (cf. \cite[Lemma 4 (2)]{Jan2}).
Note that the subgroup of $\mcg(M_1)$ generated by $D_1'$ and $D_2'$ is a free group of rank 2.
We see that
$$\Delta=\langle (D_1',id),(D_2',id)\rangle\rtimes\langle (h_1,h_1),(h_2,h_2)\rangle$$
and the subgroup of $\Delta$ generated by $(D_1',id)$ and $(D_2',id)$ is also a free group of rank 2.
Hence, we see from the exact sequence (\ref{exact}) that the subgroup of $\mcg(M)$ generated by $D_1'$ and $D_2'$ is a free group of rank 2.
On the other hand, the subgroup of $\mcg(M)$ generated by $D_l$ is an infinite cyclic group by Lemma \ref{lem-d} (3).
Since we can easily see that $D_l$ commutes with $D_1'$ and $D_2'$,
we obtain the desired result.
\end{proof}
\begin{remark}\label{rmk-mcg}
{\rm
Let $M$ be a manifold in Theorem \ref{thm-mcg}.
(1) If $M$ satisfies the condition (M3-1-1),
then $\mcg(M)=\langle B\rangle\rtimes(\langle D_1,D_2\rangle\rtimes\langle G_1,G_2\rangle)$, and has a group presentation
\begin{align*}
\begin{array}{rll}
\mcg(M)=\langle D_1,D_2,G_1,G_2,B
&|&
G_i^2, [G_1,G_2], G_1D_jG_1=D_j^{-1}, G_2D_1G_2=D_2^{-1},\\
&&B^2=D_m, [G_1,B]=D_m^{-1}, [G_2,B], [D_j,B]
\ (i,j\in\{1,2\})\rangle.
\end{array}
\end{align*}
(2) If $M$ satisfies the condition (M3-2-1),
then $\mcg(M)=\langle D_m,D_l\rangle\rtimes(\langle D_1',D_2'\rangle\rtimes\langle H_1,H_2\rangle)$, and has a group presentation
\begin{align*}
\begin{array}{rll}
\mcg(M)=\langle D_1',D_2',H_1,H_2,D_m, D_l&|&
H_i^2, [H_1,H_2], H_1D_j'H_1=D_j'^{-1}, H_2D_1'H_2=D_2'^{-1},\\
&&D_j'D_mD_j'^{-1}=D_m, D_j'D_lD_j'^{-1}=D_l,\\
&&H_1D_mH_1=D_m^{-1}, H_2D_mH_2=D_m, H_iD_lH_i=D_l^{-1}\\
&&(i,j\in\{1,2\})\rangle.
\end{array}
\end{align*}
}
\end{remark}
\section{Proof of Theorem \ref{thm-main}}\label{sec-proof}
Since the if part is already proved in \cite{Jan},
we prove the only if part.
Namely, we show that any prime, unsplittable 3-bridge link
which admits infinitely many 3-bridge spheres up to isotopy
is equivalent to a link $L(q/2p;\beta_1/\alpha_1,\beta_2/\alpha_2)$ in Figure \ref{fig-3b-inf} with $q\not\equiv 1\pmod{p}$ and $|\alpha_1|>1$ (or $|\alpha_2|>1$).
Let $L$ be a prime, unsplittable 3-bridge link in $S^3$, and
assume that $L$ admits infinitely many 3-bridge spheres up to isotopy.
Let $M=M_2(L)$ be the double branched cover of $S^3$ branched along $L$
and $\tau_L$ the covering transformation.
By Proposition \ref{prop-hs-3b},
$M$ admits infinitely many genus-2 Heegaard surfaces, up to isotopy,
whose hyper-elliptic involutions are $\tau_L$.
By \cite[Theorem 1.1]{Li}, $M$ is toroidal, and hence,
either $M$ is a Seifert fibered space or $M$ admits a nontrivial torus decomposition.
\begin{case}\label{case-1}
$M$ is a Seifert fibered space.
\end{case}
By the orbifold theorem \cite{Boi7, Coo} together with \cite[Section 5]{Dun},
$L$ is a {\it generalized Montesinos link} or a {\it Seifert link},
that is, either $L$ is equivalent to a link in Figure \ref{fig-gen-mont} or
$S^3\setminus L$ admits a Seifert fibration.
\begin{figure}\label{fig-gen-mont}
\end{figure}
Assume first that $L$ is a generalized Montesinos link.
By \cite[Theorem 2.1]{Boi3},
$L$ is equivalent to one of the links in Figure \ref{fig-case1-1} since $L$ is a 3-bridge link.
\begin{figure}\label{fig-case1-1}
\end{figure}
By \cite{Jan2,Jan3},
$L$ admits at most six 3-bridge spheres, at most two 3-bridge spheres or a unique 3-bridge sphere up to isotopy
according as $L$ is equivalent to the link in Figure \ref{fig-case1-1} (1), (2) or (3).
This contradicts the assumption that $L$ admits infinitely many 3-bridge spheres up to isotopy.
Next, assume that $L$ is a Seifert link.
By \cite{Bur2} and by the assumption that $L$ is a 3-bridge link,
we see that $L$ is equivalent to a (nontrivial) $(3,n)$-torus link
or the union of a $(2,n)$-torus knot and its core of index 2.
If $L$ is equivalent to a $(3,n)$-torus knot
or the union of a $(2,n)$-torus knot and its core of index 2,
then $M$ is a small Seifert fibered space
and admits at most four genus-2 Heegaard surfaces up to isotopy by \cite{Boi},
which is a contradiction.
We also prove the following proposition in Section \ref{sec-toruslink}.
\begin{proposition}\label{prop-toruslink}
If $L$ is a $(3,3n')$-torus link for some nonzero integer $n'$,
then $L$ admits a unique 3-bridge sphere up to isotopy.
\end{proposition}
Hence, $M$ cannot be a Seifert fibered space.
\begin{case}\label{case-2}
$M$ admits a nontrivial torus decomposition.
\end{case}
If $L$ is an arborescent link,
then $L$ admits at most four 3-bridge spheres by \cite{Jan3}.
Hence, we assume that $L$ is not an arborescent link.
Then, by Theorem \ref{thm-kob} and \cite[Proof of Theorem 1]{Jan2},
$M$ satisfies one of the conditions (M1), (M2), (M3-1-1), (M3-1-2), (M3-2-1), (M3-2-2) and (M4)
introduced at the end of Section \ref{sec-hs}.
Let $T$ be the union of tori as in Theorem \ref{thm-kob}.
\begin{subcase2}\label{case2-1}
$M$ satisfies the condition (M1).
\end{subcase2}
Note that $M$ is obtained by gluing $M_1\in D[2]$ and $M_2=L(p,q)\setminus N(K)$,
where $K$ is a 1-bridge knot in a lens space $L(p,q)$,
and that $M_2$ is hyperbolic.
Since $M_1$ is also simple,
we can see that $T=\partial M_1=\partial M_2$ is the only essential torus in $M$ up to isotopy.
By \cite[Theorem 4]{Joh2},
there exist genus-2 Heegaard surfaces $F_1,F_2,\dots,F_n$ of $M$ such that
any genus-2 Heegaard surface $F$ can be obtained from some $F_i$
by applying Dehn twists along $T$.
Recall from Theorem \ref{thm-kob} that
$F\cap M_1$ is an essential saturated annulus of $M_1$ and
$F\cap M_2$ is a 2-hold torus which gives a 1-bridge presentation of $K$.
Let $\mu$ and $\lambda$ be the meridian and a longitude of $K$,
and denote the Dehn twist along $T$ in the direction of $\mu$ and $\lambda$
by $D_{\mu}$ and $D_{\lambda}$, respectively.
Then $F$ is isotopic to $D_{\mu}^{n_1}D_{\lambda}^{n_2}(F_i)$ for some integers $n_1$ and $n_2$.
Note that any genus-2 Heegaard surface meets $T$ in the union of two meridians of $K$.
Hence, $D_{\mu}^{n_1}D_{\lambda}^{n_2}(F_i)=D_{\lambda}^{n_2}(F_i)$.
Note also that $\tau_{F_j}D_{\lambda}\tau_{F_j}=D_{\lambda}^{-1}$
because $\tau_{F_j}$ reverses the orientation of $\lambda$ (see Figure \ref{fig-case2-1}).
\begin{figure}\label{fig-case2-1}
\end{figure}
Thus,
$$
\tau_F=\tau_{D_{\lambda}^{n_2}(F_j)}=D_{\lambda}^{n_2}\tau_{F_j}D_{\lambda}^{-n_2}=D_{\lambda}^{2n_2}\tau_{F_j}.
$$
Since $\D$ is an infinite cyclic group generated by $D_{\lambda}$ by Lemma \ref{lem-d} (1),
$\{D_{\lambda}^{2n_2}\tau_{F_j}\}_{n_2\in\Z}$ are mutually distinct,
and hence, there is at most one $n_2\in\Z$ such that
$D_{\lambda}^{2n_2}\tau_{F_j}=\tau_{L}$.
So, for each Heegaard surface $F_j$,
the hyper-elliptic involution associated with $D_{\lambda}^{n_2}(F_j)$
is strongly equivalent to $\tau_L$
for at most one $n_2\in\Z$.
Hence, the number of genus-2 Heegaard surfaces whose hyper-elliptic involutions are $\tau_L$
is finite.
This contradicts the assumption.
\begin{subcase2}\label{case2-2}
$M$ satisfies the condition (M2).
\end{subcase2}
Note that $M$ is obtained by gluing $M_1\in D[2]\cup D[3]$ and $M_2=E(K)$, where $K$ is a hyperbolic 2-bridge knot,
so that the regular fiber of $M_1$ is identified with the meridian loop of $K$.
Recall from Theorem \ref{thm-kob} that,
for a genus-2 Heegaard surface $F$,
$F\cap M_1$ is the union of two essential annuli which cuts $M_1$ into three solid tori and
$F\cap M_2$ is the 2-bridge sphere of $K$.
Let $D_{\lambda}$ denote the Dehn twist along $T=\partial M_1=\partial M_2$ in the direction of a longitude of $K$, and
note that $\D$ is an infinite cyclic group generated by $D_{\lambda}$ (see Lemma \ref{lem-d} (1)).
First assume that $M_1\in D[2]$.
Note that $K$ admits a unique 2-bridge sphere up to isotopy by \cite{Sch},
and that $M_1$ contains a unique essential annulus up to isotopy.
By \cite[Lemma 6]{Jan2},
there exist genus-2 Heegaard surfaces $F_0$, $F_1$, $F_2$ and $F_3$ of $M$ such that
any genus-2 Heegaard surface of $M$ is isotopic to $D_{\lambda}^n(F_i)$ for some integer $n$ and for some $i=0,1,2,3$.
(We remark that $F_i=D_{\lambda}^{i/4}(F_0)$.)
Recall that $\tau_{F_i}D_{\lambda}\tau_{F_i}=D_{\lambda}^{-1}$.
By an argument similar to that in Case 2.\ref{case2-1},
we can see that the number of genus-2 Heegaard surfaces whose hyper-elliptic involutions are $\tau_L$ is finite, a contradiction.
Next, assume that $M_1\in D[3]$.
Note that $F\cap M_1$ is homeomorphic to one of $G_1,G_2$ and $G_3$ in Figure \ref{fig-annuli}.
\begin{figure}\label{fig-annuli}
\end{figure}
To be precise, $F\cap M_1$ is isotopic to $f_1(G_i)$ for some $f_1\in\mcg(M_1)$ and for some $i=1,2,3$.
(We may assume that $f_1|_{\partial M_1}=id$.)
For each $i=1,2,3$,
let $F_i$ be a genus-2 Heegaard surface such that $F_i\cap M_1=G_i$ and $F_i\cap M_2$ is the 2-bridge sphere of $K$.
By \cite[Lemma 6 (1)]{Jan2},
any genus-2 Heegaard surface $F$ is isotopic to $D_{\lambda}^{n/4}f(F_i)$
for some integer $n$ and for some $i=1,2,3$ and for some homeomorphism $f$ of $M$
which is obtained from some $f_1\in\mcg(M_1)$ by the rule $f|_{M_1}=f_1\in\mcg(M_1)$ and $f|_{M_2}=id$.
Let $F_i^j$ ($j=0,1,2,3$) be the Heegaard surface $D_{\lambda}^{j/4}(F_i)$.
Let $\mcg_0(M)$ be the subgroup of $\mcg(M)$ consisting of all elements whose restrictions to $M_2$ are the identity.
Then the above argument implies that
$F$ is isotopic to $g(F_i^j)$ for some $g\in\mcg_0(M)$ and for some $F_i^j$.
\begin{claim}
For each $F_i^j$, at most one of $\{g(F_i^j)\}_{g\in\mcg_0(M)}$
can have $\tau_L$ as hyper-elliptic involution.
\end{claim}
\begin{proof}
We show this only for $F_1^0$. (The other cases can be treated similarly.)
Put $\tau:=\tau_{F_1^0}$.
Then $\tau_{g(F_1^0)}=g\tau g^{-1}$.
Recall by \cite[Proof of Theorem 2 (3)]{Jan2} that
$$
\mcg(M_1)\cong (P_3/\langle (xy)^3\rangle)\rtimes \langle \tau\rangle
< (B_3/\langle (xy)^3\rangle)\rtimes \langle \tau\rangle,
$$
where $P_3$ and $B_3$ are the pure 3-braid group and the 3-braid group, respectively.
Let $\mcg_0(M_1)$ be the subgroup of $\mcg(M_1)$ consisting of all elements whose restrictions to $T$ are the identity.
Then we have $\mcg_0(M_1)\cong P_3/\langle (xy)^3\rangle$.
Recall from \cite[Theorem 15.1]{Bon} and \cite{Sak3} (cf. \cite[Section 5]{Jan2})
that there is an exact sequence
$$
1\rightarrow \D\rightarrow \mcg(M)\rightarrow \Delta\rightarrow 1,
$$
where $\Delta$ is the subgroup of $\mcg(M_1)\times \mcg(M_2)$ consisting of all elements $(f_1,f_2)$
such that $f_1|_T$ is isotopic to $f_2|_T$.
Since $\mcg_0(M)$ is the subgroup of $\mcg(M)$ consisting of all elements whose restrictions to $M_2$ are the identity,
we obtain an exact sequence
$$
1\rightarrow \D\rightarrow \mcg_0(M)\rightarrow \mcg_0(M_1)\rightarrow 1.
$$
Recall from \cite[Claim 1 (2)]{Jan2} that
the \lq\lq centralizer\rq\rq\ $$Z(\tau,\mcg_0(M_1))=\{f\in\mcg_0(M_1)\mid f\tau=\tau f\}$$ of $\tau$ in $\mcg_0(M_1)$ is $\{1\}$.
By using this fact, the identity $D_{\lambda}\tau D_{\lambda}^{-1}=D_{\lambda}^{2}\tau$ and Lemma \ref{lem-d} (1),
we can see that the \lq\lq centralizer\rq\rq\ $Z(\tau,\mcg_0(M))=\{f\in\mcg_0(M)\mid f\tau=\tau f\}$ of $\tau$ in $\mcg_0(M)$ is $\{1\}$.
This implies that
the hyper-elliptic involution associated with $g(F_1^0)$
is strongly equivalent to $\tau_L$ for at most one $g\in\mcg_0(M)$.
\end{proof}
Hence, the number of genus-2 Heegaard surfaces of $M$ whose hyper-elliptic involutions are $\tau_L$
is at most twelve,
a contradiction.
\begin{subcase2}\label{case2-3-1-1}
$M$ satisfies the condition (M3-1-1).
\end{subcase2}
Recall that $M$ is obtained by gluing $M_1\in M\ddot{o}[r]$ $(r=1,2)$ and $M_2=E(K)$, where $K$ is a $(2,n)$-torus knot,
so that the regular fiber of $M_1$ is identified with the meridian loop of $K$.
By Theorem \ref{thm-kob}, for any genus-2 Heegaard surface $F$,
$F\cap M_1$ is the union of two essential saturated annuli which cuts $M_1$ into two solid tori
and $F\cap M_2$ is a 2-bridge sphere of $K$.
Assume first that $r=1$.
Note that $M_1$ contains a unique essential saturated annulus up to isotopy
and that $K$ admits a unique 2-bridge sphere up to isotopy preserving $K$ (see \cite[Theorem 4]{Mor}).
Let $\mu$ and $\lambda$ be the meridian and a longitude of $K$, respectively.
Let $F_0$ be a genus-2 Heegaard surface of $M$.
Then, by \cite[Lemma 6 (2)]{Jan2},
any genus-2 Heegaard surface $F$ of $M$ is isotopic to
$D_{\lambda}^{n/4}(F_0)$ for some integer $n$.
Note that $\D$ is the infinite cyclic group generated by $D_{\lambda}$ (see \cite[Lemma 3]{Jan2}).
Hence, by an argument similar to that in the previous cases,
we see that the number of genus-2 Heegaard surfaces whose hyper-elliptic involutions are $\tau_L$ is finite, a contradiction.
Assume first that $r=2$.
Pick a \lq\lq standard\rq\rq\ genus-2 Heegaard surface $F_0$ of $M$,
such that $F_0\cap M_1$ is preserved by the homeomorphisms $g_1$, $g_2$ and $b$ in Lemma \ref{lem-mcg-sfs2} (1).
Then we may assume that $\tau_{F_0}(=\tau_L)=G_2$.
By Theorem \ref{thm-mcg} (1) and \cite[Lemma 6 (2)]{Jan2},
any genus-2 Heegaard surface $F$ of $M$ is isotopic to $D_1^{n_1}D_2^{n_2}\cdots D_1^{n_{2m-1}}D_2^{n_{2m}}(F_0)$
for some integers $n_1, n_2,\dots,n_{2m-1}$ and $n_{2m}$,
where $n_2,\dots,n_{2m-1}$ are nonzero.
Then $$\tau_F=D_1^{n_1}D_2^{n_2}\cdots D_1^{n_{2m-1}}D_2^{n_{2m}}G_2D_2^{-n_{2m}}D_1^{n_{2m-1}}\cdots D_2^{-n_2}D_1^{-n_1}.$$
Since $G_2D_1G_2=D_2^{-1}$ by Remark \ref{rmk-mcg},
we have $$\tau_F=D_1^{n_1}D_2^{n_2}\cdots D_1^{n_{2m-1}}D_2^{n_{2m}}D_1^{n_{2m}}D_2^{n_{2m-1}}\cdots D_1^{n_2}D_2^{n_1}G_2.$$
By Theorem \ref{thm-mcg} (1), we can see that $\tau_F=\tau_{F_0}$ implies $n_1=n_2=0$,
which means $F$ is isotopic to $F_0$.
This contradicts the assumption.
\begin{subcase2}\label{case2-3-1-2}
$M$ satisfies the condition (M3-1-2).
\end{subcase2}
Note that $M$ is obtained by gluing $M_1\in M\ddot{o}[r]$ $(r=0,1,2)$ and $M_2=E(K)$,
where $K=S(p,q)$ is a hyperbolic 2-bridge knot,
so that the regular fiber of $M_1$ is identified with the meridian loop of $K$.
We may assume that $q$ is odd.
Let $F$ be a genus-2 Heegaard surface of $M$.
By Theorem \ref{thm-kob},
$F\cap M_1$ is the union of two essential saturated annuli which cuts $M_1$ into two solid tori
and $F\cap M_2$ is a 2-bridge sphere of $K$.
Note that $M_1/\langle \tau_F\rangle$ is a solid torus and
that the image of $\fix\tau_F\cap M_1$ forms a link in it
as illustrated in Figure \ref{fig-quotient1}.
\begin{figure}\label{fig-quotient1}
\end{figure}
On the other hand, $M_2/\langle \tau_F\rangle$ is also a solid torus
and the image of $\fix\tau_F\cap M_2$ forms a knot in it
such that its exterior in $M_2$ is the exterior of a 2-bridge link $S(2p,q)$ in $S^3$ (cf. \cite[Lemma 3.2]{Jan}).
Since the meridian and the longitude of the solid torus $M_1/\langle \tau_F\rangle$ are identified with the longitude and the meridian of the solid torus $M_2/\langle \tau_F\rangle$, respectively,
we see that $L$ is equivalent to $L(q/2p;\beta_1/\alpha_1,\beta_2/\alpha_2)$.
Moreover, since $K=S(p,q)$ is a hyperbolic 2-bridge knot,
we have $q\not\equiv \pm 1\pmod{p}$ by \cite{Men}.
First assume that $r=0$.
Note that $M_1$ has a unique essential saturated annulus up to isotopy and
that $K$ admits a unique 2-bridge sphere up to isotopy.
Let $F_0$ be the pre-image of $P^0$ given in \cite{Jan} (cf. Figure \ref{fig-3b-inf})
by the covering map $M\rightarrow S^3$ (branched over $L$).
By an argument similar to that in the previous cases,
$F$ is isotopic to $D_{\lambda}^{n/4}(F_0)$ for some integer $n$,
where $D_{\lambda}$ is the Dehn twist along a component of $T=\partial M_1=\partial M_2$
in the direction of a longitude of $K$.
However, $F=D_{\lambda}^{n/4}(F_0)$ is isotopic to $F_0$ since $D_{\lambda}^{n/4}(F_0)\cap M_1$ can be isotoped to $F_0\cap M_1$ by an isotopy fixing the boundary of $M_1$ as illustrated in Figure \ref{fig-isotopy}.
Thus $M$ admits a unique genus-2 Heegaard surface up to isotopy, a contradiction.
\begin{figure}\label{fig-isotopy}
\end{figure}
Assume that $r=1$ or $r=2$.
Then, we see by \cite{Jan}
that
$L$ is equivalent to a link $L(q/2p;\beta_1/\alpha_1,\beta_2/\alpha_2)$ in Figure \ref{fig-3b-inf}
and that
$L$ admits infinitely many 3-bridge spheres $\{P^i\}_{i\in\Z}$ up to isotopy.
(Moreover, we can see that any 3-bridge sphere is isotopic to $P^i$ for some $i\in\Z$
by using an argument similar to that in the previous cases and by Lemma \ref{lem-mcg-sfs2}.)
\begin{subcase2}\label{case2-3-2}
$M$ satisfies the condition (M3-2-1) or (M3-2-2).
\end{subcase2}
Note that $M$ is obtained by gluing $M_1\in A[r]$ $(r=0,1,2)$ and $M_2=E(K)$,
where $K$ is a 2-bridge link,
so that the regular fiber of $M_1$ is identified with the meridian loop of $K$.
Let $F$ be a genus-2 Heegaard surface of $M$.
By Theorem \ref{thm-kob},
$F\cap M_1$ is the union of two essential saturated annuli which cuts $M_1$ into two solid tori
and $F\cap M_2$ is a 2-bridge sphere of $K$.
(If $M_1$ is homeomorphic to a 2-bridge knot exterior, then $F$ can intersect each $M_i$ so that $F\cap M_1$ is a 2-bridge sphere and $F\cap M_2$ is the union of two essential saturated annuli.)
Pick a \lq\lq standard\rq\rq\ genus-2 Heegaard surface $F_0$ of $M$ and assume that $\tau_{F_0}=\tau_L$.
By using an argument similar to that in the previous cases,
we see the following hold.
\begin{itemize}
\item
If $M$ satisfies the condition (M3-2-1), where $r=0$ or $1$,
then $F$ is isotopic to $D_{\lambda}^{n/2}(F_0)$ for some integer $n$,
where $\lambda$ is a longitude or a meridian of $K$ according as $F_0$ meets $T:=\partial M_1=\partial M_2$ in a meridian or a longitude of $K$.
Note that the subgroup of $\D$ generated by $D_{\lambda}$ is finite or an infinite cyclic group.
If the subgroup is finite, then $M$ admits only finitely many genus-2 Heegaard surfaces up to isotopy, a contradiction.
If the subgroup is an infinite cyclic group, then, by an argument similar to that in Cases 2.\,\ref{case2-1} together with Lemma \ref{lem-d} (4), we see that the number of genus-2 Heegaard surfaces of $M$ whose hyper-elliptic involutions are $\tau_L$ is finite, a contradiction.
\item
If $M$ satisfies the condition (M3-2-1), where $r=2$,
then we see that $F$ is isotopic to $D_l^{n_0/2}D_1'^{n_1}D_2'^{n_2}\cdots D_1'^{n_{2m-1}}D_2'^{n_{2m}}(F_0)$ for some integers $n_i$ ($i=0,1,\dots, 2m$) by using Theorem \ref{thm-mcg} (2).
By an argument similar to that in Case 2.\,\ref{case2-3-1-1} together with Theorem \ref{thm-mcg} (2), we see that $\tau_F=\tau_{F_0}(=\tau_L)$ implies $n_i=0$ for all $i=0,1,\dots, 2m$.
Hence, $F$ is isotopic to $F_0$, a contradiction.
\item
If $M$ satisfies the condition (M3-2-2), where $r=0$,
then $F$ is isotopic to $D_{\lambda}^{n/2}(F_0)$ for some integer $n$,
where $\lambda$ is a longitude of $K$.
By an an argument similar to that in Cases 2.\,\ref{case2-1} together with Lemma \ref{lem-d} (5),
we see that the number of genus-2 Heegaard surfaces of $M$ whose hyper-elliptic involutions are $\tau_L$ is finite, a contradiction.
\item
If $M$ satisfies the condition (M3-2-2), where $r=1$ or $2$,
then we see by \cite{Jan} that
the link $L$ is equivalent to a link $L(q/2p;\beta_1/\alpha_1,\beta_2/\alpha_2)$ in Figure \ref{fig-3b-inf} and that
$L$ admits infinitely many 3-bridge spheres $\{P^i\}_{i\in\Z}$ up to isotopy.
\end{itemize}
\begin{subcase2}\label{case2-4}
$M$ satisfies the condition (M4).
\end{subcase2}
Note that $M$ is obtained by gluing $M_1, M_2\in D[2]$ and $M_3=E(K_1\cup K_2)$, where $K_1\cup K_2$ is a hyperbolic 2-bridge link with components $K_1$ and $K_2$,
so that the regular fiber of $M_i$ is identified with the meridian loop of $K_i$ ($i=1,2$).
Recall from Theorem \ref{thm-kob} that,
for any genus-2 Heegaard surface $F$ of $M$,
$F\cap M_i$ ($i=1,2$) is an essential saturated annulus in $M_1$
and $F\cap M_3$ is the 2-bridge sphere of $K_1\cup K_2$.
Let $D_{\mu_i}$ and $D_{\lambda_i}$ $(i=1,2)$ denote the Dehn twists along $T=\partial M_1=\partial M_2$ in the direction of the meridian and a longitude of $K_i$, respectively.
Note that $K_1\cup K_2$ admits a unique 2-bridge sphere up to isotopy by \cite{Sch},
and that $M_i$ contains a unique essential annulus up to isotopy.
Let $F_0$ be a \lq\lq standard\rq\rq\ genus-2 Heegaard surface of $M$.
Then, by \cite[Lemma 6 (1)]{Jan2},
any genus-2 Heegaard surface $F$ of $M$ is isotopic to
$D_{\lambda_1}^{n_1/2}D_{\lambda_2}^{n_2/2}(F_0)$
for some integers $n_1$ and $n_2$ by \cite[Lemma 6 (1)]{Jan2}.
Since $\tau_{F_0}D_{\lambda_i}\tau_{F_0}=D_{\lambda_i}^{-1}$,
we have, by \cite[Lemma 5]{Jan2},
$$\tau_{D_{\lambda_1}^{n_1/2}D_{\lambda_2}^{n_2/2}(F_0)}
=D_{\lambda_1}^{n_1}D_{\lambda_2}^{n_2}\tau_{F_0}.$$
Since $\D=\langle D_{\lambda_1}, D_{\lambda_2}\rangle\cong\Z^2$ (see Lemma \ref{lem-d} (6)),
we have $D_{\lambda_1}^{n_1}D_{\lambda_2}^{n_2}\tau_{F_0}=\tau_{F_0}$ if and only if $n_1=n_2=0$.
Hence, $M$ admits a unique genus-2 Heegaard surface whose hyper-elliptic involution is $\tau_L$, a contradiction.
This completes the proof of Theorem \ref{thm-main}.
\section{Proof of Proposition \ref{prop-toruslink}}\label{sec-toruslink}
Let $L$ be a torus link $T(3,3n')$ for some nonzero integer $n'$, and
let $K_1$, $K_2$ and $K_3$ be the three components of $L$.
Let $S$ be a 3-bridge sphere of $L$.
Let $T$ be the standard torus in $S^3$ containing $L$,
and let $A_i$ $(i=1,2,3)$ be the closure of a component of $T\setminus L$
bounded by two components of $L$ different from $K_i$.
Let $V_1$ and $V_2$ be the two solid tori in $S^3$ bounded by $T$
such that the meridians of $V_1$ and $V_2$ meet $L$ in three points and $|3n'|$ points, respectively.
Since $S\cap K_i$ consists of two points for each $i=1,2,3$,
$S\cap A_i$ satisfies one of the following conditions (see Figure \ref{fig-int}).
\begin{itemize}
\item[(i)] $S\cap A_i$ contains two non-separating arcs $\gamma_i^1$ and $\gamma_i^2$.
\item[(ii)] $S\cap A_i$ contains two separating arcs $\gamma_i^1$ and $\gamma_i^2$.
\end{itemize}
\begin{figure}\label{fig-int}
\end{figure}
Thus one of the following holds.
\begin{itemize}
\item[(S1)] $S\cap A_i$ satisfies the condition (i) for every $i=1,2,3$.
\item[(S2)] Two of $S\cap A_1$, $S\cap A_2$ and $S\cap A_3$ satisfies the condition (i) and the other satisfies the condition (ii).
\item[(S3)] Two of $S\cap A_1$, $S\cap A_2$ and $S\cap A_3$ satisfies the condition (ii) and the other satisfies the condition (i).
\item[(S4)] $S\cap A_i$ satisfies the condition (ii) for every $i=1,2,3$.
\end{itemize}
Assume that $S$ satisfies the condition (S1).
Note that $\gamma:=\gamma_1^1\cup\gamma_1^2\cup\gamma_2^1\cup\gamma_2^2\cup\gamma_3^1\cup\gamma_3^2$
consists of two loops each of which contains one of the two points $S\cap K_i$ for every $i=1,2,3$.
Suppose there is a loop component, $\delta$, of $S\cap T$ other than $\gamma$.
Then $\delta$ bounds a disk in the interior of $A_i$ disjoint from $\gamma$ for some $i=1,2,3$,
and hence, $\delta\cup K_i$ is a trivial 2-component link for each $i=1,2,3$.
On the other hand, $\delta$ either bounds a disk in $S\setminus \gamma$
or is isotopic to the core of the annulus component of $S\setminus \gamma$.
In the latter case, the linking number of $\delta$ and a component of $L$ is $1$
(see Figure \ref{fig-int2}), a contradiction.
\begin{figure}\label{fig-int2}
\end{figure}
Hence, $\delta$ bounds a disk in $S\setminus \gamma$.
If we assume that $\delta$ is innermost (in $S\setminus \gamma$),
then we can eliminate it from $S\cap T$
since the union of the two disks described above is a 2-sphere bounding a 3-ball in $S^3\setminus L$.
Hence, we may assume that $S\cap T=\gamma$.
Since $\gamma$ cuts $S$ into two disks and an annulus,
it bounds two disks in $V_1$ and bounds an annulus in $V_2$.
Since such a disk is unique up to isotopy in $(V_1,L)$
and an annulus is unique up to isotopy in $(V_2,L)$,
$L$ admits a unique 3-bridge sphere satisfying the condition (S1).
(We obtain two 3-bridge spheres when $n'=\pm 1$, but it can be easily seen that they are isotopic.)
Assume that $S$ satisfies the condition (S2).
Note that $\gamma:=\gamma_1^1\cup\gamma_1^2\cup\gamma_2^1\cup\gamma_2^2\cup\gamma_3^1\cup\gamma_3^2$
is a loop containing the six points $S\cap L$.
Thus any loop component of $S\cap A_i$ except $\gamma$ bounds a disk in $S\setminus\gamma\subset S^3\setminus L$.
This implies that
any loop component of $S\cap A_i$ cannot be a core of the annulus $A_i$ for each $i=1,2,3$,
since the linking number of the core of $A_i$ and a component of $L$ is $n'(\neq 0)$.
Hence, any loop component of $S\cap A_i$ also bounds a disk in the interior of $A_i$ disjoint from $\gamma$,
and hence, we can remove all such components by an isotopy.
Thus we may assume that $S\cap T$ consists of only one loop component $\gamma$.
Since $\gamma$ itself bounds disks in $S$ on both sides,
it must be an inessential loop on $T$.
We show that this case can be reduced to the case where $S$ satisfies the condition (S1).
To this end, let $h:S^3\rightarrow [-2,2]$ be the height function
such that
$S_t:=h^{-1}(t)$ is a 3-bridge sphere of $L$ when $-1<t<1$,
that $S_t$ is a 2-sphere which meets each $K_i$ in one point when $t=\pm 1$,
that $S_t$ is a single point when $t=\pm 2$
and that $S_t$ is a 2-sphere in $S^3\setminus L$ otherwise.
Moreover, we may assume that $S_{0}=S$
and that the restriction $g:=h|_{T}$ of $h$ to $T$ has
at most one non-degenerate singular point
at every level.
Thus, for every singular value $t_0$,
$g^{-1}({t_0})$ contains a maximal point, a minimal point or a saddle point.
We represent each saddle point in $g^{-1}({t_0})$ by an arc on $T$
with endpoints on $g^{-1}({t_0-\varepsilon})$
for sufficiently small $\varepsilon >0$, as in Figure \ref{fig-saddle}.
\begin{figure}\label{fig-saddle}
\end{figure}
Such an arc, $\alpha$, is of one of the following three types
(see Figure \ref{fig-type}):
\begin{figure}
\caption{The dashed and dotted lines give all possible types of an arc $\alpha$ representing a saddle point of $g$.}
\label{fig-type}
\end{figure}
\begin{itemize}
\item
$\alpha$ is of {\it type 1}
if its endpoints are on the same component of $g^{-1}({t_0-\varepsilon})$,
and $g^{-1}({t_0+\varepsilon})$ contains a pair of parallel essential loops on $T$,
\item
$\alpha$ is of {\it type 2}
if its endpoints are on the same component of $g^{-1}({t_0-\varepsilon})$,
and $g^{-1}({t_0+\varepsilon})$ does not contain a pair of parallel essential loops on $T$,
and
\item
$\alpha$ is of {\it type 3}
if its endpoints are on different components of $g^{-1}({t_0-\varepsilon})$.
\end{itemize}
Put $X_{s}:=g^{-1}([-2,s])$ for any $s\in [-2,2]$.
Since $S(=S_{0})$ cuts $T$ into a disk and a 1-holed torus,
we may assume that $X_{0}$ is the disk.
Since $L\subset X_1$, we see that $X_1$ contains an essential loop on $T$.
Thus, there exists a singular value $s_0>0$
and a sufficiently small $\varepsilon>0$
such that
$X_{s_0-\varepsilon}$ does not contain an essential loop on $T$ and
$X_{s_0+\varepsilon}$ contains an essential loop on $T$.
Note that, if $X_{s_0-\varepsilon}$ does not contain an essential loop on $T$
and the arc representing the singular point at $s_0$ is of type 2 or of type 3,
then $X_{s_0+\varepsilon}$ cannot contain an essential loop on $T$.
Thus, the arc representing the singular point at $s_0$ must be of type 1,
and hence, $g^{-1}({s_0+\varepsilon})$ contains a pair of parallel essential loops, say $c$ and $c'$, on $T$.
Note that $c$ bounds a $k$-holed disk in $S':=S_{s_0+\varepsilon}$ disjoint from $g^{-1}({s_0+\varepsilon})\setminus (c\cup(\cup_{i=1}^{k}c_k))$
together with $k$ components $c_1,c_2,\dots,c_k$ of $g^{-1}({s_0+\varepsilon})\setminus (c\cup c')$ for some non-negative integer $k$.
We see that $c$ is null-homologous in $V_i$ for some $i=1,2$
since $c_1,c_2,\dots,c_k$ are inessential loops on $T$.
(By the choice of $s_0$, all components of $g^{-1}(s_0+\varepsilon)$ except $c$ and $c'$ are inessential on $T$.)
Hence, $c$ and $c'$ must be meridian loops of one of the solid tori $V_1$ and $V_2$.
Since each of $c$ and $c'$ meets each component of $L$ in a single point,
it intersects each of the annuli $A_1$, $A_2$ and $A_3$ in a non-separating arc.
Hence, the 3-bridge sphere $S'$, which is isotopic to $S$, satisfies the condition (S1).
Similarly, the cases where $S$ satisfies the condition (S3) or (S4) is reduced to the first case.
This implies that $L$ admits a unique 3-bridge sphere up to isotopy.
This completes the proof of Proposition \ref{prop-toruslink}.
{\bf Acknowledgment}
I would like to express my appreciation to Professor
Makoto Sakuma for his guidance, advice and encouragement.
I would also like to thank
Professor Susumu Hirose for his helpful comments, especially on Section \ref{sec-mcg}.
\end{document}
|
math
|
یہِ ژَلہٕ وٕنۍ راتھ ہے چھےٚ لونچہِ منز گنڈٕنۍ لایق
|
kashmiri
|
All songs performed live by Greg Andrew and Band.
This "Live" album features 12 of the biggest songs from the Elton John Experience show.
These are all legendary songs by Elton John and Burnie Taupin that the audience really respond to.
This concert was recorded before a large audience in the Auditorium at Twin Towns Services Club, Tweed Heads, NSW, Australia.
|
english
|
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