\section{Introduction}
Since many light scalar mesons such as $\sigma (600), \kappa (800), f_0(980)$ and $a_0(980)$
were found in experiments \cite{PDG}, a lot of theoretical studies have been devoted to
investigation of their states.
For the structure of light salar mesons, in addition to the conventional two quark state from the quark model,
several possibilities are proposed \cite{Jaffe1}; four-quark states, molecular states and scattering states.
Because the sigma meson is considered as a chiral partner of the pion in the mechanism of hadron mass generation,
it would be interesting to investigate a roll of four-quark states in the mechanism.
The study of four-quark states in light scalar mesons gives us insight of important QCD feature.

Since Alford and Jaffe showed the possibility
that scalar mesons exist as four-quark states on the lattice \cite{Jaffe},
tetraquark search on the lattice started actively.
The pioneer work of the sigma meson which is one of candidates of four-quark states
was done with full QCD by SCALAR collaboration \cite{Scalar}.
They found that disconnected diagrams which contain effectively four-quark states,
glueballs and so on make the sigma meson lighter.
Recently using tetraquark interpolators,
not only ground states but also resonance states of scalar mesons on the lattice were reported \cite{BGR}.

Following the procedure which was proposed by Alford and Jaffe \cite{Jaffe},
we explore the existence of four-quark states in the scalar channel on larger lattice with a finer lattice spacing.
We also investigate the dependence of bound energies for four-quark states
on the ratio of pion mass and rho meson mass ($m_{\pi}/m_{\rho}$)  and on the pion mass ($m_{\pi}$).

\begin{figure}[b]
\vspace{0.35cm}
\centering
\begin{picture}(250,50)
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 \put(-91,52){$D(t)$}
 \put(20,52){$C(t)$}
 \put(130,52){$A(t)$}
 \put(238,52){$G(t)$}

\end{picture}
\caption{The diagrams for four-quark correlators.}
\label{DCAG}
\end{figure}

\section{Four-quark states from Lattice QCD}
We calculate four-quark correlators in two-flavor ($N_f=2$) lattice QCD
under the assumption that four-quark states of scalar mesons are consisted of bound states of the two pions.
We employ the procedure which was proposed by Alford and Jaffe \cite{Jaffe}.
The four-quark correlators of isospin zero ($I=0$) and two ($I=2$) channels are given by,
\begin{eqnarray}
J_{I=0}(t) &=& D(t)+\frac{1}{2}C(t) -3A(t) + \frac{3}{2}G(t) \ , \\
J_{I=2}(t) &=& D(t)-C(t) \ ,
\end{eqnarray}
where $D(t)$, $C(t)$, $A(t)$ and $G(t)$ are corresponding to the diagrams in Fig.\ \ref{DCAG}.
To evaluate the four-quark states clearly,
we carry out the calculation in the quenched  approximation
where two-quark, multi-quark and glueball states do not mix with four-quark states as intermediate states.
We drop the contribution of the diagrams for $A(t)$ and $G(t)$,
ignoring the two-quark annihilation diagrams and the vacuum channels.
To obtain the bound energy $\delta E_{I}$ for the four-quark states,
we construct the ratio of the four-quark correlator $J_I(t)$ and the pion correlator $P(t)$
and fit it to an exponential at large $t$,
\begin{eqnarray}
 R_{I}(t) \ =\                                      \frac{J_{I}(t)}{\left(P(t)\right)^2}
             &\longrightarrow&               \frac{Z_{I}}{Z_{\pi}^2} \exp{(-\delta E_{I}\: t)} + \cdots \ . \\[-0.45cm]
             &{\scriptstyle {t \to \infty}}& \nonumber
\end{eqnarray}
We discuss the possibility that the four-quark states exist as bound states
from the $N_L$ dependence of the bound energies.

If a four-quark state is a bound state,
$\delta E_{I}$ is negative and
would approach to a negative constant in the large $N_L$ region.
On the other hand
if a four-quark state is a scattering state,
it is expected that $\delta E_{I}$ obeys the scattering formula of a two-particle state
in a cube box of size $L$ with periodic boundary condition in Ref.\ \cite{Luscher},
\begin{eqnarray}
 \delta E_{I}
 = E_{I} - 2m_{\pi}
 = \frac{T_{I}}{L^3} \left[ 1 + 2.8373 \left(\frac{m_{\pi}T_{I}}{4\pi L}  \right)
                                         + 6.3752 \left(\frac{m_{\pi}T_{I}}{4\pi L} \right)^2  \right] + {\cal{O}} \left(L^{-6}\right) \ ,
\label{Luscher}
\end{eqnarray}
where $E_{I}$ is the total energy, and $T_{I}$ is the scattering amplitude
which can be written by the scattering length $a_I$ as
\begin{eqnarray}
  T_{I} = - \frac{4\pi a_I}{m_{\pi}} \ .
\label{amp}
\end{eqnarray}
From Eq.(\ref{Luscher}), we find that
$\delta E_{I}$ is proportional to $L^{-3}$ in the region
where the physical spatial lattice size $L=N_L a$ ($a$ is the lattice spacing) is enough large.

\begin{figure}
\begin{center}
\includegraphics[scale=0.455]{delta_EE_mm.eps} \ \ \ \ \ \ \ \ \ \ \ \
\includegraphics[scale=0.455]{delta_EE_m.eps}
\caption{The physical lattice size $L(=N_L a)$ dependence of the energy shifts $\delta E_{I=2}$.
In left figure (a) (right figure (b)), the values of $m_{\pi}/m_{\rho}$ ($m_{\pi}$) are fixed.
In both figures the lines are proportional to $L^{-3}$.
In left figure (a), the three symbols are on the different $L^{-3}$ lines,
though they have the same values of $m_{\pi}/m_{\rho}$.
In right figure (b), the two symbols which have the same values of $m_{\pi}$ are almost on the same $L^{-3}$ line,
though they have the different values of $m_{\pi}/m_{\rho}$. }
\label{EE_mm_m}
\end{center}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[scale=0.455]{delta_EN_mm.eps} \ \ \ \ \ \ \ \ \ \ \ \
\includegraphics[scale=0.455]{delta_EN_m.eps}
\caption{The physical lattice size $L(=N_L a)$ dependence of the energy shifts $\delta E_{I=0}$.
In left figure (a) (right figure (b)), the values of $m_{\pi}/m_{\rho}$ ($m_{\pi}$) are fixed.
In both figures the lines are proportional to $L^{-3}$.
In left figure (a), the three symbols are on the different $L^{-3}$ lines,
though they have the same values of $m_{\pi}/m_{\rho}$.
In right figure (b), the two symbols which have the same values of $m_{\pi}$ are almost on the same $L^{-3}$ line,
though they have the different values of $m_{\pi}/m_{\rho}$. }
\label{mm_m}
\end{center}
\end{figure}

\section{Results}
In the work by Alford et al.\ \cite{Jaffe},
they observed that the $I=2$ channel is clearly the scattering state
but found the possibility that the $I=0$ channel is the four-quark bound state.
However to get the conclusive result for the four-quark bound state in the $I=0$ channel,
further calculations on a larger lattice with finer lattice spacing are required.

We calculate the four-quark correlators on a larger lattice with $N_f=2$ Wilson fermion
using configurations which are produced with both plaquette and Iwasaki gauge actions.
The lattice parameters are shown in Tables \ref{pa1} and \ref{pa2}.
We impose the Dirichlet boundary condition in the temporal direction on the quark fields.

The results for $I=2$ channel are shown in Fig.\ \ref{EE_mm_m}.
In Fig.\ \ref{EE_mm_m} (a),
the quark masses are set as the values of the ratio of pion mass to rho meson mass, $m_{\pi}/m_{\rho} \sim 0.74$
become close to those in Ref.\ \cite{Jaffe} (See Table \ref{para}).
In spite of the same values of $m_{\pi}/m_{\rho}$, the values of $m_{\pi}$ are different among them,
which comes from the lattice artifact due to the large lattice spacings.
In Fig.\ \ref{EE_mm_m} (b),
the bound energies are obtained under the same values of $m_{\pi} \sim 370$ MeV.
We can see that in all cases of Fig.\ref{EE_mm_m} (a) and (b) the symbols are on the $L^{-3}$ lines,
which is the same as that in Ref.\ \cite{Jaffe}.
On the other hand the results for $I=0$ channel are shown in Fig.\ \ref{mm_m}.
In Fig.\ \ref{mm_m} (a), the quark masses are set as the values of the ratio of $m_{\pi}/m_{\rho}$
become close to those in Ref.\ \cite{Jaffe} (See Table \ref{para}).
In Fig.\ \ref{mm_m} (b),
the bound energies are obtained under the same values of $m_{\pi} \sim 370$ MeV.
Again we find that in all cases of Figs.\ref{mm_m} (a) and (b) the symbols are on the $L^{-3}$ lines,
which is contradicted with the result of Ref.\ \cite{Jaffe}.
We can not observe the suggestion of the existence of the bound state in the $I=0$ channel.

In Fig.\ \ref{mm_m} (a)
the three symbols which have the same values of $m_{\pi}/m_{\rho}$ are on the different $L^{-3}$ lines,
but in Fig.\ \ref{mm_m} (b)
the two symbols which have the same values of $m_{\pi}$ are almost on the same $L^{-3}$ line.
It implies that $\delta E_{I}$ depends strongly on $m_{\pi}$ rather than $m_{\pi}/m_{\rho}$,
which is deduced from Eq.(\ref{amp}).
Since the scattering amplitude is proportional to the inversion of $m_{\pi}$,
the absolute value of $\delta E_{I}$ decreases with the increasing amount of $m_{\pi}$.

In Ref.\ \cite{Jaffe} they insisted that
they found the deviation from the $L^{-3}$ line (the black dashed line in Fig.\ \ref{pri})
in the physical lattice size dependence of energy shifts in the $I=0$ channel.
However we can give another explanation to it.
Because of the $m_{\pi}$ differences among data plots,
we can draw the three different $L^{-3}$ lines (the green, red and blue solid lines)
instead of one $L^{-3}$ line (the black dashed line) in Fig.\ \ref{pri}.
It does not suggest that the symbols deviate from the $L^{-3}$ line (the black dashed line),
but suggests that the symbols are on the three different $L^{-3}$ lines, respectively.
It means that there are no bound states in the $I=0$ channel.

\begin{figure}
\begin{center}
\includegraphics[scale=0.455]{delta_EN_pri.eps}
\caption{The physical lattice size $L(=N_L a)$ dependence of the energy shifts $\delta E_{I=0}$.
The lines are proportional to $L^{-3}$.
The color difference corresponds to $m_{\pi}$ difference.
The green, red and blue solid lines stand for $m_{\pi} = 940, 840$ and $790$ MeV, respectively.
The black dashed line was drawn in Fig.5 of Ref.\cite{Jaffe}.}
\label{pri}
\end{center}
\end{figure}

\section{Summary}
We have investigated a four-quark state as a bound state
from the spatial lattice size dependence of the bound energies for a four-quark state.
We found that
the four-quark states in the cases of both $I=0$ and $I=2$ channels exist as no bound states, i.e.,
all bound energies for the four-quark states as a function of spatial lattice size $L$ are on the $L^{-3}$ lines,
which is contradicted with the previous work \cite{Jaffe}.
The symbols which have the same values of $m_{\pi}/m_{\rho}$ are on the different $L^{-3}$ lines,
on the other hand the symbols which have the same values of $m_{\pi}$ are almost on the same $L^{-3}$ line.
This results suggest that
the absolute value of bound energy depends directly on $m_{\pi}$ rather than $m_{\pi}/m_{\rho}$.

This work is a staring point of study of four-quark states in the light scalar mesons.
To reach the conclusive results for the light scalar meson states,
further investigation is needed;
full QCD calculation with light quark masses on larger lattice size,
application of four-quark interpolators, state-of-the-art action and so on.

\begin{table}[b]
\caption{Lattice parameters.
$\kappa$ is the hopping parameter.
We adjust parameters, $\beta$ and $\kappa$ to fix the values of $m_{\pi}/m_{\rho} \sim 0.74$.
See Table \protect \ref{para}. }
\begin{center}
\begin{tabular}{c|c|c|c|c|cc|c}\hline
 \ \ \ Gauge action \ \ \ & $\beta$ & $a \ \scriptstyle{\mathrm{[fm]}}$ & $\kappa$ &$N_{L}^3\times N_{T}$ & \# Conf. \\\hline
Iwasaki    & 2.100 & 0.278(3) \cite{Iwa}    & 0.1666 & $12^3 \times  20\ $ & 150 \\
                &          &                                   &             & $16^3 \times  20\ $ & 120 \\
                &          &                                   &             & $18^3 \times  20\ $ &  78 \\
                &          &                                   &             & $20^3 \times  20\ $ & 120 \\
                &          &                                   &             & $24^3 \times  20\ $ &  78 \\\hline
Iwasaki    & 2.300 &  0.172(4) \cite{Iwa}   & 0.1565 & $12^3 \times  20\ $ & 120 \\
                &          &                                   &             & $16^3 \times  20\ $ &  60 \\
                &          &                                   &             & $20^3 \times  20\ $ &  60 \\
                &          &                                   &             & $24^3 \times  20\ $ &  54 \\\hline
Plaquette & 5.700 & 0.140(4) \cite{Fuku} & 0.1640 & $ \ 8^3 \times  20$ & 1008 \\
                &          &                                   &             & $12^3 \times  20\ $ & 240  \\
                &          &                                   &             & $16^3 \times  20\ $ & 102  \\
                &          &                                   &             & $18^3 \times  20\ $ & 210  \\
                &          &                                   &             & $20^3 \times  20\ $ & 192 \\
                &          &                                   &             & $24^3 \times  20\ $ & 300 \\
                &          &                                   &             & $36^3 \times  20\ $ & 66 \\\hline
\end{tabular}
\end{center}
\label{pa1}
\end{table}

\begin{table}[b]
\caption{Lattice parameters.
We adjust parameters, $\beta$ and $\kappa$ to fix the values of $m_{\pi} \sim 370$ MeV.
See   Table \protect \ref{para}.}
\begin{center}
\begin{tabular}{c|c|c|c|c|cc|c}\hline
 \ \ \ Gauge action \ \ \ & $\beta$ & $a \ \scriptstyle{\mathrm{[fm]}}$ & $\kappa$ &$N_{L}^3\times N_{T}$ & \# Conf. \\\hline
Iwasaki & 2.100 & 0.278(3) \cite{Iwa} & 0.1690  & $12^3 \times  20\ $ & 150   \\
             &           &               &               & $16^3 \times  20\ $ & 120  \\
             &           &               &               & $20^3 \times  20\ $ & 120 \\
             &           &               &               & $24^3 \times  20\ $ &  60 \\\hline
Iwasaki & 2.300 & 0.172(4) \cite{Iwa} & 0.1594  & $12^3 \times  20\ $ &  300 \\
             &           &               &               & $16^3 \times  20\ $ &  180 \\
             &           &               &               & $20^3 \times  20\ $ &  180 \\
             &           &               &               & $24^3 \times  20\ $ &   60 \\\hline
\end{tabular}
\end{center}
\label{pa2}
\end{table}

\begin{table}
\caption{Lattice parameters in Ref.\ \cite{Jaffe}.
They adjust parameters, $\beta$ and $\kappa$ to fix the values of $m_{\pi}/m_{\rho} \sim 0.74$.
See Table \protect \ref{para}. }
\begin{center}
\begin{tabular}{c|c|c|c|c|c|cc|c}\hline
Group & Gauge action &$\beta$ & $a \ \scriptstyle{\mathrm{[fm]}}$ & $\kappa$ &$N_{L}^3\times N_{T}$ & \# Conf. \\\hline
 Gupta et al. \cite{Gupta} &Plaquette               & 6.000 & 0.0762(8)$\!\!$ \cite{Davies} & 0.1540 & $16^3 \times  80\ $ & 35 \\ \hline
 Fukugita et al. \cite{Fuku} &Plaquette             & 5.700 & 0.162(6) \ \cite{Davies} & 0.1640 & $12^3 \times  20\ $ & 70 \\ \hline
 Alford et al. &L$\ddot{\rm{u}}$scher \& Weisz & 1.719 &0.249(5)    & -----      & $\ 6^3 \times  40$   & ----- \\
                    &                                                   &           &                 &             & $\ 8^3 \times  40$   & ----- \\\hline
 Alford et al. &L$\ddot{\rm{u}}$scher \& Weisz & 1.157 & 0.400(4)   &   -----    & $\ 5^3 \times  40$   & ----- \\
                    &                                                   &           &                 &             & $\ 6^3 \times  40$   & ----- \\
                    &                                                   &           &                 &             & $\ 8^3 \times  40$   & ----- \\\hline
\end{tabular}
\end{center}
\label{pa0}
\end{table}

\begin{table}
\caption{The values of $m_{\pi}/m_{\rho}$ and $m_{\pi}$.
In the upper table results of Ref.\ \cite{Jaffe} are shown.
In the lower table results of our calculation are shown.}
\begin{center}
\begin{tabular}{c|c|c}\hline
 data  & $m_{\pi}/m_{\rho}$ &$m_{\pi} \ \scriptstyle{\mathrm{[MeV]}}$ \\\hline
 Gupta                                                                & -----        & 940(30)    \\
 Fukugita                                                            & 0.740(8) & 620(90)    \\
 Alford ($a=0.25$ [fm])                                       & 0.756(5) & 840(11)   \\
 Alford ($a=0.40$ [fm])                                       & 0.756(4) & 790(6) \ \       \\\hline\hline
 Iwasaki ($\beta=2.1, \kappa=0.1666$)             & 0.736(3) & 451(6) \ \    \\
 Iwasaki ($\beta=2.3, \kappa=0.1565$)             & 0.746(6) & 593(16)    \\
 Plaquette ($\beta=5.7, \kappa=0.1640$) \ \ \   & 0.748(4) & 754(12)      \\\hline
 Iwasaki ($\beta=2.1, \kappa=0.1690$)             & 0.646(4) & 368(5) \ \       \\
 Iwasaki ($\beta=2.3, \kappa=0.1594$)             & 0.545(7) & 374(11)    \\\hline
\end{tabular}
\end{center}
\label{para}
\end{table}

\section*{Acknowledgments}
This work is supported in part by
Nagoya University Global COE Program (G07).
Numerical calculations were performed on
the cluster system "$\varphi$" at KMI, Nagoya University.
Grant-in-Aid for Young Scientists (B) (22740156),
Grant-in-Aid for Scientific Research (S) (22224003) and the Kurata Memorial Hitachi
Science and Technology Foundation.


\section{Automunge}

Automunge is an open source python library, available now for pip install, built on top of Pandas \citep{McKinney:10}, SciKit-Learn \citep{Pedregosa:11}, Scipy \citep{2020SciPy-NMeth}, and Numpy \citep{Walt:11}. It takes as input tabular data received in a tidy form \citep{Wickham:14}, meaning one column per feature and one row per observation, and returns numerically encoded sets with infill to missing points, thus providing a push-button means to feed raw tabular data directly to machine learning algorithms. The complexity of numerical encodings may be minimal, such as automated normalization of numerical sets and encoding of categorical, or may include more elaborate feature engineering transformations applied to distinct columns. Generally speaking, the transformations are performed based on a ``fit'' to properties of a column in a designated train set (e.g. based on a set's mean, standard deviation, or categorical entries), and then that same basis is used to consistently and efficiently apply transformations to subsequent designated test sets, such as may be intended for use in inference or for additional training data preparation.

The library consists of two master functions, automunge(.) and postmunge(.). The automunge(.) function receives a raw train set and if available also a consistently formatted test set, and returns a collection of encoded sets intended for training, validation, and inference. The function also returns a populated python dictionary, which we call the postprocess\_dict, capturing all of the steps and parameters of transformations. This dictionary may then be passed along with subsequent test data to the postmunge(.) function for consistent processing on the train set basis, such as for instance may be applied sequentially to streams of data for inference. Because it makes use of train set properties evaluated during a corresponding automunge(.) call instead of directly evaluating properties of the test data, processing of subsequent test data in the postmunge(.) function is very efficient.

Included in the platform is a library of feature engineering methods, which in some cases may be aggregated into sets to be applied to distinct columns. For such sets of transformations, as may include generations and branches of derivations, the order of implementation is designated by passing transformation categories as entries to a set of ``family tree'' primitives described further below.

\section{Categoric Encodings}

In tabular data applications, such as are the intended use for Automunge data preparations, a common tactic for practitioners is to treat categoric sets in a coarse-grained representation for presentation to a training operation. Such aggregations transform each unique categoric entry with distinct numeric encoding constructs, such as one-hot encoding in which unique entries are each represented with their own column for boolean activations, or ordinal encoding of single column integer representations. The Automunge library offers some further variations on categoric encodings [Fig. 1]. The default ordinal encoding `ord3', activated when the number of unique entries exceeds some heuristic threshold, has encoding integers sorted by frequency of occurrence followed by alphabetical, where frequency in general may be considered more useful to a training operation. For sets with a number of unique entries below this threshold, the categoric defaults instead make use of `1010' binary encodings, meaning multi-column boolean activations in which representations of distinct categoric entries may be achieved with multiple simultaneous activations, which has benefits over one-hot encoding of reduced memory bandwidth. Special cases are made for categoric sets with 2 unique entries, which are converted by `bnry' to a single column boolean representation, and sets with 3 unique entries are somewhat arbitrarily applied with `text' one-hot encoding. Note that each transform has a default convention for infill to missing values which may be updated for configure to distinct columns.

\begin{figure}[h]
  \centering

  \includegraphics[width=0.8\linewidth]{"Figure_1_052820.png}
  \caption{Categoric encoding examples}
\end{figure}

The categoric encoding defaults in general are based on assumptions of training models in the decision tree paradigms (e.g. Random Forest, Gradient Boosting, etc.), particularly considering that a binary representation may sacrifice compatibility for an entity embedding of categorical variables [4] as may be applied in the context of a neural network (for which we recommend a seeded representation of `ord3' for ordinal encoding by frequency). The defaults for automation are all configurable, and alternate root categories of transformations may also be specified to distinct columns to overwrite the automation defaults.

The characterization of these encodings as a coarse graining is meant to elude to the full discarding of the structure of the received representations. When we convert a set of unique values \{`circle', `square', `triangle'\} to \{1, 2, 3\}, we are hiding from the training operation any information that might be inferred from grammatical structure, such as the recognition of the prefix ``tri'' meaning three. Consider the word ``Automunge''. You can probably infer from the common prefix ``auto'' that there might be some automation involved, or if you are a data scientist you might recognize the word ``munge'' as referring to data transformations. Naturally we may then ask how we might incorporate practices from NLP into our tabular encodings, such as vectorized representations in a model like Word2Vec \citep{Mikolov:13}. While this is a worthy line for further investigation, some caution is deserved that the validity of such representations is built on a few assumptions, most notably the consistency of vocabulary interpretations between any pre-trained model and the target tabular application. In NLP practice it is common to fine-tune \citep{Howard:16} a pre-trained model to accommodate variation in a target domain. In tabular applications obstacles to this type of approach may arise from the limited context surrounding the categoric entries - which in practice is not uncommon to find entries as single words, or sometimes character sets that aren't even words such as e.g. serial numbers or addresses. We may thus be left without the surrounding corpus vocabulary necessary to fine tune a NLP model for some tabular target, and thus the only context available to extract domain properties may be the other entries shared in a categoric set or otherwise just the surrounding corresponding tabular features.

\section{String Parsing}

Automunge offers a novel means to extract some additional grammatical context from categoric entries prior to encoding by way of comparisons between entries in the unique values of a feature set. The operation is conducted by a kind of string parsing, in which the set of unique entries in a column are parsed to identify character subset overlaps between entries. In its current form the implementation is probably not at an optimal computational efficiency, which we estimate complexity scaling on the order of $\mathcal{O}((LN)\textsuperscript{2})$, where N is the number of unique entries and L is the average character length of those entries. Thus the intended use case is for categoric sets with a bounded range of unique entries. In general, the application of comparable transformations to test data in the postmunge(.) function is materially more computationally efficient than the initial fitting of the transformations to the train set in the automunge(.) function, particularly in variations with added assumptions for test set composition in relation to the corresponding train data.

The implementation to identify character subset overlaps is performed by first inspecting the set of unique entries from the train set, determining the longest string length (-1), then for each entry comparing each subset of that length to every equivalent length subset of the other entries, where if composition overlaps are identified, the results are populated in a data structure matching identified overlaps to their corresponding source column unique entries, and after each overlap inspection cycle incrementing that inspection length by negative steps until a configurable minimum length overlap detection threshold is reached. To keep things manageable, in the base configuration overlaps are limited to a single identification per unique entry, prioritized by longest length. Note also that transformation function parameters may be activated to exclude from overlap detections any subsets with space, punctuation, or other designated character sets such as to promote single word activations.

\begin{figure}[h]
  \centering

  \includegraphics[width=0.9\linewidth]{"Figure_2_053120.png}
  \caption{String parsing for bounded sets}
\end{figure}

There are a few options of these methods to choose from [Fig. 2]. In the first version `splt' any identified substring overlap partitions are given unique columns for boolean activations. In a second version `spl2' the full entries with identified overlaps are replaced with the overlap partitions, resulting in a reduced number of unique entries - such as may be intended as a reduced information content supplement to the original set, which we speculate could be beneficial in the context of a curriculum learning regime \citep{Bengio:09}. A third version `spl5' is similar to the second with distinction that those entries not replaced with identified overlap partitions are instead replaced with an infill plug value, such as to avoid too much redundancy between different configurations derived from the same set. The fourth shown version `sp15' is comparable to `splt' but with the allowance for multiple concurrent activations to each entry, demonstrating a tradeoff between information retention and dimensionality of returned data. Each of these versions have corresponding variants with improved computational efficiency to test set processing with the postmunge(.) function based on incorporating the assumption that the set of unique entries in the test set will be the same or a subset of those found in the train set.

Some further variations include `sp19' in which concurrent activation sets are collectively consolidated with a binary transform to reduce dimensionality. Another variation is available as `sbst' which instead of comparing string character subsets of entries to string character subsets of other entries, only compares string character subsets of entries to complete character sets of other entries.

\section{Parsing Unbounded Sets}

The transformations of the preceding section were somewhat constrained toward use against categoric sets with a bounded range of unique entries in the training data due to the complexity scaling on train set implementation. For cases where categoric sets may be presented with an unbounded range of unique entries, such as in cases for all unique entries or otherwise unbounded, Automunge offers a few more string parsing transformations to extract grammatical structure prior to categoric encodings.

As a first example, categoric entries may be passed to a search function `srch' in which entries are string parsed to identify presence of user-specified substring partitions, which are then presented to machine learning by way of recorded activations in returned boolean columns specific to each search term or alternatively with the set of activations collected into a single ordinal encoded column. Some variations on the search functions include the allowance to aggregate multiple search terms into common activations or variations for improved processing efficiency based on added assumptions of whether the target set has a narrow range of entries. Note that these methods are supported by the Automunge infrastructure allowing a user to pass transformation function specific parameters to those applied to distinct columns or also to reset transformation function parameter defaults in the context of an automunge(.) call.

Another string parsing option suitable for application to both bounded and unbounded categoric sets is intended for purposes of detecting numeric portions of categoric string entries, which are extracted and returned in a dedicated numeric column. Some different numeric formats are supported, including entries with commas via `nmcm', and using the family tree primitives the returned sets may in the same operation be normalized such as with a z-score or min-max scaling. As with other string-parsing methods priority of extracts are given to identified partitions with the longest character length. A comparable operation may instead be performed to extract string partitions from majority numeric entries, although we suggest caution of applying these methods towards numerical sets which include units of measurement for instance, as we recommend reserving engineering domain evaluations for oversight by human intelligence.

\begin{figure}[h]
  \centering

  \includegraphics[width=0.8\linewidth]{"Figure_3_052820.png}
  \caption{Numeric extraction and search for unbounded sets}
\end{figure}

\section{Family Tree Aggregations}

An example composition of string parsing transformation aggregations, including generations and branches of derivations by way of entries to the family tree primitives, are now demonstrated for the root transformation category `or19' [Fig. 4], which is available for assignment to source columns in context of an automunge(.) call. This transformation set is intended to automatically extract grammatical context from tabular data categoric features with a bounded range of unique entries.

\begin{figure}[h]
  \centering

  \includegraphics[width=0.8\linewidth]{"or19_and_or20_treesandcolumns_rev052520.jpg}
  \caption{Example family tree aggregations for bounded categoric sets}
\end{figure}

The sequence of four character keys represent transformation functions based on the transformation categories applied to a column. Note that each key represents a set of functions which may include one for application to train/test set(s) in an automunge(.) call for initial fitting of transformations to properties of a train set or a corresponding function for processing of a comparably formatted test set in postmunge(.) on the basis of properties from the train set. The steps of transformations for each returned column are logged by way of transformation function specific suffix appenders to the original column headers. Note also that some of the intermediate steps of transformations may not be retained in the returned set based on presence of downstream replacement primitive entries to family tree primitives as described further below.

The upstream application of an `UPCS' transform serves the purpose of converting categoric entry strings to all uppercase characters, thus consolidating entries with different case configurations, e.g. a received set with unique entry strings {`usa', `Usa', `USA'} would all be considered an equivalent entry (there may be some domains where this convention is not preferred in which case a user may deactivate a parameter to exclude this step). Adjacent to the `UPCS' transform is a `NArw' which returns boolean activations for those rows corresponding to infill based on the root category ``processdict'' defined type of source column values that will be target for infill (processdict is a data structure discussed further below, whose entries may either be a default or user specified). The included `1010' transform for binary encoding distinguishes all distinct entries prior to any string parsing aggregations to ensure full information retention after the `UPCS' transform both for purposes of ML training and also to support any later inversion operation to recover the original format of data pre-transforms as is supported in the postmunge(.) function. An alternate configuration could replace `1010' with another categoric transform such as one-hot or ordinal encodings. The `nmc7' transformation function is similar to `nmcm' discussed earlier, but only string parses those unique entries which were not found in the train set for a more efficient application in the postmunge(.) function. These numeric extractions are followed by a `nmbr' function for z-score normalization. Note that in some cases a numerical extract, such as those derived here from zip codes of passed addresses, may in fact be more suitable to a categoric encoding instead of numeric normalization. Such alternate configurations may easily be defined with the family tree primitives.

The remaining branches downstream of the `UPCS' start with a `spl9' function performing string parsing replacement operations comparable to the `spl2' demonstrated above, but with the assumption that the set of unique entries in the test data will be the same or a subset of the train set for efficiency considerations. The `spl9' parses through unique entry strings to identify character subset overlaps and replaces entries with the longest identified overlap, which results in returned columns with a fewer number of unique entries by aggregating entries with shared overlaps into a common representation, which may then be numerically encoded such as shown here with an `ord3' transform. The `or19' family tree also incorporates a second tier of string parsing with overlap replacement by use of the `sp10' transform (comparable to `spl5' with similar added assumptions for efficiency considerations as `spl9'). `sp10' differs from `spl9' in that unique entries without overlap are replaced in aggregate with an infill plug value to avoid unnecessary redundancy between encodings, again resulting in a reduced number of unique entries which may then be numerically encoded such as with `ord3'. Note that Fig. 4 also demonstrates an alternate root category configuration as `or20' in which an additional tier of `spl9' is incorporated prior to the `sp10'.

\begin{figure}[h]
  \centering

  \includegraphics[width=0.95\linewidth]{"Figure_5_092020.png}
  \caption{Demonstration of `or19' returned data}
\end{figure}

Fig. 5 demonstrates the numerical encodings as would be returned from the application of the `or19' root category to a small example feature set of categoric strings. It might be worth restating that due to the complexity scaling of the string parsing operation this type of operation is intended preferably for categoric sets with a bounded range of unique entries in the train set. The composition of returned sets are derived based on properties of the source column received in a designated train set, and these same bases are applied to consistently prepare data for test sets, such as sets that may be intended for an inference operation. In other words, when preparing corresponding test data, the same type and order of columns are returned, with equivalent encodings for corresponding entries and equivalent activations for specific string subset overlap partitions that were found in the train set.

\section{Specification}

The specification of transformation set compositions for these methods are conducted by way of transformation category entries to a set of family tree primitives, which distinguish for each transformation category the upstream transformation categories for when that category is applied as a root category to a source column and also the downstream transformation categories for when that category is found as an entry in an upstream primitive with offspring. Downstream primitive entries are treated as upstream primitive entries for the successive generation, and primitives further distinguish source column retention and generation of offspring.

\begin{figure}[h]
  \centering

  \includegraphics[width=0.56\linewidth]{"family_tree_rev_101719.png}
  \caption{Family tree primitives}
\end{figure}

Transformation category family tree sets may be passed to an automunge(.) call by way of a ``transformdict'' data structure, which is complemented by a second ``processdict'' data structure populated for each transformation category containing entries for the associated transformation functions and data properties. The transformdict with transformation category entries to family tree primitives and corresponding processdict transformation function entries associated with various columns returned from the `or19' root category set are demonstrated here [Fig. 7]. Here the single processdict entry of the transformation function associated with a transformation category is an abstraction for the set of corresponding transformation functions to be directed at train and/or test set feature sets.

\begin{figure}[h]
  \centering

  \includegraphics[width=0.9\linewidth]{"String_Thery_Fig_7.png}
  \caption{`or19' specifications}
\end{figure}

\section{Experiments}

Some experiments were run to evaluate comparison between standard tabular categoric representation techniques and parsed categoric encodings. The data set from the IEEE-CIS Kaggle competition \citep{IEEE:19} was selected based on known instances of feature sets containing serial number entries which were expected as a good candidate for string parsing.

The experiments were supported by the Automunge library's feature importance evaluation methods, in which a model is trained on the full feature set to determine a base accuracy on a partitioned validation set, and metrics are calculated by shuffle permutation \citep{Parr:18}, where the target feature has it's entries shuffled between rows to measure damping effect on the resulting accuracy, the delta of which may serve as a metric for feature importance. Automunge actually aggregates two metrics for feature importance, the first metric specific to a source feature by shuffling all columns derived from the same feature, in which a higher metric represents more importance, and the second metric specific to each derived column by shuffling all but the target of the columns derived form the same feature, in which a lower metric represents more relative importance between columns derived from the same feature. The experiment findings discussed below are based on the first metric. Although other auto ML options are supported in the library, this experiment used the base configuration of Random Forest \citep{Breiman:01} for the model.

For the experiment, the training data was paired down to the top ten features based on feature importance in addition to two features selected as targets for the experiments, identified in the data set by `id\_30' and `id\_31'. These two features contained entries corresponding to operating system serial numbers and browser serial numbers associated with origination of financial transactions. By inspection, there were many cases where serial numbers shared portions of grammatical structure, as for example the entries \{`Mac OS X 10\_11\_6', `Mac OS X 10\_7\_5'\} or \{`chrome 62.0', `chrome 49.0'\}. Scenarios were run in which both of these features were treated to different types of encodings, including `text' one-hot encoding, `ord3' ordinal, `1010' binary [Fig 1], and two string parse scenarios, the first with `or19' [Fig 4, 5, 7]  and the second with an aggregation of: `sp19' (string parse with concurrent activations consolidated by binary encoding) supplemented by `nmcm' (numeric extraction) [Fig 3] and `ord3' (ordinal encoding). The feature importance was then evaluated corresponding to each of these encoding scenarios [Table 1].

\begin{table}[h]
\caption{Feature Importance Metric Results}
\label{sample-table}
\begin{center}
\begin{tabular}{lllll}
\multicolumn{1}{c}{\bf Encoding}  &\multicolumn{1}{c}{\bf Category} &\multicolumn{1}{c}{\bf Accuracy} &\multicolumn{1}{c}{\bf `id\_30'} &\multicolumn{1}{c}{\bf `id\_31'}
\\ \hline \\
one-hot       & `text'                              & 0.98029    & 0.00135 & 0.00490 \\
ordinal         & `ord3'                            & 0.98040    & 0.00193 & 0.00581 \\
binary          & `1010'                           & 0.98045    & 0.00245 & 0.00699 \\
string parse & `or19'                            & 0.98082    & 0.00295 & 0.00914 \\
string parse & `sp19'    & 0.98081    & 0.00279 & 0.00924 \\
\end{tabular}
\end{center}
\end{table}

The experiments illustrate some important points about the impact of categoric encodings even outside of string parsing. Here we see that binary encoding materially outperforms one-hot encoding and ordinal encoding. We believe that one-hot encoding is best suited for labels or otherwise just used for interpretability purposes. We believe ordinal encoding is best suited for very high cardinality when a set has large number of entries. We believe binary is the best default for categoric encodings outside of vectorization, and thus serves as our categoric default under automation.

The string parsing was found to have a material benefit to both of our target features. It appears the `or19' version of string parsing was more beneficial to the `id\_30' feature and the `sp19' version to the `id\_31' feature.

Part of the challenge of benchmarking parsed categoric encodings is the nature of the application, in that performance impact of string parsing is highly dependent on data set properties, and not necessarily generalizable to a metric that would be relevant for comparison between different features or data sets. We believe this experiment has successfully demonstrated that string parsing has the potential to train better performing models in tabular applications based on improved model accuracy and feature importance in comparisons for these specific features.

\section{Discussion}

To be clear, we believe the family tree primitives [Fig 6] represent a scientifically novel framework, serving as a fundamental reframing of command line specification for multi-transform sets as may include generations and branches of derivations applied by recursion. They are built on assumptions of tidy data and that derivations are all downstream of a specific target feature set, and are well suited for the final data preprocessing steps applied prior to the application of machine learning in tabular applications. We consider these primitives a universal language for univariate data transformation specification and an improvement on mainstream practice.

Although this paper is being submitted under the subject of NLP, it should be noted that the string parsing methods as demonstrated are kind of a compromise from vocabulary vectorization, intended for tabular applications in esoteric domains with limited context or surrounding language such as could be used to fine-tune a pre-trained model, and thus not suitable for mainstream NLP models like BERT. We have attempted in this work a comprehensive overview of various permutations of string parsing that may be applied for scenarios excluding vectorization. That is not to say that a vectorization may not still be achievable - for instance each of the returned categoric encodings of varying information content returned from `or19' could be fed as input to an entity embedding layer [4] when the returned sets are used to train a model.

Further, this paper is not just intended to propose theory and methods. Automunge is a downloadable software toolkit, and the methods demonstrated here are available now in the library for push-button operation. It really is just as simple as passing a dataframe and designating a root category of `or19' to a target column. We believe the automation of string parsing for categoric encodings is a novel invention that will prove very useful for machine learning researchers and practitioners alike.

The value of the library extends well beyond string parsing. For instance, Automunge is an automated solution to missing data infill. In addition to the infill defaults for each transformation, a user can select for each column other infill options from the library, including ``ML infill'' in which column specific Random Forest models \citep{Breiman:01} are trained from partitioned subsets of the training data to predict infill to train and test sets. For example, when ML infill is applied to the `or19' set, each of the returned subsets will have their own trained infill model.

An important point of value is not just the transformations themselves, but the means by which they are applied between train and test sets. In a traditional numerical set normalization operation for instance, it is not uncommon that each of these sets is evaluated individually for a mean and standard deviation, which runs a risk of inconsistency of transformations between sets, or alternate methods to measure prior to validation set extraction runs the risk of data leakage between training and validation operations. In a postmunge(.) test set application, all of the transformation parameters are derived from corresponding columns in the train set passed to automunge(.) after partitioning validation sets, which in addition to solving these problems of inconsistency and data leakage, we speculate that at data center scale could have material benefit to computational overhead and associated carbon intensity of inference, perhaps also relevant to edge device battery constraints.

Another key point of value for this platform is simply put the reproducibility of data preprocessing. If a researcher wants to share their results for exact duplication to the same data or similar application to comparable data, all they have to do is publish the simple python dictionary returned from an automunge(.) call, and other researchers can then exactly duplicate. The same goes for archival of preprocessing experiments - a source data set need only be archived once, and every performed experiment remains accessible and reproducible with this simple python dictionary.

Beyond the core points of feature engineering and infill, the Automunge library contains several other push-button methods. The goal is to automate the full workflow for tabular data for the steps between receipt of tidy data and returned sets suitable for machine learning application. Some of the options include feature importance evaluation (by shuffle permutation \citep{Breiman:01}), dimensionality reduction (including by means of PCA \citep{Jolliffe:16}, feature importance, and binary encodings), preparation for oversampling in cases of label set class imbalance \citep{Buda:18}, evaluation of data distribution drift between initial train sets and subsequent test sets, and perhaps most importantly the simplest means for consistently and efficiently processing subsequent data with postmunge(.).

Oh, and once you try it out, please let us know.

\subsubsection*{Acknowledgments}
A thank you owed to: the Kaggle IEEE-CIS competition which helped me recognize the potential for string parsing. Mark Ryan who shared a comment in Deep Learning with Structured Data that was inspiration for the ‘UPCS’ transform. Thanks to Stack Overflow, Python, PyPI, GitHub, Colaboratory, Anaconda, and Jupyter. Special thanks to Scikit-Learn, Numpy, Scipy Stats, and Pandas.


\section{Introduction}

A {\em $\lambda$-fold} triple system of order $v$, denoted by TS$_{\lambda}(v)$, is a pair $(V,\cal{B})$ where
$V$ is a $v$-set of points and $\cal{B}$ is a set of $3$-subsets {\em (blocks)} such
that any $2$-subset of $V$ appears in exactly $\lambda$ blocks. An {\em automorphism group} of an TS$_{\lambda}(v)$ is a permutation on $V$ leaving $\cal{B}$ invariant. An TS$_{\lambda}(v)$ is {\em cyclic} if its automorphism group contains a $v$-cycle. If $\lambda=1$, an TS$_{\lambda}(v)$ is called {\em Steiner triple system} and is denoted by STS$(v)$. A cyclic STS$(v)$ is denoted by CSTS$(v)$.

An TS$_\lambda(v)$ is {\em simple} if it contains no repeated blocks. An TS$_\lambda(v)$ is called {\em indecomposable} if its blocks set $\cal{B}$ cannot be partitioned into sets $\mathcal{B}_1$, $\mathcal{B}_2$ of blocks of the form TS$_{\lambda_1}(v)$ and TS$_{\lambda_2}(v)$, where $\lambda_1+\lambda_2=\lambda$ with $\lambda_1,\lambda_2\geq 1$. A cyclic TS$_\lambda(v)$ is called {\em cyclically indecomposable} if its block set $\cal{B}$ cannot be partitioned into sets $\mathcal{B}_1$, $\mathcal{B}_2$ of blocks to form a cyclic TS$_{\lambda_1}(v)$ and TS$_{\lambda_2}(v)$, where $\lambda_1+\lambda_2=\lambda$ with $\lambda_1,\lambda_2\geq 1$.

The constructions of triple systems with the properties cyclic, simple, and indecomposable, were studied by many researchers for one property at a time; for example,
 cyclic triple systems for all $\lambda$s were constructed in \cite{colbourn2, silvesan}, simple for $\lambda=2$
 in \cite{stinson} and simple for every
 $v$ and $\lambda$ satisfying the necessary conditions in \cite{dehon}. Also, some of the properties were combined in studies. For example, in \cite{wang},
 cyclic and simple two-fold triple systems for all admissible orders were constructed, while in \cite{archdeacon, colbournrosa, dinitz, kramer, milici, zhang2}, simple and
 indecomposable designs for $\lambda=2,3,4,5,6$ and all admissible $v$ were constructed. In \cite{zhang}, simple and indecomposable designs
were constructed for all $v\geq 24\lambda -5$ satisfying the necessary conditions. For the general case of $\lambda > 6$, Colbourn and Colbourn \cite{colbourn}
constructed a single indecomposable TS$_{\lambda}(v)$ for each odd $\lambda$. Shen \cite{shen} used Colbourn and Colbourn result and some recursive constructions
to prove the necessary conditions are asymptotically sufficient. Specifically, if $\lambda$ is odd, then there exists a constant $v_0$ depending on $\lambda$ with an
 indecomposable simple TS$_\lambda(v)$ design for all $v\geq v_0$ satisfying the necessary conditions. In \cite{gruttmuller}, the authors constructed two-fold
cyclically indecomposable triple systems for all admissible orders. The authors also checked exhaustively the cyclic triple systems TS$_\lambda(v)$ for
 $\lambda=2,\;v\leq 33$ and $\lambda=3,\;v\leq21$ that are cyclically indecomposable and determined if they are decomposable (to non cyclic) or not.

In $2000$, Rees and Shalaby \cite{reesshalaby} constructed simple indecomposable two-fold cyclic triple systems for all $v\equiv 0,~1,~3,~4,~7,~\textup{and}~9~(\textup{mod}~12)$
 where $v=4$ or $v\geq12$ using Skolem-type sequences. They acknowledged that the analogous problem for $\lambda >2$ is more difficult.

In $1974$, Kramer \cite{kramer} constructed indecomposable three-fold triple systems for all admissible orders. We noticed that Kramer's construction for
$v\equiv 1\;\textup{or}\;5\,(\textup{mod}\;6)$ gives also cyclic and simple designs.

In this paper, we construct three-fold triple systems having the properties of being cyclic, simple, and indecomposable for all admissible orders
$v\equiv 3~(\textup{mod}~6)$, except for $v=9$ and $v=24c+57$, $c\geq 2$.

\section{Preliminaries}

Let $D$ be a multi set of positive integers with $|D|=n$. A {\em Skolem-type sequence of order $n$} is a sequence $(s_1,\ldots,s_t),t\geq 2n$ of $i\in D$ such that
 for each $i\in D$ there is exactly one $j\in \{1,\ldots,t-i\}$ such that $s_j=s_{j+i}=i$. Positions in the sequence not occupied by integers $i\in D$ contain null
 elements. The null elements in the sequence are also called {\em hooks}, {\em zeros} or {\em holes}. As examples, $(1,1,6,2,5,2,1,1,6,5)$ is a Skolem-type sequence
of order $5$ and $(7,5,2,0,2,0,5,7,1,1)$ is a Skolem-type sequence of order $4$.

Some special Skolem-type sequences are described below.

A {\it Skolem sequence of order $n$} is a sequence $S_n=(s_{1},s_{2},\ldots, s_{2n})$ of $2n$ integers which satisfies the conditions:
\begin{enumerate}
\item for every $k\in \{1,2,\ldots,n\}$ there are exactly two elements $s_{i},s_{j}\in S$ such that $s_{i}=s_{j}=k$, and
\item if $s_{i}=s_{j}=k,\;i<j$, then $j-i=k.$
\end{enumerate}
Skolem sequences are also written as collections of ordered pairs $\{(a_i,b_i):1\leq i\leq n,\;b_i-a_i=i\}$ with $\cup_{i=1}^{n}\{a_i,b_i\}=\{1,2,\ldots,2n\}$.

For example, $S_5=(1,1,3,4,5,3,2,4,2,5)$ is a Skolem sequence of order $5$ or, equivalently, the collection $\{(1,2),(7,9),(3,6),(4,8),(5,10)\}$.

Equivalently, a {\it Skolem sequence of order $n$} is a Skolem-type sequence with $t=2n$ and $D=\{1,\ldots,n\}$.

A {\em hooked Skolem sequence of order $n$} is a sequence $hS_n=(s_{1},\ldots, s_{2n-1}, s_{2n+1})$ of $2n+1$ integers which satisfies the above definition, as well as $s_{2n}=0.$

As an example, $hS_6=(1,1,2,5,2,4,6,3,5,4,3,0,6)$ is a hooked Skolem sequence of order $6$ or, equivalently, the collection $\{(1,2),(3,5),(8,11),
(6,10), (4,9),(7,13)\}$.

A {\em (hooked) Langford sequence of length $n$ and defect $d$, $n>d$} is a sequence $L_d^n=(l_{i})$ of $2n$ $(2n+1)$ integers which satisfies:
\begin{enumerate}
\item for every $k\in \{d,d+1,\ldots,d+n-1\}$, there exist exactly two elements $l_{i},l_{j}\in L$ such that $l_{i}=l_{j}=k$,
\item if $l_{i}=l_{j}=k$ with $i<j$, then $j-i=k$,
\item in a hooked sequence $l_{2n}=0$.
\end{enumerate}

We noticed that Kramer's construction \cite{kramer} can be obtained using the canonical starter
 $v-2,v-4,\ldots,3,1,1,3,\ldots,v-4,v-2$ and taking the base blocks $\{0,i,b_i\}(\textup{mod}\;v)|i=1,2,\ldots,\frac{1}{2}(v-1)$. So, Kramer's construction can
 be obtained using Skolem-type sequences.

We prove next, that Kramer's construction for indecomposable triple systems produces simple designs.

\begin{theorem}\cite{kramer}{\label{kramer}}
The blocks $\{\{0,\alpha,-\alpha\}(\textup{mod}\;v)|\alpha=1,\ldots,\frac{1}{2}(v-1)\}$ for $v\equiv 1\;\textup{or}\;5\,(\textup{mod}\;6)$ form a cyclic, simple, and indecomposable
three-fold triple system of order $v$.
\end{theorem}

\begin{proof}
Let $v=6n+1$. The design is cyclic and indecomposable \cite{kramer}. We prove that the cyclic three-fold triple systems produced by
 $\{\{0,\alpha,-\alpha\}(\textup{mod}\;v)|\alpha=1,\ldots,\frac{1}{2}(v-1)\}$  is also simple.

Suppose that the construction above produces $\{x,y,z\}$ as a repeated block. Any block $\{x,y,z\}$ is of the form $\{0,i,6n+1-i\}+k$ for some
 $i=1,2,\ldots,\frac{1}{2}(v-1)$ and $k \in \mathbb {Z}_{6n+1}$. Hence, if $\{x,y,z\}$ is a repeated block we have
$$\{0,i_1,6n+1-i_1\}+k_1=\{0,i_2,6n+1-i_2\}+k_2$$
whence,
$$\{0,i_2,6n+1-i_2\}=\{0,i_1,6n+1-i_1\}+k$$
for some $i_1,i_2\in\{1,2,\ldots,\frac{1}{2}(v-1)\}$ and some $k\in \mathbb{Z}_{6n+1}$.

If $k=0$, we have $i_2=6n+1-i_1$ and $i_1=6n+1-i_2$, which is impossible since $6n+1-i_1> i_2$ and $6n+1-i_2> i_1$ by definition (i.e., $i_1,i_2\in\{1,2,\ldots,3n\}$
 while $6n+1-i_1,\,6n+1-i_2\in\{3n+1,\ldots,6n\}$.

If $k=i_2$, we have $\begin{cases} i_1+i_2=6n+1 \\ 6n+1-i_1+i_2=6n+1-i_2 \end{cases}$ or

$\begin{cases} i_1+i_2=6n+1-i_2 \\ 6n+1-i_1+i_2=6n+1. \end{cases}$

Since both $i_1$ and $i_2$ are at most $3n$, it is impossible to have $i_1+i_2=6n+1$. Also $i_1\neq i_2$.

If $k=6n+1-i_2$ we have $\begin{cases} i_1+6n+1-i_2=6n+1 \\ 6n+1-i_1+6n+1-i_2=i_2+6n+1 \end{cases}  \Leftrightarrow  \linebreak \begin{cases} i_1+i_2=6n+1-i_1 \\ 6n+1-i_2+i_1=6n+1 \end{cases}$ or $\begin{cases} i_1+6n+1-i_2=i_2 \\ 6n+1-i_1+6n+1-i_2=6n+1. \end{cases}$

Since $6n+1-i_2 > i_2$ , it is impossible to have $i_1+6n+1-i_2=i_2$.

It follows that our design is simple. The case for $v=6n+5$ is similar. \end{proof}

In order to completely solve the case $\lambda=3$, we have new constructions that give cyclic, simple, and indecomposable three-fold triple systems for $v\equiv 3~(\textup{mod}\;6)$, $v\neq 9$ and
$v\neq 24c+57$, $c\geq 2$.

\section{Simple Three-fold Cyclic Triple Systems}

\begin{lemma} \label{lem1} For every $n\equiv 0 ~ \textup{or} ~1 ~(\textup{mod}\,4)$, $n\geq 8$, there is a Skolem sequence of order $n$ in
which $s_1=s_2=1$ and $s_{2n-2}=s_{2n}=2$.

\end{lemma}

\begin{proof}

To get a Skolem sequence of order $n$ for $n\equiv 0 ~ \textup{or} ~1 ~(\textup{mod}\,4)$, $n\geq 8$, take $(1,1,hL_3^{n-2})$, replace the hook with
a $2$ and add the other $2$ at the end of the sequence.

For $n=8$, take $hL_3^6=(8,3,5,7,3,4,6,5,8,4,7,0,6)$, for $n=12$ take $hL_3^{10}=(9,11,3,12,\linebreak 4, 3,7, 10,4,9,8,5,11,7,6,12,5,10,8,0,6)$ and for the
 remaining $hL_3^{n-2}$ hook a $hL_4^{n-3}$ (see \cite{simpson}, Theorem 2, Case 1) to $(3,0,0,3)$.

For $n\equiv 1~(\textup{mod}~4)$, $n\geq9$, take $hL_3^{n-2}$ (see \cite{simpson}, Theorem 2, Case 1).
\end{proof}

\begin{ex}
From the above lemma we have $S_8=(1,1,8,3,5, 7,3,4,6,5,8,4,7,2,6,2)$, $S_{12}=(1,1,9,11,3,12,4,3,7, 10,4,9,8,5,11,7,6, 12,5,10,8,2,6,
2)$ and $S_{16}=(1,1,9,6,4,14,\linebreak 15,11,4,6,13,9,16,7,12,10,8, 5,11,14,7,15,5,13,8,10,12,3,16,2,3,2)$.
\end{ex}

We use the following construction to get cyclic TS$_3(2n+1)$ for $n\equiv 0$ or $1$ (mod $4$):

\begin{construction} \cite{silvesan} {\label {c5}}
Let $S_n=(s_1,s_2,\ldots,s_{2n})$ be a Skolem sequence of order $n$ and let $\{(a_i,b_i)|1\leq i\leq n\}$ be the pairs of positions
 in $S_n$ for which $b_i-a_i=i$. Then the set $\cal{F}$=$\{\{0,i,b_i\}|1\leq i\leq n\}(\textup{mod}\; 2n+1)$ is a $(2n+1,3,3)-DF$. Hence, the set of triples in $\cal{F}$ form the base blocks of a cyclic TS$_3(2n+1)$.
\end{construction}

Then, we apply Construction \ref{c5} to the Skolem sequences given by Lemma \ref{lem1} to get cyclic three-fold triple systems that are simple and indecomposable.

\begin{construction}{\label{c7}}
Let $S_n=(s_1,s_2,\ldots,s_{2n})$ be a Skolem sequence of order $n$ given by Lemma \ref{lem1}, and let $\{(a_i,b_i)|1\leq i\leq n\}$ be the pairs of positions
 in $S_n$ for which $b_i-a_i=i$. Then the set $\cal{F}$=$\{\{0,i,b_i\}|1\leq i\leq n\}(\textup{mod}\; 2n+1)$ form the base blocks of a cyclic, simple, and
 indecomposable TS$_3(2n+1)$.
\end{construction}

\begin{ex}
If we apply Construction \ref{c7} to the Skolem sequence of order $8$: $(1,1,8,3,5,7,\linebreak 3,4,6,5,8,4,7,2,6,2)$ we get the base blocks
$\{\{0,1,2\}, \{0,2,16\},
\{0,3,7\},\{0,4,12\},\{0,5,10\},\linebreak \{0,6,15\},\{0,7,13\},\{0,8,11\}\} (\textup{mod}~17)$. These base blocks form a cyclic TS$_3(17)$ by Construction \ref{c5}.
We are going to prove next that this design is also indecomposable and simple.
\end{ex}

\begin{theorem} \label{th1}
The TS$_3(6n+3)$, $n\geq2$ produced by applying Construction \ref{c7} are simple, except for $v=24c+57$, $c\geq 2$.
\end{theorem}

\begin{proof}

Let $v=2n+1$, $n\equiv 0 ~ \textup{or} ~1\,(\textup{mod}\,4)$, $n\geq 8$.

Suppose that the construction above produces $\{x,y,z\}$ as a repeated block. With regards to Construction \ref{c7}, any block $\{x,y,z\}$ is of the
form $\{0,i,b_i\}+k$ for some $i=1,2,\ldots,n$ and $k \in \mathbb {Z}_{2n+1}$. Hence, if $\{x,y,z\}$ is a repeated block we have
$$\{0,i_1,b_{i_1}\}+k_1=\{0,i_2,b_{i_2}\}+k_2$$
whence,
$$\{0,i_2,b_{i_2}\}=\{0,i_1,b_{i_1}\}+k$$
for some $i_1,i_2\in\{1,2,\ldots,n\}$ and some $k\in \mathbb{Z}_{2n+1}$.

If $k=0$, we have $i_2=b_{i_1}$ and $i_1=b_{i_2}$ which is impossible since $b_{i_1}\geq i_1+1$ and $b_{i_2}\geq i_2+1$ from the definition of a Skolem sequence.

If $k=i_2$, we have $\begin{cases} i_1+i_2=2n+1 \\ b_{i_1}+i_2=b_{i_2} \end{cases}$ or $\begin{cases} i_1+i_2=b_{i_2} \\ b_{i_1}+i_2=2n+1. \end{cases}$

Since both $i_1$ and $i_2$ are at most $n$, it is impossible to have $i_1+i_2=2n+1$.

If $k=b_{i_2}$, we have $\begin{cases} i_1+b_{i_2}=2n+1 \\ b_{i_1}+b_{i_2}=i_2+2n+1 \end{cases} \Leftrightarrow \begin{cases} i_1+i_2=b_{i_1} \\ b_{i_2}+i_1=2n+1 \end{cases}$ or $\begin{cases} i_1+b_{i_2}=i_2 \\ b_{i_1}+b_{i_2}=2n+1. \end{cases}$

Since $b_{i_2} > i_2$ , it is impossible to have $i_1+b_{i_2}=i_2$.

So, to prove that a system has no repeated blocks is enough to show that $\begin{cases} i_1+i_2=b_{i_2}
\\ b_{i_1}+i_2=2n+1 \end{cases}$ or $\begin{cases} i_1+i_2=b_{i_1} \\ b_{i_2}+i_1=2n+1 \end{cases}$ are not satisfied.
Also, we show that $i=\frac{v}{3}$ and $b_{i}=\frac{2v}{3}$ is not allowed, which means that our systems has no short orbits.

For $n=8$ and $n=12$, it is easy to see that the Skolem sequences of order $n$ given by Lemma \ref{lem1} produce simple designs.

For $n\equiv 0(\textup{mod}~4),~n\geq 16$, let $S_n$ be the Skolem sequence given by Lemma \ref{lem1}. This Skolem sequence is constructed using
the hooked Langford sequence $hL_4^{n-3}$ from \cite{simpson}, Theorem 2, Case 1. Since $d=4$, will use only lines $(1)-(7),(14),(8*),(10*)$ and
$(11*)$ in Simpson's Table. Note that $n-3=9+4r$ in Simpson's Table, so $n=12+4r$ and $v=25+8r$, $r\geq 1$ in this case. Because we add the pair
$(1,1)$ at the beginning of the Langford sequence $hL_4^{n-3}$, $a_i$ and $b_i$ will be shifted to the right by two positions. To make it easier
for the reader, we give in Table \ref{hL4n-3} the $hL_4^{n-3}$ taken from Simpson's Table and adapted for our case.

\begin{table}[!h]
\begin{center}
\begin{tabular}{ccccc}
\hline
& $a_i+2$ & $b_i+2$ & $i=b_i-a_i$ & $0\leq j\leq $\\
\hline
$(1)$ & $2r+3-j$ & $2r+7+j$ & $4+2j$ & $r$\\
$(2)$ & $r+2-j$ & $3r+9+j$ & $2r+7+2j$ & $r-1$\\
$(3)$ & $6r+12-j$ & $6r+17+j$ & $5+2j$ & $r-1$\\
$(4)$ & $5r+12-j$ & $7r+18+j$ & $2r+6+2j$ &$r$\\
$(5)$ & $3r+8$ & $7r+17$ & $4r+9$ &-\\
$(6)$ & $4r+9$ & $8r+21$ & $4r+12$ &-\\
$(7)$ & $2r+6$ & $6r+13$ & $4r+7$ & -\\
$(14)$ & $2r+5$ & $6r+16$ & $4r+11$ & -\\
$(8*)$ & $4r+11$ & $8r+19$ & $4r+8$ & -\\
$(10*)$ & $4r+10$ & $6r+15$ & $2r+5$ & -\\
$(11*)$ & $2r+4$ & $6r+14$ & $4r+10$ & -\\
\hline

\end{tabular}
\caption{$hL_4^{n-3}$} \label{hL4n-3}
\end{center}
\end{table}
So, the base blocks of the cyclic designs
produced by Construction \ref{c7} are $\{0,1,2\},\{0,2,v-1\},\{0,3,v-2\}$ and $\{0,i,b_i+2\}$ for $i=4,\ldots,n$ and $i=b_i-a_i$.

We show first that $i=\frac{v}{3}$ and $b_{i}+2=\frac{2v}{3}$ is not allowed in the above system. In the first three base blocks is obvious
 that $i\neq \frac{v}{3}$. For the remaining base blocks we check lines $(1)-(7),(14),(8*),(10*)$ and $(11*)$ in Table \ref{hL4n-3}.

Line $(1)$ $\begin{cases} 4+2j=\frac{25+8r}{3} \\ 2r+7+j=\frac{2(25+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(2)$ $\begin{cases} 2r+7+2j=\frac{25+8r}{3} \\ 3r+9+j=\frac{2(25+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(3)$ $\begin{cases} 5+2j=\frac{25+8r}{3} \\ 6r+17+j=\frac{2(25+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(4)$ $\begin{cases} 2r+6+2j=\frac{25+8r}{3} \\ 7r+18+j=\frac{2(25+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(5)$ $\begin{cases} 4r+9=\frac{25+8r}{3} \\ 7r+17=\frac{2(25+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(6)$ $\begin{cases} 4r+12=\frac{25+8r}{3} \\ 8r+21=\frac{2(25+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(7)$ $\begin{cases} 4r+7=\frac{25+8r}{3} \\ 6r+13=\frac{2(25+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(14)$ $\begin{cases} 4r+11=\frac{25+8r}{3} \\ 6r+16=\frac{2(25+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(8*)$ $\begin{cases} 4r+8=\frac{25+8r}{3} \\ 8r+19=\frac{2(25+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(10*)$ $\begin{cases} 2r+5=\frac{25+8r}{3} \\ 6r+15=\frac{2(25+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(11*)$ $\begin{cases} 4r+10=\frac{25+8r}{3} \\ 6r+14=\frac{2(25+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Therefore, this systems has no short orbits.

Next, we have to check that $\begin{cases} i_1+i_2=b_{i_2}
\\ b_{i_1}+i_2=2n+1 \end{cases}$ or $\begin{cases} i_1+i_2=b_{i_1} \\ b_{i_2}+i_1=2n+1 \end{cases}$ are not satisfied.

Lines $(1)-(1)$: $\begin{cases} 4+2j_1+4+2j_2=2r+7+j_2 \\ 4+2j_2+2r+7+j_1=25+8r \end{cases} \Leftrightarrow \begin{cases} j_1=\frac{-2r-15}{3} \\
j_2=\frac{10r+27}{3} \end{cases}$ which is impossible since $j_1\geq 0$ and also integer.

Lines $(1)-(2)$: $\begin{cases} 4+2j_1+2r+7+2j_2=3r+9+j_2 \\ 2r+7+2j_2+2r+7+j_1=25+8r \end{cases} \Leftrightarrow \begin{cases} j_1=\frac{-2r-15}{3} \\ j_2=r-2-2j_1
\end{cases}$
which is impossible since $j_1\geq 0$ and also integer.

Lines $(1)-(3)$: $\begin{cases} 4+2j_1+5+2j_2=6r+17+j_2 \\ 5+2j_2+2r+7+j_1=25+8r \end{cases} \Leftrightarrow \begin{cases} j_1=j_2-5 \\ j_2=2r+9\end{cases}$
which is impossible since $j_2\leq r-1$.

Lines $(1)-(4)$: $\begin{cases} 4+2j_1+2r+6+2j_2=7r+18+j_2 \\ 2r+6+2j_2+2r+7+j_1=25+8r \end{cases} \Leftrightarrow \begin{cases} j_1=j_2+r-4 \\ j_2=\frac{3r+16}{3} \end{cases}$
which is impossible since $j_2\leq r$.

Lines $(1)-(5)$: $\begin{cases} 4r+2j+13=7r+17 \\ j+6r+16=25+8r \end{cases} \Leftrightarrow \emptyset$.

Lines $(1)-(6)$: $\begin{cases} 4r+2j+16=8r+21 \\ j+6r+19=25+8r \end{cases} \Leftrightarrow \emptyset$.

Lines $(1)-(7)$: $\begin{cases} 4r+2j+11=6r+13 \\ j+6r+14=25+8r \end{cases} \Leftrightarrow \emptyset$.

Lines $(1)-(14)$: $\begin{cases} 4r+2j+15=6r+16 \\ j+6r+18=25+8r \end{cases} \Leftrightarrow \emptyset$.

Lines $(1)-(8*)$: $\begin{cases} 4r+2j+12=8r+19 \\ j+6r+15=25+8r \end{cases} \Leftrightarrow \emptyset$.

Lines $(1)-(10*)$: $\begin{cases} 2r+2j+9=6r+15 \\ j+4r+12=25+8r \end{cases} \Leftrightarrow \emptyset$.

Lines $(1)-(11*)$: $\begin{cases} 4r+2j+14=6r+14 \\j+6r+17=25+8r \end{cases} \Leftrightarrow \emptyset$.

We implemented a program in Mathematica that checks all the pairs of rows in Simpson's table using the above approach. The code for the program and the results can be found in Appendix.
From the results, we can easily see that if we check any combination of two lines in Simpson's Table the conditions are not satisfied in almost all of the cases. There are
two cases where this conditions are satified. The first case is when we check line 3 with line 1, and we get that for $r=4+3c$, $j_1=2c$, and $j_2=6+2c$, $c\geq 2$ the system is
not simple. This implies that our system is not simple when $v=24c+57$, $c\geq2$. The second case is when we check line 3 with line 2. Here, we get $r=5$ and therefore $v=59$.
But $v=59$ is not congruent to $3$ (mod $6$). A TS$_3(59)$ is simple, cyclic, and indecomposable by Theorem \ref{kramer}.

For $n\equiv 1(\textup{mod}~4), n\geq 9$, let $S_n$ be the Skolem sequence given by Lemma \ref{lem1}. This Skolem sequence is constructed
 using a $hL_3^{n-2}$ (\cite{simpson}, Theorem 2, Case 1). Since $d=3$, will use only lines
 $(1)-(6),(14),(7'),(8')$ and $(10')$ in Simpson's Table. Note that $n-2=7+4r$ in Simpson's Table, so $n=9+4r$ and $v=19+8r$ in this case.
 Because we add the pair $(1,1)$ at the beginning of the Langford sequence $hL_3^{n-2}$, $a_i$ and $b_i$ will be shifted to the right
 by two positions.

Table \ref{hL3n-2} gives the $hL_3^{n-2}$ from Simpson's Table adapted to our case.

\begin{table}[!h]
\begin{center}
\begin{tabular}{ccccc}
\hline
& $a_i+2$ & $b_i+2$ & $i=b_i-a_i$ & $0\leq j\leq $\\
\hline
$(1)$ & $2r+3-j$ & $2r+6+j$ & $3+2j$ & $r$\\
$(2)$ & $r+2-j$ & $3r+8+j$ & $2r+6+2j$ & $r-1$\\
$(3)$ & $6r+10-j$ & $6r+14+j$ & $4+2j$ & $r-1$\\
$(4)$ & $5r+10-j$ & $7r+15+j$ & $2r+5+2j$ &$r$\\
$(5)$ & $3r+7$ & $7r+14$ & $4r+7$ &-\\
$(6)$ & $4r+8$ & $8r+17$ & $4r+9$ &-\\
$(14)$ & $2r+4$ & $6r+12$ & $4r+8$ & -\\
$(7')$ & $2r+5$ & $6r+11$ & $4r+6$ & -\\
$(10')$ & $4r+9$ & $6r+13$ & $2r+4$ & -\\
\hline

\end{tabular}
\caption{$hL_3^{n-2}$} \label{hL3n-2}
\end{center}
\end{table}

So, the base blocks of the cyclic designs produce by Construction \ref{c7} are $\{0,1,2\},\{0,2,v-1\}$ and $\{0,i,b_i+2\}$ for
$i=3,\ldots,n$. Using the same argument as before, we show that these designs are simple.

First, we show that $i=\frac{v}{3}$ and $b_{i}+2=\frac{2v}{3}$ is not allowed in the above system. In the first two base blocks is obvious
 that $i\neq \frac{v}{3}$. For the remaining base blocks we check lines $(1)-(6),(14),(7')$ and $(10')$ in Table \ref{hL3n-2}.

Line $(1)$ $\begin{cases} 3+2j=\frac{19+8r}{3} \\ 2r+6+j=\frac{2(19+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(2)$ $\begin{cases} 2r+6+2j=\frac{19+8r}{3} \\ 3r+8+j=\frac{2(19+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(3)$ $\begin{cases} 4+2j=\frac{19+8r}{3} \\ 6r+14+j=\frac{2(19+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(4)$ $\begin{cases} 2r+5+2j=\frac{19+8r}{3} \\ 7r+15+j=\frac{2(19+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(5)$ $\begin{cases} 4r+7=\frac{19+8r}{3} \\ 7r+14=\frac{2(19+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(6)$ $\begin{cases} 4r+9=\frac{19+8r}{3} \\ 8r+17=2\frac{19+8r}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(14)$ $\begin{cases} 4r+8=\frac{19+8r}{3} \\ 6r+12=\frac{2(19+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(7')$ $\begin{cases} 4r+6=\frac{19+8r}{3} \\ 6r+11=\frac{2(19+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(10')$ $\begin{cases} 2r+4=\frac{19+8r}{3} \\ 6r+13=\frac{2(19+8r)}{3} \end{cases} \Leftrightarrow \emptyset$.

Next, we have to check that $\begin{cases} i_1+i_2=b_{i_2}
\\ b_{i_1}+i_2=2n+1 \end{cases}$ or $\begin{cases} i_1+i_2=b_{i_1} \\ b_{i_2}+i_1=2n+1 \end{cases}$ are not satisfied. As with the previous case, the results can be found in Appendix. As before, when we check line 3 and line 1 the conditions are satisfied. But, in this case $v=24c+35$, $c\geq1$ which is not congruent to $3$ (mod $6$).
So, a TS$_3(24c+35)$ for $c\geq 1$ is cyclic, simple, and indecomposable by Theorem \ref{kramer}.\end{proof}

\begin{lemma} \label{lem2} For every $n\equiv 2 ~ \textup{or} ~3 ~(\textup{mod}\,4)$, $n\geq 7$, there is a hooked Skolem sequence of order $n$ in which $s_1=s_2=1$ and
 $s_{2n-1}=s_{2n+1}=2$.
\end{lemma}

\begin{proof}

For $n\equiv 2 ~ \textup{or} ~3 ~(\textup{mod}\,4)$, $n\geq 7$, take hS$_n=(1,1,L_3^{n-2},2,0,2)$.

When $n\equiv 2~(\textup{mod}~4)$, take $L_3^{n-2}$ (\cite{simpson}, Theorem 1, Case 3). When $n\equiv 3~(\textup{mod}~4)$, take
$L_3^5=(6,7,3,4,5,3,6,4,7,5) $ and for $n\geq 11$ take  $L_3^{n-2}$ (see \cite{bermond}, Theorem 2).
\end{proof}

We are going to use the following construction to get cyclic TS$_3(2n+1)$ for $n\equiv 2$ or $3$ (mod $4$):

\begin{construction} \cite{silvesan} {\label{c6}}
Let $hS_n=(s_1,s_2,\ldots,s_{2n-1},s_{2n+1})$ be a hooked Skolem sequence of order $n$ and let $\{(a_i,b_i)|1\leq i\leq n\}$ be
 the pairs of positions in $hS_n$ for which $b_i-a_i=i$. Then the set $\cal{F}$=$\{\{0,i,b_i+1\}|1\leq i\leq n\}(\textup{mod}\;2n+1)$ is a $(2n+1,3,3)-DF$. Hence, the set of triples in $\cal{F}$ form the base blocks of a cyclic TS$_3(2n+1)$.
\end{construction}

Then, we apply Construction \ref{c6} to the hooked Skolem sequences given by Lemma \ref{lem2} to get cyclic TS$_3(2n+1)$ for $n\equiv 2$ or $3$ (mod $4$) that are simple and indecomposable.

\begin{construction} {\label{c8}}
Let $hS_n=(s_1,s_2,\ldots,s_{2n-1},s_{2n+1})$ be a hooked Skolem sequence of order $n$ given by Lemma \ref{lem2}, and let $\{(a_i,b_i)|1\leq i\leq n\}$ be
 the pairs of positions in $hS_n$ for which $b_i-a_i=i$. Then, the set $\cal{F}$=$\{\{0,i,b_i+1\}|1\leq i\leq n\}(\textup{mod}\;2n+1)$ form the base
 blocks of a cyclic, simple, and indecomposable TS$_3(2n+1)$.
\end{construction}

\begin{ex}
If we apply Construction \ref{c8} to the hooked Skolem sequence of order $7$: $(1,1,6,7,3,4,5,3,6,4,7,5,2,0,2)$ we get the base blocks $\{\{0,1,3\},\linebreak \{0,2,1\},
\{0,3,9\},\{0,4,11\}, \{0,5,13\},\{0,6,10\},\{0,7,12\}\} (\textup{mod}~15)$. These base blocks form a cyclic TS$_3(15)$ by Construction
\ref{c6}. We are going to prove next, that this design is indecomposable and simple.
\end{ex}

\begin{theorem} \label{th2}
The TS$_3(6n+3)$, $n\geq 2$, produced by applying Construction \ref{c8}, are simple.
\end{theorem}

\begin{proof}
The proof is similar to Theorem \ref{th1}. Let $v=2n+1$, $n\equiv 2 ~ \textup{or} ~3\,(\textup{mod}\,4)$, $n\geq 10$.

For $n\equiv 2(\textup{mod}~4),~n\geq 10$, let $hS_n$ be the hooked Skolem sequence given by Lemma \ref{lem2}. This hooked Skolem sequence is
constructed using the Langford sequence $L_3^{n-2}$ from \cite{simpson}, Theorem 1, Case 3. Since $d=3$, will use only lines $(1)-(4),(6),(9),(11)$ and
$(13)$ in Simpson's Table. Note that $m=n-2=4r$ in Simpson's Table, so $n=4r+2$, $v=8r+5$, $r\geq 2$, $d=3$, $s=1$, in this case. Because we add the pair $(1,1)$ at the
beginning of the hooked Langford sequence $hL_3^{n-2}$, $a_i$ and $b_i$ will be shifted to the right by two positions. To make it easier for the reader, we
give in Table \ref{L3n-2}, the $L_3^{n-2}$ taken from Simpson's Table and adapted for our case (omit row $(1)$ when $r=2$).

\begin{table}[!h]
\begin{center}
\begin{tabular}{ccccc}
\hline
& $a_i+2$ & $b_i+2$ & $i=b_i-a_i$ & $0\leq j\leq $\\
\hline
$(1)$ & $2r-j$ & $2r+4+j$ & $4+2j$ & $r-3$\\
$(2)$ & $r+2-j$ & $3r+3+j$ & $2r+1+2j$ & $r-1$\\
$(3)$ & $6r+1-j$ & $6r+4+j$ & $3+2j$ & $r-2$\\
$(4)$ & $5r+2-j$ & $7r+4+j$ & $2r+2+2j$ &$r-2$\\
$(6)$ & $2r+3$ & $4r+3$ & $2r$ &-\\
$(9)$ & $3r+2$ & $7r+3$ & $4r+1$ &-\\
$(11)$ & $2r+1$ & $6r+3$ & $4r+2$ & -\\
$(13)$ & $2r+2$ & $6r+2$ & $4r$ & -\\
\hline
\end{tabular}
\caption{$L_3^{n-2}$} \label{L3n-2}
\end{center}
\end{table}
So, the base blocks of the cyclic designs
produce by Construction \ref{c8} are $\{0,1,3\},\{0,2,1\}$ and $\{0,i,b_i+2+1\}$ for $i=3,\ldots,n$ and $i=b_i-a_i$.

First, we show that $i=\frac{v}{3}$ and $b_{i}+2+1=\frac{2v}{3}$ is not allowed in the above system. In the first two base blocks is obvious
 that $i\neq \frac{v}{3}$. For the remaining base blocks we check lines $(1)-(4),(6),(9),(11)$ and $(13)$ in Table \ref{L3n-2}.

Line $(1)$ $\begin{cases} 4+2j=\frac{8r+5}{3} \\ 2r+5+j=\frac{2(8r+5)}{3} \end{cases} \Leftrightarrow r=\frac{1}{4}$ which is impossible since
$r\geq 2$ and also integer.

Line $(2)$ $\begin{cases} 2r+1+2j=\frac{8r+5}{3} \\ 3r+4+j=\frac{2(8r+5)}{3} \end{cases} \Leftrightarrow r=\frac{1}{2}$ which is impossible since
$r\geq 2$ and also integer.

Line $(3)$ $\begin{cases} 3+2j=\frac{8r+5}{3} \\ 6r+5+j=\frac{2(8r+5)}{3} \end{cases} \Leftrightarrow r=-\frac{1}{2}$ which is impossible since
$r\geq 2$ and also integer.

Line $(4)$ $\begin{cases} 2r+2+2j=\frac{8r+5}{3} \\ 7r+5+j=\frac{2(8r+5)}{3} \end{cases} \Leftrightarrow r=-\frac{3}{4}$ which is impossible since
$r\geq 2$ and also integer.

Line $(6)$ $\begin{cases} 2r=\frac{8r+5}{3} \\ 4r+4=\frac{2(8r+5)}{3} \end{cases} \Leftrightarrow \emptyset$. Line $(9)$ $\begin{cases} 4r+1=\frac{8r+5}{3} \\ 7r+4=2\frac{8r+5}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(11)$ $\begin{cases} 4r+2=\frac{8r+5}{3} \\ 6r+4=\frac{2(8r+5)}{3} \end{cases} \Leftrightarrow \emptyset$. Line $(13)$ $\begin{cases} 4r=\frac{8r+5}{3} \\ 6r+3=\frac{2(8r+5)}{3} \end{cases} \Leftrightarrow \emptyset$.

Next, we have to show that $\begin{cases} i_1+i_2=b_{i_2}
\\ b_{i_1}+i_2=2n+1 \end{cases}$ or $\begin{cases} i_1+i_2=b_{i_1} \\ b_{i_2}+i_1=2n+1 \end{cases}$ are not satisfied. The results for this can be found Appendix. As before, when we check line $3$ with line $1$, the conditions are satisfied. But $v=24c+5$, $c\geq2$ in this case which is not congruent to $3$ (mod $6$). So,
by Theorem \ref{kramer}, there exists a cyclic, simple, and indecomposable TS$_3(24c+5)$ for $c\geq 2$.

For $n\equiv 3(\textup{mod}~4), n\geq 11$, let $hS_n$ be the hooked Skolem sequence given by Lemma \ref{lem2}. This hooked Skolem sequence is constructed
 using a $L_3^{n-2}$ (\cite{bermond}, Theorem 2). Since $d=3$ will use only lines
 $(1)-(4),(6)-(10)$ in \cite{bermond}. Note that $m=n-2=4r+1, r\geq 2$, $e=4$ in \cite{bermond}, so $n=4r+3$ and $v=8r+7$ in this case.
 Because we add the pair $(1,1)$ at the beginning of the Langford sequence $L_3^{n-2}$, $a_i$ and $b_i$ will be shifted to the right
 by two positions.

Table \ref{L3n-2germa} gives the $L_3^{n-2}$ from \cite{bermond} adapted to our case.

\begin{table}[!h]
\begin{center}
\begin{tabular}{ccccc}
\hline
& $a_i+2$ & $b_i+2$ & $i=b_i-a_i$ & $0\leq j\leq $\\
\hline
$(1)$ & $2r+2-j$ & $2r+6+j$ & $4+2j$ & $r-2$\\
$(2)$ & $r+2-j$ & $3r+5+j$ & $2r+3+2j$ & $r-2$\\
$(3)$ & $3$ & $4r+4$ & $4r+1$ & -\\
$(4)$ & $2r+4$ & $4r+5$ & $2r+1$ &-\\
$(6)$ & $r+3$ & $5r+5$ & $4r+2$ &-\\
$(7)$ & $2r+5$ & $6r+5$ & $4r$ &-\\
$(8$ & $2r+3$ & $6r+6$ & $4r+3$ & -\\
$(9)$ & $6r+4-j$ & $6r+7+j$ & $3+2j$ & $r-2$\\
$(10)$ & $5r+4-j$ & $7r+6+j$ & $2r+2+2j$ & $r-2$\\
\hline

\end{tabular}
\caption{$L_3^{n-2}$} \label{L3n-2germa}
\end{center}
\end{table}

So, the base blocks of the cyclic designs produce by Construction \ref{c8} are $\{0,1,3\},\{0,2,1\}$ and $\{0,i,b_i+2+1\}$ for
$i=3,\ldots,n$. Using the same argument as before, we show that these designs are simple.

First we show that $i=\frac{v}{3}$ and $b_{i}+2+1=\frac{2v}{3}$ is not allowed in the above system. In the first two base blocks is obvious
 that $i\neq \frac{v}{3}$. For the remaining base blocks we check lines $(1)-(4),(6)-(10)$ in Table \ref{L3n-2germa}.

Line $(1)$ $\begin{cases} 4+2j=\frac{8r+7}{3} \\ 2r+7+j=\frac{2(8r+7)}{3} \end{cases} \Leftrightarrow r=\frac{3}{4}$ which is impossible since
$r\geq 2$ and also integer.

Line $(2)$ $\begin{cases} 2r+3+2j=\frac{8r+7}{3} \\ 3r+6+j=\frac{2(8r+7)}{3} \end{cases} \Leftrightarrow r=\frac{1}{2}$ which is impossible since
$r\geq 2$ and also integer.

Line $(3)$ $\begin{cases} 4r+1=\frac{8r+7}{3} \\ 4r+5=\frac{2(8r+7)}{3} \end{cases} \Leftrightarrow \emptyset$. Line $(4)$ $\begin{cases} 2r+1=\frac{8r+7}{3} \\ 4r+6=\frac{2(8r+7)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(6)$ $\begin{cases} 4r+2=\frac{8r+7}{3} \\ 5r+6=\frac{2(8r+7)}{3} \end{cases} \Leftrightarrow \emptyset$. Line $(7)$ $\begin{cases} 4r=\frac{8r+7}{3} \\ 6r+6=\frac{2(8r+7)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(8)$ $\begin{cases} 4r+3=\frac{8r+7}{3} \\ 6r+7=\frac{2(8r+7)}{3} \end{cases} \Leftrightarrow \emptyset$. Line $(9)$ $\begin{cases} 3+2j=\frac{8r+7}{3} \\ 6r+8+j=
\frac{2(8r+7)}{3} \end{cases} \Leftrightarrow \emptyset$.

Line $(10)$ $\begin{cases} 2r+2+2j=\frac{8r+7}{3} \\ 7r+7+j=\frac{2(8r+7)}{3} \end{cases} \Leftrightarrow \emptyset$.

Next, we have to check that $\begin{cases} i_1+i_2=b_{i_2}
\\ b_{i_1}+i_2=2n+1 \end{cases}$ or $\begin{cases} i_1+i_2=b_{i_1} \\ b_{i_2}+i_1=2n+1 \end{cases}$ are not satisfied. The results for this can be found in Appendix. Here, for $v=3c-1$, $c\geq 4$ and for $v=55$ the conditions are satisfied but these orders are not congruent to $3$ (mod $6$). Therefore, by Theorem \ref{kramer},
 there exists cyclic, simple, and indecomposable TS$_3(3c-1)$ for $c\geq 4$ and cyclic, simple, and indecomposable TS$_3(55)$.
\end{proof}

\section{Indecomposable Three-fold  Cyclic Triple Systems}

In this section, we prove that the Constructions \ref{c7} and \ref{c8} produce indecomposable three-fold triple systems for $v\equiv~3~(\textup{mod}~6)$,
$v\geq 15$.

\begin{theorem}\label{ind1}
The TS$_3(v)$ produced by Constructions \ref{c7} and \ref{c8} are indecomposable for every $v\equiv~3~(\textup{mod}~6)$, $v\geq 15$.
\end{theorem}

\begin{proof}
Suppose that $v\equiv~3~(\textup{mod}~6)$ and write $v=2n+1$, $n\equiv 0 ~ \textup{or} ~1\,(\textup{mod}\,4)$, $n\geq 8$.

Now, for an TS$_3(2n+1)$ to be decomposable, there must be a Steiner triple system STS($2n+1$) inside the TS$_3(2n+1)$.

If $2n+1\equiv ~3~(\textup{mod}\,6)$, let $\{x_i,x_j,x_k\}$ be a triple using symbols from N$_{2n+1}=\{0,1,\ldots,2n\}$. Let
 $d_{ij}=\textup{min}~\{|x_i-x_j|,2n+1-|x_i-x_j|\}$ be the difference between $x_i$ and $x_j$. An STS($2n+1$) on N$_{2n+1}$ must
 have a set of triples with the property that each difference $d$, $1\leq d\leq n$, occurs exactly $2n+1$ times. Assume there is an
 STS($2n+1$) inside our TS$_3(2n+1)$ and let $f_{\alpha}$ be the number of triples inside the STS($2n+1$) which are a cyclic shift of $\{0,\alpha,b_{\alpha}\}$.

It is enough to look at the first two base blocks of our TS$_3(2n+1)$. These are $\{0,1,2\}$ $(\textup{mod} ~ 2n+1)$ and $\{0,2,2n\}~(\textup{mod} ~ 2n+1)$.
Then the existence of an STS($2n+1$) inside our TS$_3(2n+1)$ requires that the equation $2f_1+f_2=2n+1$ must have a solution in nonnegative integers (we need
 the difference $1$ to occur exactly $2n+1$ times).

{\bf Case \mbox {\boldmath $1$}: \mbox {\boldmath $f_1=1$}}

Suppose we choose one block from the orbit $\{0,1,2\}(\textup{mod}\;2n+1)$. Since this orbit uses the difference $1$ twice and the difference $2$, and the
 orbit $\{0,2,2n\}(\textup{mod} ~ 2n+1)$ uses the differences $1,\,2$ and $3$, whenever we pick one block from the first orbit we cannot choose three blocks from
 the second orbit (i.e., those blocks where the pairs $(0,1)$, $(0,2)$ and $(1,2)$ are included). So, we just have $2n-2$ blocks in the second orbit to choose from.
But we need $2n-1$ blocks from the second orbit in order to cover difference $1$ exactly $2n+1$ times.

Therefore, we have no solution in this case.

{\bf Case \mbox {\boldmath$2$}: \mbox {\boldmath $f_1=2$}}

Since $f_2=\frac{2n-3}{2}$ is not an integer, we have no solution in this case.

{\bf Case \mbox {\boldmath$3$}: \mbox {\boldmath $f_1=3,5,\ldots, n\,(\textup{or}\,n-1)$}}

Similar to Case $1$. So, there is no solution in this case.

{\bf Case \mbox {\boldmath$4$}: \mbox {\boldmath $f_1=4,6,\ldots, n\,(\textup{or}\,n-1)$}}

Similar to Case $2$. So, there is no solution in this case.

{\bf Case \mbox {\boldmath$5$}: \mbox {\boldmath $f_1=0$}}

Note that our cyclic TS$_3(v)$ has no short orbits (Theorem \ref{th1}), while a cyclic STS$(v)$ will have a short orbit. Therefore, if a design
exists inside our TS$_3(v)$, that design is not cyclic.

Now, we choose no block from the first orbit and all the blocks in the second orbit (i.e., $f_1=0,\;f_2=2n+1$). Therefore differences $1$, $2$ and $3$ are all
 covered each exactly $2n+1$ in the STS$(v)$. From the remaining $n-2$ orbits $\{0,i,b_i\},\,i\geq 3$ there will be two or three orbits which will use differences $2$ and $3$.
Since differences $1$, $2$ and $3$ are already covered, we cannot choose any block from those orbits that uses these three differences. So, we are left with $n-4$
 or $n-5$ orbits to choose from.
We need to cover differences $4,5,\ldots,n$ ($n-3$ differences) each exactly $v=2n+1$ times.

We form a system of $n-3$ equations with $n-4$ or $n-5$ unknowns in the following way: when a difference appears in different orbits,
 the sum of the blocks that we choose from each orbit has to equal $v$, i.e., if difference $4$ appears in $\{0,5,b_5\}$, $\{0,7,b_7\}$ and $\{0,10,b_{10}\}$ we have
 $f_5+f_7+f_{10}=v$ or
 if difference $4$ appears in $\{0,7,b_7\}$ twice and in $\{0,9,b_9\}$ once, we have $2f_7+f_9=v$. The system that we form has two or three entries in each row non-zero
 while all the others entries will equal zero. The rows in the system can be rearranged so that we get an upper triangular matrix. Therefore, the system of equations is non-singular and it has the unique solution
$f_{i_1}=f_{i_2}=\ldots =f_{i_k}=v$ for some $4\leq i_1,i_2,\ldots,i_k\leq n$ and $f_{j_1}=f_{j_2}=\ldots =f_{j_k}=0$ for some $4\leq j_1,j_2,\ldots,j_k\leq n$. But this
 implies that the STS$(v)$ inside our TS$_3(v)$ is cyclic, which is impossible.

Therefore, we have no solution in this case. It follows that our TS$_3(2n+1)$ is indecomposable.

Now, suppose that $v=2n+1$, $n\equiv 2 ~ \textup{or} ~3 ~(\textup{mod} ~4)$, $n\geq 7$. Let $f_{\alpha}$ be the number of triples inside the STS($2n+1$) which are a cyclic shift of $\{0,\alpha,b_{\alpha}+1\}$.
 Using the same argument as before, it is easy to show that the equation $2f_2+f_1=2n+1$ has no solution. Therefore our TS$_3(2n+1)$ is indecomposable.
\end{proof}

\section{Cyclic, Simple, and Indecomposable Three-fold Triple Systems}

\begin{theorem}
There exists cyclic, simple, and indecomposable three-fold triple systems, TS$_3(v)$, for every $v\equiv 1~(\textup{mod}~2)$, $v\geq 5, v\neq 9$ and $v\neq 24c+57$, $c\geq 2$.
\end{theorem}

\begin{proof}

Let $v\equiv 1~\textup{or}~5~(\textup{mod}~6)$ and take the base blocks $\{0,\alpha,-\alpha\}(\textup{mod}\;v)|\alpha=0,1,\ldots,\frac{1}{2}(v-1)$. By Theorem
\ref{kramer}, these will be the base blocks of a cyclic, simple, and indecomposable three-fold triple system.

Let $v\equiv 3~(\textup{mod}~6)$, and write $v=2n+1$, $n\equiv 0 ~ \textup{or} ~1\,(\textup{mod}\,4)$, $n\geq 8$. Apply Construction \ref{c7} to the Skolem sequence
of order $n$ given by Lemma \ref{lem1}. These designs are cyclic by Construction \ref{c5}, simple for all $v$ except $v=24c+57$, $c\geq2$ by Theorem \ref{th1} and indecomposable by Theorem \ref{ind1}.

Let $v\equiv 3~(\textup{mod}~6)$, and write $v=2n+1$, $n\equiv 2 ~ \textup{or} ~3\,(\textup{mod}\,4)$, $n\geq 7$. Apply Construction \ref{c8} to the hooked Skolem
 sequence of order $n$ given by Lemma \ref{lem2}. These designs are cyclic by Construction \ref{c6}, simple by Theorem \ref{th2} and indecomposable by
 Theorem \ref{ind1}.
\end{proof}

\section{Conclusion and Open Problems}

We constructed, using Skolem-type sequences, three-fold triple systems having all the properties of being cyclic, simple and indecomposable,
 for $v\equiv~3~(\textup{mod}~6)$ except for $v=9$ and $v=24c+57$, $c\geq2$. Our results, together with Kramer's results \cite{kramer}, completely solve the problem of finding
three-fold triple systems having three properties: cyclic, simple, and indecomposable with some possible exceptions for $v=9$ and $v=24c+7$, $c\geq2$.

In our approach of finding cyclic, simple, and indecomposable three-fold triple systems, proving the simplicity of the designs was a
tedious and long task. Another approach that we tried was in constructing three disjoint (i.e. no two pairs in the same
positions) sequences of order $n$ and taking the
 base blocks $\{0,i,b_i+n\}, i=1,2,\ldots,n$. These base blocks form a cyclic TS$_3(6n+1)$. Also, someone can take three disjoint
hooked sequences of order $n$ and taking the base blocks
 $\{0,i,b_i+n\}, i=1,2,\ldots,n$ together with the short orbit $\{0,\frac{v}{3},\frac{2v}{3}\}$. These base blocks form a cyclic
TS$_3(6n+3)$.

\begin{ex}
For $n=5$, take the three disjoint hooked sequences of order $n$:
\begin{center}
$(1,1,4,1,1,0,4,2,3,2,0,3)$

$(2,3,2,3,3,0,3,4,1,1,0,4)$

$(4,5,5,5,4,0,5,5,5,2,0,2)$
\end{center}

Then the base blocks $\{0,1,7\},\{0,1,10\},\{0,1,15\},\{0,2,8\},\{0,2,15\},\linebreak \{0,2,17\},\{0,3,10\},
\{0,3,12\},\{0,3,17\},\{0,4,10\},\{0,4,17\},\{0,4,12\},\linebreak \{0,5,12\},\{0,5,13\},\{0,5,14\},\{0,11,22\}$ are the base blocks of a cyclic, simple, and indecomposable TS$_3(33)$.
\end{ex}

The simplicity is easy to prove here since the three hooked sequences that we used share no pairs in the same positions. On the other
hand, on first inspection, to prove the indecomposability of such designs appears to be more difficult. Also, someone needs to find
three disjoint such sequences for all admissible orders $n$.

\begin{problem}
Can the above approach of finding cyclic, simple, and indecomposable three-fold triple systems of order $v$ be generalized for all admissible orders?
\end{problem}

\begin{problem}
Are there cyclic, simple, and indecomposable TS$_3(24c+57)$, $c\geq 2$?
\end{problem}

\begin{problem}
Are there cyclic, simple, and indecomposable designs for $\lambda\geq 4$ and all admissible orders?
\end{problem}

\begin{problem}
For $\lambda\geq 3$ what is the spectrum of those $v$ for which there exists a cyclically indecomposable but decomposable cyclic TS$_\lambda(v)$?
\end{problem}

\begin{ex}

Let $V=\{0,1,2,3,4,5,6,7,8\}$. Then the blocks of a cyclic TS$_3(9)$ are:

$\{0,1,2\}, \{0,2,7\}, \{0,3,6\}, \{0,4,8\}$

$\{1,2,3\}, \{1,3,8\}, \{1,4,7\}, \{1,5,0\}$

$\{2,3,4\}, \{2,4,0\}, \{2,5,8\}, \{2,6,1\}$

$\{3,4,5\}, \{3,5,1\}, \{3,6,0\}, \{3,7,2\}$

$\{4,5,6\}, \{4,6,2\}, \{4,7,1\}, \{4,8,3\}$

$\{5,6,7\}, \{5,7,3\}, \{5,8,2\}, \{5,0,4\}$

$\{6,7,8\}, \{6,8,4\}, \{6,0,3\}, \{6,1,5\}$

$\{7,8,0\}, \{7,0,5\}, \{7,1,4\}, \{7,2,6\}$

$\{8,0,1\}, \{8,1,6\}, \{8,2,5\}, \{8,3,7\}$

This design is cyclic, simple, and is decomposable. The blocks $\{0,1,2\},\linebreak \{3,4,5\},\{6,7,8\}, \{1,3,8\},\{4,6,2\},\{7,0,5\},\{0,3,6\},\{1,4,7\},\{2,5,8\},\linebreak \{0,4,8\},
 \{3,7,2\},\{6,1,5\}$ form an STS$(9)$. The designs is cyclically indecomposable since no CSTS$(9)$ exists.
\end{ex}