Finding a Length-Constrained Maximum-Density Path in a Tree Rung-Ren Lin, Wen-Hsiung Kuo, and Kun-Mao Chao.

Presentation on theme: "Finding a Length-Constrained Maximum-Density Path in a Tree Rung-Ren Lin, Wen-Hsiung Kuo, and Kun-Mao Chao."— Presentation transcript:

Finding a Length-Constrained Maximum-Density Path in a Tree Rung-Ren Lin, Wen-Hsiung Kuo, and Kun-Mao Chao

Overview  Introduction of the problem.  A brief discussion of one-dimensional sequence.  Two approaches of finding the maximum- density path in a tree.

Input & Output  Input ： A weighted tree with n edges and a lower bound L.  The lower bound is necessary.  Output ： A maximum-density path with length at least L.

The Density of a Path  The density of a path is defined as follow ：  Given a path with k edges E={e 1, e 2, …, e k }, and w(e) is an edge weight function. The density of such path is defined as

Lower Bound L = 4 3 2 3 3 5 6 8 7 5 6 3 1 5 1 4 2 1 8 4 2 7 4 8 Max-Density = ( 7+4+6+5+8 ) / 5 = 6

The Efficiency  We propose two efficient algorithms reach O(nL) time.  One of them is further modified to solve some special cases such as full m-ary trees in linear time.

One-Dimensional Sequence  The maximum-average segment problem arises naturally in several areas of sequence analysis.  For example, given a DNA sequence, which segment of the sequence of length at least L has the highest GC ratio.  The efficiency of the naïve algorithm of one- dimensional sequence is O(n 2 ).

Lower Bound L  There exists an optimal segment of length at most 2L-1.  It can be proved by a counter argument.  The naïve algorithm is O(nL) now. AB

The Relation between Two Overlapped Sequences  Let P[x] denotes the maximum-density of those segments that start from x.  For a given j { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4183782/slides/slide_9.jpg", "name": "The Relation between Two Overlapped Sequences  Let P[x] denotes the maximum-density of those segments that start from x.", "description": " For a given j

Right-Skew Segment  A sequence S={s 1, s 2, …, s k } is right-skew if and only if the average of any prefix {s 1, s 2, …, s i ) is never larger than the average of the remaining subsequence {s i+1, …, s k ).  According to the tricks mentioned above, Goldwasser et al. proposed a linear time algorithm for one-dimensional sequence.

Right-skew Decomposition 5397738918635 4 5 6 7 Decreasing right-skew decomposition is unique

Finding Maximum-Density Path In a Tree  The path in a tree is similar to the segment of an one-dimensional sequence. That is, there exists a maximum-density path of length at most 2L-1.  We propose two approaches reach the time complexity of O(nL).

Downward & Upward Paths  We classify the paths that start from node K into two types called downward and upward paths.  One is to stretch downward to its children only, called downward paths. And the other is to include at least its parent, called upward paths.

An upward path of node K K A downward path of node K A general tree

Notation  Let denotes the maximum-density downward path of node K of length i.  Let denotes the second-best one. If there is a tie, choose an arbitrary path.  Similarly, represent the maximum upward path of node K of length i.

3 5 3 4 3 6 9 2 7 6 3 5 8 4 2 1 8 4 2 7 4 8 8 A

Contributor  The node which determines its parent’s maximum-density downward path is called contributor.  Each maximum-density downward path of a given length has its own contributor.  The upward path has no contributor because each node has at most one parent.

3 5 3 4 3 6 9 2 7 6 3 5 8 4 2 1 8 4 2 7 4 8 8 A B C D node D node C node D node B

Downward & Upward Table  The downward table of node K is composed of and, where 1 ≦ i ≦ 2L-1.  Note that there exists an optimal path with length at most 2L-1  Similarly, the upward table of node K is composed of, where 1 ≦ i ≦ 2L-1.

Constructing Downward Table  For a given internal node K with m children {K 1, K 2, …, K m }. Let e j denotes the edge (K, K j ).  We can construct the downward table of node K by bottom-up dynamic programming.  All contributors should be recorded.

Constructing Upward Table  Suppose K’ is the parent of node K, and e’ denote the edge (K’, K).  Upward Table of node K can be constructed by top-down dynamic programming.

Time Complexity  Deciding a given length of downward and upward path of any node is O(1).  Thus, constructing downward and upward table of a node takes O(L).  There are totally n nodes in a tree, so it take O(nL) time to complete all downward and upward tables.

Approach I ： Finding the Path from its End Node  Once the downward and upward table of each node are constructed, then we can determine the maximum-density path of a given node with length from L to 2L-1 in O(L).  Therefore, it takes O(nL) time to check all nodes since there are n nodes in such tree.

Approach II ： Finding the Path from its LCA Node  LCA stands for least common ancestor.  We are now combining two downward paths together.

Download ppt "Finding a Length-Constrained Maximum-Density Path in a Tree Rung-Ren Lin, Wen-Hsiung Kuo, and Kun-Mao Chao."

Similar presentations
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}