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Randomized Algorithms for the Loop Cutset Problem

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reynold
12 diapositives · 246 Views · Feb 24, 2013

Explore the Loop Cutset Problem and its applications, deterministic vs. randomized algorithms, FVS and WFVS problems, and their respective algorithms. Understand the basic ideas and definitions related to Feedback Vertex Set (FVS) and Weighted Feedback Vertex Set (WFVS) problems. Learn about the efficient randomized algorithms for solving these challenging problems.

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Randomized Algorithms for the Loop Cutset Problem
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  1. Randomized Algorithms for the Loop Cutset Problem Class presentation for CPSC 689-605 Author: Ann Becker, Beuven Bar-Yehuda Dan Geiger Presented by: Songjian Lu

  2. Content • Introduction • FVS problem • Algorithm for FVS problem • WFVS problem • Algorithm for WFVS problem

  3. Introduction • FVS or WFVS problems have many applications. Such as solving dead-lock. • Deterministic algorithm for FVS • O(17k4)!n) —Bodlaender 1990. • O((2k+1)kn2) —Downey & Fellows 1995. • O(10.567kn3) — Dehne, Fellows et al. 2005. • Randomized algorithm • O(4kkn) for FVS. —Becker et al. 2000. • O(6kkn) for WFVS. —Becker et al. 2000.

  4. Definition of FVS problem • Definition: Given a graph G=(V,E), decide if there is a subset of vertices FV such that each cycle in G passes through at least one vertex in F.

  5. Basic idea for the algorithm of FVS problem • Degree 1 vertices in G can be deleted. • Degree 2 vertices in G can be bypassed (delete degree 2 vertex and connect its two neighbors with an edge). • A vertex has loop must be in F.

  6. Basic idea for the algorithm of FVS problem • Given G=(V,E) that each vertex has degree at lease 3 and has no self loops, let F be the feedback set, X=V-F. EX be the set of edges that both ends are in X, EF,X be the set of edges that one end is in X and another end is in F. Then, |Ex|≤|EF,X| (i.e. |Ex|≤|E|/2). X=V-F Feedback set F

  7. Basic idea for the algorithm of FVS problem • Given a graph G=(V,E) that each vertex has degree at least 3 and no self loops. Suppose F is a feedback set of size K. • Randomly choose an edge e from E, the probability that eEF,X is 1/2. Then randomly choose one vertex u from e, the probability that uF is 1/2. So the total probability that uF is 1/4. • After choosing k vertices randomly, the probability that all chosen k vertices are in F is 1/4k. So if we try c4ktimes, the probability that we find a feedback set F is 1-(1-1/4k)c4k.

  8. Definition of WFVS problem • Definition: Given a graph G=(V,E) and each vertex has a positive weight, decide if there is a subset of vertices FV such that each cycle in G passes through at least one vertex in F. And if G has a feedback set of size k, find a feedback set of size k with minimum weight (The weight of a feedback set is the sum of weight of all vertices in the feedback set).

  9. Basic idea for the algorithm of WFVS problem • Degree 1 vertices in G can be deleted. • A vertex has loop must be in F. • Degree 2 vertices in G can be bypassed ?-----NO. • If two degree 2 vertices are connected, we can bypass the one with larger weight. • A vertex has loop must be in F.

  10. Basic idea for the algorithm of WFVS problem • Given G=(V,E) that each vertex has degree at lease 2 , no two degree 2 vertices are connected and no self loops, let F be the feedback set, X=V-F. EX be the set of edges that both ends are in X, EF,X be the set of edges that one end is in X and another end is in F. Then, |Ex|≤2|EF,X| (i.e. |Ex| ≤ 2|E|/3). • |Ex| ≤ 2|Ex’| ≤2|EF,X’|= 2|EF,X|. X=V-F X’=V’-F Feedback set F Feedback set F

  11. Basic idea for the algorithm of WFVS problem • Given a graph G=(V,E) that each vertex has degree at least 2, no two degree 2 vertices are connected, no self loops and each vertex has a positive weight. Suppose F is a feedback set of size K with minimum weight. • Randomly choose an edge e from E, the probability that eEF,X is 1/3. Then randomly choose one vertex u from e, the probability that uF is 1/2. So the total probability that uF is 1/6. • After choosing k vertices randomly, the probability that all chosen k vertices are in F is 1/6k. So if we try c6ktimes, the probability that we find a feedback set F is 1-(1-1/6k)c6k.

  12. Question? Thank you very much!