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Randomized Algorithms for the Loop Cutset Problem

Randomized Algorithms for the Loop Cutset Problem. Class presentation for CPSC 689-60 5. Author: Ann Becker, Beuven Bar-Yehuda Dan Geiger. Presented by: Songjian Lu . Content. Introduction FVS problem Algorithm for FVS problem WFVS problem Algorithm for WFVS problem.

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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!

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