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

Content Introduction FVS problem Algorithm for FVS problem WFVS problem Algorithm for WFVS problem

Introduction FVS or WFVS problems have many applications. Such as solving dead- lock. Deterministic algorithm for FVS –O(17k 4 )!n) —Bodlaender –O((2k+1) k n 2 ) —Downey & Fellows –O( k n 3 ) — Dehne, Fellows et al Randomized algorithm –O(4 k kn) for FVS. —Becker et al –O(6 k kn) for WFVS. —Becker et al

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.

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.

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. E X be the set of edges that both ends are in X, E F,X be the set of edges that one end is in X and another end is in F. Then, |E x |≤|E F,X | (i.e. |E x |≤|E|/2). Feedback set F X=V-F

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E F,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 1/4 k. So if we try c4 k times, the probability that we find a feedback set F is 1-(1-1/4 k ) c4 k.

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).

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.

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. E X be the set of edges that both ends are in X, E F,X be the set of edges that one end is in X and another end is in F. Then, |E x |≤2|E F,X | (i.e. |E x | ≤ 2|E|/3). |E x | ≤ 2|E x’ | ≤2|E F,X’ |= 2|E F,X |. Feedback set F X=V-F Feedback set F X’=V’-F

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E F,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 1/6 k. So if we try c6 k times, the probability that we find a feedback set F is 1-(1-1/6 k ) c6 k.

Question? Thank you very much!