Class presentation for CPSC 689-605
Randomized Algorithms
for the Loop Cutset
Problem
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
Introduction
• FVS or WFVS problems have many
applications. Such as solving deadlock.
• 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.
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. 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.
X=V-F
|Ex|≤|E|/2).
Feedback
set F
| E X || X |, 3 | X |  d (v) | EF , X | 2 | E X | .
vX
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 c4k times, the probability that
k
we find a feedback set F is 1-(1-1/4k)c4 .
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. 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).
X=V-F
Feedback
set F
• |Ex| ≤ 2|Ex’| ≤2|EF,X’|= 2|EF,X|.
X’=V’-F
Feedback
set 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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 c6k times, the probability
that
k
we find a feedback set F is 1-(1-1/6k)c6 .
Question?
Thank you very much!