Approximating the undirected Multicut problem A folklore algorithm. Based on a result of Karloff et al for Multiway cut. Problem definition Input: Given connected G(V,E) w(e)0 and a collection si, ti of vertex pairs. Required: a minimum
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Presentation Transcript
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Approximating the undirected Multicut problem A folklore algorithm.
Based on a result of Karloff et al for
Multiway cut.<br>
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Problem definition Input: Given connected G(V,E) w(e)≥0 and a collection {si, ti} of vertex pairs.
Required: a minimum cost E’ so that in
G(V,E-E’), si and ti are not in the same
connected component
In the fractional program: put fractions on edges so that the distance between si and ti
is at least 1 for every i.<br>
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The LP Minimize e w(e) xe
Such that:
x,e, eP xe ≥ 1
For every P from si to ti<br>
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Separating oracle Is there
x,eP xe <1
For some P from si to ti
This is a shortest-path problem
From now on the xe are treated as
distances<br>
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A distance cut A distance cut for s toward t is taking a number 1/2 and taking out of the graph all edges with one vertex with xe distance less than from s, and for the other the distance from s is at least
No other pairs remain inside the sphere as if so dist(s, si)+dist(s, ti)<1. Contradiction.<br>
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The algorithm 1) Choose once R (0,1/2).
2) Order the pairs randomly
3) For i=1 to k do.
3.1 Add to the solution all edges (w, z)
so that dist(si, w)< but dist(si, z)≥
3.2 All edges (w,z) so that dist(si, w)≤
dist(si, z)≤ will not be cut edges.<br>
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The algorithm is randomized Let cut(e) be the random variable set to w(e) if e is in the solution and 0 otherwise.
Note that Cut=e cut(e)
E(Cut)=E(e cut(e))= e Pr(e is in the cut)w(e).
The question is: how high is the probability that e is in the cut?<br>
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The algorithm is randomized The question is: how high is the probability that e is in the cut?
We will see that the probability that e is in the cut is at most 2(ln k+2) xe (more precisely, the minimum between this and 1). This implies that E(Cut)≤2(ln k+2)optf which means a 2(ln k+1) approximation.<br>
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From now on we bound the probability that e is in the cut Let d(i) be the distance of si from e (the minimum between the distances to u and v). Assume by renaming the pairs that
d(1)≤d(2)≤…. ≤d(k)<br>
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Definition We say that si adds (u, v) to the solution
si is the first pair so that dist(si, u)< and
dist(si, v)≥
It can be that dist(ti, v)< for example, but we can rename the vertices.<br>
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What is the probability that s13 added e? u v s7 This can’t happen<br>
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If this happens (u,v) no longer can be a cut edge u v s7 This can’t happen<br>
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And this can’t happens u v s7 s7 added the edge first before s13<br>
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This must happen u v s7<br>
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What happens if s7 is before s13 in the random order? u v s7 s13<br>
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This can’t happen since s7 is closer to the edge u v s7 s13<br>
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A condition For s13 to have a chance to add e to the cut, s13 must be before s1, s2…,s12 in the cut. Probability 1/13.
In general 1/i<br>
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The algorithm We have disjoint events. e was taken into the cut by s1, by s2 or by s3 to e was taken by sk
Pr(e is in the cut)=Pr(e was taken by s1)+Pr(e was taken by s2)+…+pr(e was taken by sk).<br>
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The first condition for si to add e is that si will be in the ordering before sj for all j<i Probability 1/i<br>
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The first condition for si to add e is that si will be in the ordering before sj for all j<i But even then it is not sure<br>
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˄ 1/2 Say that si is before s1,s2,…,si-1 x e n si<br>
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˄ 1/2 The blue line doesn’t separate, the red does x e n si<br>
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You have an interval of at most x(e) length out of ½ that implies that e is taken by si What is the probability that e=uv is cut by pair i?<br>
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In total, the probability that si takes e is at most 2xe/i What is the probability that e=uv is cut by pair i?<br>
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Pr(e is in the cut)=
Pr(e is taken by s1)+Pr(e is taken by s2)+… +Pr(e is taken by sk) The probability that e is in the solution<br>
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Pr(e is cut)
≤2x(e)/1+2x(e)/3+…+2x(e)/k<(2ln k+2)x(e)
This is what we claimed Probability that e is in the cut<br>
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There is an O(log n) expected ratio approximation algorithm for the Undirected Multicut problem. We have shown<br>
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The integrality gap It is Ω(log n) on a random graph
and expanders.
We don’t have a good hardness of
approximation.<br>
