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Given connected undirected graph , capacity ,find with such that is minimum.

Def: of are separated by a cut , or is a -cut if exactly one of is in .

To find min -cut in an undirected graph, we replace each edge by two oppositely directed arcs and give same capacities as the edge, then solve max flow problem.

If we fix some node , then any cut is -cut for some node Solve min -cut problem for and take min of them.

But other direct method exists with running time.

3.5 Minimum Cuts in Undirected Graphs

3.5.1 Global Minimum Cuts

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Definition: : capacity of a min cut of . : capacity of a minimum -cut of is a graph obtained by identifying node with . ( distinct nodes)( , where is a new node, ; for each and end of in , is an end of if , otherwise is an end of in . capacities are the same)

(If multiple edges created, may replace them with an edge with total capacities. For ease of exposition, we do not do this.)

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Identifying nodes f, g

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Prop 3.35: Every cut of is a cut of . Every cut of that does not separate from is a cut of .

In , each cut corresponds to a cut in having and in one side of the cut.Hence, if we can find the min cut among the cuts that separate and , we can compare the value with the min cut value in .

Straightforward implementation: solve min -cut for any , . Then identity with and solve min cut on again.

Continue times on smaller graphs.

Small improvement, but not significant.make smart choice of .

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rationale: min Need to find a min cut separating (using max flow algorithm after doubling

arcs, need times ).However, choosing carefully reduces the work.

Def: A legal ordering of is an ordering of the nodes of such that, where denotes

for Choose any node as and at step choose that has the largest total capacity of edges joining it to the previously chosen nodes.

Can be found in (refer Prim’s or Dijkstra’s algorithm)

Thm 3.36: If is a legal ordering of , then is a minimum -cut of .

Node Identification Algorithm

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Legal ordering beginning with a is : a, b, c, d, e, h, g, f

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Node Identification Minimum Cut Algorithm

Initialize to be , to be undefined

While has more than one node

find a legal ordering of ;

If

Replace by , by ;

Replace by ;

Return .

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In proving the theorem, we do not assume is connected.

Lemma 3.37: If , then min

(pf) Choose a min -cut with .If , then is an -cut so .Otherwise, is a -cut, so .

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(Proof of Thm 3.36)Need to show . Use induction on the number of edges and nodes. (trivial if or )Consider two cases: (1) is an edge of (2) not adjacent in .(1) Let denote with edge deleted. Then still legal ordering of

(2) In this case, is a -cut, need to show that it is min -cut. From lemma 3.37, enough to show that

and

For the first case, apply induction on . is a legal ordering of . Hence

For the second case, apply induction on is a legal ordering of . Hence

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randomized algorithm for minimum cut problem

Random Contraction Algorithm

Random Contraction Algorithm

While has more than two nodes

Choose an edge of with probability ;

Where , replace by (contraction);

Return , the unique cut of .

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Thm 3.38: Let be a minimum cut of . Then the random contraction algorithm returns with probability at least .(pf) Algorithm will return provided that none of its edges is chosen to be con-tracted. (some edges might be lost even though not contracted, but it causes no problem)Suppose edges have been chosen, none from . Let the current graph be , with .Since is min cut of , its capacity is at most the average of capacities , ( ). Thus, .Then the probability that an edge from is chosen at step is

The probability that no edge of is chosen at step is at least ,so the probability that no edge of is ever chosen is at least .

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Cor 3.39: Let be a minimum cut of and let be a positive integer.The probability that the random contraction algorithm does not return in one of runs is at most .(pf) By thm 3.38, the probability is at most (since )

ex: run the algorithm for times :

Reference: Randomized Algorithms, Rajeev Motwani and Prabhakar Raghavan, Cambridge, 1995

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3.5.2 Cut-Trees

Multiterminal Cut Problem:Given an undirected graph , for all , and a set of terminals , find a mini-mum -cut for each pair of nodes .

Procedure:Choose some pair of terminal nodes . Find min -cut and denotes the two node sets in each side of the cut as and .

p

q

𝑅𝑆 𝑃

𝑄

𝑆𝑓 (𝑟 , 𝑠)

𝑓 (𝑟 , 𝑠)𝑓 (𝑝 ,𝑞)

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General Procedure

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Thm 3.40: For any , the capacity of min -cut equals the minimum label in the -path in tree . Moreover, min -cut is obtained by the bipartition of the tree obtained after eliminating the minimum weight edge from .

Tree satisfying the first property: K-flow-equivalent to .Tree that also satisfies the second property : Gomory-Hu K-cut-tree ( Gomory-Hu cut-tree)

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Lemma 3.41: Let be a minimum -cut for some node , and let . Then there exists a minimum -cut such that .(pf) Let be a min -cut. Assume that . Consider 2 cases: (1) (2) (1) From submodular inequality, we have

Now since is an -cut So is a min -cut.Similarly for (2).

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Let be a tree produced at some stage of Gomory-Hu procedure and is an edge joining the sets(identified nodes) and .We say that terminal nodes and are representatives for if .

Lemma 3.42: At every stage of the Gomory-Hu procedure, there exists represen-

tatives for each edge of the tree (pf) True for the initial two node tree.Suppose we split a node into and , based on an -cut. (, ) and are representatives for the edge joining and and other edges, except the ones incident to earlier, are not affected.Suppose is a node joined to (using edge ) before the iteration is executed and assume is joined to node in the new tree. Then there exist , which are representatives for such that . Now consider 2 cases: (1) (2) .(1): and continue to be representatives for .(2): can show that and are representatives for in the new tree.

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Proof of Lemma 3.42

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Thm 3.40: For any , the capacity of min -cut equals the minimum label in the -path in tree . Moreover, min -cut is obtained by the bipartition of the tree obtained after eliminating the minimum weight edge from .(Proof of Theorem 3.40)The second assertion follows from the first.Let be the path that joins and .Then min since each of the cuts corresponding to the edges separates and .Let be the unique terminal node in for . Have , .By Lemma 3.42, , for By exercise 3.69, min

min Each edge in corresponds to a cut in specified by the bipartition of obtained by deleting from , and this cut has capacity .

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A variant of Gomory-Hu procedure can be used to identify the violated odd set constraint for the matching problem(Ref: M. W. Padberg and M. R. Rao (1982), Odd Minimum Cut-Sets and b-Matchings, Mathematics of Operations Research 7, 67-80.)

More efficient implementation: D. Gusfield, "Very simple methods for all pairs network flow analysis," SIAM Journal on Computing 19 (1990) 143-155