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Let G = (V,E) be an arbitrary flow network, with a source s and target t and positive integer capacity c(e) for every edge e. Let (S,T) be a minimum s-t cut with respect to these capacities. Now suppose we increase the capacity of every edge by one, i.e. c_new(e) = c(e) + 1 for all edges, then is (S, T) still a minimum s-t cut with respect to these new capacities {c_new} ?

My intuition is, had G contained edges of different capacities, increased capacity might have resulted in different minimum cut. But when all edges have same capacity then minimum cut would remain same.

Am I correct? How to prove this?

Sanket Achari
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3 Answers3

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Yes, your intuition is correct.

When G contains edges of different capacities, increasing the capacity of every edge by 1 might change the minimum cut. This is easily demonstrated by example, as shown below. The minimum cut (in red) has capacity 3. Increasing the capacity of each edge increases that cut to 6. So the connection from S to A becomes the new minimum cut with a capacity of 5.

enter image description here

When all the edges have the same capacity, increasing the capacity of every edge by 1 will not change the minimum cut. The basic idea behind the proof is that the capacity of a cut is nc where n is the number of edges cut, and c is the capacity of each edge. Since c is a constant, the minimum cut is the cut with minimum n. We'll refer to that minimum as N.

Now if the capacity of each edge is increased by 1, then the new capacity of each cut is n(c+1). Hence, the new capacity of the cut that used to be the minimum cut is N(c+1). Suppose a cut with capacity larger than N(c+1) exists: since all edges have weight c+1, such a cut must be an M-edge cut, for some M > N. But then in the original graph this same cut would have capacity Mc > Nc, contradicting the assumption that the N-edge cut is optimal there, so no such M-edge cut can exist, meaning that the N-edge cut (now with capacity N(c+1)) remains optimal in the new graph.

j_random_hacker
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user3386109
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If ALL edge capicities are increased by a constant, then the min-cut is the same. Provided the edge capacities in the graph are same. Otherwise it may change.

A simple explanation would be -

We calculate min-cut/max-flow by dinic algorithm, using BFS. We apply bfs from source to sink and substact the least capacity/bottle-neck capacity edge in the bfs path. We add this edge-wt. to flow. We continue this until there is no path from source to sink.

If you were to increse edge capacities by a constant, the cut would always remain same. As BFS paths in all iterations of this algorithm would be the same. Only the max-flow value would change.

adisticated
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If all the edges have the same capacity,then the problem is equal to: "If all edge capacities are multiplied by a positive number, the minimum cut remains unchanged. (A)" It's easy to prove (A).And this question should be a special case of (A) where all edge capacities are multiplied by c+1/c.

uohz
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