given an undirected graph G, find a minimum cut (A,A')

such that A has odd cardinality,

and A does not contain the special vertex s.

such that A has odd cardinality,

and A does not contain the special vertex s.

We showed that if the weight of the edges crossing such a min-cut was strictly less than 1, then A would correspond to a violated constraint in the (exponentially-sized) LP. And one could find the min-cut (B, B') in the graph in polynomial time. But what if the side B not containing the special vertex s in this min-cut turned out to be of even cardinality? Well, It turns out we need a little more work.

To handle this, we do the following: first, we recurse on G/B (which is the graph obtained by shrinking the set B to a single node), and on G/B' obtained by shrinking B' to a single node. In the latter, we treat the new node obtained from shrinking B' as the "special" node.

Why would one of the two recursions succeed in finding a minimum odd cut? Suppose (A, A') was a minimum odd cut we were looking for (with s not in A). Recall that (B, B') was the mincut but B was even. And s is in neither of A or B,

- If A lay completely inside B or B', we'd find it in one of the recursive calls.
- If A contains B, we'd find it the recursive call on G/B.
- Suppose A intersects both B and B' but does not contain B. Then consider the 4 parts: A cap B, A cap B', A' cap B, and A' cap B'. |A| is odd, |B| is even. So either |A cap B| is odd, or |A cap B'| is odd.
- Say it's the former: |A cap B| is odd. A' cap B' contains s, so it's non-empty.

Now we bust out the following fact (which is called submodularity of the cut function): if cut(X) is the weight of all edges going out of some set X,cut(X) + cut(Y) >= cut(X cap Y) + cut(X cup Y).

So cut(A) + cut(B) >= cut(A cap B) + cut(A cup B) = cut(A cap B) + cut(A' cap B').

cut(A) = min-odd-cut.

cut(B) = min-cut.

cut(A' cap B') >= min-cut, since A' cap B' is non-empty.

cut(A cap B) >= min-odd-cut.

Putting these together, the only way to satisfy submodularity is when we have equality everywhere. so cut(A cap B) = min-odd-cut, and we will find it recursively. - If it's the latter: |A cap B'| is odd. Since A does not contain all of B, A' cap B is non-empty. Now apply submodularity to A and B', and repeat the same argument.

This shows that at least one of the recursive calls to G/B or G/B' still has an odd-cut with the same value as cut(A), and hence we will find it recursively. - Say it's the former: |A cap B| is odd. A' cap B' contains s, so it's non-empty.

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