A comment about HW3 #5

Pranav and Sidhanth asked me for the motivation behind that problem: why did we just not find a perfect matching in the graph (say by the blossom algorithm), put weight 0 on the matching edges, and 1s elsewhere?

The reason is that we want to find perfect matchings in parallel. If we had a single perfect matching in the graph, we could check (in parallel) for each edge if it was in this perfect matching. (How? Use Lovasz’s algorithm via Tutte’s theorem on G and then on G-e. This requires computing determinants, which is doable in parallel.) And then (in parallel) output all edges that belong to the matching.

But if there are many perfect matchings, maybe all edges may find that G-e still has a perfect matching. Which of these should we output? We don’t want to do things sequentially, so we seem stuck.

This is why Ketan Mulmuley, Umesh Vazirani, and Vijay Vazirani came up with the isolation lemma. Suppose we choose random weights for all edges. Whp there is a unique min-weight matching. So now each edge e’s subproblem is: does e belong to the unique min-weight matching? This can also be done in parallel, using a slight extension of the Lovasz idea. See the MVV paper for details.

The next question arises: do we really need randomization? And this has been open for some time, and people have tried to reduce the number of random bits you need. The naive approach uses O(log m) bits per edge, so O(m log m) random bits over all. In this question you proved a bound of O(log^2 m) bits, which is much better. It still remains an open problem how to remove the randomness altogether.

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