We consider the Bipartite Matching Problem on a bipartite graph G = (V, E). As usual, we s...

We consider the Bipartite Matching Problem on a bipartite graph G = (V, E). As usual, we say that V is partitioned into sets X and Y, and each edge has one end in X and the other in Y.

If M is a matching in G, we say that a node y ε Y is covered by M if y is an end of one of the edges in M.

(a) Consider the following problem. We are given G and a matching M in G. For a given number k, we want to decide if there is a matching M' in G so that

(i) M' has k more edges than M does, and

(ii) every node y e Y that is covered by M is also covered by M'. We call this the Coverage Expansion Problem, with input G, M, and k. and we will say that M' is a solution to the instance.

Give a polynomial-time algorithm that takes an instance of Coverage Expansion and either returns a solutionM' or reports (correctly) that there is no solution. (You should include an analysis of the running time and a brief proof of why it is correct.)

Note: You may wish to also look at part (b) to help in thinking about this.

Example. Consider Figure 1, and suppose M is the matching consisting of the edge (x1, y2). Suppose we are asked the above question with k = 1.

Then the answer to this instance of Coverage Expansion is yes. We can let M' be the matching consisting (for example) of the two edges (x1, y2) and (x2, y4); M' has one more edge thanM, and y2 is still covered by M'.

(b) Give an example of an instance of Coverage Expansion, specified by G, M, and k, so that the following situation happens.

The instance has a solution; but in any solution M', the edges of M do not form a subset of the edges of M'.

(c) Let G be a bipartite graph, and let M be any matching in G. Consider the following two quantities.

- K1 is the size of the largest matching M' so that every node y that is covered by M is also covered by M'.

- K2 is the size of the largest matching M" in G.

Clearly K12, since K2 is obtained by considering all possible match-ings in G.

Prove that in fact K1 = K2; that is, we can obtain a maximum matching even if we're constrained to cover all the nodes covered by our initial matching M.

Figure 1 An instance of Coverage Expansion.