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I am reading on how solving maximum flow (Ford-Fulkerson) can be also used to solve unweighted bipartite graph matching problem. I think I don't understand the essence of this problem, because to me it seems trivial.

The method of solving the problem says to convert the original bipartite graph into a network, by creating a Source and Sink vertices, directing all edges towards the Sink and setting all edges' capacity to 1. Then I should run Ford-Fulkerson. Fair enough.

My question is, can't I just do this linearly? (Obviously not, but I don't see why). The goal of the problem seems to be to find a maximum matching in a complete bipartite graph - i.e. the maximum number of edges between the two "sections" of the graph that do not share any vertices.

To illustrate, see this picture

enter image description here

  • In the first graph the maximum matching will be 2 - any of the two vertices on the right may only be connected to a single vertex on the left. Since the edges are unweighted, it doesn't really matter which?
  • Similarly in the second graph, the matching will again be just 2.

Can you tell me where am I thinking wrong? I don't understand where the complexity of the problem comes from.

And sorry if any of the terms were used incorrectly, I am not studying CS in English. Thanks

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  • $\begingroup$ Do you realize that not all vertices on the right need to be connected to all vertices on the left? (then it will still be a bipartite graph). edit: I see now that you say "complete bipartite graph". Then your solution would be correct. $\endgroup$ – Albert Hendriks Jun 1 '16 at 19:34
  • $\begingroup$ It might also be that the input is guaranteed to be a (complete) bipartite graph, but that your algorithm just needs to figure out how many nodes are on the left and how many on the right. $\endgroup$ – Albert Hendriks Jun 1 '16 at 19:56
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I am reading on how [...] to solve unweighted bipartite graph matching problem. [...] The goal of the problem seems to be to find a maximum matching in a complete bipartite graph

No, the goal of the problem is to find a maximum matching in any unweighted bipartite graph.

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  • $\begingroup$ Then it seems my professor's text has an error in it. This explains it then, thanks! $\endgroup$ – Martin Melka Jun 2 '16 at 8:04
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    $\begingroup$ @MartinMelka maybe the confusion comes from the problem of finding a maximum-weight matching in a bipartite graph with weighted edges. Then, it is indeed without loss of generality to assume that the graph is complete, since you can add the missing edges with weight 0. $\endgroup$ – Erel Segal-Halevi Feb 11 '19 at 17:09

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