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Therefore, the first √''n'' phases, in a graph with ''n'' vertices and ''m'' edges, take time O(''m'' √''n'').
It can be shown that each phase increases the length of the shortest augmenting path by at least one: the phase finds a maximal set of augmenting paths of the given length, so any remaining augmenting path must be longer. Therefore, once the initial √''n'' phases of the algorithm are complete, the shortest remaining augmenting path has at least √''n'' edges in it. However, symmetric difference of the eventual optimal matching
Since the algorithm performs a total of at most 2√''n'' phases, it takes a total time of O(''m'' √''n'').
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