Section 14.4 391 are edge disjoint. Thus, there are at least m edge disjoint paths from s to t. But there cannot be more since deg s = m and edge disjoint paths must start with different edges. We claim that m is also the minimum number of edges which must be removed to sever all paths from s to t (in accordance with Menger's Theorem). To sever all such paths, we must remove at least m edges since for every v E V2, either sv or vt must be removed. On the other hand, removing all m edges incident with s clearly severs all paths from s to t so the minimum number of edges that have to be removed is m. (b) Employing the notation of (a), suppose s is in VI and t is in V2. The maximum number of edge disjoint paths between s and t and the minimum number of edges that must be removed in order to sever all paths from s to t is mini m, n} (in accordance with Menger's Theorem). Without loss of generality, we assume that m ::; n, so that the minimum of m and n is m. Let VI, V2, . .. , Vm-I be the vertices of
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