Data Structures and Algorithms 2: Graph Algorithms and by K. Mehlhorn

By K. Mehlhorn

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C = v is impossible i* T* 37 since z (V. In the first case we are done. So suppose c v. Since ~* T G" - FATHER[v] is connected (FATHER[v] might not even be a node of G") there must be a path vo, •• ,vk from v = Vo to c = v k avoiding FATHER[v]. Let vi be the first node of that path which is not a descendant of v. As in the proof of part b) one shows LOWPT[v] ~ DFSNUM[v i ] < DFSNUM[FATHER[v]], a contradiction. This completes the proof of part c). d) Immediate from the definition of LOWPT. o Lemma 2 directly leads to an algorithm for finding the biconnected components of an undirected graph; in fact we can use our algorithm for strongly connected components with only three minor changes.

In addition, they require us to change priority queue PQ. If v is not in U then we have to add COST[v] reduce COST[~ from its old value to its new value. The latter operation requires that we can quickly find is to have an array P[1 •• p[~ = nil if v to PQ,if v is in U then we have to ~ ru U and p[v] COST[~ in PQ given v. A good solution of pOinters to elements of PQ. We have points to COST[~ in PQ if v E U. In summary, the following operations (cf. 1. that Min, Deletemin take time O(a log n/log a) and Insert, Demote* take time O(log n/log a) if PQ is realized as an unordered (a,2a)-tree for integer a choice of a?

What set V. should we choose in ~ line 3, how do we find (v,w) in line (4) and how do we represent sets Vi. Let us resolve the latter problem first. 3. to represent sets Vi. Then line (6) is a Union operation (and we do n - 1 of those) and testing whether both endpoints of edge (v,w) E E belong to the same Vi corresponds to two Finds. Since this test has to be done at most once for every edge (v,w) E E the number of Finds is O(e). 3 •• The former questions are harder to resolve. We discuss three strategies: considering edges in order of increasing weight, always growing component V1 and growing components uniformly.

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