ps8sol - T that end at C 2 respectively It is not hard to...

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6.889: Algorithms for Planar Graphs and Beyond Problem Set 8 - Solutions 1. Solution: (a) The Monge property cannot be satisfied because shortest paths need not cross. (b) This is essentially the bipartite case treated in class. The Monge property holds. Choose, say, a clockwise ordering of the nodes on both cycles. (c) Every path in G 00 corresponds to a path in G 0 , so shortest paths in G 0 are at least as short as those in G 00 . Hence A 0 i,j A 00 i,j . Equality holds if a shortest i -to- j path in G 0 does not cross the path P . (d) Pick an arbitrary node x C 1 and compute a shortest path tree T in G 0 rooted at x . Let P and P r be the leftmost and rightmost paths among all paths of
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Unformatted text preview: T that end at C 2 , respectively. It is not hard to see that for every i ∈ C 1 and j ∈ C 2 , there exists a shortest i-to-j path in G that does not cross both P ‘ and P r . Let A ‘ and A r be the dense distance matrices that correspond to the graphs obtained by cutting G open along P ‘ and P r , respectively. Both A ‘ and A r are Monge, and A i,j ≤ A ‘ i,j and A i,j ≤ A r i,j . It follows that the column minima of A are the minimum between the column minima of A ‘ and A r which can be found in O ( | C 1 | + | C 2 | ) time each using SMAWK....
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