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Unformatted text preview: AMS/MBA 546 (Spring, 2009) Estie Arkin Network Flows  Final, sketch of solutions Average 85.67, high 95! 1). (a). A Feedback edge set of an undirected graph G = ( V NE ) is a subset of edges E ′ ⊆ E whose removal results in an acyclic graph G ′ = ( V NE \ E ′ ). Given a graph with positive edge weights w ij > 0, describe a polynomial time algorithm that finds a minimum weight feedback edge set. Recall, this problem is NP Complete for directed graphs. The feedback edge set is a set of edges to be removed so that the result is a tree. Since we want to remove the smallest weight edge set, this is equivalent to finding the set of edges not to be removed of max weight, namely, the max spanning tree, which can be computed in polynomial time. (b). For a graph G = ( N,E ) define conn ( G ) the number of connected components of G . Prove that conn ( G ) + m ≥ n for every graph G .  N  = n , and  E  = m . For each connected components, we have m i ≥ n i − 1, since each connected component has a spanning tree. Adding over all i , we get ∑ i ( m i + 1) ≥ ∑ i n i , and ∑ i m i = m , ∑ i n i = n , ∑ i 1 = conn ( G ). 2). We say that an undirected graph has a strongly connected orientation if its edges can be directed so that the resulting graph is strongly connected (meaning that for every two nodes i , j there is a directed path from i to j and from j to i ). A bridge in a graph G is an edge whose removal disconnects G . Prove: A graph G = ( V NE ) has a strongly connected orientation if and only if G is connected and has no bridge. (Ideally, your proof should be constructive. It should describe an efficient algorithm that gives a strongly connected orientation,if one exists). Clearly the condition is necessary. If the graph is not connected, then nodes in different connected components will not have a directed path connecting them regardless of how the edges are oriented. Also, if the graph has a bridge, say ( i,j ) then without loss of generality say the edge is directed from i to j , there is no directed path from j to i . One such algorithm: Find the 2 connected components of the graph. Note that by our assumption that no bridges exist, each 2 connected component must be more than a single edge. We show how to direct the edges of each such 2 connected component to get that component strongly connected: Do DFS and direct the tree edges down, and non tree edges up. This works, for the following reason: The resulting graph will have a path from the root of the DFS tree to every node i . Also, since the graph under consideration is 2connected, using the non tree arc (and possibly some tree arcs) from every node i there is a directed path to the root. The subtree rooted at i has a “back edge” to a node between i and the root (a node of level less than i ’s level) say node j . If j negationslash = the root, then the subtree rooted at...
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 Fall '08
 Arkin,E
 Graph Theory, TSP, Estie Arkin, DTSP

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