5.4 Minimum cost arborescences
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1 5.4 Minimum cost arborescences Let G = (V,A) be a directed graph and let r V be a node called the root. An arborescence with root r is a subgraph T = (V,F) of G which does not contain a pair of opposite edges and such that the following conditions hold. a) If the directions of the edges are ignored, then T is a spanning tree. b) There is a path from r to every other v V. Lemma A subgraph T = (V,F) of G is an arborescence if and only if T does not contain any cycle and each vertex v r satisfies δ in (v) = 1. Proof. Suppose that T does not contain any cycle and each vertex v r satisfies δ in (v) = 1. Let v V be a node. Since δ in (v) = 1 there exists a node u with (u,v) A. If u is not the root, then there exists a node x v with (x,u) A. Continuing this way we eventually obtain r, since T does not contain a cycle. Since F = V 1 we thus have a spanning tree, after ignoring the directions of the edges. Suppose now that T is an arborescence. Then T does not contain a cycle, since such a cycle would also induce an undirected cycle. The only edge which enters v is the last edge on a directed path from r to v. If each node is reachable from r, then G contains an arborescence which is constructed by breadth-first-search. Lemma G contains an arborescence if and only if each node in G is reachable from r. Let F A be the subset of the edges of G which is constructed as follows. Set F = /0. For each v V, v r choose a cheapest edge e δ in (v) and set F := F {e}. Thus F is a cheapest set of edges such that δ in (v) F = 1 for each v r. An arborescence F also satisfies δ in (v) F = 1. Therefore one has for each arborescence F. c(f ) c(f) (5.10)
2 62 If F does not contain a cycle, then it follows from Lemma 5.14 that F is a minimum weight arborescence. Let δ v = min{c(e) e δ in (v)}. We now consider the edge weights c (u,v) = c(u,v) δ v. Theorem An arborescence F is minumum-weight w.r.t. c if and only if it is minumumweight w.r.t. c. Proof. Let F be an arborescence. One has c(f) c (F) = (u,v) F = δ v. v V ( c(u,v) c (u,v) ) The last term is independent from F. Therefore the claim follows. Suppose now that we have replaced c by c. Then we have the following facts. i) Each edge-weight is non-negative. ii) δ v = min{c(e) e δ in (v)} = 0 iii) F contains only edges of weight 0. The crucial idea is now in the following Lemma. Remember that we are done with our search for a minimum weight arborescence if F does not contain a cycle. If F does contain a cycle, then we can essentially shrink this cycle into one node, recursively continue our search for a minimum-cost arborescence in the new graph and construct an minumumweight arborescence for the original graph from an minumum-weight arborescence of the smaller graph. Theorem Let G = (V,A) be a directed graph containing an arborescence and let c : A R 0 be edge-weights. Further let C be a cycle not containing r with c(c) = 0. Then there exists a minimum-weight arborescence T = (V, F) which enters C exactly once. Proof. Let T = (V, F) be an minimum-weight arborescence. There exists at least one edge (a,b) F entering C, meaning a / C and b C. Suppose (a,b) is such an edge minimizing the number of edges on the unique r b-path in T. We now change F as follows. For each node v C, v b replace the edge e δ in T (v) with the edge e δc in (v). Let F be the thereby obtained edge-set, then the following facts hold.
3 5.4 MINIMUM COST ARBORESCENCES F = V 1 2. c(f ) c(f), since each edge on C has weight 0 and all other edge-weights are nonnegative. 3. The path from r to b in T is untouched 4. Each node in C is reachable from r It remains to show that each node v V is reachable using edges in F. Let P be the path in T = (V,F) from r to v. If P does not contain a node of C, then P is still present in F. Otherwise, let w be the last node on P which is also a node of C. By 4) above we have that w is reachable from r in F. The edge leaving w in P is still present in F since its end-node is not in C. The same holds for all other edges on the path from w to v in T. Therefore v is reachable from r.
4 64
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