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Unformatted text preview: 2.6 Minimum Diameter of Orientations of Multigraphs 63 2.6 Minimum Diameter of Orientations of Multigraphs The same complexity result holds for the following problem: find a minimum diameter orientation of a graph. Indeed, the following assertion holds. Theorem 2.6.1 (Chv´ atal and Thomassen) [164] It is NPcomplete to decide whether an undirected graph admits an orientation of diameter 2. For a bridgeless multigraph G , let diam min ( G ) denote the minimum di ameter of an orientation of G . We will present a minor modification of the original proof of Theorem 2.6.1 by Chv´ atal and Thomassen [164]. The main difference is in the use of Lemma 2.6.2 (which is applied to two different results in this section). Define a bipartite tournament BT s , with partite sets U,W , each of cardinality s , as follows. Let U = { u 1 ,u 2 ,...,u s } and W = { w 1 ,w 2 ,...,w s } . The vertex u i dominates only vertices w i ,w i +1 ,..., w i + b s/ 2 c 1 (the subscripts are taken modulo s ) for every i = 1 , 2 ,...,s . Lemma 2.6.2 Let s ≥ 2 . The diameter diam( BT s ) equals 3. In particular, dist( U,U ) = dist( W,W ) = 2 . Proof: Clearly, it suffices to show that dist( U,U ) = dist( W,W ) = 2 . This follows from the fact that, for every i 6 = j , we have N + ( u i ) N + ( u j ) 6 = ∅ and, hence, there is a vertex w ∈ W such that u i → w → u j . ut Lov´ asz [520] proved that it is NPhard to decide whether a hypergraph of rank 3 3 is 2colourable. By the result of Lov´ asz, Theorem 2.6.1 follows from the next theorem. Theorem 2.6.3 Given a hypergraph H of rank 3 and order n , one can con struct in polynomial time (in n ) a graph G such that diam min ( G ) = 2 if and only if H is 2colourable. Proof: Let k be the integer satisfying 8 ≤ k ≤ 11 and n + k is divisible by 4. Let H be a hypergraph obtained from H by adding k new vertices v 1 ,...,v k . Moreover, append three new edges {{ v i ,v i +1 } : i = 1 , 2 , 3 } to H if H has an odd number of edges, and add four new edges {{ v i ,v i +1 } : i = 1 , 2 , 3 , 4 } to H otherwise. Observe that H has an even number of edges, which is at least four. To construct G , take disjoint sets R and Q such that the elements of R ( Q ) are in a onetoone correspondence with the vertices (the edges) of H . Let G h R i and G h Q i be complete graphs, and p ∈ R and q ∈ Q be adjacent if and only if the vertex corresponding to p belongs to the edge corresponding to q (in H ). Append four new vertices w 1 ,w 2 ,w 3 ,w 4 and join each of them to all the vertices in R ∪ Q . Finally, add a new vertex x and join it to all the vertices 3 Recall that the rank of a hypergraph is the cardinality of its largest edge. 64 2. Distances in R . We show that the obtained graph G has the desired property. (Clearly, G can be constructed in polynomial time.) Assume that G admits an orientation G * of diameter 2. For a vertex u ∈ R , set f ( u ) = 0 if and only if x → u in G * ; otherwise, f ( u ) = 1. Since dist G * ( x,q...
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 Spring '11
 Algorithms

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