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Unformatted text preview: MATH 682 Problem Set #3 Solutions 1. (15 points) Demonstrate the following facts about a directed graph D . (a) (5 points) Prove that ∑ v ∈ V ( D ) d D ( v ) = ∑ v ∈ V ( D ) d + D ( v ) . Recall that d and d + represent the indegree and outdegree respectively. Note that d ( v ) = { ( v,u ) ∈ E ( G ) } = ∑ u ← v 1, and similarly with opposite directionality for d + ( v ). Thus: X v ∈ V ( D ) d D ( v ) = X v ∈ V ( D ) X u ← v 1 = X u,v ∈ V ( G ) ( v,u ) ∈ E ( G ) 1 =  E ( G )  Likewise, ∑ v ∈ V ( D ) d + D ( v ) =  E ( G )  . (b) (10 points) Prove that a directed Eulerian tour (i.e. a directed closed trail traversing every edge) on a strictly connected graph D exists if and only if d ( v ) = d + ( v ) for all v ∈ V ( D ) . Let us start by supposing D has an eulerian tour: we shall show that all vertices of D have equal indegree and outdegree. Let us denote our eulerian tour by v 1 → v 2 → v 3 → v 4 → ··· → v m → v 1 , and let e i = ( v i ,v i +1 ), with the special case e m = ( v m ,v 1 ). Then, by the definition of an eulerian tour, all the e i are distinct and E ( G ) = { e 1 ,e 2 ,...,e m } . Now, consider a vertex u in G . Some number k of the v i are equal to u : let us denote u = v i 1 = v i 2 = ··· = v i k . Then, for any j , note that v i j ± 1 (using the wraparound convention that v = v k and v k +1 = v 1 where applicable) must not be identical to u , since v i j ± 1 is adjacent to u and G is a simple digraph. Thus, since none of the i j are consecutive, the indices { i 1 1 ,i i ,i 2 1 ,i 2 ,...,i k 1 ,i k } are all distinct. Since u is the head of e i if and only if either i = i j and u is the tail if and only if i = i j 1, it follows that the above set of indices is exactly the indices of the edges incident on u , and exactly half of them are incoming, and half of them outgoing. Since every element of E ( G ) is associated with exactly one index, the set of edges incident on u is equinumerous with this index set, so d + ( u ) = d ( u ) = k . To prove that equal indegree and outdegree at every vertex is sufficient, we dis pense with the trivial case where D consists only of an isolated vertex and prove the nontrivial digraphs are eulerian by contradiction. Consider a nontrivial conthe nontrivial digraphs are eulerian by contradiction....
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 Spring '09
 WILDSTROM
 Math

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