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PS03-100225-solutions

# PS03-100225-solutions - MATH 682 Problem Set#3 Solutions...

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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 0 = 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 con- nected digraph D in which every vertex has equal indegree and outdegree. Since D is strictly connected and nontrivial, δ + ( D ) 1 and δ - ( D ) 1, since there

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