exam-1-100224-solutions - MATH 682 Exam #1 1. (6 of 12...

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Unformatted text preview: MATH 682 Exam #1 1. (6 of 12 students attempted this) Answer the following questions related to bound- subverting examples: (a) It is known that the connectivity ( G ) of a graph G is bounded above by the minimum degree ( G ) . Describe a method of constructing a connected graph with arbitrarily large but with a fixed, small . One easy example is 2 K n s connected by a single edge. This is connected, and has minimum degree n , but it has connectivity 1, since removing either of the endpoints of the bridge will disconnect the graph. (b) Let a flow f on a weighted digraph D be called addition-maximal if there is no valid flow g 6 = f such that g ( e ) f ( e ) for all edges; in other words, there is no valid flow resulting solely from adding flow to f . Describe (with an example, if such is useful) the construction of a weighted digraph and addition-maximal flow f such that | f | = 1 but D has maximal flow k for an arbitrarily large integer k . Consider the following construction: we have source s , sink t , and intermediate vertices u 1 ,...,u k and v 1 ,...v k . Our edges will be ( s,u i ), ( u i ,v i ), ( v i ,u i +1 ), and ( v i ,t ) for all i , and we will have capacity 1 on each edge. Suppose f has flow of 1 on the edges ( s,u 1 ), ( u i ,v i ), ( v i ,u i +1 ), ( v k ,t ), and flow 0 on every other edge. Then | f | = 1, but the maximum flow on this graph is clearly k , since we could route flow 1 through each of the k paths s u i v i t . Furthermore, f cannot be improved simply by adding flow, since all of the forward paths ( u i ,v i ) are fully utilized. Below is an exhibit of how this graph would look in the specific case of k = 4: s u1 u2 u3 u4 v1 v2 v3 v4 t 1(1) 0(1) 0(1) 0(1) 1(1) 1(1) 1(1) 1(1) 1(1) 1(1) 1(1) 0(1) 0(1) 0(1) 1(1) 2. (4 of 12 students attempted this) Let G be a graph containing a cycle C , such that there is a path P of length k between two vertices of G . Show that G contains a cycle of length k . (Hint: give a name to the number of times the P intersects C , and find cycles whose size is dictated by this parameter) Let us consider the aforementioned path of length k : P clearly intersects C at least twice, at its endpoints, but let us say that it has r + 1 points of intersection with C . We shall see that G must thus contain a cycle of length at least r and a cycle of length at least k r . Page 1 of 5 February 25, 2010 MATH 682 Exam #1 The first of the above assertions is easy: since P intersects C in r +1 points, C contains at least r + 1 points, so the length of the cycle C is atleast r (actually, at least r + 1). The second assertion is subtler: let us denote the points of P s intersection with C , sequentially along the length of P , as v , v 1 , v 2 ,... , v r . Now let us denote the fragment of P connecting v i- 1 to v i as P i . So P , a path of length k , is constructed by splicing together the r paths P 1 ,P 2 ,...,P r ; thus, at least one of these P i has length k r . Now, since...
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This note was uploaded on 01/12/2012 for the course MATH 682 taught by Professor Wildstrom during the Spring '09 term at University of Louisville.

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exam-1-100224-solutions - MATH 682 Exam #1 1. (6 of 12...

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