exam-1-100224-solutions

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

### Page1 / 5

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online