PS01-100121-solutions

PS01-100121-solutions - MATH 682 Problem Set #1 Solutions...

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Unformatted text preview: MATH 682 Problem Set #1 Solutions 1. (10 points) Prove that if graph G is connected and contains a cycle, then there is an edge e in G such that G- e is still connected. Let a cycle in G be denoted by the adjacencies v 1 v 2 v 3 v 4 v k v 1 . Let e be the edge { v 1 ,v 2 } (in principle any edge in the cycle will work; we choose a specific one to simplify the argument). For arbitrary vertices u and v in G , connectivity of G means there is a path u = u 1 u 2 u 3 u 4 u = v . If this path does not traverse the edge e , then u and v are connected in G- e by this path; on the other hand, if the edge e occurs on this path, then we know that there is exactly one i such that { u i ,u i +1 } = e = { v 1 ,v 2 } . There are two possibilities: either u i = v 1 and u i +1 = v 2 , or vice versa. These are handled nearly identically. In the first case, we consider the walk: u = u 1 u 2 u i = v 1 v k v k- 1 v 2 = u i +1 u i +2 u = v And in the second: u = u 1 u 2 u i = v 2 v 3 v k v 1 = u i +1 u i +2 u = v In both cases we replace the traversal of the edge e in a path from u to v with a traversal of the path around the cycle the long way. Note that this is not necessarily a path; splicing together paths does not guarantee non-self-intersection, but a walk is sufficient to demonstrate connectedness. It is easy to see that e does not appear in this walk: the first and third sections consist of all edges in the path from u to v except for e , and the middle section consists of every edge in the cycle except for e . Thus, u is connected to v in G- e , and since u and v were arbitrary, G- e is connected. 2. (15 points) Recall that ( G ) , ( G ) , ( G ) , and ( G ) are the independence number, clique number, minimum degree, and maximum degree of G respectively: (a) (5 points) Prove that ( G ) ( G ) + 1 and that ( G ) | G | - ( G ) ....
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PS01-100121-solutions - MATH 682 Problem Set #1 Solutions...

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