so9 - Math 55 Homework 9 Solutions Monday August 8 2011 1...

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Math 55 - Homework 9 Solutions Monday, August 8, 2011 1. Suppose the G is a simple, undirected, planar graph and every vertex in G has degree at least 5. Prove that G has at least 12 vertices of degree exactly 5. We will prove this for when G is connected. If G is not connected, then each connected component must have at least 12 such vertices, so there are at least 12 overall. Since G is connected and planar, by Euler’s formula r = e - v +2 . Since G is simple, each region has at least 3 edges on it. By counting edge-region touchings, we then know 2 e 3 r . Let v 5 be the number of vertices of degree 5, and let w be the number of vertices of degree 6 or more. So v = v 5 + w . By counting vertex-edge touchings, we get 2 e 5 v 5 + 6 w . Putting this all together, we get: v 5 = v 5 + 6 ( v - v 5 - w ) = 6 v - (5 v 5 + 6 w ) 6( e - r + 2) - 2 e = 2(2 e - 3 r ) + 12 12 2. Suppose that T is a tree and e is an edge which goes between two vertices of T , but is not in T . Prove that if you add e to T , then the new graph has exactly one cycle (which doesn’t repeat a vertex). Let ( v,w ) be an edge not in T . Since T is connected, T contains a path from ( v,w ) . By adding ( v,w ) to this path, we get a cycle. So T + ( v,w ) has at least one cycle. Suppose T + ( v,w ) had two cycles C 1 and C 2 . Both C 1 and C 2 must contain ( v,w ) , since T is a tree and has no cycles on its own. If C 1 6 = C 2 , let ( s,t ) be some edge in one but not the other. Without loss of generality, let ( s,t ) be in C 1 . In C 1 , pick the longest
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This note was uploaded on 03/14/2012 for the course MATH 55 taught by Professor Strain during the Summer '08 term at Berkeley.

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so9 - Math 55 Homework 9 Solutions Monday August 8 2011 1...

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