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Unformatted text preview: EECS 310, Fall 2011 Instructor: Nicole Immorlica Problem Set #6 Due: November 8, 2011 1. (25 points) Consider the following false claim: Let G be a simple graph with maximum degree at most k . If G has a vertex of degree less than k , then G is kcolorable. (a) (5 points) Give a counterexample to the false claim when k = 2. (b) (15 points) Consider the following proof of the false claim: Proof. Proof by induction on the number of vertices n . Inductive Hypothesis : P ( n ) is defined to be: Let G be a graph with n vertices and maximum degree at most k . If G has a vertex of degree less than k , then G is kcolorable. Base Case : ( n = 1) G has only one vertex and is 1colorable, so P (1) holds. Inductive Step : We may assume P ( n ). To prove P ( n +1), let G n +1 be a graph with n + 1 vertices and maximum degree at most k . Also, suppose G n +1 has a vertex v of degree less than k . We need only prove that G n +1 is kcolorable. To do this, first remove the vertex v to produce a graph G n with n vertices. Removing v reduces the degree of all vertices adjacent to v by 1. So in G n , each of these vertices has degree less than k . Also the maximum degree of G n is at most k . So G n satisfies the conditions of the inductive hypothesis P ( n ). We conclude that G n is kcolorable. Now a kcoloring of G n gives a coloring of all vertices of G n +1 except for v . Since v has degree less than k , there will be fewer than k colors assigned to the nodes adjacent to v . So among the k possible colors, there will be a color not used to color these adjacent nodes, and this color can be assigned to v to form a kcoloring of G n +1 . Identify the exact sentence where the proof goes wrong and explain the error. (c) (5 points) Add a condition to the theorem statement to make it a true claim, and explain how to revise the above proof to give a proof of this true claim. Solution: (a) Consider the triangle graph and then another disconnected node. This is a counterexample. (b) As we might expect from our experiences with induction, the error in this proof comes from the shrinking down process. In particular, our shrinking down process will not guarantee that we have a graph which satisfies the conditions for the predicate. Consider the lines Removing v reduces the degree of all nodes adjacent to v by 1. So in G n , all of these vertices have degree less than k. This doesnt cover the case where v was adjacent to no other nodes. In this case, it could be that every node has degree k , v had degree 0, and then the removal of v from the graph results in a graph that has no vertex of degree less than k. If this is the case, then we cannot apply the inductive hypothesis....
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This note was uploaded on 01/15/2012 for the course ECON 101 taught by Professor N/a during the Fall '11 term at Middlesex CC.
 Fall '11
 N/A
 Economics

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