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
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 k-colorable. (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 k-colorable. Base Case : ( n = 1) G has only one vertex and is 1-colorable, 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 k-colorable. 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 k-colorable. Now a k-coloring 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 k-coloring 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 counter-example. (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....
View Full Document
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