Winter 2009 Assignment8 Soln

Winter 2009 Assignment8 Soln - MATH 239 Assignment 8 DUE:...

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MATH 239 Assignment 8 DUE: NOON Friday 3 April 2009 in the drop boxes opposite the Math Tutorial Centre MC 4067. 1. Prove that a planar graph without a 3-cycle has a vertex of degree 3 or less. Solution: Let G be a planar graph without a 3-cycle. There are two cases to consider. Case 1: G has no cycles. Thus the components of G are trees. If there are any com- ponents with no edges (i.e. single vertices), then we have a vertex of degree 0, and the result follows. Otherwise, we know from Theorem 5.1.5 that every tree with at p 2 has at least two vertices of degree 1, and the result also follows. Case 2: G has a cycle, so its girth is defined. Let us suppose to the contrary that G has all vertices of degree at least 4 (i.e. deg( v ) 4 for any vertex v ). From Theorem 4.1.7, we know that 2 q = v deg( v ) 4 p , and thus q 2 p . Since G has no 3-cycles, we know the girth is at least 4 (and G has at least four vertices). Thus from Theorem 6.4.6, we know that q 2 p - 4, which contradicts q 2 p . Thus we know that G cannot have all vertices of degree 4 or more (i.e. has at least one vertex of degree at most 3). The result follows. 2. Prove that if G is a planar graph with no cycles of length less than 4, then G is 4- colourable. (Do not use the 4-colour theorem). Solution: We can prove this result by induction ( note that this closely mimics the proof of the 6 -colour theorem ). Any graph with p 4 is trivially 4-colourable (one can assign a different colour to every vertex). (Inductive Hypothesis:) Suppose that any planar graph G on p vertices with no cycles of length less than 4 is 4-colourable. Consider a planar graph on p + 1 vertices with no cycles of length less than 4. From question 1, we know that such a graph has a vertex v of degree at most 3. Consider the graph G 0 = G - v , the graph formed by removing v and its incident edges from G . Note that G - v is still planar, and still has no cycles of length less than 4 (i.e. we still satisfy the assumptions of the inductive hypothesis); thus G - v is 4-colourable. Consider a proper 4-colouring of G - v , and use the same colour assignment to colour all the vertices except v in G ; in this case, all edges not containing v have each endpoint coloured differently. We only need to colour v in such a way that v is coloured with a colour different from each of its neighbours. Since v has at most 3 neighbours, then there will be at least one colour available to colour v so that the complete colouring is proper (i.e. no adjacent vertices are coloured the same). Thus G is also 4-colourable, and the induction follows.
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3. Prove that a graph with maximum degree Δ can be (Δ + 1)-coloured. Solution 1:
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Winter 2009 Assignment8 Soln - MATH 239 Assignment 8 DUE:...

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