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Unformatted text preview: MATH 4171 Graph Theory SPRING 2006 Homework Set VII  Solutions 1. Prove without using Theorem 8.24 that every planar graph is 6colorable. Proof. Let G be a planar simple graph. Then every induced subgraph G of G is also a planar simple graph. By Theorem 6.5, G has a vertex of degree at most 5, which means that ( G ) 5. Therefore, min { ( G ) : G is an induced subgraph of G } 5. Now we deduce from Theorem 8.6 that ( G ) 1 + min ( G ) 6. 2. For the given graph G , prove that ( G ) = 4 by showing: (a) ( G ) 4; and (b) ( G ) &gt; 3. Proof. (a) The coloring given in 1 2 2 3 1 3 4 1 4 x G= the middle graph proves that ( G ) 4. (b) Let H be the last graph shown above, which is obtained from G by deleting the bottom three vertices. Let C = H x , which is a cycle of length five. Since x is adjacent to all vertices on C , we must have ( H ) &gt; ( C ). Therefore, ( G ) ( H ) &gt; ( C ) = 3....
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This note was uploaded on 02/17/2009 for the course MATH 4171 taught by Professor Lax,r during the Spring '08 term at LSU.
 Spring '08
 Lax,R
 Graph Theory

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