# sol7 - MATH 239 ASSIGNMENT 7 Suggested solutions 1. For...

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Unformatted text preview: MATH 239 ASSIGNMENT 7 Suggested solutions 1. For each n 1, let T n be the graph whose vertices are the { , 1 , 2 }-strings of length n , in which two strings are adjacent if they dier in exactly one position. For example, in T 2 , 00 is adjacent to 10, 20, 01 and 02. (a) How many vertices and edges does T n have? (b) Show that T n is non-planar for n 2. (c) Show that T n is 3-colourable. Solution. (a) The number of vertices in T n is the number { , 1 , 2 }-strings of length n , which is 3 n . Each vertex has degree 2 n (each of the n positions may be changed in two dierent ways) so, if q is the number of edges in T n , 2 q = X v V ( T n ) deg ( v ) = X v V ( T n ) 2 n = 2 n 3 n , and therefore q = n 3 n . (b) Planar graphs with p vertices and q edges satisfy q 3 p-6. For n 3, q = n 3 n 3 n +1 &gt; 3 p-6, so T n is non-planar. However, T 2 has 9 vertices and 18 edges. Therefore, to show T 2 is non-planar, we need to nd an edge subdivision of K 3 , 3 or K 5 . There are several options here, here is one example. 12 22 20 01 02 21 11 10 00 (c) For each vertex v in T n , let sum( v ) be the sum of the digits of v . Colour the vertices of T n as follows: colour the vertex v red if sum( v ) is a multiple of 3, blue if sum( v ) is 1 more than a multiple of 3, and yellow if sum( v ) is 2 more than a multiple of 3. Notice that, if u and v are adjacent, sum( u ) and sum( v ) are dierent and dier by at most 2, so u and v are coloured dierently. 2. Without using the four colour theorem, show that every planar graph without a triangle is 4-colourable. Solution. The proof is similar to the proof of the six colour theorem given in class. Planar graphs with p vertices, q edges and girth at least 4 satisfy q 2 p-4 (by Theorem 6.4.6). Therefore, every such graph has a vertex of degree at most 3 (graphs where every vertex has degree at least 4 satisfy 2 q 4 p ). Now we proceed by induction on the number p of vertices. All graphs on at most 4 vertices are 4-colourable, so the result is true for p 4. For the induction hypothesis, assume the result is true for all planar graphs on p k vertices, where k 4. Now consider a planar graph G on p = k + 1 vertices. By the discussion above, G has a vertex v with deg( v ) 3. Let G be the graph obtained by removing v and all its incident edges from G . By the induction hypothesis, G is 4-colourable. We can 4-colour is 4-colourable....
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## sol7 - MATH 239 ASSIGNMENT 7 Suggested solutions 1. For...

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