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Unformatted text preview: 1. Proof required. 2. x 10 1x 10 x 20 1x 20 x 50 1x 50 3. (a) { } * ( { 1 }{ 1 } * { }{ } * \ { 1 }{ 11 } * { }{ 00 } * ) * { 1 } * (b) a n = a n1 + a n2 , n 2 . a = 1, a 1 = 2. 4. b n = (2n )2 n . 5. (a) a connected graph with no cycles. (b) Let n i be the number of vertices of degree i . Then using sum of degrees = 2 q (Handshake theorem), and also q = p1 for trees, we can see that n 1n 32 n 43 n 5 =2 . This is true for all trees. Substituting n 1 = 7 and n 4 = 2 we get 7n 32 43 n 5 =2 . Thus 1 = n 3 + 3 n 5 + 4 n 6 + . Since the n i are nonnegative integers, the only possibility is n 3 = 1 (and n 5 = n 6 = = 0). (c) Should be easy. 6. (a) The vertices at levels are as follows (if these are correct, you probably have the right tree): level 0: a level 1: b e s 2: c i f o r t 3: g j d p n u 4: h k q 5: m (b) The graph is not bipartite because there is an edge between t wo vertices at the same level (gu). Tracing back up the tree we nd the cycle dfguqmhd. level (gu)....
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This note was uploaded on 08/09/2009 for the course MATH 239 taught by Professor M.pei during the Spring '09 term at Waterloo.
 Spring '09
 M.PEI
 Combinatorics

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