Winter 2009 Sample Final Soln

Winter 2009 Sample Final Soln - 1. Proof required. 2. x 10...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1. Proof required. 2. x 10 1-x 10 x 20 1-x 20 x 50 1-x 50 3. (a) { } * ( { 1 }{ 1 } * { }{ } * \ { 1 }{ 11 } * { }{ 00 } * ) * { 1 } * (b) a n = a n-1 + a n-2 , n 2 . a = 1, a 1 = 2. 4. b n = (2-n )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 = p-1 for trees, we can see that n 1-n 3-2 n 4-3 n 5- =-2 . This is true for all trees. Substituting n 1 = 7 and n 4 = 2 we get 7-n 3-2 4-3 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)....
View Full Document

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.

Page1 / 2

Winter 2009 Sample Final Soln - 1. Proof required. 2. x 10...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online