HW1Soln

# HW1Soln - CSCI 2670 Fall 2005 HW 1 August 30, 2005 0.3 a)...

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CSCI 2670 Fall 2005 HW 1 August 30, 2005 0.3 a) no b) yes c) {x,y,z} d) {x,y} e) {(x,x),(x,y),(y,x),(y,y),(z,x),(z,y)} f) { ,{x},{y},{x,y}} 0.7 a) {(x,y) | x,y are stings of 0’s and 1’s the same number of 0’s OR 1’s} b) {(x,y) | x ≥ y} c) 0.12 Proof by contradiction. Let G be a graph with n≥ 2 nodes and assume all nodes of G have different degrees. Since G contains n nodes, the degree of each node must be some value in the set D={1,2,3,…,n-1}. Note that the number of values in D is n. Therefore, for each value d D, there is exactly one node v in G such that the degree of v is d. In particular, there is some node with degree equal to n-1. This node must have an edge to every other node in G. But this means that there can be no node with degree equal to 0. This is a contradiction since 0 D. Therefore, our assumption that all nodes have different degrees must be incorrect. I.e., there are at least two nodes with the same degree.

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1.3 1.4 c) L(M 1 ) = {w | w has an even number of a’s}
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## This note was uploaded on 02/07/2011 for the course CS 501 taught by Professor Sm during the Spring '11 term at Indiana.

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HW1Soln - CSCI 2670 Fall 2005 HW 1 August 30, 2005 0.3 a)...

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