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Unformatted text preview: CS 173: Discrete Structures, Spring 2010 Homework 11 Solutions This homework contains 3 problems worth a total of 42 regular points and 4 bonus points. It is due on Friday, April 30th at 4pm. Put your homework in the appropriate dropbox in the Siebel basement. 1. Partial Orders [16 points] (a) Suppose that A is the set of finite nonzero length sequences of integers . E.g. 5 , 2 is an element of A . So is 3 , 4 , 3 , 3, and so is 3 , 1 , 37 , 67 , , 3. A generic element of A would be x 1 ,x 2 ,...,x n where n 1 (since each sequence contains at least one integer.) Let S be the subsequence of relation on A defined as follows: x 1 ,...,x m S y 1 ,...,y n if and only if m n and there is an integer k such that x 1 = y k , x 2 = y k +1 , ..., x m = y k + m 1 . Prove that S is a partial order. Solution: To prove that S is a partial order, we show that it is reflexive, antisymmetric and tran sitive: i. Reflexive: x 1 ,...,x m S x 1 ,...,x m holds if we set k = 1 since m m and x 1 = x k = x 1 , x 2 = x k +1 = x 2 , ... x m = x k + m 1 = x m . ii. Antisymmetric: Suppose x 1 ,...,x m S y 1 ,...,y n and y 1 ,...,y n S x 1 ,...,x m . If we take the first relation, m n (by the definition of S ) and there must be an integer k 1 such that: x 1 = y k 1 , x 2 = y k 1 +1 , ..., x m = y k 1 + m 1 . For the second relation, n m and there must be an integer k 2 such that: y 1 = x k 2 , y 2 = x k 2 +1 , ... y n = x k 2 + n 1 . By the antisymmetric property of the operator, m n and n m must imply that n = m . If we substitute m for n in the second relation, we get: y 1 = x k 2 , y 2 = x k 2 +1 , ... y n = x k 2 + m 1 . Therefore, k 1 = k 2 = 1 , which implies: x 1 = y 1 , x 2 = y 2 , ..., x m = y m . iii. Transitive: Suppose x 1 ,...,x m S y 1 ,...,y n and y 1 ,...,y n S z 1 ,...,z p such that m n p . We want to prove that x 1 ,...,x m S z 1 ,...,z p . Note that m n p implies that m p because is a transtive operator. Let k 1 and k 2 be two integers such that: x 1 = y k 1 ,x 2 = y k 1 +1 ,...,x m = y k 1 + m 1 and y 1 = z k 2 ,y 2 = z k 2 +1 ,...,y n = z k 2 + n 1 We can deduce the k 1 th element in the y sequence (by substituting k 1 for n ) is y k 1 = z k 2 + k 1 1 . Thus, x 1 = z k 2 + k 1 1 ,x 2 = z k 2 + k 1 ,...,x m = z k 2 + k 1 + m 2 Therefore, x 1 ,x 2 ,...x m S z 1 ,z 2 ,...z p because m p , which implies that S is transitive....
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This note was uploaded on 11/09/2011 for the course CS 173 taught by Professor Erickson during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Erickson

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