hw11-soln - CS 173: Discrete Structures, Spring 2010...

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: 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 non-zero 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....
View Full Document

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.

Page1 / 5

hw11-soln - CS 173: Discrete Structures, Spring 2010...

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