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Unformatted text preview: Comp201: Discrete Computational Structures Fall Term, 2010 Homework 4: Solutions 1 Answers for Section 8.1 8.1.56 We prove this by induction on n . It is very easy to show that this is true for the base case n = 1, since R 1 = R and R is symmetric. At the inductive step, assume that R n is symmetric, and show that R n +1 is symmetric. Let ( a,c ) ∈ R n +1 = R n ◦ R . Then ∃ b ∈ A such that ( a,b ) ∈ R and ( b,c ) ∈ R n . Since R n and R are symmetric, ( b,a ) ∈ R and ( c,b ) ∈ R n . Thus by definition ( c,a ) ∈ R ◦ R n = R n +1 . Note that R ◦ R n = R n ◦ R since R and R n are both symmetric square matrices. 2 Answers for Section 8.4 8.4.26 b) M R = 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 M 2 R = 0 0 0 0 0 0 1 1 0 1 0 1 1 0 0 0 0 0 0 0 0 0 1 0 1 M 3 R = 0 0 0 0 0 0 1 1 0 1 0 0 1 0 1 0 0 0 0 0 0 1 1 0 1 M 4 R = 0 0 0 0 0 0 1 1 0 1 0 1 1 0 1 0 0 0 0 0 0 1 1 0 1 = M 5 R B = M R ∨ M 2 R ∨ M 3 R ∨ M 4 R ∨ M 5 R = 0 0 0 0 0 0 1 1 0 1 0 1 1 0 1 1 0 0 0 0 0 1 1 0 1 B is the matrix representation of the connectivity relation R * . 1 3 ANSWERS FOR SECTION 8.5 2 d) M R = 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 1 0 0 1 1 1 0 1 M 2 R = 1 1 1 0 1 1 0 1 0 1 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 M 3 R = 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 1 1 1 1 1 M 4 R = 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 M 5 R = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 = B That is, every pair of elements in the set are connected.That is, every pair of elements in the set are connected....
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This note was uploaded on 04/13/2011 for the course ENGR 301 taught by Professor Raultekin during the Spring '10 term at Boğaziçi University.
 Spring '10
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