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### Homework 10 Solutions W11

Course: EECS 203, Spring 2011
School: Michigan
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Word Count: 213

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203: EECS Homework 10 Solutions Section 8.4 1. (E) 2 When we add the pairs (x, x) to the given relation we have all of Z Z; in other words, we have the relation that always holds. 2. (E) 6 We form the reflexive closure by taking the given directed graph and appending loops at all vertices at which there are not already loops. a b c d 3. (E) 14 Suppose that the closure C exists. We must show that C is the...

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203: EECS Homework 10 Solutions Section 8.4 1. (E) 2 When we add the pairs (x, x) to the given relation we have all of Z Z; in other words, we have the relation that always holds. 2. (E) 6 We form the reflexive closure by taking the given directed graph and appending loops at all vertices at which there are not already loops. a b c d 3. (E) 14 Suppose that the closure C exists. We must show that C is the intersection I of all the relations S that have property P and contain R. Certainly, I C, since C is one of the sets in the intersection. Conversely, by definition of closure, C is a subset of every relation S that has the property P and contains R; therefore C is contained in their intersection. 4. (M) 22 Since R R , clearly if R, then R . 5. (C) 24 It is certainly possible for R2 to contain some pairs (a, a). For example , let R = {(1, 2), (2, 1)} 6. (M) 26 (a) We show the various 0 0 1 0 0 0 A= 1 0 0 0 1 0 0 0 0 matrices that are invoked. First, 0 0 1 0 0 0 1 0 1 0 0 0 , A[2] = 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 ,A [3] = 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 (1) It follows that A[4] = A[2] and A[5] = A[3] . Therefore 1 namely 0 1 0 1 0 1 0 1 0 1 0 0 1 0 the answer B, the meet of all the A's, is A A[2] , 0 1 0 1 1 0 0 0 0 0 (c) A= A[4] = 7. 28 (b) W0 = W5 = (d) W0 = W3 = 0 1 0 1 1 0 1 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 1 1 1 1 1 0 0 0 1 1 1 0 1 1 , W1 = , W4 = 0 1 0 1 1 0 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 1 0 1 1 1 1 0 1 1 0 0 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 , W2 = , W5 = 0 1 0 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 (4) 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 1 1 0 1 0 1 0 0 1 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 1 0 1 = W1 , (3) W2 = 0 0 0 1 0 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 1 1 0 1 = W3 = W4 0 1 1 1 0 1 1 1 1 1 1 0 1 0 0 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 0 1 1 0 0 0 0 1 1 1 1 0 , A[2] = , A[5] = 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 , A[3] = , A[3] = 1 1 1 1 0 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 0 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 (2)
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Michigan - EECS - 203
Homework 10Section 8.11.(M) 48a) Since R and S are both having all pairs like (x,x) where x is an element of set A the union will also have it. Thus R S is also reflexive. 48b) R S will be reflexive as both R and S have pairs like (x,x). 48c) R S will b
Michigan - EECS - 203
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Michigan - EECS - 203
Homework 11 SolutionsSection 8.4 Initial Matrix Using cfw_a as an interior vertex Using cfw_a,b as interior vertices Using cfw_a,b,c as interior vertices Using cfw_a,b,c,d as interior verticesUsing cfw_a,b,c,d,e as interior verticesFinal AnswerSection
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Michigan - EECS - 203
EECS 203: Homework 7Instructions: The questions below come from the course textbook and are rated as (E)asy, (M)edium or (C)hallenging. Answer them as completely, concisely, and legibly as possible and your answers will be scored and weighted according t
Michigan - EECS - 203
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Michigan - EECS - 203
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Michigan - EECS - 203
Michigan - EECS - 203
Michigan - EECS - 203
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