Unformatted text preview: Homework 11 Solutions
Section 8.4 Initial Matrix Using {a} as an interior vertex Using {a,b} as interior vertices Using {a,b,c} as interior vertices Using {a,b,c,d} as interior vertices Using {a,b,c,d,e} as interior vertices Final Answer Section 8.5 14. We must show that the relation satisfies each of the following three properties: Reflexivity: Every character of a word W is obviously equal to itself, so the nth character of W is always the same letter as the nth character of W. . Symmetry: If , then the nth character of is the same letter as the nth character of . Since the characters remain unchanged depending on which string we consider first, . Transitivity: If and , then the nth character of is the same letter as the nth character of and the nth character of is the same letter as the nth character of , then we can also conclude that the nth character of is the same letter as the nth character of , and that . 60. Recall that iff and . We must show that this relation satisfies each of the following three properties: Reflexivity: We must show that . Choose C=1 and k=1. Since the definition of bigO notation allows for equality, this statement holds. Symmetry: If , then and which are exactly the conditions necessary to conclude that . Transitivity: If and , then , , and . We showed in Section 3.2 that if and , then . Using this we can conclude both and , which is equivalent to saying . Section 8.6 22. a) 4 6 2 3 5 1
c) 10 15 25 2 3 5 11 40. a) Suppose for the sake of contradiction that there is more than one greatest element in a poset. Call two distinct greatest elements a and b. By the definition of greatest element, and . Since a is in the poset, and . This is not possible unless (since a partial order is antisymmetric), but we assumed that a and b are distinct. This is a contradiction. Therefore a poset must have exactly one greatest element if such an element exists. b) Suppose for the sake of contradiction that there is more than one least element in a poset. Call two distinct least elements a and b. By the definition of greatest element, and . Since a is in the poset, and . This is not possible unless (since a partial order is antisymmetric), but we assumed that a and b are distinct. This is a contradiction. Therefore a poset must have exactly one greatest element if such an element exists. 60. Following the same reasoning as in the proof of Lemma 1, choose an arbitrary element of the poset S. If is not a minimal element, then there is some . Continue this process of choosing lesser elements, creating a decreasing chain . Because partial orders are antisymmetric, it is impossible for an element to be repeated in this chain. And because the poset contains a finite number of elements, it must be possible to reach a with no lesser elements. By definition, this will be a minimal element. 64. There are many possible correct answers to this problem. One such answer is Section 9.2 24. The graph is bipartite. can be colored red and can be colored blue. 34. This is possible as the graph below illustrates: 36. a) This is not graphic because the sum of the degrees of all the nodes is odd. c) Yes, this is graphic: 46. a) 2e is equal to the sum of the degrees of all the vertices in the graph. 2e/v is the average degree of each vertex. It is impossible for the average to be less than the minimum. b) 2e is equal to the sum of the degrees of all the vertices in the graph. 2e/v is the average degree of each vertex. It is impossible for the average to be greater than the maximum. 60. Start by choosing an arbitrary vertex and color it red. Then color all of its neighbors blue. For each blue vertex, color all of its uncolored neighbors red. For each red vertex color all of its uncolored neighbors blue. Repeat this process until every vertex has been colored. If any vertex neighbors a vertex of the same color, then the graph is not bipartite. Otherwise it is. Section 9.3 38. Yes, the two graphs are isomorphic. This is a bijection between the two: Edges between vertices and are transformed according to the above. 54. a) There are only two possible graphs with two vertices: the one with an edge and the one without. b) A graph with three vertices can contain 0, 1, 2, or 3 edges. Excluding isomorphisms, there is only one graph for each number of edges. Therefore the answer is 4. c) There is one graph with zero edges and one graph with one edge. If there are two edges, the edges can either be adjacent or nonadjacent, yielding two graphs. If there are three edges, the edges may form a triangle, a star, or a path, giving three possible graphs. To save calculating this for higher numbers of edges, we realize that graphs with four, five, or six edges are actually just complements of graphs with zero, one, or two edges, so the number of possible graphs must be the same for four, five, and six edges. Adding all these possibilities together yields: 2 * (1 + 1 + 2) + 3 = 11 ...
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This note was uploaded on 02/12/2012 for the course EECS 203 taught by Professor Yaoyunshi during the Spring '07 term at University of Michigan.
 Spring '07
 YaoyunShi

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