Exam (elaborations) TEST BANK FOR Introduction to the Theory of Computation 3rd Edition By M. (Michael) Sipser (Solution Manual)-Converted

0.1 a. The odd positive integers. b. The even integers. c. The even positive integers. d. The positive integers which are a multiple of 6. e. The palindromes over {0,1}. f. The empty set. 0.2 a. {1, 10, 100}. b. {n| n > 5}. c. {1, 2, 3, 4}. d. {aba}. e. {ε}. f. ∅. 0.3 a. No. b. Yes. c. A. d. B. e. {(x, x), (x, y), (y, x), (y, y), (z, x), (z, y)}. f. {∅, {x}, {y}, {x, y}}. 0.4 A × B has ab elements, because each element of A is paired with each element of B, so A × B contains b elements for each of the a elements of A. 0.5 P(C) contains 2c elements because each element of C may either be in P(C) or not in P(C), and so each element of C doubles the number of subsets of C. Alternatively, we can view each subset S of C as corresponding to a binary string b of length c, where S contains the ith element of C iff the ith place of b is 1. There are 2c strings of length c and hence that many subsets of C. 0.6 a. f(2) = 7. b. The range = {6, 7} and the domain = {1, 2, 3, 4, 5}. c. g(2,10) = 6. d. The range = {1, 2, 3, 4, 5} × {6, 7, 8, 9, 10} and the domain = {6, 7, 8, 9, 10}. e. f(4) = 7 so g(4, f(4)) = g(4, 7) = 8. 1 c 2013 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part. 2 Theory of Computation, third edition 0.7 The underlying set is N in these examples. a. Let R be the “within 1” relation, that is, R = {(a, b)| |a − b| ≤ 1}. b. Let R be the “less than or equal to” relation, that is, R = {(a, b)| a ≤ b}. c. Finding a R that is symmetric and transitive but not reflexive is tricky because of the following “near proof” that R cannot exist! Assume that R is symmetric and transitive and chose any member x in the underlying set. Pick any other member y in the underlying set for which (x, y) ∈ R. Then (y, x) ∈ R because R is symmetric and so (x, x) ∈ R because R is transitive, hence R is reflexive. This argument fails to be an actual proof because y may fail to exist for x. Let R be the “neither side is 1” relation, R = {(a, b)| a = 1 and b = 1}. 0.10 Let G be any graph with n nodes where n ≥ 2. The degree of every node in G is one of the n possible values from 0 to n − 1. We would like to use the pigeon hole principle to show that two of these values must be the same, but number of possible values is too great. However, not all of the values can occur in the same graph because a node of degree 0 cannot coexist with a node of degree n − 1. Hence G can exhibit at most n − 1 degree values among its n nodes, so two of the values must be the same. 0.11 The error occurs in the last sentence. If H contains at least 3 horses, H1 and H2 contain a horse in common, so the argument works properly. But, if H contains exactly 2 horses, then H1 and H2 each have exactly 1 horse, but do not have a horse in common. Hence we cannot conclude that the horse in H1 has the same color as the horse in H2. So the 2 horses in H may not be colored the same. 0.12 a. Basis: Let n = 0. Then, S(n) = 0 by definition. Furthermore, 1 2n(n + 1) = 0. So S(n) = 1 2n(n + 1) when n = 0. Induction: Assume true for n = k where k ≥ 0 and prove true for n = k + 1. We can use this series of equalities: S(k + 1) = 1 + 2 + · · · + k + (k + 1) by definition = S(k) + (k + 1) because S(k) = 1 + 2 + · · · + k = 1 2k(k + 1) + (k + 1) by the induction hypothesis = 1 2 (k + 1)(k + 2) by algebra b. Basis: Let n = 0. Then, C(n) = 0 by definition, and 1 4 (n4 + 2n3 + n2) = 0. So C(n) = 1 4 (n4 + 2n3 + n2) when n = 0. Induction: Assume true for n = k where k ≥ 0 and prove true for n = k + 1. We can use this series of equalities: C(k + 1) = 13 + 23 + · · · + k3 + (k + 1)3 by definition = C(k) + (k + 1)3 C(k) = 13 + · · · + k3 = 1 4 (n4 + 2n3 + n2) + (k + 1)3 induction hypothesis = 1 4 ((n + 1)4 + 2(n + 1)3 + (n + 1)2) by algebra 0.13 Dividing by (a − b) is illegal, because a = b hence a − b = 0 and division by 0 is undefined. c 2013 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1 1.12 Observe that D ⊆ b∗a∗ because D doesn’t contain strings that have ab as a substring. Hence D is generated by the regular expression (aa)∗b(bb)∗. From this description, finding the DFA for D is more easily done. 1.14 a. LetM be the DFAM with the accept and non-accept states swapped. We show thatM recognizes the complement of B, where B is the language recognized by M. Suppose M accepts x. If we run M on x we end in an accept state of M. Because M and M have swapped accept/non-accept states, if we runM on x, we would end in a non-accept state. Therefore, x ∈ B. Similarly, if x is not accepted by M, then it would be accepted byM. SoM accepts exactly those strings not accepted byM. Therefore,M recognizes the complement of B. Since B could be any arbitrary regular language and our construction shows how to build an automaton to recognize its complement, it follows that the complement of any regular language is also regular. Therefore, the class of regular languages is closed under complement. b. Consider the NFA in Exercise 1.16(a). The string a is accepted by this automaton. If we swap the accept and reject states, the string a is still accepted. This shows that swapping the accept and non-accept states of an NFA doesn’t necessarily yield a new NFA recognizing the complementary language. The class of languages recognized by NFAs is, however, closed under complement. This follows from the fact that the class of languages recognized by NFAs is precisely the class of languages recognized by DFAs which we know is closed under complement from part (a). 1.18 LetΣ = {0, 1}. a. 1Σ∗0 b. Σ∗1Σ∗1Σ∗1Σ∗ c. Σ∗0101Σ∗ d. ΣΣ0Σ∗ e. (0 ∪ 1Σ)(ΣΣ)∗ f. (0 ∪ (10)∗)∗1∗ g. (ε ∪ Σ)(ε ∪ Σ)(ε ∪ Σ)(ε ∪ Σ)(ε ∪ Σ) h. Σ∗0Σ∗ ∪ 1111Σ∗ ∪ 1 ∪ ε i. (1Σ)∗(1 ∪ ε) j. 0∗(100 ∪ 010 ∪ 001 ∪ 00)0∗ k. ε ∪ 0 l. (1∗01∗01∗)∗ ∪ 0∗10∗10∗ 3 c 2013 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part. 4 Theory of Computation, third edition m. ∅ n. Σ+ 1.20 a. ab, ε; ba, aba b. ab, abab; ε, aabb c. ε, aa; ab, aabb d. ε, aaa; aa, b e. aba, aabbaa; ε, abbb f. aba, bab; ε, ababab g. b,