TEST BANK FOR Introduction to the Theory of Computation 3rd Edition By M. (Michael) Sipser

Chapter 0
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
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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
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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, ab; ε, bb
h. ba, bba; b, ε
1.21 In both parts we first add a new start state and a new accept state. Several solutions are
possible, depending on the order states are removed.
a. Here we remove state 1 then state 2 and we obtain
a∗b(a ∪ ba∗b)∗
b. Here we remove states 1, 2, then 3 and we obtain
ε ∪ ((a ∪ b)a∗b((b ∪ a(a ∪ b))a∗b)∗(ε ∪ a))
1.22 b. /#(#∗(a ∪ b) ∪ /)∗#+/
1.24 a. q1, q1, q1, q1; 000.
b. q1, q2, q2, q2; 111.
c. q1, q1, q2, q1, q2; 0101.
d. q1, q3; 1.
e. q1, q3, q2, q3, q2; 1111.
f. q1, q3, q2, q1, q3, q2, q1; 110110.
g. q1; ε.
1.25 A finite state transducer is a 5-tuple (Q,Σ, Γ, δ, q0), where
i) Q is a finite set called the states,
ii) Σ is a finite set called the alphabet,
iii) Γ is a finite set called the output alphabet,
iv) δ : Q × Σ−→Q × Γ is the transition function,
v) q0 ∈ Q is the start state.
Let M = (Q,Σ, Γ, δ, q0) be a finite state transducer, w = w1w2 · · ·wn be a string
over Σ, and v = v1v2 · · · vn be a string over the Γ. Then M outputs v if a sequence of
states r0, r1, . . . , rn exists in Q with the following two conditions:
i) r0 = qo
ii) δ(ri, wi+1) = (ri+1, vi+1) for i = 0, . . . , n − 1.
1.26 a. T1 = (Q,Σ, Γ, δ, q1), where
i) Q = {q1, q2},
ii) Σ = {0, 1, 2},
iii) Γ = {0, 1},
iv) δ is described as
0 1 2
q1 (q1,0) (q1,0) (q2,1)
q2 (q1,0) (q2,1) (q2,1)
v) q1 is the start state.
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.
Instructor’s Manual 5
b. T2 = (Q,Σ, Γ, δ, q1), where
i) Q = {q1, q2, q3},
ii) Σ = {a, b},
iii) Γ = {0, 1},
iv) δ is described as
a b
q1 (q2,1) (q3,1)
q2 (q3,1) (q1,0)
q3 (q1,0) (q2,1)
v) q1 is the start state.
1.29 b. Let A2 = {www| w ∈ {0,1}∗}. We show that A2 is nonregular using the pumping
lemma. Assume to the contrary that A2 is regular. Let p be the pumping length given by
the pumping lemma. Let s be the string apbapbapb. Because s is a member of A2 and
s has length more than p, the pumping lemma guarantees that s can be split into three
pieces, s = xyz, satisfying the three conditions of the lemma. However, condition 3
implies that y must consist only of as, so xyyz ∈ A2 and one of the first two conditions
is violated. Therefore A2 is nonregular.
1.30 The error is that s = 0p1p can be pumped. Let s = xyz, where x = 0, y = 0 and
z = 0p−21p. The conditions are satisfied because
i) for any i ≥ 0, xyiz = 00i0p−21p is in 0∗1∗.
ii) |y| = 1 > 0, and
iii) |xy| = 2 ≤ p.
1.31 We construct a DFA which alternately simulates the DFAs for A and B, one step at a time.
The new DFA keeps track of which DFA is being simulated. LetM1 = (Q1,Σ, δ1, s1, F1)
and M2 = (Q2,Σ, δ2, s2, F2) be DFAs for A and B. We construct the following DFA
M = (Q,Σ, δ, s0, F) for the perfect shuffle of A and B.
i) Q = Q1 × Q2 × {1, 2}.
ii) For q1 ∈ Q1, q2 ∈ Q2, b ∈ {1, 2}, and a ∈ Σ:
δ((q1, q2, b), a) =

(δ1(q1, a), q2, 2) b = 1
(q1, δ1(q2, a), 1) b = 2.
iii) s0 = (s1, s2, 1).
iv) F = {(q1, q2, 1)| q1 ∈ F1 and q2 ∈ F2}.
1.32 We construct an NFA which simulates the DFAs for A and B, nondeterministically
switching back and forth from one to the other. Let M1 = (Q1,Σ, δ1, s1, F1) and
M2 = (Q2,Σ, δ2, s2, F2) be DFAs for A and B. We construct the following NFA
N = (Q,Σ, δ, s0, F) for the shuffle of A and B.
i) Q = Q1 × Q2.
ii) For q1 ∈ Q1, q2 ∈ Q2, and a ∈ Σ:
δ((q1, q2), a) = {(δ1(q1, a), q2), (q1, δ2(q2, a))}.
iii) s0 = (s1, s2).
iv) F = {(q1, q2)| q1 ∈ F1 and q2 ∈ F2}.
1.33 Let M = (Q,Σ, δ, q0, F) be a DFA that recognizes A. Then we construct NFA N =
(Q,Σ, δ, q
0, F) recognizing DROP-OUT(A). The idea behind the construction is
that N simulates M on its input, nondeterministically guessing the point at which the
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different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.
6 Theory of Computation, third edition
dropped out symbol occurs. At that point N guesses the symbol to insert in that place,
without reading any actual input symbol at that step. Afterwards, it continues to simulate
M.
We implement this idea in N by keeping two copies ofM, called the top and bottom
copies. The start state is the start state of the top copy. The accept states of N are the
accept states of the bottom copy. Each copy contains the edges that would occur in M.
Additionally, include ε edges from each state q in the top copy to every state in the
bottom copy that q can reach.
We describe N formally. The states in the top copy are written with a T and the
bottom with a B, thus: (T, q) and (B, q).
i) Q = {T,B} × Q,
ii) q
0 = (T, q0),
iii) F = {B} × F,
iv) δ((T, q), a) =

{(T, δ(q, a))} a ∈ Σ
{(B, δ(q, b))| b ∈ Σ} a = ε
δ((B, q), a) =

{(B, δ(q, a))} a ∈ Σ
∅ a = ε
1.35 Let M = (Q,Σ, δ, q0, F) be a DFA that recognizes A. We construct a new DFA M =
(Q,Σ, δ, q0, F) that recognizes A/B. Automata M and M differ only in the sets of
accept states. Let F = {r| starting at r and reading a string in B we get to an accept
state of M}. Thus M accepts a string w iff there is a string x ∈ B where M accepts
wx. Hence M recognizes A/B.
1.36 For any regular language A, let M1 be the DFA recognizing it. We need to find a DFA
that recognizes AR. Since any NFA can be converted to an equivalent DFA, it suffices to
find an NFA M2 that recognizes AR.
We keep all the states in M1 and reverse the direction of all the arrows in M1. We
set the accept state of M2 to be the start state in M1. Also, we introduce a new state q0
as the start state for M2 which goes to every accept state in M1 by an -transition.
1.39 The idea is that we start by comparing the most significant bit of the two rows. If the
bit in the top row is bigger, we know that the string is in the language. The string does
not belong to the language if the bit in the top row is smaller. If the bits on both rows
are the same, we move on to the next most significant bit until a difference is found. We
implement this idea with a DFA having states q0, q1, and q2. State q0 indicates the result
is not yet determined. States q1 and q2 indicate the top row is known to be larger, or
smaller, respectively. We start with q0. If the top bit in the input string is bigger, it goes
to q1, the only accept state, and stays there till the end of the input string. If the top bit
in the input string is smaller, it goes to q2 and stays there till the end of the input string.
Otherwise, it stays in state q0.
1.40 Assume language E is regular. Use the pumping lemma to a get a pumping length p
satisfying the conditions of the pumping lemma. Set s = [01
]p [ 10
]p. Obviously, s ∈ E
and |s| ≥ p. Thus, the pumping lemma implies that the string s can be written as xyz
with x = [01
]a , y = [01
]b , z = [01
]c [ 10
]p, where b ≥ 1 and a + b + c = p. However, the
string s = xy0z = [01
]a+c [ 10
]p ∈ E, since a + c < p. That contradicts the pumping