Instructor’s Solutions Manual for Introduction to the Theory
of Computation third edition
Michael Sipser Mathematics Department MIT
Preface
This Instructor’s Manual is designed to accompany the textbook, Introduction to the Theory
of Computation, third edition, by Michael Sipser, published by Cengage, 2013. It contains
solutions to almost all of the exercises and problems in Chapters 0–9. Most of the omitted
solutions in the early chapters require figures, and producing these required more work that
we were able to put into this manual at this point. A few problems were omitted in the later
chapters without any good excuse.
Some of these solutions were based on solutions written by my teaching assistants and
by the authors of the Instructor’s Manual for the first edition.
This manual is available only to instructors.
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 (C) contains 2c elements because each element of C may either be in (C) or not in
, (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.
0.7 The underlying set is U 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
∈ set. Pick any other
underlying ∈ 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.
1
0.12 a. Basis: Let n = 0. Then, S(n) = 0 by definition. Furthermore, n(n + 1) = 0. So
2
1
S(n) = n(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
= k(k + 1) + (k + 1) by the induction hypothesis
1
= (k + 1)(k + 2) by algebra
1 4 3 2
b. Basis: Let n = 0. Then, C(n) = 0 by definition, and (n + 2n + n ) = 0. So
4
, 1 4 3 2
C(n) = (n + 2n + n ) 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) = 1 + 2 + · · · + k + (k + 1)
3 3 3 3
by definition
C(k) = 1 + · · · + k
3 3 3
= C(k) + (k + 1)
1 4 3 2 3
= (n + 2n + n ) + (k + 1) induction hypothesis
1 4 3 2
= ((n + 1) + 2(n + 1) + (n + 1) ) by algebra
0.13 Dividing by (a b) is illegal, because a = b hence a b = 0 and division by 0 is
undefin
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. Let M ′ be the DFA M with the accept and non-accept states swapped. We show that M ′
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 run M 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
by M . So M ′ accepts exactly those strings not accepted by M . 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 rec-
ognizing 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 Let Σ = {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
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
o ∪ ((a ∪ b)a∗b((b ∪ a(a ∪ b))a∗b)∗(ε ∪ a))
b. /#(#∗(a ∪ b) ∪ /)∗# /
+
1.22
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, r 1 , ... , 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
q1 (q1,0) (q1,0) (q2,1)
q2 (q1,0) (q2,1) (q2,1)
v) q1 is the start state.
b. T2 = (Q, Σ, Γ, δ, q1), where
i) Q = {q1, q2, q3},
of Computation third edition
Michael Sipser Mathematics Department MIT
Preface
This Instructor’s Manual is designed to accompany the textbook, Introduction to the Theory
of Computation, third edition, by Michael Sipser, published by Cengage, 2013. It contains
solutions to almost all of the exercises and problems in Chapters 0–9. Most of the omitted
solutions in the early chapters require figures, and producing these required more work that
we were able to put into this manual at this point. A few problems were omitted in the later
chapters without any good excuse.
Some of these solutions were based on solutions written by my teaching assistants and
by the authors of the Instructor’s Manual for the first edition.
This manual is available only to instructors.
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 (C) contains 2c elements because each element of C may either be in (C) or not in
, (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.
0.7 The underlying set is U 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
∈ set. Pick any other
underlying ∈ 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.
1
0.12 a. Basis: Let n = 0. Then, S(n) = 0 by definition. Furthermore, n(n + 1) = 0. So
2
1
S(n) = n(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
= k(k + 1) + (k + 1) by the induction hypothesis
1
= (k + 1)(k + 2) by algebra
1 4 3 2
b. Basis: Let n = 0. Then, C(n) = 0 by definition, and (n + 2n + n ) = 0. So
4
, 1 4 3 2
C(n) = (n + 2n + n ) 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) = 1 + 2 + · · · + k + (k + 1)
3 3 3 3
by definition
C(k) = 1 + · · · + k
3 3 3
= C(k) + (k + 1)
1 4 3 2 3
= (n + 2n + n ) + (k + 1) induction hypothesis
1 4 3 2
= ((n + 1) + 2(n + 1) + (n + 1) ) by algebra
0.13 Dividing by (a b) is illegal, because a = b hence a b = 0 and division by 0 is
undefin
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. Let M ′ be the DFA M with the accept and non-accept states swapped. We show that M ′
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 run M 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
by M . So M ′ accepts exactly those strings not accepted by M . 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 rec-
ognizing 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 Let Σ = {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
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
o ∪ ((a ∪ b)a∗b((b ∪ a(a ∪ b))a∗b)∗(ε ∪ a))
b. /#(#∗(a ∪ b) ∪ /)∗# /
+
1.22
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, r 1 , ... , 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
q1 (q1,0) (q1,0) (q2,1)
q2 (q1,0) (q2,1) (q2,1)
v) q1 is the start state.
b. T2 = (Q, Σ, Γ, δ, q1), where
i) Q = {q1, q2, q3},
