Answer Problem1(5pts)

• Please give details of your answer. A direct answer without explanation is not counted.

• Your answers must be in English.

• Please carefully read problem statements.

• During the exam you are not allowed to borrow others’ class notes.

• Try to work on easier questions first.

Problem 1 (5 pts)

Give the formal definition of the following NFA.

q₁

start q₂ q₃ q₄

0,1

1 0,1, 0,1,

Answer

The formal definition of the NFA is (Q, Σ, δ, q₁, F ), where

• Q = {q₁, q₂, q₃, q₄},

• Σ = {0, 1},

• δ is given as

0 1 

q₁ {q₁} {q₁, q₂} ∅ q₂ {q₃} {q₃} {q₃} q₃ {q₄} {q₄} {q₄}

q₄ ∅ ∅ ∅

,

• q₁ is the start state, and

• F = {q₄}.

Common Mistakes

• Σ 6= {0, 1, }

• start state is q₁ (not {q₁}).

• elements in δ should be sets (not states).

Problem 2 (20 pts)

Use the procedure in section 1.2 of the textbook to convert the NFA in Problem 1 to DFA.

Simplify the obtained DFA to have only 4 states. (You need to first draw ALL possible states and then eliminate unnecessary states.)

Answer

∅

q₂

q3 q4

q12

q₁₃

q₂₃ q₂₄

q₃₄ q123

q₁₂₄ q₂₃₄

q₁₄ q134

q₁₂₃₄

q₁

start

0,1

0 1

0,1

0,1 0,1

0

1

0 1

0 1

0,1 0,1 0,1 0

1 0

1 1

0

0,1 0

1

the simplified diagram:

q14

q₁₃₄

q1234

q1

start 0

1 0

1 1

0

0

1

Problem 3 (20 pts)

Is the following language regular?

L = {a^mbⁿ| m = kn, where k is positive integer } Explain your answer clearly.

Answer

No. Assume L is regular. By pumping lemma, there is a positive integer p ≥ 1 such that for all s ∈ L and |s| ≥ p, s = xyz satisfies the following conditions:

• |y| ≥ 1,

• |xy| ≤ p, and

• for all i ≥ 0, xyⁱz ∈ L.

Let s = a^pb^p. Because p = 1p, s ∈ L. For i = 0, xy⁰z = a^(p−|y|)b^p, then p − |y| < kp for any positive integer k. We have proved xy⁰z /∈ L and got a contradiction, so the language L is not regular.

Common Mistakes

• Some of you choose s = a^2pb^p. However, for i = 0, |y| = p, xy⁰z = a^2p−|y|b^p = a^pb^p ∈ L.

• When you choose a string s, you cannot sepcify x, y, or z there. For example, you cannot choose s to be x = , y = a^p, z = b^p and say that xy⁰z /∈ L.

Problem 4 (20 pts)

Is the following language regular?

L = {a^mbⁿ | m + n is prime}

Explain your answer clearly.

Answer

No. Assume L is regular. By pumping lemma, there is a positive integer p ≥ 1 such that for all s ∈ L and |s| ≥ p, s = xyz satisfies the following conditions:

• |y| ≥ 1,

• |xy| ≤ p, and

• for all i ≥ 0, xyⁱz ∈ L.

Let s = a^pb^q, where q is the smallest positive integer so that p + q is prime. Choose y = a^t for t ≥ 1, then xyⁱz = a^p−t+itb^q = a^p+(i−1)tb^q. Select i = p + q + 1, xy^p+q+1z = a^p+(p+q)tb^q, then (p + (p + q)t) + q = (p + q)(t + 1) is not prime. We have proved xy^p+q+1z /∈ L and got a contradiction. L is not regular.

Problem 5 (25 pts)

Consider the following language

A = {}, where the alphabet Σ = {0}.

1. Give a 1-state NFA to recognize this language (diagram and formal definition).

2. Give a 2-state NFA to recognize this language (diagram and formal definition).

3. Convert the 2-state NFA to DFA using the procedure in section 1.2 of the textbook. (You need to first draw ALL possible states and then simplify it to a 2-state DFA.)

4. Convert the DFA to GNFA and follow the procedure in section 1.3 of the textbook to obtain the corresponding regular expression.

5. Consider the DFA of subproblem 3. Is the 2-state DFA unique? Why? If you use 3 states to recognize the language, is there an unique DFA? Why?

Answer

1.

q₁ start

The formal definition of the NFA is (Q, Σ, δ, q₁, F ), where

• Q = {q1},

• Σ = {0},

• δ is given as

0  q₁ ∅ ∅

,

• q₁ is the start state, and

• F = {q1}.

2.

q₁

start q₂

0 0

The formal definition of the NFA is (Q, Σ, δ, q₁, F ), where

• Q = {q₁, q₂},

• Σ = {0},

0  q₁ {q₂} ∅ q₂ {q₂} ∅

,

• q₁ is the start state, and

• F = {q₁}.

3.

q₁

start q₂ q₁₂ q∅

0 0

0

0

the simplified diagram:

q₁

start q₂

0 0

4.

• add q_s and q_a:

qs

start q1

qa

q2

0 0





• remove q2:

q_s

start q₁

q_a





• remove q₁:

qs

start  qa

The regular expression is .

5.

Yes. The DFA must accept at first state. For the alphabet Σ = {0}, it requires another state for transition. The 2-state DFA is unique.

No. The one more state can offer another state for transition or be unconnected.

q₁

start q₂ q₃

0

0 0

q₁

start 0 q₂ q₃

0

0

q1

start q2 q3

0 0

0

The 3-state DFA is not unique.

Problem 6 (10 pts)

Give a counterexample to show that the following construction fails to prove the closure of the class of regular languages under the star operation. Let N₁ = (Q₁, Σ, δ₁, q₁, F₁) recognize A₁, where Σ = {0, 1}. Construct N = (Q₁, Σ, δ, q₁, F ) as follows. N is supposed to recognize A^∗₁.

• The states of N are the states of N1.

• F = {q₁} ∪ F₁.

The accept states F are the old accept states plus its start state.

• Define δ so that for any q ∈ Q and any a ∈ Σ ∪ {},

δ(q, a) =

(δ₁(q, a) q /∈ F₁ or a 6=  δ₁(q, a) ∪ {q₁} q ∈ F₁ and a = .

The counterexample must be a NFA diagram with the smallest number of states, and explain your counterexample clearly.

Answer

Here is a counterexample: A₁ = ∅ and N₁ =

q₁ start

0

Then N =

q₁ start

0

N accepts 0, but 0 /∈ A^∗₁. Therefore, N does not recognize A^∗₁.

Common Mistakes

• One state is enough.

• Your NFA should recognize your langauge.