Cs133 lec03

Regular Expressions

Non-Regular Languages

CS 133 : Automata Theory and Computability Regular Languages and Finite Automata Henry N. Adorna Nestine Hope S. Hernandez Algorithms and Complexity Laboratory Department of Computer Science University of the Philippines, Diliman {hnadorna,nshernandez}@dcs.upd.edu.ph

LEC 3

Regular Expressions

Non-Regular Languages

Regular Languages and Finite Automata

Regular Expressions Converting Regular Languages to Regular Expressions Converting Regular Expressions to Finite Automata

Non-Regular Languages

Regular Expressions

Non-Regular Languages

Regular Languages and Finite Automata

Regular Expressions Converting Regular Languages to Regular Expressions Converting Regular Expressions to Finite Automata

Non-Regular Languages

Regular Expressions

Non-Regular Languages

Regular Expressions

Definition R is a regular expression if R is 1. a for some a in the alphabet Σ 2. ε 3. ∅ 4. (R1 ∪ R2 ), where R1 and R2 are regular expressions 5. (R1 ◦ R2 ), where R1 and R2 are regular expressions 6. (R1∗ ), where R1 is a regular expression

Regular Expressions

Example (Let Σ = {0, 1}) 1. 1∗ 01∗

Non-Regular Languages

Regular Expressions

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

Non-Regular Languages

Regular Expressions

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗

Non-Regular Languages

Regular Expressions

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

Non-Regular Languages

Regular Expressions

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗

Non-Regular Languages

Regular Expressions

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗ = {w|w is a string over Σ excluding ε} = Σ+ .

Non-Regular Languages

Regular Expressions

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗ = {w|w is a string over Σ excluding ε} = Σ+ .

4. (ΣΣ)∗

Non-Regular Languages

Regular Expressions

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗ = {w|w is a string over Σ excluding ε} = Σ+ .

4. (ΣΣ)∗ = {w|w is a string of even length }.

Non-Regular Languages

Regular Expressions

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗ = {w|w is a string over Σ excluding ε} = Σ+ .

4. (ΣΣ)∗ = {w|w is a string of even length }.

5. 1(01+ )∗

Non-Regular Languages

Regular Expressions

Non-Regular Languages

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗ = {w|w is a string over Σ excluding ε} = Σ+ .

4. (ΣΣ)∗ = {w|w is a string of even length }.

5. 1(01+ )∗ = {w|w begins with 1 and every 0 in w is followed by at least one 1}.

Regular Expressions

Non-Regular Languages

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗ = {w|w is a string over Σ excluding ε} = Σ+ .

4. (ΣΣ)∗ = {w|w is a string of even length }.

5. 1(01+ )∗ = {w|w begins with 1 and every 0 in w is followed by at least one 1}.

6. (1 ∪ ε)0∗

Regular Expressions

Non-Regular Languages

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗ = {w|w is a string over Σ excluding ε} = Σ+ .

4. (ΣΣ)∗ = {w|w is a string of even length }.

5. 1(01+ )∗ = {w|w begins with 1 and every 0 in w is followed by at least one 1}.

6. (1 ∪ ε)0∗ = 10∗ ∪ 0∗ .

Regular Expressions

Non-Regular Languages

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗ = {w|w is a string over Σ excluding ε} = Σ+ .

4. (ΣΣ)∗ = {w|w is a string of even length }.

5. 1(01+ )∗ = {w|w begins with 1 and every 0 in w is followed by at least one 1}.

6. (1 ∪ ε)0∗ = 10∗ ∪ 0∗ .

7. ∅∗

Regular Expressions

Non-Regular Languages

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗ = {w|w is a string over Σ excluding ε} = Σ+ .

4. (ΣΣ)∗ = {w|w is a string of even length }.

5. 1(01+ )∗ = {w|w begins with 1 and every 0 in w is followed by at least one 1}.

6. (1 ∪ ε)0∗ = 10∗ ∪ 0∗ .

7. ∅∗ = {ε}.

Regular Expressions

Non-Regular Languages

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗ = {w|w is a string over Σ excluding ε} = Σ+ .

4. (ΣΣ)∗ = {w|w is a string of even length }.

5. 1(01+ )∗ = {w|w begins with 1 and every 0 in w is followed by at least one 1}.

6. (1 ∪ ε)0∗ = 10∗ ∪ 0∗ .

7. ∅∗ = {ε}.

8. 1∗ ∅

Regular Expressions

Non-Regular Languages

Example (Let Σ = {0, 1}) 1. 1∗ 01∗ = {w|w contains a single 0}.

2. Σ∗ 0Σ∗ = {w|w has at least one 0}.

3. ΣΣ∗ = {w|w is a string over Σ excluding ε} = Σ+ .

4. (ΣΣ)∗ = {w|w is a string of even length }.

5. 1(01+ )∗ = {w|w begins with 1 and every 0 in w is followed by at least one 1}.

6. (1 ∪ ε)0∗ = 10∗ ∪ 0∗ .

7. ∅∗ = {ε}.

8. 1∗ ∅ = ∅.

Regular Expressions

Non-Regular Languages

Theorem A language is regular if and only if some regular expression describes it.

Regular Expressions

Non-Regular Languages

Theorem A language is regular if and only if some regular expression describes it.

Lemma If a language is described by a regular expression, then it is regular.

Regular Expressions

Non-Regular Languages

Theorem A language is regular if and only if some regular expression describes it.

Lemma If a language is described by a regular expression, then it is regular.

Lemma If a language is regular, then it is described by a regular expression.

Regular Expressions

Non-Regular Languages

Generalized Nondeterministic Finite Automaton

Definition A generalized nondeterministic finite automaton(GNFA) is a 5-tuple (Q, Σ, δ, qstart , qaccept ), where 1. Q is the finite set of states, 2. Σ is the input alphabet, 3. δ : (Q − {qaccept }) × R → (Q − {qstart }) is the transition function, 4. qstart is the start state, and 5. qaccept is the final state.

Regular Expressions

Non-Regular Languages

Generalized Nondeterministic Finite Automaton

Definition A generalized nondeterministic finite automaton(GNFA) is a 5-tuple (Q, Σ, δ, qstart , qaccept ), where 1. Q is the finite set of states, 2. Σ is the input alphabet, 3. δ : (Q − {qaccept }) × R → (Q − {qstart }) is the transition function, 4. qstart is the start state, and 5. qaccept is the final state.

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example1: Step1

a start

1

2

a b

a

b 3

b

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example1: Step1

a start

1

2

a b

b

a

b 3 b a start

s

ε

1

ε

2

a b

ε

a

b 3

a

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example1: Step2

b a start

s

ε

1

b b

ε

2

a

ε

a 3

a

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example1: Step2

b a start

s

ε

1

ε

2

a b

ε

a

b

a

3 aa ∪ b

start

a

s

b

ε

2 a ∪ ba

ab

3

a

ε bb

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example1: Step3

aa ∪ b

start

s

a

2

ba ∪ ba

ε

a

ab ε 3

bb

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example1: Step3

aa ∪ b

start

s

a

2

ba ∪ ba

ε

a

ab ε 3

bb a(aa ∪ b)∗

start

s

b ∪ a(aa ∪ b)∗ ab

(a ∪ ba)(aa ∪ b)∗ ∪ ε

3

bb ∪ (a ∪ ba)(aa ∪ b)∗ ab

a

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example1: Step4

a(aa ∪ b)∗

start

s

b ∪ a(aa ∪ b)∗ ab

(a ∪ ba)(aa ∪ b)∗ ∪ ε

3

bb ∪ (a ∪ ba)(aa ∪ b)∗ ab

a

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example1: Step4

a(aa ∪ b)∗

start

s

b ∪ a(aa ∪ b)∗ ab

(a ∪ ba)(aa ∪ b)∗ ∪ ε

3

a

bb ∪ (a ∪ ba)(aa ∪ b)∗ ab

(a(aa ∪ b)∗ ) ∪ (b ∪ a(aa ∪ b)∗ ab)(bb ∪ (a ∪ ba)(aa ∪ b)∗ ab)∗ ((a ∪ ba)(aa ∪ b)∗ ∪ ε)

start

s

a

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example2: Step1

start

b

1

a

2 Îľ

a,b

a 3

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example2: Step1

start

b

1

a

2 ε

a,b

a 3 start

ε

s

b

1

ε

ε a

a

2 a∪b

a 3

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example2: Step2

start

ε

s

b

1

ε

ε a

a

2 a∪b

a 3

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example2: Step2

start

ε

s

b

1

ε

ε a

a∪b

a 3

start

b

s

ε

a

2

a

a

a∪b

ab

ε

3

a

2

a

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example2: Step3

start

b

s

ε

a

2

a∪b

ab

ε

a

a

3

a

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example2: Step3

start

b

s

ε

a start

a∪b

ab

ε

a

a

2

3

a

s ε ∪ ba∗ (a ∪ b)

ε

a

a

3

a ∪ aba∗ (a ∪ b)

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example2: Step4

start

s ε ∪ ba∗ (a ∪ b)

ε a

a

3

a ∪ aba∗ (a ∪ b)

Regular Expressions

Non-Regular Languages

Regular Language → Regular Expression Example2: Step4

start

s ε ∪ ba∗ (a ∪ b)

ε a

start

s

a

3

a ∪ aba∗ (a ∪ b)

ε ∪ ((ε ∪ ba∗ (a ∪ b))(a ∪ aba∗ (a ∪ b))∗ a)

a

Regular Expressions

Non-Regular Languages

Convert the following finite automata to regular expressions. start

a

1

start

a,b

1

a

2

b b

b 2 (a)

a

b 3

a (b)

Regular Expressions

Non-Regular Languages

Convert the following finite automata to regular expressions. a b

start

start

a

1

2

2 a

a

b

b

b

a

a,b 3

(c)

b

1

3

a (d)

4

b

Regular Expressions

Non-Regular Languages

Algebraic Laws for Regular Expressions • Commutativity

B for Union • Associativity

B for Union B for Concatenation • Identity (is a value such that when the operator is applied to it

and some other value, the result is the other value) B ∅ is the identity for union.

B ε is the identity for concatenation.

• Annihilator (is a value such that when the operator is applied to

it and some other value, the result is the annihilator) B ∅ is the annihilator for concatenation.

• Distributivity B of Concatenation over Union • L(M ∪ N ) = LM ∪ LN • (M ∪ N )L = M L ∪ N L

Regular Expressions

Non-Regular Languages

Algebraic Laws for Regular Expressions • Idempotence

B for Union • Laws involving ∗

B B B B B

(L∗ )∗ = L∗ ∅∗ = ε ε∗ = ε LL∗ = L∗ L = L+ L∗ = L+ ∪ ε

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ 2. ε

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ 2. ε

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ 2. ε 3. ∅

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ 2. ε 3. ∅

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ 2. ε 3. ∅ 4. (R1 ∪ R2 )

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ 2. ε 3. ∅ 4. (R1 ∪ R2 )

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ 2. ε 3. ∅ 4. (R1 ∪ R2 )

5. (R1 ◦ R2 )

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ 2. ε 3. ∅ 4. (R1 ∪ R2 )

5. (R1 ◦ R2 )

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ 2. ε 3. ∅ 4. (R1 ∪ R2 )

5. (R1 ◦ R2 )

6. (R1∗ )

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata R is a regular expression if R is 1. a, for some a in the alphabet Σ 2. ε 3. ∅ 4. (R1 ∪ R2 )

5. (R1 ◦ R2 )

6. (R1∗ )

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata Convert the regular expression (ab ∪ a)∗ to an NFA:

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata Convert the regular expression (ab ∪ a)∗ to an NFA:

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata Convert the regular expression (ab ∪ a)∗ to an NFA:

Note that this has an equivalent NFA with only 2 states. Can you find it?

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata Convert the regular expression (a ∪ b)∗ aba to an NFA:

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata Convert the regular expression (a ∪ b)∗ aba to an NFA:

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata Convert the regular expression (a ∪ b)∗ aba to an NFA:

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata Convert the regular expression (a ∪ b)∗ aba to an NFA:

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata Convert the regular expression (a ∪ b)∗ aba to an NFA:

Regular Expressions

Non-Regular Languages

Regular Expression → Finite Automata Convert the regular expression (a ∪ b)∗ aba to an NFA:

Regular Expressions

Non-Regular Languages

Regular Languages and Finite Automata

Regular Expressions Converting Regular Languages to Regular Expressions Converting Regular Expressions to Finite Automata

Non-Regular Languages

Regular Expressions

Non-Regular Languages

Is B = {0n 1n |n ≼ 0} regular?

Regular Expressions

Non-Regular Languages

Pumping Lemma for Regular Languages

Theorem (Pumping Lemma) If A is a regular language, then there is a number p (the pumping length) where, if s is any string in A of length at least p, then s may be divided into three pieces, s = xyz, satisfying the following conditions:

Regular Expressions

Non-Regular Languages

Pumping Lemma for Regular Languages

Theorem (Pumping Lemma) If A is a regular language, then there is a number p (the pumping length) where, if s is any string in A of length at least p, then s may be divided into three pieces, s = xyz, satisfying the following conditions: 1. for each i ≼ 0, xy i z ∈ A,

Regular Expressions

Non-Regular Languages

Pumping Lemma for Regular Languages

Theorem (Pumping Lemma) If A is a regular language, then there is a number p (the pumping length) where, if s is any string in A of length at least p, then s may be divided into three pieces, s = xyz, satisfying the following conditions: 1. for each i ≼ 0, xy i z ∈ A, 2. |y| > 0, and

Regular Expressions

Non-Regular Languages

Pumping Lemma for Regular Languages

Theorem (Pumping Lemma) If A is a regular language, then there is a number p (the pumping length) where, if s is any string in A of length at least p, then s may be divided into three pieces, s = xyz, satisfying the following conditions: 1. for each i ≥ 0, xy i z ∈ A, 2. |y| > 0, and 3. |xy| ≤ p.

Regular Expressions

Non-Regular Languages

Proof: Let M = (Q, Σ, δ, q1 , F ) be a DFA recognizing A and p be the number of states of M .

Let s = s1 s2 · · · sn be a string in A of length n, where n ≥ p.

Let r1 , · · · , rn+1 be the sequence of states that M enters while processing s, so ri+1 = δ(ri , si ) for 1 ≤ i ≤ n.

This sequence has length n + 1, which is at least p + 1. Among the first p + 1 elements in the sequence, 2 must be of the same state, by the pigeonhole principle. We call the first of these rj and the second rl , we have l ≤ p + 1.

Now let x = s1 · · · sj−1 , y = sj · · · sl−1 , z = sl · · · sn .

As x takes M from r1 to rj , y takes M from rj to rj , and z takes M from rj to rn+1 , which is an accept state, M must accept xy i z for i ≥ 0.

We know that j 6= l, so |y| > 0; and l ≤ p + 1, so |xy| ≤ p.

Thus we have satisfied all the conditions of the pumping lemma.

Regular Expressions

Non-Regular Languages

Proof: Let M = (Q, Σ, δ, q1 , F ) be a DFA recognizing A and p be the number of states of M .

Let s = s1 s2 · · · sn be a string in A of length n, where n ≥ p.

Let r1 , · · · , rn+1 be the sequence of states that M enters while processing s, so ri+1 = δ(ri , si ) for 1 ≤ i ≤ n.

This sequence has length n + 1, which is at least p + 1. Among the first p + 1 elements in the sequence, 2 must be of the same state, by the pigeonhole principle. We call the first of these rj and the second rl , we have l ≤ p + 1.

Now let x = s1 · · · sj−1 , y = sj · · · sl−1 , z = sl · · · sn .

As x takes M from r1 to rj , y takes M from rj to rj , and z takes M from rj to rn+1 , which is an accept state, M must accept xy i z for i ≥ 0.

We know that j 6= l, so |y| > 0; and l ≤ p + 1, so |xy| ≤ p.

Thus we have satisfied all the conditions of the pumping lemma.

Note: If no strings in A are at least p then the theorem becomes vacuously true. Obviously the 3 conditions hold for all strings of length at least p if there are no such strings.

Regular Expressions

Non-Regular Languages

We use the pumping lemma to prove that a language L is not regular.

(How???)

Regular Expressions

Non-Regular Languages

We use the pumping lemma to prove that a language L is not regular.

(How???) ♣ Proof by contradiction:

Regular Expressions

Non-Regular Languages

We use the pumping lemma to prove that a language L is not regular.

(How???) ♣ Proof by contradiction: • Assume that L is regular.

Regular Expressions

Non-Regular Languages

We use the pumping lemma to prove that a language L is not regular.

(How???) ♣ Proof by contradiction: • Assume that L is regular.

B the pumping lemma guarantee the existence of a pumping length p such that all strings of length p or greater in L can be pumped.

Regular Expressions

Non-Regular Languages

We use the pumping lemma to prove that a language L is not regular.

(How???) ♣ Proof by contradiction: • Assume that L is regular.

B the pumping lemma guarantee the existence of a pumping length p such that all strings of length p or greater in L can be pumped. • Find a string s in L that has length p or greater but that cannot be pumped.

Regular Expressions

Non-Regular Languages

We use the pumping lemma to prove that a language L is not regular.

(How???) ♣ Proof by contradiction: • Assume that L is regular.

B the pumping lemma guarantee the existence of a pumping length p such that all strings of length p or greater in L can be pumped. • Find a string s in L that has length p or greater but that cannot be pumped.

B involves grouping the various ways of dividing s into several cases and analyzing them individually.

Regular Expressions

Non-Regular Languages

We use the pumping lemma to prove that a language L is not regular.

(How???) ♣ Proof by contradiction: • Assume that L is regular.

B the pumping lemma guarantee the existence of a pumping length p such that all strings of length p or greater in L can be pumped.

• Find a string s in L that has length p or greater but that cannot be pumped.

B involves grouping the various ways of dividing s into several cases and analyzing them individually.

• The existence of s contradicts the pumping lemma if L were

regular.

Regular Expressions

Non-Regular Languages

We use the pumping lemma to prove that a language L is not regular.

(How???) ♣ Proof by contradiction: • Assume that L is regular.

B the pumping lemma guarantee the existence of a pumping length p such that all strings of length p or greater in L can be pumped.

• Find a string s in L that has length p or greater but that cannot be pumped.

B involves grouping the various ways of dividing s into several cases and analyzing them individually.

• The existence of s contradicts the pumping lemma if L were

regular.

• Hence L is not regular.

Regular Expressions

Non-Regular Languages

Some Non-Regular Languages

Example 1. B = {0n 1n |n ≼ 0}

Regular Expressions

Non-Regular Languages

Some Non-Regular Languages

Example 1. B = {0n 1n |n ≼ 0} 2. C = {w|w has an equal number of 0s and 1s }

Regular Expressions

Non-Regular Languages

Some Non-Regular Languages

Example 1. B = {0n 1n |n ≥ 0} 2. C = {w|w has an equal number of 0s and 1s } 3. F = {ww|w ∈ {0, 1}∗ }

Regular Expressions

Non-Regular Languages

Some Non-Regular Languages

Example 1. B = {0n 1n |n ≥ 0} 2. C = {w|w has an equal number of 0s and 1s } 3. F = {ww|w ∈ {0, 1}∗ } 2

4. G = {1n |n ≥ 0}

Regular Expressions

Non-Regular Languages

Some Non-Regular Languages

Example 1. B = {0n 1n |n ≥ 0} 2. C = {w|w has an equal number of 0s and 1s } 3. F = {ww|w ∈ {0, 1}∗ } 2

4. G = {1n |n ≥ 0} 5. E = {0i 1j |i > j}

Regular Expressions

Non-Regular Languages

Questions?

See you next meeting!