1

Closure Properties of Regular Languages

L1 and L2 are regular.

How about L1L2, L1L2 , L1L2 , L1, L1* ?

2

Theorem 1

If L1 and L2 are regular, then so are L1L2, L1L2 , L1L2 ,

L1, L1*.

(The family of regular languages is closed under intersection, union, concatenation, complement, and star-closure.)

3

Proof

• L1 = L(r1) L2 = L(r2)

L(r1 + r2) = L(r1)L(r2)

L(r1 . r2) = L(r1)L(r2)

L(r1*) = (L(r1))*

4

Proof

• M = (Q, , , q0, F) accepts L1.

M = (Q, , , q0, Q – F) accepts L1.

5

Proof

• M1 = (Q, , 1, q0, F1) accepts L1.

M2 = (P, , 2, p0, F2) accepts L2.

q0 qf

a1 an

p0 pf

a1 an

2(pj, a) = pl

1(qi, a) = qk

1((qi, pj), a) = (qk, pl)

6

Proof

• Example:

L1 = {abn | n 0}

L2 = {anb | n 0}

L1L2 = {ab}

7

Theorem 2

The family of regular languages is closed under reversal:

If L is regular, then so is LR.

8

Proof

?

9

Homomorphism

Suppose and are alphabets.

h: *

is called a homomorphism

10

Homomorphism

Extended definition:

w = a1a2 ... an

h(w) = h(a1)h(a2) ... h(an)

11

Homomorphism

If L is a language on , then its homomorphic image is defined as:

h(L) = {h(w): w L}

12

Example

= {a, b}

= {a, b, c}

h(a) = ab h(b) = bbc

h(aba) = abbbcab

L = {aa, aba} h(L) = {abab, abbbcab}

13

Homomorphism

If L is the language on of a regular expression r,

then the regular expression for h(L) is obtained by

applying the homomorphism to each symbol of r.

14

Example

= {a, b}

= {b, c, d}

h(a) = dbcc h(b) = bdc

r = (a + b*)(aa)* L = L(r)

h(r) = (dbcc + (bdc)*)(dbccdbcc)*

15

Theorem 3

The family of regular languages is closed under homomorphism:

If L is regular, then so is h(L).

16

Proof

Let L(r) = L for some regular expression r.

Prove: h(L(r)) = L(h(r)).

17

Homework

• Exercises: 2, 4, 6, 8, 9, 11, 18, 22 of Section 4.1 - Linz’s book.

• Presentation: Section 4.3 - Linz’s book (pigeonhole principle & pumping lemma and examples).

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Right Quotient

Let L1 and L2 be languages on the same alphabet.

Then the right quotient of L1 with L2 is defined as:

L1/L2 = {x | xy L1 and y L2}

19

Example

L1 = {anbm | n 1, m 0}{ba}

L2 = {bm | m 1}

L1/L2 = ?

20

Example

L1 = {anbm | n 1, m 0}{ba}

L2 = {bm | m 1}

L1/L2 = {anbm | n 1, m 0}

21

Example

q0 q1 q2

b

b

b

a

q3

a

a, b

a

q4

q5

a, b

a

L1 = {anbm | n 1, m 0}{ba}

22

Example

q0 q1 q2

b

b

b

a

q3

a

a, b

a

q4

q5

a, b

a

L1 = {anbm | n 1, m 0}{ba}

L2 = {bm | m 1}

*(q0, x) = qi

*(qi, y) F and y L2

23

Example

q0 q1 q2

b

b

b

a

q3

a

a, b

a

q4

q5

a, b

a

L1 = {anbm | n 1, m 0}{ba}

L2 = {bm | m 1}

*(q0, x) = qi

*(qi, y) F and y L2

24

Theorem

The family of regular languages is closed under right quotient:

If L1 and L2 are regular, then so is L1/L2.

25

Proof

• M = (Q, , , q0, F) accepts L1.

M^ = (P, , , q0, F^) accepts L1/L2.

If y L2 and *(qi, y) F add qi to F^

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Proof

• M = (Q, , , q0, F) accepts L1.

M^ = (P, , , q0, F^) accepts L1/L2.

If y L2 and *(qi, y) F add qi to F^

Mi = (P, , , qi, F) and L(Mi) L2 .

27

Example

L1 = L(a*baa*)

L2 = L(ab*)

L1/L2 = ?

28

Example

q0 q1 q2

a

a

b

b

b

a, b

q3

a

L1 = L(a*baa*)

L(M0) L2 =

L(M1) L2 = {a}

L(M2) L2 = {a}

L(M3) L2 =

L1 = L(ab*)

29

Example

q0 q1 q2

a

a

b

b

b

a, b

q3

a

L1 = L(a*baa*)

L(M0) L2 =

L(M1) L2 = {a}

L(M2) L2 = {a}

L(M3) L2 =

L1 = L(ab*)

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Questions about RLs

1. Given a regular language L on and any w *, is there an algorithm to determine whether or not w L?.

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Questions about RLs

1. Given a regular language L on and any w *, is there an algorithm to determine whether or not w L?.

Yes.

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Questions about RLs

1. Is there an algorithm to determine whether or nota regular language is empty, finite, or infinite?.

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Questions about RLs

1. Is there an algorithm to determine whether or nota regular language is empty, finite, or infinite?.

Yes.

34

Questions about RLs

1. Given two regular languages L1 and L2, is there an algorithm to determine whether or not L1 = L2?.

35

Questions about RLs

1. Given two regular languages L1 and L2, is there an algorithm to determine whether or not L1 = L2?.

Yes.

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Homework

• Exercises: 1, 2, 3, 5, 9 of Section 4.2 - Linz’s book.

• Exercises: 3, 4, 5, 6, 8, 10, 12 of Section 4.3 - Linz’s book.