Chapter 4

• Pumping Lemma

• Properties of Regular Languages

• Decidable questions on Regular Languages

Theorem 4.1: Pumping lemma for regular languages

Let L be a regular language. Then there is a constant n (which depends on L) such that for every string w in L such that |w| n, we can break w into three strings, w = xyz, such that y , |xy| n, and for all i 0, xyiz is also in L.

q qlf

Proof: Let m be the number of states in the smallest FA accepting L, and let w = a1a2…an where n m.

If w is in L, then we have

(q, a1a2…aj) = ql, (ql, aj+1..ak) = ql, and

(ql, ak+1…an) = f.

 

choose x = a1a2…aj

y = aj+1..ak

z = ak+1…an

It is obvious that (ql, vi) = ql for i 0.

So, if the FA accepts w = xyz, it also accepts zyiz.

Applications of the pumping lemma:

 

Useful to prove a language L is not a regular set

Method

    select an arbitrary ‘n’

choose a string w in L where |w| n

for any partition of w = xyz such that

|xy| n and |y| 1, show a contradiction;

i.e. show that there is a string xykz not in L;

k will depend on n, x, y, and z

Example L = {0i1i | k > 0}

given arbitrary n, choose w = 0n1n

for any partition of w as xyz,

y is one of 0j, 1j, or 0j1l

in all cases xy2z is not in L

So, L is not a regular set

Example: L = { w in (a+b)* | # a’s = # b’s}

Example: L = {w | w = ap where p is a prime}

Definition of concatenation

Let u, v *. The concatenation of u and v, written as uv, is a binary operation on the strings of * defined as follows:

1. Basis: If lenghth(v) = 0, then v = and uv = u

2. Recursive step: Let v be a string of length n > 0 and v = wa where

a and lenghth(w) = n-1. Then uv = (uw)a.

Definition of uR (reversal of u)

1. Basis: length(u) = 0, then u = and = R

2. Recursive step: If length(u) = n > 0, then u = wa for some string w such that length(w) = n-1 and some a , and uR = awR.

u is a substring of v if v = xuy

u is a prefix of v if v = ux

u is a suffix of v if v = xu

Theorem 1: (uv)w = u(vw). Concatenation is associative.

Theorem 2: (uv)R = vRuR

A language L such that no string in L is a proper prefix (suffix) of any other string L is said to have the prefix(suffix) property.

Example:

{a}* does not have prefix property

{aib| i 0} does have the prefix property

Definition of homomorphism:

Let 1 and 2 be alphabets. A homomorphism is a mapping h: 1 2*.

It is extended to 1* as follows:

h: 1* 2

* is defined as h() =

and h(xa) = h(x)h(a) for all x1* and a1.

Homomorphism applied to a language

Let L be a language and h be a homomorphism.

h(L) = {h(w) | w is in L}

Definition of inverse homomorphism:

If h: 1 2* is a homomorphism, then the relation h-1: 2

* (1*),

called an inverse homomorphism and is defined as follows:

If y 2* then h-1(y) = {x | h(x) = y}

h-1(L) = h-1(y) = {x | y = h(x) L}

Examples:

Let 1 = {0, 1}, 2 = {a, b}, L = {0n1n | n > 0}.

Let h(0) = a, h(1) = bb

Then, h(L) = {anb2n | n > 0}

h-1(abb) = {01}

Let h(0) = ab and h(1) =

Then, h(L) = (ab)n

h-1(ab) = {1n01n}

Closure Properties of Regular Languages

The class of regular sets is closed under– Union

– Intersection

– Complement

– Difference

– Reversal

– Closure (*)

– Concatenation

– Homomorphism

– Inverse homomorphism

Theorem (4.5): If L is a regular language over an alphabet , then L = * – L is also regularProof: Let L = L(A) for some DFA

A = (Q, , , q0, F). Let B be the DFA

B = (Q, , , q0, Q – F). Then L(B) = L.

Therefore L is regular.

Corollary: If L and M are regular

languages, then L M is regular.

Theorem (4.11): If L is a regular language, so is LR.

Proof: (Use structural induction on the size of the regular expression E representing L.)

Claim: ER represents LR

Basis: If L is , {} or {a} then claim is true

Induction step:

1)If E = E1 + E2, then ER = E1R + E2

R

2)If E = E1 + E2, then ER = E2RE1

R

3)If E = E1*, then ER = (E1

R)*

Theorem (4.14): If L is a regular language over an alphabet , and h is a homomorphism on , then h(L) is also regular.

Proof: Let r be a regular expression, representing L, and s be a regular expression obtained from r by replacing every symbol a of by h(a), then s represents h(L) (use structural induction to prove this).

s is a regular expression.

So, h(L) is regular language.

Theorem (4.16): if h is a homomorphism from an alphabet to * and L *, then h-1(L) is also a regular set.

Proof: Let M=(Q, , , q0, F) be such that L = L(M).

Let M1 = (Q, , 1, q0, F) where (q, a) in

M1 = (q, h(a)) in M.

Then L(M1) = h-1(L).

Example: Show that {anban | n 1} is not regular.

Example: Show that {0,1}* - {0n1n | n > 0} is not regular

Note:

The Class of Regular sets included

in * is a Boolean algebra of sets for any

alphabet .