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
ql
f
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 x1* and a1.
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
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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
regular
Proof: 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 + E2R
2)If E = E1 + E2, then ER = E2RE1R
3)If E = E1*, then ER = (E1R)*
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 .