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Pumping Lemma for Regular Languages

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Title: Pumping Lemma for Regular Languages

1
Pumping Lemma for Regular Languages

Pumping Lemma relates the size of string accepted
with the number of states in a DFA
What is the largest string accepted by a DFA with
n states?
Suppose there is no loop?
Now, if there is a loop, what type of strings are
accepted via the loop(s)?

2
Pumping Lemma for Regular Languages

Lemma (the pumping lemma)
Let M be a DFA with Q n states. If there
exists a string x in L(M), such that x ? n,
then there exists a way to write it as x uvw,
where u,v, and w are all in S and
1? uv ? n
v ? 1
such that, the strings uviw are also in L(M), for
all i ? 0

3
Pumping Lemma for Regular Languages

Proof
Let x a1a2 am where m ? n, x is in L(M), and
d(q0, a1a2 ap) qjp
a1 a2 a3 am
qj0 qj1 qj2 qj3 qjm m ? n and qj0 is q0
Consider the first n symbols, and first n1
states on the above path
a1 a2 a3 an
qj0 qj1 qj2 qj3 qjn
Since Q n, it follows from the pigeon-hole
principle that js jt for some 0 ? sltt ? n,
i.e., some state appears on this path twice
(perhaps many states appear more than once, but
at least one does).

same
loop

Let
u a1as
v as1at
Since 0 ? sltt ? n and uv a1at it follows that
1 ? v and therefore 1 ? uv
uv ? n and therefore 1 ? uv ? n
In addition, let
w at1am
It follows that uviw a1as(as1at)iat1am is
in L(M), for all i ? 0.
In other words, when processing the accepted
string x, the loop was traversed once, but could
have been traversed as many times as desired, and
the resulting string would still be accepted.

Example
n 5
x 0001000 is in L(M)
0 0 0 1 0 0 0
q0 q1 q2 q3 q1 q2 q3 q4
first n1 states
u 0
v 001
w 000 uviw is in L(M), i.e., 0(001)i000 is
in L(M), for all i ? 0

Note that this does not mean that every string
accepted by the DFA has this form
001 is in L(M) but is not of the form 0(001)i000
Similarly, this does not mean that every long
string accepted by the DFA has this form
0011111 is in L(M), is very long, but is not of
the form 0(001)i000
Note, however, in this latter case 0011111 could
be similarly decomposed.

Note It may be the case that no x in L(M) has
x ? n.

Example
n 4
x bbbab is in L(M) b b b a
b
x 5
q0 q0 q0 q0 q1 q3
u e u b u bb
v b or v b v b
w bbab w bab w ab
(b)ibbab is in L(M), for all i ? 0 b(b)ibab is
in L(M), for all i ? 0

a
b
a
q0
q1
b
b
a
q2
a/b
10
NonRegularity Example

Theorem The language
L 0k1k k ? 0 (1)
is not regular.
Proof (by contradiction) Suppose that L is
regular. Then there exists a DFA M such that
L L(M) (2)
We will show that M accepts some strings not in
L, contradicting (2).
Suppose that M has n states, and consider a
string x0m1m, where mgtgtn.
By (1), x is in L.
By (2), x is also in L(M), note that the machine
accepts a language not just a string

Since x m gtgt n, it follows from the pumping
lemma that
x uvw
1 ? uv ? n
1 ? v, and
uviw is in L(M), for all i ? 0
Since 1 ? uv ? n and nltltm, it follows that 1 ?
uv ? m.
Also, since x 0m1m it follows that uv is a
substring of 0m.
In other words v0j, for some j ? 1.
Since uviw is in L(M), for all i ? 0, it follows
that 0mcj1m is in L(M), for all c ? 1.
But by (1) and (2), 0mcj1m is not in L(M), for
any c ? 1, a contradiction.
Note that L basically corresponds to balanced
parenthesis.

12
NonRegularity Example

Theorem The language
L 0k1k2k k ? 0 (1)
is not regular.
Proof (by contradiction) Suppose that L is
regular. Then there exists a DFA M such that
L L(M) (2)
We will show that M accepts some strings not in
L, contradicting (2).
Suppose that M has n states, and consider a
string x0m1m2m, where mgtgtn.
By (1), x is in L.
By (2), x is also in L(M), note that the machine
accepts a language not just a string

Since x m gtgt n, it follows from the pumping
lemma that
x uvw
1 ? uv ? n
1 ? v, and
uviw is in L(M), for all i ? 0
Since 1 ? uv ? n and nltltm, it follows that 1 ?
uv ? m.
Also, since x 0m1m2m it follows that uv is a
substring of 0m.
In other words v0j, for some j ? 1.
Since uviw is in L(M), for all i ? 0, it follows
that 0mcj1m2m is in L(M), for all c ? 1.
But by (1) and (2), 0mcj1m2m is not in L(M),
for any c ? 1, a contradiction.
Note that the above proof is almost identical to
the previous proof.

14
NonRegularity Example

Theorem The language
L 0m1n2mn m,n ? 0 (1)
is not regular.
Proof (by contradiction) Suppose that L is
regular. Then there exists a DFA M such that
L L(M) (2)
We will show that M accepts some strings not in
L, contradicting (2).
Suppose that M has n states, and consider a
string x0m1n2mn, where mgtgtn.
By (1), x is in L.
By (2), x is also in L(M).

Since x m gtgt n, it follows from the pumping
lemma that
x uvw
1 ? uv ? n
1 ? v, and
uviw is in L(M), for all i ? 0
Since 1 ? uv ? n and nltltm, it follows that 1 ?
uv ? m.
Also, since x 0m1n2mn it follows that uv is a
substring of 0m.
In other words v0j, for some j ? 1.
Since uviw is in L(M), for all i ? 0, it follows
that 0mcj1m2mn is in L(M), for all c ? 1. In
other words v can be pumped as many times as we
like, and we still get a string in L(M).
But by (1) and (2), 0mcj1n2mn is not in L(M),
for any c ? 1, because the acceptable expression
should be 0mcj1n2mcjn, a contradiction.
Note that the above proof is almost identical to
the previous proof.

Theorem Let M (Q, S, d, q0, F) be a DFA. Then
L(M) is finite iff x lt Q for all x in L(M).
Proof
(if) Suppose that x lt Q for all x in L(M).
Since the number of states Q and the number of
input symbols S are both fixed, it follows that
there are at most a finite number of strings of
length less than Q. It follows that L(M) is
finite (exercise give an upper bound on the
number of such strings).
(only if) By contradiction. Suppose that L(M) is
finite, but that x ? Q for some x in L(M).
From the pumping lemma it follows that xuvw, v
? 1 and uviw is in L(M) for all i ? 0. But then
L(M) would be infinite, a contradiction.

Theorem Let M (Q, S, d, q0, F) be a DFA. Then
L(M) is infinite iff there exists an x in L(M)
such that x ? Q.
Proof
(if) Suppose there exists an x in L(M) such that
x ? Q. From the pumping lemma it follows
that xuvw, v ? 1 and uviw is in L(M) for all i
? 0. Therefore L(M) is infinite.
(only if) By contradiction. Suppose that L(M) is
infinite, but that there is no x in L(M) such
that x ? Q. It follows that each x in L(M)
has length less than Q. Since the number of
states Q and the number of input symbols S
are both fixed, it follows that there are at most
a finite number of strings of length less than
Q. It follows that L(M) is finite. A
contradiction.
Note that the above also follows directly from
the previous theorem.

Theorem Let M (Q, S, d, q0, F) be a DFA. Then
L(M) is non-empty iff there exists an x in L(M)
such that x lt Q.
Proof
(if) Suppose there exists a string x in L(M)
such that x lt Q. Then clearly L(M) is
non-empty.
(only if) By contradiction. Suppose that L(M) is
non-empty, but that there exists no string x in
L(M) such that x lt Q. It follows that y ?
n, where n Q, for all y in L(M). Let z be a
string of shortest length in L(M). Then z ? n
and, by the pumping lemma zuvw, v ? 1 and uviw
is in L(M) for all i ? 0. But then uv0w uw is
in L(M) and
uw z - v
? z - 1
lt z
Since uw is in L(M), it follows that z is not a
string of shortest length in L(M), a
contradiction.

Corollary Let M (Q, S, d, q0, F) be a DFA.
Then there is an algorithm to determine if L(M)
is empty.
Proof
From the theorem it follows that if L(M) is
non-empty then there exists a string x where x
lt n and n Q such that M accepts x. We can
try running M on each string of length lt n to see
if any are accepted. If one is accepted then
L(M) is non-empty, otherwise L(M) is empty. Since
the number of states Q and the number of input
symbols S are both fixed, it follows that there
are at most a finite number of strings (of length
less than Q) that need to be tested.
Note that Dijkstras algorithm works here too

Theorem Let M (Q, S, d, q0, F) be a DFA and
let n Q. Then L(M) is infinite iff there
exists an x in L(M) such that n ? x lt 2n.
Proof (left as an exercise similar to the
previous theorem).
(Contradiction) Suppose there is no string of
length between n and 2n in the language accepted
by DFA M with n states.
Suppose a string z is the shortest string
accepted by the machine M, such that z ? 2n.
Since z gt n, by pumping lemma z uvw, and 1 ?
v ? n, because uv ? n. Then zuw must be
accepted by the machine M. So, z z - v
cannot be lt n. As per assumption, z ? 2n is
not possible as z is the shortest such string and
zgtz. And we presumed that there is no
string of size between n and 2n, so, n ? z ?
2n cannot be true either. Contradiction.
Corollary Let M (Q, S, d, q0, F) be a DFA.
Then there is an algorithm to determine if L(M)
is infinite.
Proof (left as an exercise).

21
Closure Properties of Regular Languages

Consider various operations on languages
x x is in S and x is not in L
x x is in L1 or L2
x x is in L1 and L2
x x is in L1 but not in L2
L1L2 xy x is in L1 and y is in L2
L Li L0 U L1 U L2 U
L Li L1 U L2 U

Theorem Let M (Q, S, d, q0, F) be a DFA. Then
M (Q, S, d, q0, Q-F) is a DFA where L(M)
S - L(M).
Proof (left as an exercise).
Corollary The regular languages are closed under
complementation, i.e., if L is a regular language
then so is .
Does the above construction work for NFAs?

Theorem The regular languages are closed with
respect to union, concatenation and Kleene
closure.
Proof Let L1 and L2 be regular languages. By
definition there exist regular expressions r1 and
r2 such that L(r1) L1 and L(r2) L2. But then
r1 r2 is a regular expression representing
. Similarly for concatenation and Kleene
closure.