MELJUN CORTES Automata Lecture the Pumping Lemma 1

8/21/2019 MELJUN CORTES Automata Lecture the Pumping Lemma 1

1/21

Theory of Computation (With Automata Theory)

* Property of STI

Page 1 of 21

The Pumping Lemma

The Pumping Lemma

The Pumping Lemma

Introduction

• Recall that all regular

languages have this property

that if they have strings that

are long enough, then each of

these strings has a substring

that can be repeated an

arbitrary number of times, and

the resulting string will still be

in the language.

• This property can be used to

prove the non-regularity of

languages and is the basis of

the pumping lemma.


2/21


* Property of STI

Page 2 of 21

The Pumping Lemma

• Consider the following DFA:

Assume that the input string w

= 001100. Let m = 6 be the

number of states of the DFA.

Since w ≥ m, then at least

one state will be traversed at

least twice.

The states that will be

traversed are

q 0 q 5q 1 q 2 q 3 q 40 0 1 1 1

1 1

0

0 0, 1

0

q 0

0

q 1

0

q 2

1

q 3

1

q 4

0

q 2

0

q 5


3/21


* Property of STI

Page 3 of 21

The Pumping Lemma

• In this example, state q2 was

traversed twice therebyforming a loop or cycle

consisting of states q2, q3, and

q4. Specifically,

• The part of the input string

001100 that caused the loop is

the substring 110.

• Take note that the first

repeated state appeared within

the first m symbols of the input

string.

q 2 q 3 q 4

0 0 1 1 0

this part

caused theloop

0


4/21


* Property of STI

Page 4 of 21

The Pumping Lemma

• Therefore, the substring 110can be repeated an arbitrary

number of times (pumped) and

the resulting strings will still be

part of the language. These

include strings such as001101100, 001101101100,

and 001101101101100.

• The input string can actually

be viewed as being composed

of three parts (called x , y , and

z) as shown below:

0 0 1 1 0

y

0

x z


5/21


* Property of STI

Page 5 of 21

The Pumping Lemma

• Substring x is the first part of

the input string that iscomposed of the symbols

before the 1st appearance of

the repeated state which is q2.

• Substring y is the second partof the input string that contains

the input symbols from the first

occurrence of the repeated

state up to its second

occurrence. This is the loopwithin the input string.

• Substring z is composed of the

remaining symbols of the input

string that lead to the finalstate.

• The input string w can then be

written as

w = xyz


6/21


* Property of STI

Page 6 of 21

The Pumping Lemma

• Since substring y can be

repeated an arbitrary number

of times and the resulting

string is still part of thelanguage, then w can be

generalized as

w = xy iz where i ≥ 0

Examples:

w = xy 1z = 001100

w = xy 2z = 001101100

w = xy 3z = 001101101100


7/21


* Property of STI

Page 7 of 21

The Pumping Lemma

• To summarize this specialproperty of regular languages:

1. Let w be any string of

regular language L whoselength is greater than or

equal to the number of

states of the DFA that

recognizes language L ( w

≥ m).

2. The string w can be

divided into three parts xyz

where y contains the inputsymbols from the first

occurrence of a repeated

state to its second

occurrence.


8/21


* Property of STI

Page 8 of 21

The Pumping Lemma

3. The length of substring xy

will be less than or equal to

the number of states of the

DFA ( xy ≤ m).

4. The length of y ≥ 1.

5. Substring y can berepeated or pumped any

number of times and the

resulting string is still a

member of language L.

• The summary given is an

informal statement of the

pumping lemma.


9/21


* Property of STI

Page 9 of 21

The Pumping Lemma

The Pumping Lemma

• If L is a regular language, then

there is a number p (the

pump ing leng th ) where, if w

is any string in L whose length

is greater than or equal to p,

then w may be divided into

three parts, w = xyz , satisfying

the following conditions:

1. for each i ≥ 0, xy iz L,

2. y ≥ 1, and

3. xy p.

• Take note that the pumpinglength p refers to the number of

states of the DFA recognizing

the language. Hence, w ≥ p in

order for the pumping lemma to

be used.


10/21


* Property of STI

Page 10 of 21

The Pumping Lemma

• Major points:

1. The string w = xy iz L for

each i ≥ 0. This states that

the loop or cycle within the

input string w can be

pumped any number oftimes and w will still be a

member of L.

2. y ≥ 1 refers to that fact

that y cannot be an empty

string. There has to be aloop or cycle that can be

pumped.

3. xy p refers to the fact

that the loop or cycle that

will be pumped is the firstone encountered from the

start of the input string. In

other words, the part to be

pumped is within the first p

symbols of the input string.


11/21


* Property of STI

Page 11 of 21

The Pumping Lemma

• The pumping lemma can be

used to prove the non-

regularity of languages.

• Pumping lemma proofs often

use proof by contradiction:

1. First, assume that the

language L is regular.

2. Since L is regular, the

pumping lemma states thatall strings of L that are long

enough can be pumped.

3. Find a string w that is long

enough that cannot bepumped.

4. If there is such a string,

then the pumping lemma

was contradicted. Hence,

language L is not regular.


12/21


* Property of STI

Page 12 of 21

The Pumping Lemma

• Example 1: Prove that the

language L1 = {0 x 1 x x > 0} is

not regular.

Assume first that L1 is regular.

Let p be the pumping length of

L1.

Let the string to be pumped bew = 0p1p.

Since w ≥ p, the string chosen

is long enough to be pumped.

Consider now the various

ways in which w can be

divided into x , y , and z

substrings.


13/21


* Property of STI

Page 13 of 21

The Pumping Lemma

Option 1: Let y be all 0s.

If y = all 0s, pumping it

would create a string in

which there are more 0s

than 1s. Hence, the

resulting string will not be

in L1. This violates

condition 1 of the pumping

lemma.

Option 2: Let y be all 1s.

If y = all 1s, pumping it

would create a string in

which there are more 1s

than 0s. Hence, theresulting string will not be

in L1. This violates


lemma. Condition 3 is also

violated.


14/21


* Property of STI

Page 14 of 21

The Pumping Lemma

Option 3: Let y be composedof 0s and 1s.

If y is composed of 0s and

1s, pumping it would

create a string that mayhave the same number of

0s and 1s, but they will be

out of order with some 1s

before 0s. Hence, the

resulting string will not bein L1. This violates


lemma. Condition 3 is also

violated.

Since L1 has a string that

cannot be pumped, then this

contradicts the original

assumption that L1 is regular.

Therefore, L1 is not regular.


15/21


* Property of STI

Page 15 of 21

The Pumping Lemma


language L2 = {ww w {0,1}*}is not regular.

Assume first that L2 is regular

and let p be the pumping

length of L2.

Let the string to be pumped be

w = 0p10p1. Since w ≥ p, the

string can pumped.

From condition 3, substring y

can only be 0s that belong to

the first half of the input string.

If y is pumped, then there will

be more 0s in the first half of the string compared to its

second half, thereby creating

an imbalance. This contradicts

the original assumption that L2is regular. Therefore, L

2 is not

regular.


16/21


* Property of STI

Page 16 of 21

The Pumping Lemma


language L3 = {w {0,1}* w =wR } is not regular.



length of L3.


w = 0p10p. Since w ≥ p, the

string can pumped.

From condition 3, the substring

y can only be 0s that belong to

the first block of consecutive

0s (to the left of the single 1) of

the input string. If y is pumped,then the string is no longer a

palindrome.

This contradicts the original




17/21


* Property of STI

Page 17 of 21

The Pumping Lemma


language L4 = {0i1 j i > j } is not

regular.



length of L4.


w = 0p+11p. Since w ≥ p, the

string chosen is long enough

to be pumped.

From condition 3, substring y

can only be 0s that belong to

the block of consecutive 0s of

the input string. If y is pumped,then there will still be more 0s

than 1s. Hence, the string is

still in language L4.

Is the language regular?


18/21


* Property of STI

Page 18 of 21

The Pumping Lemma

Condition 1 of the pumping

lemma states that w = xy iz L

for each i ≥ 0. This means that

i can be equal to 0 making w =

xy i

z = xy 0

z = xz . This is calledpump ing down .

If y is pumped down, then

there will be more consecutive1s than 0s. Hence, the

resulting string is not in L4.

Since L4 has a string thatcannot be pumped, then this

contradicts the original




19/21


* Property of STI

Page 19 of 21

The Pumping Lemma


language L5 = { 0i 2 i ≥ 0} is

not regular.



length of L5.

Consider two strings xy 1z and

xy 2z . These two strings differ

in length by y .

Condition 3 states that xy p.

This means that y p.

If the string to be pumped is w = 0 p

2, then w = xyz = p2. If y

is pumped once, then w =

xy 2z . The length of w will then

be w = xy 2z p2 + p.


20/21


* Property of STI

Page 20 of 21

The Pumping Lemma

If xy 2z p2 + p, then this

inequality can be rewritten as

xy 2z < ( p + 1)2

Observe that replacing p2 + p

with ( p + 1)2 does not

invalidate the inequalitybecause p2 + p < ( p + 1)2.

Since the length of xy 1z is

equal to p2 (a perfect square),the length of the next string

should be ( p + 1)2, which is the

next perfect square after p2.


21/21


* Property of STI The Pumping Lemma

But the next string after xy 1z is

xy 2z and its length is less than

( p + 1)2

. Therefore, the lengthof xy 2z is not a perfect square

because there is no perfect

square number between p2

and ( p + 1)2. Hence, xy 2z

does not belong to L5.

Since L5 has a string that

cannot be pumped, then this

contradicts the originalassumption that L5 is regular.


MELJUN CORTES Automata Lecture the Pumping Lemma 1

Documents

Transcript of MELJUN CORTES Automata Lecture the Pumping Lemma 1