s30 pumping lemma

8/2/2019 s30 Pumping Lemma

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Formal Methods in Computer Science

CS1502

Pumping Lemma

Patchrawat Uthaisombut

University of Pittsburgh


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What do we know about NFAs and DFAs?

Describe the concept of closure.

What are closure properties of regular

languages that we know of?

Are all languages regular?


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Is language { 0n10n | n >= 0 } regular?Lets try to find an DFA or NFA

recognizing it.

What do we need to show to proof that

a language is NOT regular?

How?


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Using some other property

All regular languages share some property Q.

x (Reg(x) Q(x)) (All Ps are Qs)

Note, some non-regular languages may also have property Q.

If we can show that a language L doesnt have this property,then L cannot be regular.

~Q(L) ~Reg(L)

What property Q is a good choice to use? All regular languages must have property Q. If a language does not have property Q (and thus not regular), it

should be easy to prove this.


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Goals

Pumping lemma

Understanding the Pumping condition

Shared by all regular languages.

Proof of Pumping lemma

(Negation of Pumping condition)

(Using pumping lemma to show that alanguage is not regular.)

(Existence of non-regular languages)


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Pumping Condition

A language L satisfies thepumping condition if there exists an integer

p > 0 such that for all strings s in L of length atleast p

there exist strings x, y, z suchthat

s = xyzand

|xy| p and |y| > 0 and for all i0, xyiz is in L.

p(p>0 s((sL |s|p)

xyz(( s= xyz|xy| p |y| > 0

) i(i 0 xyizL)

))

)

Focus on the RED part first


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Let x = abcdefg be in L. Then there exists a substring y in s such that y

can be repeated (pumped) in place any number

of times and the resulting string is still in L. xyiz is in L for all i 0.

Example y = cde

xy0

z = xz =abfg is in L xy1z = xyz = abcdefgis in L xy2z = xyyz = abcdecdefgis in L xy3z = xyyyz = abcdecdecdefg is in L

1) s in L

2) s = xyz

3)for all i0, xyiz is in Ly can be pumped


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What the other parts mean

A language L satisfies the pumping condition if there exists an integer p > 0 such that

will discuss later

for all strings s in L of length at least p s must be in L and have sufficient length

there exist strings x, y, z such that s = xyz and |xy| n and

y occurs in the first p characters of s.

|y| > 0 and y is not the empty string.

for all i 0, xyiz is in L.


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Example

Let L be the set of even length strings over{a,b}.

Consider p = 2and s = abaa. What are the possibilities for y? abaa, abaa

abaa

Which one satisfies the pumping condition? abaa

Do all strings of L satisfy this?


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Example

Let L be the set of strings over {a,b} wherethe number of as mod 3 is 1.

Consider p = 3 and s = abbaaa. What are the possibilities for y?

abbaaa, abbaaa, abbaaa abbaaa, abbaaa abbaaa

Which ones satisfy the pumping condition? abbaaa, abbaaa, abbaaa

Do all strings of L satisfy this?


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Pumping Lemma

A language L satisfies the pumping condition if: there exists an integer p > 0 such that

for all strings s in L of length at least p

there exist strings x, y, z such that

s = xyzand

|xy| = 1 and

for all i >= 0, xyiz is in L

Pumping Lemma: All regular languages satisfy thepumping condition.

Now we proof the Pumping Lemma


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High Level Outline

Let L be an arbitrary regular language.

Let M be a DSA such that L(M) = L.

M exists by definition of L being regular.

Show that L satisfies the pumpingcondition

Use M in this part

Pumping Lemma follows.


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First step: n+1 prefixes of x

Let p be the number of states in M.

Let s be any string in L of length at least p.

Let si

denote the ith character of string s.

There are at least p+1 distinct prefixes of s

length 0:

length 1: s1

length 2: s1s2 ...

length i: s1s2 si ...

length n: s1s2 si sn


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Example

Let p = 8

Let s = abcdefgh

There are 9 distinct prefixes of s. length 0:

length 1: a

length 2: ab ...

length 8: abcdefgh


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Second step: Pigeon-hole Principle

As M processes string s, it processes each prefix of s

In particular, each prefix of s must end up in some state of M

Situation

There are p+1 distinct prefixes of s.

There are only p states in M.

Conclusion

At least two prefixes of s must end up in the same state of M

Pigeon-hole principle

If there are more birds than holes, then some hole has two or more birds.

Name these two prefixes w1 and w2.


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Third step: Forming x, y, z

Setting:

Prefix w1 has length i

Prefix w2 has lengthj > i

prefix w1

of length i: s1

s2

si

prefix w2 of length j: s1s2 si si+1 sj

Forming x, y, z

Set x = w1 = s1s2 si

Set y = si+1 sj Set z = sj+1 s|s|

xyz = s1s2 sisi+1 sjsj+1 s|s|x y z


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Example

Let M be a 5-state FSA that accepts all strings over{a,b,c,,z} whose length mod 5 = 3.

Consider x = abcdefghijklmnopqr, a string in L.

What are the two prefixes w1 and w2? w1 =

w2 = abcde

What are x, y, z?

x = y = abcde

z = fghijklmnopqr

xyz =abcdefghijklmnopqr

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Fourth step: Showing x, y, z

satisfy all the conditions |xy| p

xy = w2

w2 is one of the first p+1 prefixes of string s

|y| 1 y consists of the characters in w2 after w1 Since w2 and w1 are distinct prefixes of s, y is not

For all i 0, xyiz in L x=w1 and xy=w2 end up in the same state qof M

This is how we defined w1 and w2 Thus for all i 0, xyi ends up in state q

The string z causes M to go from state qto an acceptingstate.


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Example

Let M be a 5-state FSA that accepts all strings over{a,b,c,,z} whose length mod 5 = 3.

Consider x = abcdefghijklmnopqr, a string in L.

What are x, y, z? x =

y = abcde

z = fghijklmnopqr

|xy| = 5 p |y| = 5 1

For all i 0, xyiz = (abcde)ifghijklmnopqr is in L.

1 2 3 40


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Pumping Lemma

A language L satisfies the pumping condition if: there exists an integer p > 0 such that

for all strings s in L of length at least p

there exist strings x, y, z such that

s = xyzand

|xy| = 1 and

for all i >= 0, xyiz is in L

Pumping Lemma: All regular languages satisfy thepumping condition.

Now we proof the Pumping Lemma


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Rephrasing pumping lemma

Pumping Lemma: All regular languages satisfy thepumping condition. x ( Reg(L) Pump(L) )

A language is regular if and only if it satisfies thepumping condition. Equivalent? x ( ~Reg(L) ~Pump(L) ) Not Equivalent

All non-regular languages do not satisfy the pumpingcondition. Equivalent? x ( ~Reg(L) ~Pump(L) ) Not Equivalent

All languages that do not satisfy the pumpingcondition are not regular. Equivalent? x ( ~Pump(L) ~Reg(L) ) Equivalent


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How to use the pumping lemma

Contrapositive of Pumping Lemma:All languages that do not satisfy thepumping condition are not regular.

x ( ~Pump(L) ~Reg(L)

To prove that a language L is not regular1. Show that L does NOT satisfy the pumping

condition.2. Cite pumping lemma and conclude that L is not

regular.

Need to find negation of Pumping Condition


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Pumping Condition

A language L satisfies thepumping condition if there exists an integer

p > 0 such that for all strings s in L of length at

least p there exist strings x, y, z such

that s = xyzand

|xy|

p and |y| > 0 and for all i0, xyiz is in L.

p(p>0 s((sL |s|p)xyz(( s= xyz

|xy| p |y| > 0

) i(i 0 xyizL)

))

)

Need to find negation of Pumping Condition


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x ( P(x) Q(x) ) Not all Ps are Qs x ( P(x) Q(x) ) x (P(x) Q(x) )

x ( P(x) Q(x) ) Some Ps are not Qs

x ( P(x) Q(x) ) No Ps are Qs. x ( P(x) Q(x) )

x (

P(x)

Q(x) ) x ( P(x) Q(x) )All Ps are not Qs


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Negation of Pumping condition

p(p>0 s((sL |s|p)xyz(

( s= xyz

|xy| p |y| > 0

) i(i 0 xyizL)

))

)

p( p> 0 s( (sL |s| p) xyz(

( s= xyz

|xy| p|y| > 0

) i(i 0 xyizL)

))

)


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Negation of Pumping condition A language L does NOT

satisfy the pumping condition

if: for all integers p > 0

there exists a string s in L oflength at least p such that

for all strings x, y, z such that s = xyz,

|xy| p, and

|y| > 0, there exists an i 0 suchthat xyiz is not in L

p( p> 0 s( (sL |s| p) xyz(

( s= xyz

|xy|p

|y| > 0) i(i 0 xyizL)

)

))


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Pumping condition and its negation A language L satisfiesthe

pumping condition if:

there existsan integer p > 0such that

for allstrings s in L of length at

least p

there existstrings x, y, z suchthat

s = xyz,

|xy| p, and

|y| > 0, for alli 0, xyiz is inL

A language L does not satisfythe

pumping condition if: for allintegers p > 0

there existsa string s in L oflength at least p such that

for allstrings x, y, z such that

s = xyz,

|xy| p, and

|y| > 0,

there existsan i 0 such

that xyiz is not inL

s30 pumping lemma

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