Formal Methods in Computer Science CS1502 Pumping Lemma

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Formal Methods in Computer ScienceCS1502Pumping
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 gt 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 a language is
  not regular.)
• (Existence of non-regular languages)

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Pumping Condition

• ?p (pgt0 ??s ((s?L ? s?p)? ?x ?y ?z (
  ( s xyz ? xy ? p ? y gt
  0 ) ? ?i (i ? 0 ? xyiz ?L) ))
• )

• A language L satisfies the pumping condition if
• there exists an integerp gt 0 such that
• for all strings s in L of length at least p
• there exist strings x, y, z such that
• s xyz and
• xy ? p and
• y gt 0 and
• for all i ? 0, xyiz is in L.

Focus on the RED part first
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y can be pumped
1) s in L 2) s xyz 3) for all i ? 0, xyiz is
in L

• 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
• xy0z xz abfg is in L
• xy1z xyz abcdefg is in L
• xy2z xyyz abcdecdefg is in L
• xy3z xyyyz abcdecdecdefg is in L

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What the other parts mean

• A language L satisfies the pumping condition if
• there exists an integer p gt 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 gt 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 2 and 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 where the
  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 gt 0 such that
• for all strings s in L of length at least p
• there exist strings x, y, z such that
• s xyz and
• xy lt p and
• y gt 1 and
• for all i gt 0, xyiz is in L
• Pumping Lemma All regular languages satisfy the
  pumping 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 pumping condition
• Use M in this part
• Pumping Lemma follows.

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First step n1 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 p1 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 p1 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 length j gt i
• prefix w1 of length i s1s2 si
• prefix w2 of length j s1s2 si si1 sj
• Forming x, y, z
• Set x w1 s1s2 si
• Set y si1 sj
• Set z sj1 ss
• xyz s1s2 si si1 sj sj1 ss
• x y
  z

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Example
3
1
2
4
0

• 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 p1 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
• xw1 and xyw2 end up in the same state q of 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 q to an
  accepting state.

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Example
3
1
2
4
0

• 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.

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Pumping Lemma

• A language L satisfies the pumping condition if
• there exists an integer p gt 0 such that
• for all strings s in L of length at least p
• there exist strings x, y, z such that
• s xyz and
• xy lt p and
• y gt 1 and
• for all i gt 0, xyiz is in L
• Pumping Lemma All regular languages satisfy the
  pumping condition.

Now we proof the Pumping Lemma
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Rephrasing pumping lemma

• Pumping Lemma All regular languages satisfy the
  pumping condition.
• ?x ( Reg(L) ? Pump(L) )
• A language is regular if and only if it satisfies
  the pumping condition. Equivalent?
• ?x ( Reg(L) ? Pump(L) ) Not Equivalent
• All non-regular languages do not satisfy the
  pumping condition. Equivalent?
• ?x ( Reg(L) ? Pump(L) ) Not Equivalent
• All languages that do not satisfy the pumping
  condition are not regular. Equivalent?
• ?x ( Pump(L) ? Reg(L) ) Equivalent

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How to use the pumping lemma

• Contrapositive of Pumping LemmaAll languages
  that do not satisfy the pumping condition are not
  regular.
• ?x ( Pump(L) ? Reg(L)
• To prove that a language L is not regular
• Show that L does NOT satisfy the pumping
  condition.
• Cite pumping lemma and conclude that L is not
  regular.

Need to find negation of Pumping Condition
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Pumping Condition
Need to find negation of Pumping Condition

• ?p (pgt0 ??s ((s?L ? s?p)? ?x ?y ?z (
  ( s xyz ? xy ? p ? y gt
  0 ) ? ?i (i ? 0 ? xyiz ?L) ))
• )

• A language L satisfies the pumping condition if
• there exists an integerp gt 0 such that
• for all strings s in L of length at least p
• there exist strings x, y, z such that
• s xyz and
• xy ? p and
• y gt 0 and
• for all i ? 0, xyiz is in L.

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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 (pgt0 ??s ((s?L ? s?p)? ?x ?y ?z (
  ( s xyz ? xy ? p ? y gt
  0 ) ? ?i (i ? 0 ? xyiz ?L) ))
• )

• ?p ( p gt 0 ? ?s ( (s?L ? s ? p) ?
  ?x ?y ?z ( ( s xyz ?
  xy ? p ? y gt 0 ) ?
  ?i (i ? 0 ? xyiz ? L) ) ))

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Negation of Pumping condition

• A language L does NOT satisfy the pumping
  condition if
• for all integers p gt 0
• there exists a string s in L of length at least p
  such that
• for all strings x, y, z such that
• s xyz,
• xy ? p, and
• y gt 0,
• there exists an i ? 0 such that xyiz is not in L

• ?p ( p gt 0 ? ?s ( (s?L ? s ? p) ? ?x
  ?y ?z ( ( s xyz ? xy
  ? p ? y gt 0 ) ?
  ?i (i ? 0 ? xyiz ? L) ) ))

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Pumping condition and its negation

• A language L satisfies the pumping condition if
• there exists an integer p gt 0 such that
• for all strings s in L of length at least p
• there exist strings x, y, z such that
• s xyz,
• xy ? p, and
• y gt 0,
• for all i ? 0, xyiz is in L

• A language L does not satisfy the pumping
  condition if
• for all integers p gt 0
• there exists a string s in L of length at least p
  such that
• for all strings x, y, z such that
• s xyz,
• xy ? p, and
• y gt 0,
• there exists an i ? 0 such that xyiz is not in L