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Formal Methods in Computer Science CS1502 Pumping Lemma

2. . 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?. 3. . Is language { 0n10n | n >= 0 } regular? Let's try to find an DFA or NFA recognizing it.What do we need to show to pro

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Formal Methods in Computer Science CS1502 Pumping Lemma

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    1. 1 Formal Methods in Computer Science CS1502 Pumping Lemma Patchrawat Uthaisombut University of Pittsburgh

    2. 2 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?

    3. 3 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?

    4. 4 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.

    5. 5 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)

    6. 6 Pumping Condition 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 = xyz and |xy| ? p and |y| > 0 and for all i ? 0, xyiz is in L. ?p (p>0 ? ?s ((s?L ? |s|?p)? ?x ?y ?z ( ( s = xyz ? |xy| ? p ? |y| > 0 ) ? ?i (i ? 0 ? xyiz ?L) ) ) )

    7. 7 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 y can be pumped

    8. 8 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.

    9. 9 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?

    10. 10 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?

    11. 11 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 = xyz and |xy| <= p and |y| >= 1 and for all i >= 0, xyiz is in L Pumping Lemma: All regular languages satisfy the pumping condition.

    12. 12 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.

    13. 13 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

    14. 14 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

    15. 15 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.

    16. 16 Third step: Forming x, y, z Setting: Prefix w1 has length i Prefix w2 has length j > i prefix w1 of length i: s1s2 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 si si+1 sj sj+1 s|s| x y z

    17. 17 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

    18. 18 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 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.

    19. 19 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.

    20. 20 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 = xyz and |xy| <= p and |y| >= 1 and for all i >= 0, xyiz is in L Pumping Lemma: All regular languages satisfy the pumping condition.

    21. 21 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

    22. 22 How to use the pumping lemma Contrapositive of Pumping Lemma: All 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.

    23. 23 Pumping Condition 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 = xyz and |xy| ? p and |y| > 0 and for all i ? 0, xyiz is in L. ?p (p>0 ? ?s ((s?L ? |s|?p)? ?x ?y ?z ( ( s = xyz ? |xy| ? p ? |y| > 0 ) ? ?i (i ? 0 ? xyiz ?L) ) ) )

    24. 24 ??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

    25. 25 Negation of Pumping condition ??p (p>0 ? ?s ((s?L ? |s|?p)? ?x ?y ?z ( ( s = xyz ? |xy| ? p ? |y| > 0 ) ? ?i (i ? 0 ? xyiz ?L) ) ) ) ?p ( p > 0 ? ?s ( (s?L ? |s| ? p) ? ?x ?y ?z ( ( s = xyz ? |xy| ? p ? |y| > 0 ) ? ?i (i ? 0 ? xyiz ? L) ) ) )

    26. 26 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 of length 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 such that xyiz is not in L ?p ( p > 0 ? ?s ( (s?L ? |s| ? p) ? ?x ?y ?z ( ( s = xyz ? |xy| ? p ? |y| > 0 ) ? ?i (i ? 0 ? xyiz ? L) ) ) )

    27. 27 Pumping condition and its negation 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 = xyz, |xy| ? p, and |y| > 0, for all i ? 0, xyiz is in L A language L does not satisfy the pumping condition if: for all integers p > 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| > 0, there exists an i ? 0 such that xyiz is not in L

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