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- 1. Theory of Computation Pumping Lemma for Regular Languages, Proof by Contradiction, Pumping Lemma Examples, Context-Free Grammars, Backus-Naur Notation Vladimir Kulyukin
- 2. Outline ● Pumping Lemma for Regular Languages ● Proof by Contradiction ● Pumping Lemma Examples ● Context-Free Grammars ● Backus-Naur Notation
- 3. Pumping Lemma for Regular Languages
- 4. Review: Processing State Sequence q0 q1 1 0 A processing state sequence is a sequence of states that an FA uses (enters and leaves or just enters in the case of final states) to process a given input. Let x = 001. Then the processing path of the above FA is q0, q0, q0, q1: from q0 to q0 on the first 0, on to q0 on the second 0, and on to q1 on 1. q0 q0 q0 q1 0 0 1
- 5. Review: Pumping Lemma ( ) .0for,and1where ,can writeThen we.andLet states.DFA withaiswhere,Let ≥∈≤ =≥∈ = iLwuvv uvwxnxLx nMMLL i
- 6. Review: Pictorial Insight q0 q’q M u v w nxuvwx ≥= ,
- 7. Pumping Lemma Use & Proof by Contradition
- 8. Pumping Lemma Use ● Pumping Lemma is not used to prove that languages are regular ● Pumping Lemma states that if a language is regular, then its strings have a specific property (pumping property) ● Pumping Lemma is used to show that languages are not regular: assume that a language is regular and then show that a specific string does not satisfy the pumping property
- 9. Proof by Contradiction ● Pumping Lemma applications are examples of a proof technique known as proof by contradiction ● Proof by contradiction is used to prove statements of the form IF A THEN B ● A typical use of this technique consists of two steps: – Assume that A is true and B is false – Find a contradiction, i.e., a statement that contradicts another statement known to be true; that statement can be A itself, which is assumed to be true
- 10. Proof by Contradiction ● Why does proof by contradiction works? ● It works because the statement IF A THEN B is true in all cases except when A is true and B is false ● Proof by contradiction allows us to rule out that case by finding a contradiction ● It is this contradiction that proves out the original statement IF A THEN B because it excludes the only case under which this statement can be false
- 11. Three Questions about Proof by Contradiction ● What contradictions can/should we look for? – Experience, intuition, and insight ● How can we use the assumptions that A is true and B is false to react a contradiction? – Work forward from both assumptions and see what shakes loose ● When can/should we use proof by contradiction? – A reliable indicator is the word not in the statement B
- 12. Proof by Contradiction: Example If r is a real number such that r2 = 2, then r is not rational (irrational)
- 13. Proof by Contradiction: Example A: r is a real number such that r2 = 2 B: r is not rational (irrational)
- 14. Proof by Contradiction: Example Assume A and NOT B: A: r is a real number such that r2 = 2 NOT B: r is rational Now we have to find a contradiction
- 15. Proof by Contradiction: Example If r is rational (remember that NOT B is assumed to be true), then r = p/q, q != 0
- 16. Proof by Contradiction: Example We can assume that p and q have no common divisors (aka mutually prime); if they had, they could be divided out
- 17. Proof by Contradiction: Example ● Since r = p/q, r2 = (p/q)2 ● Since r2 = 2, then 2 = (p/q)2 ● Since 2 = (p/q)2 , 2q2 = p2 ● Since p2 = 2q2 , p2 is even ● Since p2 is even, p is even
- 18. Proof by Contradiction: Example ● Since p is even, p = 2k, where k is an integer ● Since 2q2 = p2 , 2q2 = (2k)2 = 4k2 ● Since 2q2 = (2k)2 = 4k2 , q2 = 2k2 ● Since q2 = 2k2 , then q is even ● But then, since both p and q are even, they have at least one common divisor ● We found a contradiction because p and q were assumed to have no common divisors
- 19. Pumping Lemma Examples
- 20. Pumping Lemma Example: Proof 1 { } regular.notisHence,s.'ofnumberthe angreater thstrictlyisyins'ofnumberthebecause, ,Since.Consider.forchoice onlytheisThus,states.1ofsequenceprocessinga needsstatesDFA withthe,processTo.1, Lemma,PumpingBy the.||,Then.Let Lemma.PumpingtheofconstantthebeLet:2Proof regular.notis|:Claim 2 Lb aLy avuvvwwuvyv avn navuvwz nzLzbaz n NkbaL n nn kk ∉ === =+ ≤= ≥∈= ∈= + +
- 21. Pumping Lemma Example: Proof 2 { } ion.contradictaachievewecases,3theofeachIn. Thus,.,,If.Thus,s.' follows'wherestringais,If. Thus,.,,If.; ;:forchoices3areThere.0, Then.1,Then.||,Let Lemma.PumpingtheofconstantthebeLet:1Proof regular.notis|:Claim 0 02 20 0 Lwuv kmbawuvbvLwuvb awuvbavLwuv mkbawuvavbvbav avvmbaz vuvwznzLz n NkbaL km mk mm kk ∉ >==∉ =∉ <==== =>= ≤=≥∈ ∈= + ++ ++++ +
- 22. Main Implication of the Pumping Lemma There are languages that are not regular, i.e., languages for which it is impossible to construct DFAs, NFAs or regular expressions
- 23. Pumping Lemma & Finite Languages ● Any finite language is regular ● We can construct a finite automaton for each individual string (that automaton will accept that string and only that string) ● The individual automata can then be compiled into one union automaton that accepts all strings in that language
- 24. Pumping Lemma & Finite Languages ● Does the Pumping Lemma contradict the regularity of finite languages? ● No, because for any finite language we can find the length of the longest string (say n) ● The Pumping Lemma's statement is vacuously true for all strings whose length is greater than n, because there are no such strings in the language ● So we do not get any contradiction out of the Pumping Lemma for finite languages
- 25. Context-Free Grammars
- 26. Context-Free Grammar (CFG): Definition ( ) . overstringaisandwhere,formtheof isproductioneachs;productionofsetfiniteais symbol;starttheis alphabet;terminaltheis alphabet;lnonterminatheis where,,,,tuple-4aisCFGA VΣ yVXyX P VS Σ V PSVGG ∪ ∈→ ∈ Σ=
- 27. Example 01 { } .fromderivedisthatshow,oninduction By.|:CFGfollowingheConsider t:Proof free.-contextisLthatShow.0Let:Claim Sban aSbS |nbaL nn nn ε→ ≥=
- 28. Example 02 { } .fromderivedisthatshow,oninduction By.|:CFGfollowingheConsider t:Proof free.-contextisthatShow.0|Let:Claim 3 3 Sban aSbbbS LnbaL nn nn ε→ ≥=
- 29. Example 03 { } { } { } ε ε εε | | :LforGCFGaconstructcan weNow.|:forCFGabeLet.|:for CFGabeLet.thatObservefree.-contextareandboth 01,ExampleBy.0|and0|Let:Proof .free-contextis0,0|Let:Claim 22 11 21 2222111 12121 21 dcSS baSS SSS dcSSLGbaSSL GLLLLL mdcLnbaL mndcbaL mmnn mmnn → → → →→ = ≥=≥= ≥≥=
- 30. Example 04 .andof rulestheto|addingbyforCFGaconstructcan Then wely.respective,andofsymbolsstartingtwothebe andLetly.respective,andforandgrammars free-contextexist twotherefree,-contextareandSince:Proof free.-contextisThen .Letlanguages.free-contextbeandLet:Claim 21 21 21 212121 21 2121 GG SSSLG GG SSLLGG LL L LLLLL → ∪=
- 31. Example 05 { }{ } ( ) .,oflengthon theinductionbyThen, | | :CFGfollowingthebeLet:Proof .,12|,:Claim * LGLz baX XXXSS G NkkzbazL = → → ∈+=∈=
- 32. Example 06 { }{ } ( ) .Then .|| :CFGfollowingthebeLet:Proof .|,:Claim * LGL bSbaSaS G xxzbazL R = → =∈= ε
- 33. Backus-Naur Notation
- 34. Backus-Naur Notation (BNN) ● BNN is frequently used in R&D on compilers and programming languages ● BNN is a CFG specification language ● The main difference is that BNN uses ::= instead of and puts non-terminals inside <>
- 35. BNN Example ( ) cba |||exp |expexp |expexp |expexp |exp*exp |expexp::exp < = + −=
- 36. Two Derivations of Expression a-b*c <exp> <exp> * <exp> <exp> <exp>- a b c <exp> <exp> - <exp> <exp> * <exp> a b c
- 37. References & Reading Suggestions ● Davis, Weyuker, Sigal. Computability, Complexity, and Languages, 2nd Edition, Academic Press ● Brooks Webber. Formal Language: A Practical Introduction, Franklin, Beedle & Associates, Inc ● Hopcroft and Ullman. Introduction to Automata Theory, Languages, and Computation, Narosa Publishing House ● D. Solow. How to Read & Do Proofs, Wiley & Sons

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