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Theory of Computation, GNFA, Pigeonhole, Pumping Lemma

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- 1. Equivalence with FA * Any Regex can be converted to FA and vice versa, because: * Regex and FA are equivalent in their descriptive power ** Regular language is one that is recognized by some FA, remember? Hence: (Lemma) if a language is described by a RegEx, then it is a regular language. See Proof: P.67, example 1.56 11/25/13
- 2. Lemma: If a language is regular, then it is descried by a RegEx. Because A is regular, it is accepted by a DFA. We describe a procedure for converting DFAs into equivalent RegEx. Breaking the procedure into 2 parts, using a new type of FA called Generalized Nondeterministic Finite Automaton (GNFA) Hence, 1st convert DFA to GNFA, then GNFA to RegEx 11/25/13
- 3. GNFA is NFA, but Regex could be used as labels on arrows. Figure 1.61 P.70 shows an example of GNFA, which is on the special form that it should ALWAYS have. 11/25/13
- 4. To convert a NFA to GNFA (and keep the special form) Add a new start state with ε arrow to the old start Add a new accept state with ε arrow coming from the old accept state. If any arrow has more than one label, or there are more than one arrow between two states on the same direction, replace with one arrow and “union” the labels If two states have no arrows between them, add an arrow with Φ label 11/25/13
- 5. Convert the GNFA to RegEx In GNFA, the start and accept states are must, and they're different, hence no. of states (k) of any GNFA is always >=2. If k>2, then GNFA is constructed by reducing the number to k=2 (an arrow form start to accept states) whose label is the RegEx. Example, converting a 3 state DFA to RegEx. 3k DFA > 5k GNFA > 4k GNFA > 3k GNFA > 2k GNFA > RegEx 11/25/13
- 6. Convert the GNFA to RegEx The Challenge in reducing the states if greater than 2, by Ripping the a state out of the machine (other than the start, accept states), we call it q rip , we repair the machine by altering the removed qrip with labeled arrow. The label should cover the absence of qrip .The new label form start to accept states is the RegEx. Figure 1.63 P. 72 11/25/13
- 7. Formal definition of GNFA Just like the DFA, it's a 5-tuple with same parameters except for δ, which can be given as: δ: (Q - {qaccept }) X (Q – {qstart}) → R R is the collection of all RegEx over the alphabet ∑, The domain of the transition function is (Q - {qaccept }) X (Q – {qstart}) simply because in GNFA, an arrow connects every state to every other state, except no arrows are coming from qaccept or going to qstart. Read: P.73 - 76 11/25/13
- 8. Nonregular Languages Some languages cannot be recognized by any FA (a limitation of FA) Assume the language B = {0n1n | n>=0}. No DFA can recognize it because the resulting number of 0s is unlimited, and therefore no DFA can remember this number of possibilities. But just because the language appears to require unbounded memory doesn't mean it's nonregular. So we need a method to prove the nonregularity 11/25/13
- 9. Nonregular Languages Assume we have 2 languages C and D, over the alphabet ∑ = {0,1}, where: C = { w | w has an equal no. of 0s and 1s}, and D = { w | w has an equal number of occurrences of 01 and 10 as substrings} Which one is nonregular? See problem 1.48 >> Difficult? We need a mathematical proof for certainty. 11/25/13
- 10. The pumping lemma for regular languages Theorem (Pumping Lemma): “All regular languages have a special property, if a language doesn't have that property, it is not regular”. The property states that all strings in the language can be “pumped” if they are at least as long as a certain special value called the “pumping length”. In other words: Such string contains a section that can be repeated any number of times with the resulting string remaining in the language 11/25/13
- 11. Pumping lemma: if A is a regular language, then there is a number p (the pumping length) where, if s is any string in A of length at least p, then s may be divided into three pieces, s=xyz, satisfying the conditions: 1- for each i>=0, xyiz ∈ A 2- |y| >0, and 3- |xy| <= p. Figure 1.71 P.78 11/25/13
- 12. Nonregular languages n {a b : n≥ 0} vv R : Regular languages a∗b b∗c + a b + c ( a + b )∗¿ ¿ etc . .. 11/25/13 n v ∈ {a , b } ∗¿ ¿ ¿ ¿
- 13. How can we prove that a language L is not regular? Prove that there is no DFA or NFA or RE that accepts L Difficulty: this is not easy to prove (since there is an infinite number of them) Solution: use the Pumping Lemma !!! 11/25/13
- 14. The Pigeonhole Principle 11/25/13
- 15. 4 3 11/25/13 pigeons pigeonholes
- 16. A pigeonhole must contain at least two pigeons 11/25/13
- 17. n pigeons ........... m pigeonholes ........... 11/25/13 n> m
- 18. The Pigeonhole Principle pigeons n pigeonholes m n> m There is a pigeonhole with at least 2 pigeons ........... 11/25/13
- 19. The Pigeonhole Principle and DFAs 11/25/13
- 20. Consider a DFA with 4 b q1 b b a q2 a a 11/25/13 states b q3 a q4
- 21. aaaab Consider the walk of a “long’’ string: (length at least 4) A state is repeated in the walk of aaaab a q1 q2 b q1 a q3 a a q3 b a q2 b q4 b a a 11/25/13 q2 q3 b a q4
- 22. The state is repeated as a result of the pigeonhole principle Walk of Pigeons: (walk states) q1 a q2 a q3 a aaaab q2 a q3 b q4 Are more than q1 Nests: (Automaton states) 11/25/13 q2 q3 Repeated state q4
- 23. aabb Consider the walk of a “long’’ string: (length at least 4) Due to the pigeonhole principle: A state is repeated in the walk of aabb a q1 q2 b a q3 b q4 b q4 b b q1 a q2 a a 11/25/13 q3 b a q4
- 24. The state is repeated as a result of the pigeonhole principle aabb Walk of Pigeons: q1 a a q2 q3 b q4 b q4 (walk states) Are more than Nests: (Automaton states) 11/25/13 q1 q2 q3 Automaton States q4 Repeated state
- 25. In General: If ∣w∣ ≥ #states of DFA, by the pigeonhole principle, a state is repeated in the walk Walk of q1 σ1 σ2 .... σi w = σ 1 σ 2 ⋯σ k qi σ i +1 σj .... qi σ j+1 .... σk Arbitrary DFA q1 11/25/13 σ1 σ2 w ...... qi ...... Repeated state σk qz qz
- 26. ∣w∣ ≥ #states of DFA =m Pigeons: (walk states) .... q1 Walk of qi .... w qi .... qz Are more than Nests: q1 q2 (Automaton states) 11/25/13 .... qi .... A state is repeated q m−1 qm
- 27. The Pumping Lemma 11/25/13
- 28. Take an infinite regular language L (contains an infinite number of strings) There exists a DFA that accepts L m states 11/25/13
- 29. Take string w ∈L ∣w∣ ≥ m with (number of states of DFA) then, at least one state is repeated in the walk of w Walk in DFA of w = σ 1 σ 2 ⋯σ k σ1 σ2 ...... q ...... σk Repeated state in DFA 11/25/13
- 30. There could be many states repeated Take q to be the first state repeated One dimensional projection of walk w : First occurrence σ1 σ2 .... σi Second occurrence q σ i +1 Unique states 11/25/13 .... σj q σ j+1 .... σk
- 31. We can write w = xyz One dimensional projection of walk w : First occurrence σ1 σ2 .... x=σ 1 ⋯σ i 11/25/13 σi Second occurrence q σ i +1 .... y =σ i +1 ⋯σ j σj q σ j+1 .... σk z=σ j +1 ⋯σ k
- 32. In DFA: w=x y z contains only first occurrence of y ... σj ... σ1 11/25/13 x σi σ i +1 q σ j+1 ... ... z σk q
- 33. Observation: length ∣x y∣ ≤ m number of states of DFA y ... Unique States σj ... σ1 11/25/13 x σi σ i +1 q Since, in xy no state is repeated (except q)
- 34. Observation: length ∣ y∣ ≥ 1 Since there is at least one transition in loop y ... σj σ i +1 q 11/25/13
- 35. We do not care about the form of string z x may actually overlap with the paths of y ... z ... 11/25/13 x q z y and
- 36. Additional string: The string is accepted Do not follow loop x z y ... σj ... σ1 11/25/13 x σi σ i +1 q σ j+1 ... ... z σk
- 37. Additional string: The string is accepted Follow loop 2 times y ... σj ... σ1 11/25/13 x y y z x σi σ i +1 q σ j+1 ... ... z σk
- 38. The string is accepted Additional string: Follow loop 3 times y ... σj ... σ1 11/25/13 x y y y z x σi σ i +1 q σ j+1 ... ... z σk
- 39. In General: The string is accepted Follow loop i times xy z i=0, 1, 2, . .. y ... σj ... σ1 11/25/13 i x σi σ i +1 q σ j+1 ... ... z σk
- 40. i Therefore: x y z ∈L i=0, 1, 2, . .. Language accepted by the DFA y ... σj ... σ1 11/25/13 x σi σ i +1 q σ j+1 ... ... z σk
- 41. We just described: The Pumping Lemma !!! 11/25/13
- 42. The Pumping Lemma: • Given a infinite regular language • there exists an integer • for any string w ∈ L • we can write (critical length) with length ∣w∣ ≥ m w=x y z • with ∣x y∣ ≤ m • such that: 11/25/13 m L i and x y z ∈L ∣ y∣ ≥ 1 i=0, 1, 2, . ..
- 43. In the book: Critical length m 11/25/13 p = Pumping length
- 44. Applications of the Pumping Lemma 11/25/13
- 45. Observation: Every language of finite size has to be regular (we can easily construct an NFA that accepts every string in the language) Therefore, every non-regular language has to be of infinite size (contains an infinite number of strings) 11/25/13
- 46. Suppose you want to prove that An infinite language L is not regular 1. Assume the opposite: L is regular 2. The pumping lemma should hold forL 3. Use the pumping lemma to obtain a contradiction 4. Therefore, L is not regular 11/25/13
- 47. Explanation of Step 3: How to get a contradiction 1. Let m be the critical length for 2. Choose a particular stringw ∈ L ∣w∣≥m the length condition 3. Write L which satisfies w=xyz 4. Show that ' i w =xy z∉L for some 5. This gives a contradiction, since from ' i pumping lemma w =xy z∈L 11/25/13 i≠1
- 48. Note: It suffices to show that only one string w ∈ L gives a contradiction You don’t need to obtain contradiction for everyw ∈ L 11/25/13
- 49. Example Theorem: L = { 0 n 1 n | n ∈ N } is not regular. Proof: By contradiction; assume L is regular. Let n be the pumping length guaranteed by the pumping lemma. Consider the string w = 0 n 1 n . Then |w| = 2n ≥ n and w ∈ L, so we can write w = xyz such that y ≠ ε and for any natural number i, xyiz ∈ L. We consider three cases:
- 50. Case 1: y consists solely of 0s. Then xyiz = xz = 0 n - |y| 1 n , and since |y| > 0, xz ∉ L. Case 2: y consists solely of 1s. Then xy i z = xz = 0 n 1 n - |y| , and since |y| > 0, xz ∉ L. Case 3: y consists of k 0s followed by m 1s. Then xy i z has the form 0 n 1 m 0 k 1 n + m , so xy i z ∉ L. In all three cases we reach a contradiction, so our assumption was wrong and L is not regular. ■
- 51. Example of Pumping Lemma application Theorem: The language n n L={a b : n≥0} is not regular Proof: 11/25/13 Use the Pumping Lemma
- 52. n n L={a b : n≥0} Assume for contradiction thatL is a regular language SinceL is infinite we can apply the Pumping Lemma 11/25/13
- 53. n n L={a b : n≥0} Let m be the critical length for L Pick a string w such that: w ∈ L and length ∣w∣ ≥m We pick 11/25/13 m w=a b m
- 54. From the Pumping Lemma: m m we can write w=a b =x y z with lengths ∣x y∣ ≤m , ∣ y∣≥1 m w= m m m xyz=a b = a ...aa ... aa .. . ab.. . b x 11/25/13 Thus: k y z y =a , 1≤k ≤m
- 55. m m k y =a , 1≤k ≤m x y z=a b From the Pumping Lemma: i x y z ∈L i=0, 1, 2, . .. Thus: 11/25/13 2 x y z ∈L
- 56. m m k y =a , 1≤k ≤m x y z=a b 2 x y z ∈L From the Pumping Lemma: m+k m 2 xy z= a...aa...aa ...aa...ab...b ∈L x Thus: 11/25/13 y a y m+k z m b ∈L
- 57. a BUT: m+k m b ∈L n n L={a b : n≥0} a m+k m b ∉L CONTRADICTION!!! 11/25/13 k≥ 1
- 58. Therefore: Our assumption thatL is a regular language is not true Conclusion: L 11/25/13 is not a regular language END OF PROOF
- 59. Non-regular language n n {a b : n≥0} Regular languages ¿ ¿ L(a b ) 11/25/13
- 60. Read P. 80 – 82 (examples) Solve Exercises and Problems (P. 83 – 98) Prepare for a quiz! 11/25/13

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