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Theory of Computer Science March 30, 2020 C4. Regular Languages: Pumping Lemma, Closure Properties and Decidability Theory of Computer Science C4. Regular Languages: Pumping Lemma, Closure Properties C4.1 Pumping Lemma and Decidability


  1. Theory of Computer Science March 30, 2020 — C4. Regular Languages: Pumping Lemma, Closure Properties and Decidability Theory of Computer Science C4. Regular Languages: Pumping Lemma, Closure Properties C4.1 Pumping Lemma and Decidability C4.2 Closure Properties Gabriele R¨ oger University of Basel C4.3 Decidability March 30, 2020 Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 1 / 35 Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 2 / 35 C4. Regular Languages: Pumping Lemma, Closure Properties and C4. Regular Languages: Pumping Lemma, Closure Properties and Decidability Pumping Lemma Decidability Pumping Lemma Overview Regular Languages Grammars & Grammars DFAs C4.1 Pumping Lemma Regular NFAs Languages Automata & Regular Formal Languages Expressions Context-free Pumping Languages Lemma Properties Context-sensitive & Type-0 Languages Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 3 / 35 Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 4 / 35

  2. C4. Regular Languages: Pumping Lemma, Closure Properties and C4. Regular Languages: Pumping Lemma, Closure Properties and Decidability Pumping Lemma Decidability Pumping Lemma Pumping Lemma: Motivation Pumping Lemma You can show that a language is regular by specifying Theorem (Pumping Lemma) an appropriate grammar, finite automaton, or regular expression. Let L be a regular language. Then there is an n ∈ N How can you you show that a language (a pumping number for L) such that all words x ∈ L with | x | ≥ n is not regular? can be split into x = uvw with the following properties: 1 | v | ≥ 1 , 2 | uv | ≤ n, and ◮ Direct proof that no regular grammar exists 3 uv i w ∈ L for all i = 0 , 1 , 2 , . . . . that generates the language � difficult in general Question: what if L is finite? ◮ Pumping lemma: use a necessary property that holds for all regular languages. Picture courtesy of imagerymajestic / FreeDigitalPhotos.net Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 5 / 35 Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 6 / 35 C4. Regular Languages: Pumping Lemma, Closure Properties and C4. Regular Languages: Pumping Lemma, Closure Properties and Decidability Pumping Lemma Decidability Pumping Lemma Pumping Lemma: Proof Pumping Lemma: Proof Theorem (Pumping Lemma) Theorem (Pumping Lemma) Let L be a regular language. Then there is an n ∈ N Let L be a regular language. Then there is an n ∈ N (a pumping number for L) such that all words x ∈ L with | x | ≥ n (a pumping number for L) such that all words x ∈ L with | x | ≥ n can be split into x = uvw with the following properties: can be split into x = uvw with the following properties: 1 | v | ≥ 1 , 1 | v | ≥ 1 , 2 | uv | ≤ n, and 2 | uv | ≤ n, and 3 uv i w ∈ L for all i = 0 , 1 , 2 , . . . . 3 uv i w ∈ L for all i = 0 , 1 , 2 , . . . . Proof. Proof (continued). For regular L there exists a DFA M = � Q , Σ , δ, q 0 , E � with Choose a split x = uvw so M is in the same state after reading u L ( M ) = L . We show that n = | Q | has the desired properties. and after reading uv . Obviously, we can choose the split in a way that | v | ≥ 1 and | uv | ≤ | Q | are satisfied. . . . Consider an arbitrary x ∈ L ( M ) with length | x | ≥ | Q | . Including the start state, M visits | x | + 1 states while reading x . Because of | x | ≥ | Q | at least one state has to be visited twice. . . . Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 7 / 35 Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 8 / 35

  3. C4. Regular Languages: Pumping Lemma, Closure Properties and C4. Regular Languages: Pumping Lemma, Closure Properties and Decidability Pumping Lemma Decidability Pumping Lemma Pumping Lemma: Proof Pumping Lemma: Application Theorem (Pumping Lemma) Let L be a regular language. Then there is an n ∈ N (a pumping number for L) such that all words x ∈ L with | x | ≥ n Using the pumping lemma (PL): can be split into x = uvw with the following properties: 1 | v | ≥ 1 , Proof of Nonregularity 2 | uv | ≤ n, and ◮ If L is regular, then the pumping lemma holds for L . 3 uv i w ∈ L for all i = 0 , 1 , 2 , . . . . ◮ By contraposition: if the PL does not hold for L , then L cannot be regular. Proof (continued). ◮ That is: if there is no n ∈ N with the properties of the PL, The word v corresponds to a loop in the DFA after reading u and then L cannot be regular. can thus be repeated arbitrarily often. Every subsequent continuation with w ends in the same end state as reading x . Therefore uv i w ∈ L ( M ) = L is satisfied for all i = 0 , 1 , 2 , . . . . Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 9 / 35 Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 10 / 35 C4. Regular Languages: Pumping Lemma, Closure Properties and C4. Regular Languages: Pumping Lemma, Closure Properties and Decidability Pumping Lemma Decidability Pumping Lemma Pumping Lemma: Caveat Pumping Lemma: Example Example The language L = { a n b n | n ∈ N } is not regular. Caveat: The pumping lemma is a necessary condition for a language Proof. to be regular, but not a sufficient one. Assume L is regular. Then let p be a pumping number for L . � there are languages that satisfy the pumping lemma The word x = a p b p is in L and has length ≥ p . conditions but are not regular Let x = uvw be a split with the properties of the PL. � for such languages, other methods are needed to show Then the word x ′ = uv 2 w is also in L . Since | uv | ≤ p , uv consists that they are not regular (e.g., the Myhill-Nerode theorem) only of symbols a and x ′ = a | u | a 2 | v | a p −| uv | b p = a p + | v | b p . Since | v | ≥ 1 it follows that p + | v | � = p and thus x ′ / ∈ L . This is a contradiction to the PL. � L is not regular. Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 11 / 35 Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 12 / 35

  4. C4. Regular Languages: Pumping Lemma, Closure Properties and C4. Regular Languages: Pumping Lemma, Closure Properties and Decidability Pumping Lemma Decidability Pumping Lemma Pumping Lemma: Another Example I Pumping Lemma: Another Example II Example The language L = { ab n ac n +2 | n ∈ N } is not regular. Example The language L = { ab n ac n +2 | n ∈ N } is not regular. Proof (continued). If | u | = 0, then word v starts with an a . Proof. Hence, uv 0 w = b p −| v | +1 ac p +2 does not start with symbol a Assume L is regular. Then let p be a pumping number for L . The word x = ab p ac p +2 is in L and has length ≥ p . and is therefore not in L . This is a contradiction to the PL. Let x = uvw be a split with the properties of the PL. If | u | > 0, then word v consists only of b s. Consider uv 0 w = ab p −| v | ac p +2 . As | v | ≥ 1, this word does not From | uv | ≤ p and | v | ≥ 1 we know that uv consists of one a contain two more c s than b s and is therefore not in language L . followed by at most p − 1 b s. This is a contradiction to the PL. We distinguish two cases, | u | = 0 and | u | > 0. . . . We have in all cases a contradiction to the PL. � L is not regular. Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 13 / 35 Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 14 / 35 C4. Regular Languages: Pumping Lemma, Closure Properties and C4. Regular Languages: Pumping Lemma, Closure Properties and Decidability Closure Properties Decidability Closure Properties Closure Properties How can you combine regular languages in a way to get another regular language C4.2 Closure Properties as a result? Picture courtesy of stockimages / FreeDigitalPhotos.net Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 15 / 35 Gabriele R¨ oger (University of Basel) Theory of Computer Science March 30, 2020 16 / 35

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