Next: Pushdown automata. Pushdown Automata. Informal Description PDA (1) Informal Description PDA (2) Formal Description PDA

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1 CE 2001: Introdution to Theor of Computation ummer 2007 uprakash Datta Net: Pushdown automata dd stak to definition of Finite utomata Can serve as tpe of memor or ounter More powerful than Finite utomata epts Contet-Free Languages (CFLs) Offie: CEB 3043 Phone: et Course page: 6/25/2007 CE 2001, ummer /25/2007 CE 2001, ummer Pushdown utomata Pushdown automata are for ontet-free languages what finite automata are for regular languages. PDs are reognizing automata that have a single stak (= memor): Last-In First-Out pushing and popping We onsider onl non-deterministi PDs. 6/25/2007 CE 2001, ummer Informal Desription PD (1) input w = internal state set Q stak z The PD M reads w and stak element. Depending on - input w i Σ ε, - stak s j Γ ε, and - state q k Q the PD M: - jumps to a new state, - pushes an element Γ ε (nondeterministiall) 6/25/2007 CE 2001, ummer Informal Desription PD (2) input w = internal state set Q stak z fter the PD has read omplete input, M will be in state Q If possible to end in aepting state F Q, then M aepts w 6/25/2007 CE 2001, ummer Formal Desription PD Pushdown utomata M is defined b a si tuple (Q,Σ,Γ,δ,q 0,F), with Q finite set of states Σ finite input alphabet Γ finite stak alphabet q 0 start state Q F set of aepting states Q δ transition funtion δ: Q Σ ε Γ ε P (Q Γ ε ) 6/25/2007 CE 2001, ummer

2 PD for L = { 0 n 1 n n 0 } Eample 2.9: The PD first pushes $ 0 n on stak. Then, while reading the 1 n string, the zeros are popped again. If, in the end, $ is left on stak, then aept q 1 q 4 ε, ε $ ε, $ ε 6/25/2007 CE 2001, ummer q 2 q 3 0, ε 0 Mahine Diagram for 0 n 1 n q 1 q 4 ε, ε $ ε, $ ε 6/25/2007 CE 2001, ummer q 2 q 3 0, ε 0 On w = (state; stak) evolution: (q 1 ; ε) (q 2 ; $) (q 2 ; 0$) (q 2 ; 00$) (q 2 ; 000$) (q 3 ; 00$) (q 3 ; 0$) (q 3 ; $) (q 4 ; ε) This final q 4 is an aepting state Mahine Diagram for 0 n 1 n q 1 q 4 ε, ε $ ε, $ ε q 2 q 3 0, ε 0 On w = 0101 (state; stak) evolution: (q 1 ; ε) (q 2 ; $) (q 2 ; 0$) (q 3 ; $) (q 4 ; ε) But we still have part of input 01. There is no aepting path. PDs and CFL Theorem 2.12: language L is ontet-free if and onl of there is a pushdown automata M that reognizes L. Two step proof: 1) Given a CFG G, onstrut a PD M G 2) Given a PD M, make a CFG G M 6/25/2007 CE 2001, ummer /25/2007 CE 2001, ummer Converting a CFL to a PD Basi idea tart with the start smbol Non-deterministiall appl rules top when a string of terminals is left. Compare with input aept if math. Detail: How do ou store the intermediate string on the stak? 6/25/2007 CE 2001, ummer Converting a CFL to a PD - 2 In q start, initialize the stak to ontain $ and make an ε-transition to q loop. In q loop, if top of stak ontains a variable, replae it with a terminal w for some rule w Else, if top of stak ontains a terminal w, ompare with net input smbol. If unequal, rejet. Else if the top of the stak is $ (empt stak), make an ε-transition to q aept. 6/25/2007 CE 2001, ummer

3 Converting a PD to a CFG Basi idea: For eah pair of states p,q use a variable pq whih generates all strings that take the PD from p to q. The start variable is q0 q aept Details: ssume that the PD Has a single aept state Empties stak before aepting Eah transition either pops or pushes. Converting a PD to a CFG - 2 p,q,r,s Q, t Γ, a,b Σ ε, if δ(p,a,ε) ontains (r,t) and if δ(s,b,t) ontains (q, ε) add the rule pq a rs b p,q,r Q, add the rule pq pr rq p Q, add the rule pp ε 6/25/2007 CE 2001, ummer /25/2007 CE 2001, ummer Converting a PD to a CFG - 3 Corretness - Easier diretion: If pq generates, then brings PD P from p with empt stak to q with empt stak. trong indution on the number of steps in the derivation of. Base ase: (1 step derivation) pp ε Indutive ase: pq a rs bor pq pr rq Converting a PD to a CFG - 4 Corretness - Harder diretion: If brings PD P from p with empt stak to q with empt stak, then pq generates. trong indution on the number of steps in the omputation of P going from p to q (both with empt staks) on input. Base ase: (0 step derivation) = ε [ pp ε] Indutive ase: stak not empt eept before and end [ pq a rs b] OR stak empt some time before the end [ pq pr rq ] 6/25/2007 CE 2001, ummer /25/2007 CE 2001, ummer Net Non-CF languages CFL pumping lemma Non-CF Languages The language L = { a n b n n n 0 } does not appear to be ontet-free. Informal: The problem is that ever variable an (onl) at b itself (ontet-free). The problem of * v : If * uz * uvz * uvz L, then * uz * uvz * * uv i i z * uv i i z L as well, for all i=0,1,2, 6/25/2007 CE 2001, ummer /25/2007 CE 2001, ummer

4 Pumping Lemma for CFLs Idea: If we an prove the eistene of derivations for elements of the CFL L that use the step * v, then a new form of v- pumping holds: * v * v 2 2 * v 3 3 * ) Observation: We an prove this eistene if the parse-tree is tall enough. 6/25/2007 CE 2001, ummer Remember Parse Trees Parse tree for bbba * bbaba bbaa b b 6/25/2007 CE 2001, ummer B a a B Pumping a Parse Tree u v z If s = uvz L is long, then its parse-tree is tall. Hene, there is a path on whih a variable repeats itself. We an pump this part. 6/25/2007 CE 2001, ummer Tree Tall Enough Let L be a ontet-free language, and let G be its grammar with maimal b smbols on the right side of the rules: X 1 X b parse tree of depth h produes a string with maimum length of b h. Long strings implies tall trees. Let V be the number of variables of G. If h = V +2 or bigger, then there is a variable on a top-down path that ours more than one. 6/25/2007 CE 2001, ummer uvz L uv 2 2 z L u v z B repeating the part we get 6/25/2007 CE 2001, ummer /25/2007 CE 2001, ummer u v R z v while removing the - gives

5 Pumping down: uz L u z In general uv i i z L for all i=0,1,2, 6/25/2007 CE 2001, ummer Pumping Lemma for CFL For ever ontet-free language L, there is a pumping length p, suh that for ever string s L and s p, we an write s=uvz with 1) uv i i z L for ever i {0,1,2, } 2) v 1 3) v p Note that 1) implies that uz L 2) sas that v annot be the empt string ε Condition 3) is not alwas used 6/25/2007 CE 2001, ummer Different parts of the parse tree u v z 6/25/2007 CE 2001, ummer Formal Proof of Pumping Lemma Let G=(V,Σ,R,) be the grammar of a CFL. Maimum size of rules is b 2: X 1 X b string s requires a minimum tree-depth log b s. If s p=b V +1, then tree-depth V +1 (has V +2 nodes), hene there is a path and variable where repeats itself: * uz * uvz * uvz It follows that uv i i z L for all i=0,1,2, Furthermore: v 1 beause tree is minimal v p beause bottom tree with p leaves has a repeating path 6/25/2007 CE 2001, ummer Pumping a n b n n (E. 2.36) ssume that B = {a n b n n n 0} is CFL Let p be the pumping length, and s = a p b p p B P.L.: s = uvz = a p b p p, with uv i i z B for all i 0 Options for v : 1) The strings v and are uniform (v=a a and =, for eample). Then uv 2 2 z will not ontain the same number of a s, b s and s, hene uv 2 2 z B 2) v and are not uniform. Then uv 2 2 z will not be a ab b Hene uv 2 2 z B 6/25/2007 CE 2001, ummer Pumping a n b n n (ont.) ssume that B = {a n b n n n 0} is CFL Let p be the pumping length, and s = a p b p p B P.L.: s = uvz = a p b p p, with uv i i z B for all i 0 We showed: No options for v suh that uv i i z B for all i. Contradition. B is not a ontet-free language. 6/25/2007 CE 2001, ummer

6 Eample 2.37 (Pumping down) Proof that C = {a i b j k 0 i j k } is not ontet-free. Let p be the pumping length, and s = a p b p p C P.L.: s = uvz, suh that uv i i z C for ever i 0 Two options for 1 v p: 1) v = a*b*, then the string uv 2 2 z has not enough s, hene uv 2 2 z C 2) v = b**, then the string uv 0 0 z = uz has too man a s, hene uv 0 0 z C Contradition: C is not a ontet-free language. 6/25/2007 CE 2001, ummer D = { ww w {0,1}* } (E. 2.38) Carefull take the strings s D. Let p be the pumping length, take s=0 p 1 p 0 p 1 p. Three options for s=uvz with 1 v p: 1) If a part of is to the left of middle ( ) in 0 p 1 p 0 p 1 p, then the seond half of uv 2 2 z starts with 1 2) If a part of v is to the right of middle of 0 p 1 p 0 p 1 p, then the first half of uv 2 2 z ends with 0, hene uv 2 2 z D 3) If is in the middle of 0 p 1 p 0 p 1 p, then uz equals 0 p 1 i 0 j 1 p D (beause i or j p) Contradition: D is not ontet-free. 6/25/2007 CE 2001, ummer Pumping Problems Using the CFL pumping lemma is more diffiult than the pumping lemma for regular languages. Net Closure properties of CFL You have to hoose the string s arefull, and divide the options effiientl. dditional CFL properties would be helpful (like we had for regular languages). What about losure under standard operations? 6/25/2007 CE 2001, ummer /25/2007 CE 2001, ummer Union Closure Properties Lemma: Let 1 and 2 be two CF languages, then the union 1 2 is ontet free as well. Proof: ssume that the two grammars are G 1 =(V 1,Σ,R 1, 1 ) and G 2 =(V 2,Σ,R 2, 2 ). Construt a third grammar G 3 =(V 3,Σ,R 3, 3 ) b: V 3 = V 1 V 2 { 3 } (new start variable) with R 3 = R 1 R 2 { }. It follows that L(G 3 ) = L(G 1 ) L(G 2 ). Intersetion & Complement? Let again 1 and 2 be two CF languages. One an prove that, in general, the intersetion 1 2, and the omplement Ā 1 = Σ* \ 1 are not ontet free languages. One proves this with speifi ounter eamples of languages. 6/25/2007 CE 2001, ummer /25/2007 CE 2001, ummer