Lecture 17 Oct 25, 2011 • Section 2.1 (push-down automata) • Section 2.2 (pumping lemma for context-free languages) Pushdown Automata Pushdown automata are for context-free languages what finite automata are for regular languages. PDAs are recognizing automata that have a single stack (= memory): Last-In First-Out pushing and popping Note: PDAs are nondeterministic. Informal Description PDA (1) input w = 00100100111100101 internal state set Q stack x y y z x The PDA M reads w and stack element. Depending on - input wi , - stack sj , and - state qk Q the PDA M: - jumps to a new state, - pushes an element (nondeterministically) Informal Description PDA (2) input w = 00100100111100101 internal state set Q After the PDA has read complete input, M will be in state Q x y y z x If possible to end in accepting state FQ, then M accepts w stack Formal Description PDA A Pushdown Automata M is defined by a six tuple (Q,,,,q0,F), with • Q finite set of states • finite input alphabet • finite stack alphabet • q0 start state Q • F set of accepting states Q • transition function : Q P (Q ) PDA for L = { 0n1n | n0 } Example 2.14: The PDA first pushes “ $ 0n ” on stack. Then, while reading the 1n string, the zeros are popped again. If, in the end, $ is left on stack, then “accept” q1 q4 , $ 0, 0 q2 1, 0 , $ q3 1, 0 Machine Diagram for 0n1n q1 q4 , $ 0, 0 q2 1, 0 , $ q3 1, 0 On w = 000111 (state; stack) evolution: (q1; ) (q2; $) (q2; 0$) (q2; 00$) (q2; 000$) (q3; 00$) (q3; 0$) (q3; $) (q4; ) This final q4 is an accepting state Machine Diagram for 0n1n q1 q4 , $ 0, 0 q2 1, 0 , $ q3 1, 0 On w = 0101 (state; stack) evolution: (q1; ) (q2; $) (q2; 0$) (q3; $) (q4; ) … But we still have part of input “01”. There is no accepting path. Another Example of a PDA Another example of PDA Consider the language over the alphabet {a, b}: L = { w | #a(w) = #b(w) } (#a(w) stands for the number of a’s in w.) PDA design intuition: push a symbol 1 on seeing a’s, pop on seeing b’s. Problem: what if we see a lot of b’s in the start, and a’s come later? Can change the role. Push on b, pop on a. Need to know which one – using two different states. Another example of PDA Consider the language over the alphabet {a, b}: L = { w | #a(w) = #b(w) } One more PDA – for even length palindromes L = { w wR | w is in {0, 1}* } PDAs versus CFL Theorem 2.20: A language L is context-free if and only if there is a pushdown automata M that recognizes L. Two step proof: 1) Given a CFG G, construct a PDA MG 2) Given a PDA M, make a CFG GM Equivalence of PDA and CFG (0) Part 1: For every CFG, we can build an equivalent PDA. General construction: each rule of CFG A w is included in the PDA’s move. Equivalence of PDA and CFG (1) Part 1: For every CFG, we can build an equivalent PDA. Example: (page 115 of text) NPDA, CFG equivalence Proof of (): L is recognized by a NPDA implies L is described by a CFG. – harder direction – first step: convert NPDA into “normal form”: • single accept state • empties stack before accepting • each transition either pushes or pops a symbol 2011 NPDA, CFG equivalence – main idea: non-terminal Ap,q generates exactly the strings that take the NPDA from state p (w/ empty stack) to state q (w/ empty stack) – then Astart, accept generates all of the strings in the language recognized by the NPDA. 2011 NPDA, CFG equivalence • Two possibilities to get from state p to q: generated by Ap,r generated by Ar,q stack height p input r q abcabbacacbacbacabacabbabbabaacab bbababaacaccaccccc string taking NPDA from p to q 2011 NPDA, CFG equivalence • NPDA P = (Q, Σ, , δ, start, {accept}) • CFG G: – non-terminals V = {Ap,q : p, q Q} – start variable Astart, accept – productions: for every p, r, q Q, add the rule Ap,q → Ap,rAr,q 2011 NPDA, CFG equivalence • Two possibilities to get from state p to q: generated by Ar,s stack height r p input push d s pop d q abcabbacacbacbacabacabbabbabaacab bbababaacaccaccccc string taking NPDA from p to q 2011 NPDA, CFG equivalence • NPDA P = (Q, Σ, , δ, start, {accept}) from state p, read a, push d, • CFG G: move to state r – non-terminals V = {Ap,q : p, q Q} – start variable Astart, accept from state s, read b, pop d, move to state q – productions: for every p, r, s, q Q, d , and a, b (Σ {ε}) if (r, d) δ(p, a, ε), and (q, ε) δ(s, b, d), add the rule Ap,q → aAr,sb 2011 NPDA, CFG equivalence • NPDA P = (Q, Σ, , δ, start, {accept}) • CFG G: – non-terminals V = {Ap,q : p, q Q} – start variable Astart, accept – productions: for every p Q, add the rule Ap,p → ε NPDA, CFG equivalence • two claims to verify correctness: 1. if Ap,q generates string x, then x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack) 2. if x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack), then Ap,q generates string x 2011 NPDA, CFG equivalence 1. if Ap,q generates string x, then x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack) – induction on length of derivation of x. – base case: 1 step derivation. must have only terminals on rhs. In G, must be production of form Ap,p → ε. 2011 NPDA, CFG equivalence 1. if Ap,q generates string x, then x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack) – assume true for derivations of length at most k, prove for length k+1. – verify case: Ap,q → Ap,rAr,q →k x = yz – verify case: Ap,q → aAr,sb →k x = ayb 2011 NPDA, CFG equivalence 2. if x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack), then Ap,q generates string x – induction on # of steps in P’s computation – base case: 0 steps. starts and ends at same state p. only has time to read empty string ε. – G contains Ap,p → ε. 2011 NPDA, CFG equivalence 2. if x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack), then Ap,q generates string x – induction step. assume true for computations of length at most k, prove for length k+1. – if stack becomes empty sometime in the middle of the computation (at state r) • y is read going from state p to r y) • z is read going from state r to q z) • conclude: Ap,q → Ap,rAr,q →* yz = x 2011 (Ap,r→* (Ar,q→* NPDA, CFG equivalence 2. if x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack), then Ap,q generates string x – if stack becomes empty only at beginning and end of computation. • • • • 2011 first step: state p to r, read a, push d go from state r to s, read string y (Ar,s→* y) last step: state s to q, read b, pop d conclude: Ap,q → aAr,sb →* ayb = x PDACFG conversion Summary of the construction: Non-CF Languages The language L = { anbncn | n0 } does not appear to be context-free. Informal: A PDA can compare #a’s with #b’s. But by the time b’s are processed, the stack is empty. Not possible to count a’s with c’s. The problem of A * vAy : If S * uAz * uvAyz * uvxyz L, then S * uAz * uvAyz * … * uviAyiz * uvixyiz L as well, for all i=0,1,2,… Pumping Lemma for CFLs Idea: If we can prove the existence of derivations for elements of the CFL L that use the step A * vAy, then a new form of ‘v-y pumping’ holds: A * vAy * v2Ay2 * v3Ay3 * …) Observation: We can prove this existence if the parsetree is tall enough. Recall Parse Trees Parse tree for S AbbcBa * cbbccccaBca cbbccccacca S A b b c B a c c a B c c A c c Pumping a Parse Tree S A A v x y u z If s = uvxyz L is long, then its parse-tree is tall. Hence, there is a path on which a variable A repeats itself. We can pump this A–A part. A Tree Tall Enough Let L be a context-free language, and let G be its grammar with maximal b symbols on the right side of the rules: A X1…Xb A parse tree of depth h produces a string with maximum length of bh. 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 occurs more than once. uvxyz L S A A u v x y z By repeating the A–A part we get… uv2xy2z L S A A u v A R x y z y x v … while removing the A–-A gives… uxz L S A x u z In general uvixyiz L for all i=0,1,2,… Pumping Lemma for CFL For every context-free language L, there is a pumping length p, such that for every string sL and |s|p, we can write s = uvxyz with 1) uvixyiz L for every i{0,1,2,…} 2) |vy| 1 3) |vxy| p Note that 1) implies that uxz L 2) says that v and y cannot be both empty strings Condition 3) is not always used. (It is not crucial part of pumping lemma, but helps to reduce the number of cases.) Formal Proof of Pumping Lemma Let G=(V,,R,S) be the grammar of a CFL. Maximum size of rules is b2: A X1…Xb A string s requires a minimum tree-depth logb|s|. If |s| p=b|V|+2, then tree-depth |V|+2, hence there is a path and variable A where A repeats itself: S * uAz * uvAyz * uvxyz It follows that uvixyiz L for all i=0,1,2,… Furthermore: |vy| 1 because tree is minimal |vxy| p because bottom tree with p leaves has a ‘repeating path’ Pumping lemma for {anbncn | n >= 0} Assume that B = {anbncn | n0} is CFL Let p be the pumping length, and s = apbpcp B P.L.: s = uvxyz = apbpcp, with uvixyiz B for all i0 Options for vxy: 1) The strings v and y are uniform (v=a…a and y=c…c, for example). Then uv2xy2z will not contain the same number of a’s, b’s and c’s, hence uv2xy2zB 2) At least one of v or y is not uniform. (i.e., it has at least two different symbols occurring in it). Then uv2xy2z will not be a…ab…bc…c Hence uv2xy2zB Pumping lemma applied to {anbncn} continued Assume that B = {anbncn | n0} is CFL Let p be the pumping length, and s = apbpcp B P.L.: s = uvxyz = apbpcp, with uvixyiz B for all i0 We showed: For every way of partitioning s into uvxyz, there is an i such that uvixyiz is not in B. Contradiction. B is not a context-free language. Another example Proof that C = {aibjck | 0ijk } is not context-free. Let p be the pumping length, and s = apbpcp C P.L.: s = uvxyz, such that uvixyiz C for every i 0 vxy can’t have a’s and c’s. Why? So only two options for vxy: 1) vxy belongs to a*b*, then the string uv2xy2z has not enough c’s, hence uv2xy2zC 2) vxy belongs to b*c*, then the string uv0xy0z = uxz has too many a’s, hence uv0xy0zC Contradiction: C is not a context-free language. D = { ww | w{0,1}* } (Ex. 2.22) Carefully take the strings sD. Let p be the pumping length, take s=0p1p0p1p. Three options for s=uvxyz with 1 |vxy| p: 1) If a part of y is to the left of | in 0p1p|0p1p, then second half of uv2xy2z starts with “1” 2) Same reasoning if a part of v is to the right of middle of 0p1p|0p1p, hence uv2xy2z D 3) If x is in the middle of 0p1p|0p1p, then uxz equals 0p1i 0j1p D (because i or j < p) Contradiction: D is not context-free. Pumping lemma for CFG - remarks Using the CFL pumping lemma is more difficult than the pumping lemma for regular languages. You have to choose the string s carefully, and divide the options efficiently. Additional CFL properties would be helpful (like we had for regular languages). What about closure under standard operations? Union Closure Properties Lemma: Let A1 and A2 be two CF languages, then the union A1A2 is context free as well. Proof: Assume that the two grammars are G1=(V1,,R1,S1) and G2=(V2,,R2,S2). Construct a third grammar G3=(V3,,R3,S3) by: V3 = V1 V2 { S3 } (new start variable) with R3 = R1 R2 { S3 S1 | S2 }. It follows that L(G3) = L(G1) L(G2). Intersection, Complement? Let again A1 and A2 be two CF languages. One can prove that, in general, the intersection A1 A2 , and the complement Ā1= * \ A1 are not context free languages. Intersection, Complement? Proof for complement: Recall that a problem in HW 5 shows that L = { x#y | x, y are in {a, b}*, x != y} IS context-free. Complement of this language is L’ = { w | w has no # symbol} U { w | w has two or more # symbols} U { w#w | w is in {a,b}* }. We can show that L’ is NOT context-free. Context-free languages are NOT closed under intersection Proof by counterexample: Recall that in an earlier slide in this lecture, we showed that L = {anbncn | n >= 0} is NOT context-free. Let A = {anbncm | n, m >= 0} and B = L = {anbmcm | n, m >= 0}. It is easy to see that both A and B are context-free. (Design CFG’s.) This shows that CFG’s are not closed under intersection.

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