Lecture 17 Oct 25, 2011 Section 2.1 (push-down automata) Section 2.2 (pumping lemma for context-free languages)

Last-In First-Out pushing and popping 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) internal state set Q input w = 00100100111100101 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) x y z stack

Informal Description PDA (2) internal state set Q input w = 00100100111100101 After the PDA has read complete input, M will be in state  Q If possible to end in accepting state FQ, then M accepts w x y z 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 | n0 } , $ , $ 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 q3 q2 q4 , $ , $ 1, 0 0, 0

Machine Diagram for 0n1n , $ , $ 0, 0 q2 q1 1, 0 q3 q4 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 , $ , $ 0, 0 q2 q1 1, 0 q3 q4 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: 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 q r abcabbacacbacbacabacabbabbabaacabbbababaacaccaccccc input 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 s p push d pop d q abcabbacacbacbacabacabbabbabaacabbbababaacaccaccccc input string taking NPDA from p to q 2011

NPDA P = (Q, Σ, , δ, start, {accept}) CFG G: 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, 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 from state p, read a, push d, move to state r from state s, read b, pop d, move to state q 2011

NPDA P = (Q, Σ, , δ, start, {accept}) CFG G: 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: if Ap,q generates string x, then x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack) 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 (Ap,r→* y) z is read going from state r to q (Ar,q→* z) conclude: Ap,q → Ap,rAr,q →* yz = x 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 if stack becomes empty only at beginning and end of computation. 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 2011

PDACFG conversion Summary of the construction:

Non-CF Languages The language L = { anbncn | n0 } 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 parse-tree is tall enough.

Recall Parse Trees Parse tree for S  AbbcBa * cbbccccaBca  cbbccccacca S b a c A B

Pumping a Parse Tree S A A u v x y 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 A R y u v x z y v x … 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 sL 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 b2: 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 | n0} is CFL Let p be the pumping length, and s = apbpcp  B P.L.: s = uvxyz = apbpcp, with uvixyiz  B for all i0 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 uv2xy2zB 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 uv2xy2zB

Pumping lemma applied to {anbncn} continued Assume that B = {anbncn | n0} is CFL Let p be the pumping length, and s = apbpcp  B P.L.: s = uvxyz = apbpcp, with uvixyiz  B for all i0 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 | 0ijk } 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 uv2xy2zC 2) vxy belongs to b*c*, then the string uv0xy0z = uxz has too many a’s, hence uv0xy0zC Contradiction: C is not a context-free language.

D = { ww | w{0,1}* } (Ex. 2.22) Carefully take the strings sD. Let p be the pumping length, take s=0p1p0p1p. Three options for s=uvxyz with 1  |vxy|  p: 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 A1A2 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.