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Today

Today. Chapter 2: (Pushdown automata) Non-CF languages CFL pumping lemma Closure properties of CFL. Pushdown Automata. Pushdown automata are for context-free languages what finite automata are for regular languages.

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Today

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  1. Today • Chapter 2: • (Pushdown automata) • Non-CF languages • CFL pumping lemma • Closure properties of CFL

  2. Pushdown Automata Pushdown automata are for context-free languages what finite automata are for regularlanguages. PDAs are recognizing automata that have a single stack (= memory):Last-In First-Out pushing and popping Difference: PDAs are inherently nondeterministic.(They are not practical machines.)

  3. x y y z x stack Informal Description PDA (1) input w = 00100100111100101 internal state set Q The PDA M reads w and stack element. Depending on - input wi , - stack sj  , and - state qk  Qthe PDA M: - jumps to a new state, - pushes an element (nondeterministically)

  4. Informal Description PDA (2) input w = 00100100111100101 internal state set Q 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 y z x stack

  5. Formal Description PDA • A Pushdown Automata M is defined by asix 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  )

  6. 0, 0 , $ q2 q1 1, 0 , $ 1, 0 q3 q4 PDA for L = { 0n1n | n0 } Example 2.9: The PDA first pushes “ $ 0n ” on stack. Then, while reading the 1n string, thezeros are popped again.If, in the end, $ is left on stack, then “accept”

  7. 0, 0 , $ q2 q1 1, 0 , $ 1, 0 q3 q4 Machine Diagram for 0n1n • 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

  8. 0, 0 , $ q2 q1 1, 0 , $ 1, 0 q3 q4 Machine Diagram for 0n1n 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.

  9. PDAs versus CFL Theorem 2.12: A language L is context-free if and only of 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

  10. Non-CF Languages The language L = { anbncn | n0 } does notappear to be context-free. Informal: The problem is that every variablecan (only) act ‘by itself’ (context-free). 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,…

  11. “Pumping Lemma for CFLs” Idea: If we can prove the existence of derivationsfor 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.

  12. S b b A c a B c A c c a B c c c Remember Parse Trees Parse tree for S  AbbcBa * cbbccccaBca  cbbccccacca

  13. 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.

  14. A Tree Tall Enough Let L be a context-free language, and letG 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 stringwith 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 ona ‘top-down path’ that occurs more than once.

  15. uvxyz L S A A u v x y z By repeating the A–A part we get…

  16. uv2xy2z L S A A A R y v u x z y x v … while removing the A–-A gives…

  17. uxz  L S A x u z In general uvixyiz  L for all i=0,1,2,…

  18. 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 with1) uvixyiz  L for every i{0,1,2,…}2) |vy|  13) |vxy|  p Note that 1) implies that uxz  L 2) says that vy cannot be the empty string Condition 3) is not always used

  19. 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’

  20. Pumping anbncn (Ex. 2.20) 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) v and y are not uniform. Then uv2xy2z will not be a…ab…bc…c Hence uv2xy2zB

  21. Pumping anbncn (cont.) Assume that B = {anbncn | n0} is CFLLet p be the pumping length, and s = apbpcp  BP.L.: s = uvxyz = apbpcp, with uvixyiz  B for all i0 We showed: No options for |vxy| such that uvixyiz  B for all i. Contradiction. B is not a context-free language.

  22. Example 2.21 (Pumping down) Proof that C = {aibjck | 0ijk } is not context-free. Let p be the pumping length, and s = apbpcp  CP.L.: s = uvxyz, such that uvixyiz  C for every i0 Two options for 1  |vxy|  p:1) vxy = a*b*, then the string uv2xy2z has not enough c’s, hence uv2xy2zC 2) vxy = b*c*, then the string uv0xy0z = uxz has too many a’s, hence uv0xy0zC Contradiction: C is not a context-free language.

  23. 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: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.

  24. Pumping Problems Using the CFL pumping lemma is more difficultthan 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?

  25. 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 areG1=(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).

  26. 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= * \ A1are not context free languages. One proves this with specific counter examples of languages (see homework).

  27. Homework • Exercise 2.2 (a, b) • Exercise 2.7 • Problem 2.18 (a, b, c, d)

  28. Practice Problems • Exercise 2.10 • Problem 2.19 • Problem 2.20 • Problem 2.23

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