Intro to Theory of Computation

Transcription

1 Intro to Theory of Computation LECTURE 9 Last time: Converting a PDA to a CFG Pumping Lemma for CFLs Today: Pumping Lemma for CFLs Review of CFGs/PDAs Sofya Raskhodnikova 2/9/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L9.1

2 I-clicker problem (frequency: AC) To prove that a language L is not context free, we can A. argue that a PDA cannot remember enough information to recognize L; B. use pumping lemma; C. give a CFG and show that it does not generate L. D. None of the above. E. More than one choice above works.

3 Pumping lemma for CFLs Let L be a context-free language Then there exists P such that For every w L with w P there exist uvxyz=w, where: 1. vy > 0 2. vxy P 3. uv i xy i z L for all i 0 L9.3

4 Negating the pumping lemma L Let is L not be context-free a language. Then If there for every exists P, P such that there for every is a w L with w P there for every exist uvxyz=w, where: 1. vy > 0 2. vxy P 3. there For is every an i 0, uv i xy i z L L8.4

5 Pumping lemma as a game 1. YOU pick the language L to be proved not CFL. 2. ADVERSARY picks p, but doesn't reveal to YOU what p is; YOU must devise a play for all possible p's. 3. YOU pick w L, which should depend on p and which must be of length at least p. 4. ADVERSARY divides w into u, v, x, y, z, obeying PL conditions: vy is not empty and vxy has length p. Again, ADVERSARY does not tell YOU what u, v, x, y, z are. 5. YOU win by picking i, which may be a function of p, u, v, x, y, z, such that uv i xy i z is not in L. 2/9/2016 L5.5

6 I-clicker problem (frequency: AC) Prove {ww w {0,1}*} is not context free. Proof: Assume pumping length p. What string can you choose in your next move? A B. 0 p 1 p C. 01 p D. 0 p 1 p 0 p 1 p E. More than one choice above works.

7 Using the pumping lemma Prove L = {ww w {0,1}*} is not context-free Assume L is context-free. Then there is a pumping length P. No matter what P is, the string s = 0 P 1 P 0 P 1 P has s P and s L. So there should be uvxyz=s with: 1) vy > 0, 2) vxy P, 3) i, uv i xy i z L. P P P P s = L8.7

8 {ww w {0,1}*} is not a CFL: proof vxy No matter what P is, string s = 0 P 1 P 0 P 1 P has Then cannot pumping be only down in must the first remove half, at since least pumping one 1 down s P and s L. or would one move zero, e.g. a 0 to uxz the = 0end P 1 i 0of j 1 P the where first half. vxyicannot p So or there j be p. only should So in uxz the be last L, uvxyz=s no half, matter since with: how pumping up they would 1) are vy chosen; move > 0, 2) a vxy 0 so to L the cannot P, end 3) of be i, the uv context-free! i xy first i z half. L. P P P P s = vxy L8.8

9 Using the pumping lemma Prove L = {w#w R w {0,1}* and #1s = #0s} is not context-free Assume L is context-free. Then there is a pumping length P. No matter what P is, the string s = 0 P 1 P #1 P 0 P has s P and s L. So there should be uvxyz=s with: 1) vy > 0, 2) vxy P, 3) i, uv i xy i z L. P P P P s = #

10 {w#w R w {0,1}* and #1s = #0s} is not a CFL: proof Assume L is context-free. vxy cannot Then be there only is in a either pumping half, length since P. pumping No would matter make what one P side is, the of string # longer s = than 0 P 1 P the #1 P 0other. P has # cannot be in v s or y P since and s pumping L. would add since # is in x and vxy P, neither v nor y can too many So #s. there So should it must be be uvxyz=s in x. have 0s, and at least one must have with: a 1. Then 1) uv vy 2 xy> 2 z 0, has 2) vxy more 1s P, than 3) i, 0s uvand i xy i zis not L. in L. P P P P s = # vxy

11 I-clicker problem (frequency: AC) S asb What is the language of this CFG? A. a n b n n 0} b n a n n 0} B. The set of strings over alphabet a, b C. The set of strings over alphabet a, b with the same number of a s and b s D. None of the above bsa SS ε

12 Give the language of this CFG S asb bsa SS ε Answer: the set of all strings over alphabet {a,b} that have the same number of a s and b s Prove that your answer is correct L9.12

13 Proof of correctness S asb bsa SS ε 1. Every generated string has the same number of a s and b s Proof: Every rule that generates terminals, generates the same number of a s and b s 2/9/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L9.13

14 Proof of correctness S asb bsa SS ε 2. Every string that has the same number of a s and b s is generated. Proof idea: Consider a string w 1 w n that has the same number of a s and b s. Plot # of a s # of b s in the first i characters of w as a function of i. 2/9/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L9.14

15 2. Every string that has the same number of a s and b s is generated. 2/9/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

16 2. Every string that has the same number of a s and b s is generated. Claim. Every string of length 2m that has the same number of a s and b s is generated, integer m 0. Proof by strong induction on m. Base case: m = 0. String ε is generated by the rule S ε. IH: Suppose the claim holds m = 0, 1,, k for some k. IS: We prove it for m = k + 1. Each string of length 2(k + 1) with the same number of a s and b s falls in one of the 3 cases on previous slide. 2/9/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

17 2. Every string that has the same number of a s and b s is generated. Case 1: w = aw b, where w is a string of length 2k with the same number of a s and b s. Then, by IH, we can derive w from S: S w (1) To get w, we use S asb and then (1): S asb aw b. We proved that w is generated. 2/9/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

18 2. Every string that has the same number of a s and b s is generated. Case 2: w = bw a, where w is a string of length 2k with the same number of a s and b s. Then, by IH, we can derive w from S: S w (2) To get w, we use S bsa and then (2): S bsa bw a. We proved that w is generated. 2/9/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

19 2. Every string that has the same number of a s and b s is generated. Case 3: w = w s w e, where w s and w e are strings of length 2k with the same number of a s and b s. Then, by IH, we can derive w s and w e from S: S w s (3) S w e (4) To get w, we use S SS, then (3) and then (4): S SS w s S w s w e. We proved that w is generated. To conclude, all strings of length 2(k + 1) with the same number of a s and b s are generated. 2/9/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

20 Context-free or not? NOT L₁ = { xy x,y {0,1}* and x=y} YES L₂ = {xy x,y {0,1}*, x = y and x y} L9.20

21 Give an algorithmic description of a PDA for {xy x, y 0, 1 and x = y but x y}. Hint: x y i j i j L9.21

22 Give an algorithmic description of a PDA for {xy x, y 0, 1 and x = y but x y}. Hint: i i j j L9.22