Non-CF Languages The language L = { a n b n c n | 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  * …  * uv i Ay i z  * uv i xy i z  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  * v 2 Ay 2  * v 3 Ay 3  * …) 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 b ac A c c ca A c c c B B

Pumping a Parse Tree A A u vxy 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. S

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  X 1 …X b A parse tree of depth h produces a string with maximum 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 occurs more than once.

uvxyz  L A A u vxy z By repeating the A–A part we get… S

uv 2 xy 2 z  L A u v x y z R A A v x y … while removing the A–-A gives… S

uxz  L A uz In general uv i xy i z  L for all i=0,1,2,… S x

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) uv i xy i z  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  X 1 …X b A string s requires a minimum tree-depth  log b |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 uv i xy i z  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 {a n b n c n | n >= 0} Assume that B = {a n b n c n | n  0} is CFL Let p be the pumping length, and s = a p b p c p  B P.L.: s = uvxyz = a p b p c p, with uv i xy i z  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 uv 2 xy 2 z will not contain the same number of a’s, b’s and c’s, hence uv 2 xy 2 z  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 uv 2 xy 2 z will not be a…ab…bc…c Hence uv 2 xy 2 z  B

Pumping lemma applied to {a n b n c n } continued Assume that B = {a n b n c n | n  0} is CFL Let p be the pumping length, and s = a p b p c p  B P.L.: s = uvxyz = a p b p c p, with uv i xy i z  B for all i  0 We showed: For every way of partitioning s into uvxyz, there is an i such that uv i xy i z is not in B. Contradiction. B is not a context-free language.

Another example Proof that C = {a i b j c k | 0  i  j  k } is not context-free. Let p be the pumping length, and s = a p b p c p  C P.L.: s = uvxyz, such that uv i xy i z  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 uv 2 xy 2 z has not enough c’s, hence uv 2 xy 2 z  C 2) vxy belongs to b*c*, then the string uv 0 xy 0 z = uxz has too many a’s, hence uv 0 xy 0 z  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=0 p 1 p 0 p 1 p. Three options for s=uvxyz with 1  |vxy|  p: 1)If a part of y is to the left of | in 0 p 1 p |0 p 1 p, then second half of uv 2 xy 2 z starts with “1” 2) Same reasoning if a part of v is to the right of middle of 0 p 1 p |0 p 1 p, hence uv 2 xy 2 z  D 3) If x is in the middle of 0 p 1 p |0 p 1 p, then uxz equals 0 p 1 i 0 j 1 p  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 A 1 and A 2 be two CF languages, then the union A 1  A 2 is context free as well. Proof: Assume that the two grammars are G 1 =(V 1, ,R 1,S 1 ) and G 2 =(V 2, ,R 2,S 2 ). Construct a third grammar G 3 =(V 3, ,R 3,S 3 ) by: V 3 = V 1  V 2  { S 3 } (new start variable) with R 3 = R 1  R 2  { S 3  S 1 | S 2 }. It follows that L(G 3 ) = L(G 1 )  L(G 2 ).

Intersection, Complement? Let again A 1 and A 2 be two CF languages. One can prove that, in general, the intersection A 1  A 2, and the complement Ā 1 =  * \ A 1 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 = {a n b n c n | n >= 0} is NOT context-free. Let A = {a n b n c m | n, m >= 0} and B = L = {a n b m c m | 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.