CS 461 – Oct. 5 Pumping lemma #2 –Understanding –Use to show a language is not a CFL Next: Applications of CFLs –Expression grammars and Compiling

Pumping lemma L is a CFL implies: There is a p, such that for any string s, |s| >= p We can break up s into 5 parts uvxyz: | v y | >= 1 | v x y | <= p and uv i xy i z  L. How does this differ from the first pumping lemma?

In other words If L is a CFL, then for practically any word w in L, we should be able to find 2 substrings within w located near each other that you can simultaneously pump in order to create more words in L. Contrapositive: If w is a word in L, and we can always create a word not in L by simultaneously pumping any two substrings in w that are near each other, then L is not a CFL.

Gist A string can grow in 2 places. –When we write recursive rules, we can add symbols to left or right of a variable. Ex.S  #A# A  ε | 1A2 Look at derivation of: #111…222# Even a complex rule like S  S1S1S2 can be “simplified”, and still only need 2 places to grow the string. –This is because any CFG can be written in such a way that there are no more than 2 symbols on the right side of .

Example L = { 1 n 2 n } should satisfy the pumping lemma. Let p = 2. Strings in L of length >= 2 are: { 12, 1122, , … }. All these words have both a 1 and a 2, which we can pump. Just make sure the 2 places to pump are within p symbols of each other. So let’s say they are in the middle. u v x y z 11111…1 1 ε ….2 Could p = 1 work?

Example 2 { w # w R }. Think about where we can grow. Let p = 3. All strings of length 3 or more look like w # w R where | w | >= 1. As a technicality, would other values of p work?

Non-CFLs It’s possible for a language to be non-CFL! Typically this will be a language that must simultaneously pump in 3+ places. Ex. How could we design a PDA for { 1 n 2 n 3 n }?

Adversary’s Revenge In Game #1, how did we win? Ex. { 0 n 1 n }. –Adversary chose p. –We chose s. –Adversary broke up s = xyz subject to constraints. –We were always able to pump and find words outside L. Game #2 strategy –Adversary’s constraints looser. The middle 3 parts have to be within p of each other. Can be anywhere in the string, not just in first p symbols of word.

{ 1 n 2 n 3 n } Let p be any number. Choose s = 1 p 2 p 3 p. Let s = uvxyz such that |vxy| = 1. Where can v and y be? –All 1’s –All 2’s –All 3’s –Straddling 1’s and 2’s –Straddling 2’s and 3’s In every case, can we find a word not in L?

More examples { 1 i 2 j 3 k : i < j < k } –What should we chose as s? –How can the word be split up? { w w } –Let s = 0 p 1 p 0 p 1 p. Where can v & y be? –Same section of 0’s and 1’s. –In neighboring sections –Either v or y straddles a border.