1. Context-Free Languages
Pumping Lemma

2. Pumping Lemma for CFL
If L is a context-free language, then there is a number p (the pumping length) where, if s is any string in L of length at least p, then s may be divided into five pieces s=uvxyz satisfying the conditions:
For each i ≥ 0, s=uvixyiz  L,
|vy| > 0, and
|vxy| ≤ p.

3. Pumping Lemma as an “Adversary Game”
We pick a language L that we want to show is not a CFL.
Our “adversary” gets to pick p.
We get to pick s, and may use p as a parameter when we do so.
Our adversary gets to break s into uvxyz, subject only to the constraints that |vy| > 0, and |vxy| ≤ p.
We “win” the game, if we can, by picking i and showing that uvixyiz is not in L.

4. Example 1: L = { 1n where n is prime}
By intuition:
Is L a context-free language?
n=2: 11
n=3: 111
n=5: 11111
n=7:
n=13:

5. Example 1: L = { 1n where n is prime}
Adversary picks a pumping length of p.
Consider some prime n ≥ p+2.
Such an n must exist since there are an infinite number of primes.
Let s= 1n
Adversary picks uvxyz.
Let |vy| = m.
Then, |uxz| = n-m.

6. Example 1: L = { 1n where n is prime}
By the pumping lemma, uvn-mxyn-mz  L
|uvn-mxyn-mz| = |uxz| + (n-m)(|v| + |y|)
= n-m + (n-m) m
= (n-m) (m+1)
Thus, |uvn-mxyn-mz| is not prime unless one of the above factors is 1.
But, (m+1) > 1 since m=|vy|, and |vy| > 0.
(n-m) > 1 since:
n ≥ p+2 was chosen, and
m ≤ p since m=|vy| ≤ |vxy| ≤ p