# Non Context-Free Languages

### Overview

We have seen that context-free languages are more powerful than regular languages.
However they have limits; there are many types of languages that are *not* context free.

We will discuss languages in this category and how we can prove a language is not context-free.

### Pumping Lemma for Context-Free Languages

Like regular languages, context-free languages have a pumping lemma which is given below:

If $A$ is a context free language, then there is a number $p$ where, if $s$ is a string in
$A$ of length at least $p$, then $s$ may be divided into *five* pieces $s = uvxyz$ such that:

- $\forall i \ge 0, uv^ixy^iz \in A$
- $|vy| > 0$
- $|vxy| \le p$

### Pumping Lemma Explanation

If $A$ is a context-free language, it has some grammar, $G$, which generates it.
The grammar $G$ has a finite number of variables, let's call this the number $p$.

If $s$ is some string in $A$, then there is a parse tree of $s$.
If the string $s$ is larger than $p$, then some variable $R$ must be repeated in the parse tree (by the pigeon-hole principle).

Because there was some sequence of the derivation that brought us from $R$ back to $R$, then
we could repeat that sequence and "pump" $R$ again, and still have a valid parse tree.

We could also remove the portion of the string that repeated $R$ and still have a valid string in $A$:

Another way of looking at the pumping lemma for context-free languages is by
considering how a grammar can "loop", which is via recursion. A recursive rule
can be generalized like this:

$A \rightarrow vAy \ |\ x$

Here, $v$, $y$, and $x$ are strings of terminals (or variables
which can reduce to terminals).
A recursive rule contains the portion of the string before the recursion ($v$),
the portion after the recursion ($y$) and also the base case ($x$).

$u$ and $z$ are the parts leading up to the recursive rule, and following
the recursive rule:

$S \rightarrow uAz$

Any context-free grammar which can generate arbitrarily-long strings *must*
have some type of recursive rule like this.

The pumping lemma just says that we can repeat that recursion any number of
times we like and still have a string in the language.

### Context Free Example

How can we show the pumping lemma holds for $\{0^n1^n | n \ge 0\}$?

### Proving A Language is Not Context-Free

To prove that a language is not context-free, we can show that, for some string in the language
with at least length $p$, there is no way to divide it such that the pumping lemma holds.

### Example: $\{a^nb^nc^n | n \ge 0\}$

### Example: $\{ww | w \in \{0,1 \}^\ast\}$

Copyright © 2018 Ian Finlayson | Licensed under a Creative Commons Attribution 4.0 International License.