Context Free Grammar for {a^mb^nc^n | m , n ε N } | CFG for {a^mb^nc^n

Context Free Grammar for {a^mb^nc^n | m , n ε N } | CFG for {a^mb^nc^n | m , n ε N }

To construct a context-free grammar (CFG) for the language

L = {a^m b^n c^n mid m, n in mathbb{N}}

, let’s break down the problem.

Language Description

The language $L$ consists of strings where:

There are $m$ occurrences of the letter $a$ ,
Followed by $n$ occurrences of the letter $b$ ,
Followed by $n$ occurrences of the letter $c$ ,
The number of $b$ ‘s and $c$ ‘s is the same, but there can be any non-negative number of $a$ ‘s.

Constructing the CFG

We need a CFG that generates strings in the form of $a^m b^n c^n$ where $m$ and $n$ are non-negative integers. Here’s a CFG that generates this language:

Start Symbol: $S$

Production Rules:

$S rightarrow aS mid B$
$B rightarrow bBc mid epsilon$

Explanation of the Rules:

This rule allows for the generation of any number of $a$ ‘s at the beginning of the string. After generating $a$ ‘s, it transitions to the non-terminal $B$ to handle the $b$ ‘s and $c$ ‘s.

This rule generates the sequence of $b$ ‘s followed by $c$ ‘s. For every $b$ introduced, a corresponding $c$ is introduced, maintaining the balance between the number of $b$ ‘s and $c$ ‘s. The rule $B rightarrow epsilon$ allows for the case when there are no $b$ ‘s or $c$ ‘s.

Example Derivations:

For $m = 2$ and $n = 3$ , the string $a^2 b^3 c^3$ would be generated as follows:
- $S Rightarrow aS Rightarrow aaS Rightarrow aaB$
- $B Rightarrow bBc Rightarrow bbBc Rightarrow bbbBc Rightarrow bbbepsilon ccc$
- Therefore, $aaB Rightarrow aabbbccc$
For $m = 0$ and $n = 2, the string$ $b^2 c^2$ would be generated as follows:
- $S Rightarrow B$
- $B Rightarrow bBc Rightarrow bbBc Rightarrow bbbc$
- $B Rightarrow epsilon$ giving $bbc$