# Context-Free Grammar (CFG)

CFG stands for context-free grammar. It is is a formal grammar which is used to generate all possible patterns of strings in a given formal language. Context-free grammar G can be defined by four tuples as:

- G = (V, T, P, S)

**Where,**

**G** is the grammar, which consists of a set of the production rule. It is used to generate the string of a language.

**T** is the final set of a terminal symbol. It is denoted by lower case letters.

**V** is the final set of a non-terminal symbol. It is denoted by capital letters.

**P** is a set of production rules, which is used for replacing non-terminals symbols(on the left side of the production) in a string with other terminal or non-terminal symbols(on the right side of the production).

**S** is the start symbol which is used to derive the string. We can derive the string by repeatedly replacing a non-terminal by the right-hand side of the production until all non-terminal have been replaced by terminal symbols.

### Example 1:

Construct the CFG for the language having any number of a’s over the set ∑= {a}.

**Solution:**

As we know the regular expression for the above language is

- r.e. = a*

Production rule for the Regular expression is as follows:

- S → aS rule 1
- S → ε rule 2

Now if we want to derive a string “aaaaaa”, we can start with start symbols.

- S
- aS
- aaS rule 1
- aaaS rule 1
- aaaaS rule 1
- aaaaaS rule 1
- aaaaaaS rule 1
- aaaaaaε rule 2
- aaaaaa

The r.e. = a* can generate a set of string {ε, a, aa, aaa,…..}. We can have a null string because S is a start symbol and rule 2 gives S → ε.

### Example 2:

Construct a CFG for the regular expression (0+1)*

**Solution:**

The CFG can be given by,

- Production rule (P):
- S → 0S | 1S
- S → ε

The rules are in the combination of 0’s and 1’s with the start symbol. Since (0+1)* indicates {ε, 0, 1, 01, 10, 00, 11, ….}. In this set, ε is a string, so in the rule, we can set the rule S → ε.

### Example 3:

Construct a CFG for a language L = {wcwR | where w € (a, b)*}.

**Solution:**

The string that can be generated for a given language is {aacaa, bcb, abcba, bacab, abbcbba, ….}

The grammar could be:

- S → aSa rule 1
- S → bSb rule 2
- S → c rule 3

Now if we want to derive a string “abbcbba”, we can start with start symbols.

- S → aSa
- S → abSba from rule 2
- S → abbSbba from rule 2
- S → abbcbba from rule 3

Thus any of this kind of string can be derived from the given production rules.

### Example 4:

Construct a CFG for the language L = a^{n}b^{2n} where n>=1.

**Solution:**

The string that can be generated for a given language is {abb, aabbbb, aaabbbbbb….}.

The grammar could be:

- S → aSbb | abb

Now if we want to derive a string “aabbbb”, we can start with start symbols.

- S → aSbb
- S → aabbbb

**Theory of Automata****Finite Automata****Transition Diagram****Transition Table****DFA (Deterministic finite automata)****Examples of DFA****NFA (Non-Deterministic finite automata)****Examples of NFA****Eliminating ε Transitions****Conversion from NFA to DFA****Conversion from NFA with ε to DFA****Minimization of DFA****Regular Expression****Examples of Regular Expression****Moore Machine****Mealy Machine****Context Free Grammar****Simplification of CFG****Chomsky’s Normal Form (CNF)****Greibach Normal Form (GNF)****Pushdown Automata(PDA)****Non-deterministic Pushdown Automata****Turing Machine****Examples of TM**