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- Automata Theory Introduction
- Deterministic Finite Automaton
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- Classification of Grammars
- Introduction to Grammars
- Language Generated by Grammars
- Chomsky Grammar Classification

- Regular Grammar
- Regular Expressions
- Regular Sets
- Arden's Theorem
- Constructing FA from RE
- Pumping Lemma for Regular Grammar
- DFA Complement

- Context-Free Grammars
- Context-Free Grammar Introduction
- Ambiguity in Grammar
- CFL Closure Properties
- CFG Simplification
- Chomsky Normal Form
- Greibach Normal Form
- Pumping Lemma for CFG

- Pushdown Automata
- Pushdown Automata Introduction
- Pushdown Automata Acceptance
- PDA & Context Free Grammar
- PDA & Parsing

- Turing Machine
- Turing Machine Introduction
- Accepted & Decided Language
- Multi-tape Turing Machine
- Multi-Track Turing Machine
- Non-Deterministic Turing Machine
- Semi-Infinite Tape Turing Machine
- Linear Bounded Automata

- Decidability
- Language Decidability
- Undecidable Language
- Turing Machine Halting Problem
- Rice Theorem
- Post Correspondence Problem

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# Language Generated by a Grammar

The set of all strings that can be derived from a grammar is said to be the language generated from that grammar. A language generated by a grammar **G** is a subset formally defined by

L(G)={W|W ∈ ∑*, S ⇒G **W**}

If **L(G1) = L(G2)**, the Grammar **G1** is equivalent to the Grammar **G2**.

### Example

If there is a grammar

G: N = {S, A, B} T = {a, b} P = {S → AB, A → a, B → b}

Here **S** produces **AB**, and we can replace **A** by **a**, and **B** by **b**. Here, the only accepted string is **ab**, i.e.,

L(G) = {ab}

### Example

Suppose we have the following grammar −

G: N = {S, A, B} T = {a, b} P = {S → AB, A → aA|a, B → bB|b}

The language generated by this grammar −

L(G) = {ab, a^{2}b, ab^{2}, a^{2}b^{2}, ………}

= {a^{m} b^{n} | m ≥ 1 and n ≥ 1}

## Construction of a Grammar Generating a Language

We’ll consider some languages and convert it into a grammar G which produces those languages.

### Example

** Problem** − Suppose, L (G) = {a

^{m}b

^{n}| m ≥ 0 and n > 0}. We have to find out the grammar

**G**which produces

**L(G)**.

*Solution*

Since L(G) = {a^{m} b^{n} | m ≥ 0 and n > 0}

the set of strings accepted can be rewritten as −

L(G) = {b, ab,bb, aab, abb, …….}

Here, the start symbol has to take at least one ‘b’ preceded by any number of ‘a’ including null.

To accept the string set {b, ab, bb, aab, abb, …….}, we have taken the productions −

S → aS , S → B, B → b and B → bB

S → B → b (Accepted)

S → B → bB → bb (Accepted)

S → aS → aB → ab (Accepted)

S → aS → aaS → aaB → aab(Accepted)

S → aS → aB → abB → abb (Accepted)

Thus, we can prove every single string in L(G) is accepted by the language generated by the production set.

Hence the grammar −

G: ({S, A, B}, {a, b}, S, { S → aS | B , B → b | bB })

### Example

** Problem** − Suppose, L (G) = {a

^{m}b

^{n}| m > 0 and n ≥ 0}. We have to find out the grammar G which produces L(G).

** Solution** −

Since L(G) = {a^{m} b^{n} | m > 0 and n ≥ 0}, the set of strings accepted can be rewritten as −

L(G) = {a, aa, ab, aaa, aab ,abb, …….}

Here, the start symbol has to take at least one ‘a’ followed by any number of ‘b’ including null.

To accept the string set {a, aa, ab, aaa, aab, abb, …….}, we have taken the productions −

S → aA, A → aA , A → B, B → bB ,B → λ

S → aA → aB → aλ → a (Accepted)

S → aA → aaA → aaB → aaλ → aa (Accepted)

S → aA → aB → abB → abλ → ab (Accepted)

S → aA → aaA → aaaA → aaaB → aaaλ → aaa (Accepted)

S → aA → aaA → aaB → aabB → aabλ → aab (Accepted)

S → aA → aB → abB → abbB → abbλ → abb (Accepted)

Thus, we can prove every single string in L(G) is accepted by the language generated by the production set.

Hence the grammar −

G: ({S, A, B}, {a, b}, S, {S → aA, A → aA | B, B → λ | bB })