Question

Find the grammar for the following language L(G) = {am bn | m>0,n > 0}​

Answer 1

Answer:

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, a2b, ab2, a2b2, ………}

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

Solution

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

Solution −

Since L(G) = {am bn | 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 })