Theory of Computation | Relationship between grammar and language

A grammar is a set of production rules which are used to generate strings of a language. In this article, we have discussed how to find the language generated by a grammar and vice versa as well.

Language generated by a grammar –

Given a grammar G, its corresponding language L(G) represents the set of all strings generated from G. Consider the following grammar,

G: S-> aSb|ε

In this grammar, using S-> ε, we can generate ε. Therefore, ε is part of L(G). Similarly, using S=>aSb=>ab, ab is generated. Similarly, aabb can also be generated.
Therefore,

L(G) = {a^nb^n, n>=0}

In language L(G) discussed above, the condition n = 0 is taken to accept ε.
Key Points –

  • For a given grammar G, its corresponding language L(G) is unique.
  • The language L(G) corresponding to grammar G must contain all strings which can be generated from G.
  • The language L(G) corresponding to grammar G must not contain any string which can not be generated from G.

Let us discuss questions based on this:



Que-1. Consider the grammar: (GATE-CS-2009)

S -> aSa|bSb|a|b 

The language generated by the above grammar over the alphabet {a,b} is the set of:
(A) All palindromes
(B) All odd length palindromes.
(C) Strings that begin and end with the same symbol
(D) All even length palindromes

Solution: Using S->a and S->b, a and b can be generated. Similarly using S=>aSa=>aba, aba can be generated. Other strings which can be generated from grammar are: a, b, aba, bab, aaa, bbb, ababa, …
Therefore, option (B) is correct.

Que-2. Consider the following context-free grammars: (GATE-CS-2016)

Which one of the following pairs of languages is generated by G1 and G2, respectively?

Solution: Consider the grammar G1:
Using S=>B=>b, b can be generated.
Using S=>B=>bB, bb can be generated.
Using S=>aS=>aB=>ab can be generated.
Using S=>aS=>aB=>abB=>abb can be generated.
As we can see, number of a’s can be zero or more but number of b is always greater than zero.
Therefore,

L(G1) = {a^mb^n| m>=0 and n>0}

Consider the grammar G2:
Using S=>aA=>a, a can be generated.
Using S=>bB=>b, b can be generated.
Using S=>aA=>aaA=>aa can be generated.
Using S=>bB=>bbB=>bb can be generated.
Using S=>aA=>aB=>abB=>abb can be generated.

As we can see, either a or b must be greater than 0.
Therefore,

L(G2) = {a^mb^n| m>0 or n>0}

Grammar generating a given language –

Given a language L(G), its corresponding grammar G represents the production rules which produces L(G). Consider the language L(G):

L(G) = {a^nb^n, n>=0}

The language L(G) is set of strings ε, ab, aabb, aaabbb….
For ε string in L(G), the production rule can be S->ε.
For other strings in L(G), the production rule can be S->aSb| ε.
Therefore, grammar G corresponding to L(G) is:

S->aSb| ε 

Key Points –

  • For a given language L(G), there can be more than one grammar which can produce L(G).
  • The grammar G corresponding to language L(G) must generate all possible strings of L(G).
  • The grammar G corresponding to language L(G) must not generate any string which is not part of L(G).

Let us discuss questions based on this:

Que-3. Which one of the following grammar generates the language L = {a^ib^j | i≠j}? (GATE-CS-2006)

Solution: The given language L contains the strings :

{a, b, aa, bb, aaa, bbb, aab, abb…}

It means either the string must contain one or more number of a OR one or more number of b OR a followed by b having unequal number of a and b.

If we consider grammar in option (A), it can generate ab as:

S=>AC=>aAC=>aC=>ab

However, ab can’t be generated by language L. Therefore, grammar in option (A) is not correct.

Similarly, grammar in option (B) can generate ab as:

S=>aS=>ab

However, ab can’t be generated by language L. Therefore, grammar in option (B) is not correct.

Similarly, grammar in option (C) can generate ab as:

S=>AC=>C=>aCb=>ab

However, ab can’t be generated by language L. Therefore, grammar in option (C) is not correct.
Therefore, using method of elimination, option (D) is correct.



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