Question 14164 – Compiler-Design
May 17, 2024Question 7934 – Theory-of-Computation
May 17, 2024Question 7933 – Theory-of-Computation
Consider the context-free grammars over the alphabet {a, b, c} given below. S and T are non-terminals.
G1: S → aSb|T, T → cT|ϵ
G2: S → bSa|T, T → cT|ϵ
The language L(G1) ∩ L(G2) is
Correct Answer: B
Question 17 Explanation:
Strings generated by G1:
{ϵ, c, cc, ccc, … ab, aabb, aaabbb….acb, accb… aacbb, aaccbb, …}
Strings generated by G2:
{ϵ, c, cc, ccc, … ba, bbaa, bbbaaa….bca, bcca… bbcaa, bbccaa, …}
The strings common in L (G1) and L (G2) are:
{ϵ, c, cc, ccc…}
So, L (G1) ∩ L (G2) can be represented by regular expression: c*
Hence the language L (G1) ∩ L (G2) is “Not finite but regular”.
{ϵ, c, cc, ccc, … ab, aabb, aaabbb….acb, accb… aacbb, aaccbb, …}
Strings generated by G2:
{ϵ, c, cc, ccc, … ba, bbaa, bbbaaa….bca, bcca… bbcaa, bbccaa, …}
The strings common in L (G1) and L (G2) are:
{ϵ, c, cc, ccc…}
So, L (G1) ∩ L (G2) can be represented by regular expression: c*
Hence the language L (G1) ∩ L (G2) is “Not finite but regular”.
Finite
Not finite but regular
Context-Free but not regular
Recursive but not context-free