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Consider the context-free grammar G below
S → aSb | X
X → aX | Xb | a | b,
where S and X are non-terminals, and a and b are terminal symbols. The starting non-terminal is S.
Which one of the following statements is CORRECT?
- The language generated by G is (a + b)*
- The language generated by G is a*(a + b)b*
- The language generated by G is a*b*(a + b)
- The language generated by G is not a regular language
Correct answer: The language generated by G is a*(a + b)b*
Solution
S -> aSb gives a^n ... b^n wrapping X, and X -> aX | Xb | a | b generates a*(a+b)b*; the a^n/b^n wrappers are absorbed, so L(G) = a*(a+b)b*, which IS regular. Hence the correct statement is option index 1, not 'not regular'.
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