MadeEasy Test Series 2019: Theory of computation - Grammer

Question

Let L be the language of all strings on [0,1] ending with 1.

Let X be the language generated by the grammar G.

$S \rightarrow 0S/1A/ \epsilon $

$A \rightarrow 1S/0A$

Then $L \cup X= $

• A. ∅
• B. ∑*
• C. L
• D. X

Ans given : B. ∑*

They said that X is a language which contains all strings that do not end with 1. But is it so?

Can’t we generate 11 from the grammar?

Please verify.

Comments

@Registered user 48

Oh it was right before only.. I suddenly got confused. See their solution..

They said that L(X) = not ending with 1. But 11 is a valid string.

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the DFA shown in the figure is of language generated by x   i.e. no of 1’s should be even

so the string 11 will be accepted by X

i hope your doubt got cleared

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now i have a question what about the string which have odd no of 1’s and end with a zero

how they are present in the language