UGC NET CSE | June 2012 | Part 3 | Question: 32

Question

The equivalent grammar corresponding to the grammar $G:S \rightarrow aA, A \rightarrow BB, B \rightarrow aBb \mid \varepsilon$ is

• A. $S \rightarrow aA, A \rightarrow BB, B \rightarrow aBb$
• B. $S \rightarrow a \mid aA, A \rightarrow BB, B \rightarrow aBb \mid ab$
• C. $S \rightarrow a \mid aA, A \rightarrow BB \mid B, B \rightarrow aBb$
• D. $S \rightarrow a \mid aA, A \rightarrow BB \mid B , B \rightarrow aBb \mid ab$

Comments

shekhar chauhan Please check the Question .now i think everything is ok. 

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Try to derive string "aab" from options only option D can

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@ srestha

In option B)    A->BB

Option D)    A->BB | B

 

Rest everything is same

Answer 1

G:S$\rightarrow$aA

A $\rightarrow$ BB

B$\rightarrow$ aBb /∈

(A) Here S can generate a (s->aA->aBB->aB->a) which can not generate by A.So we can eliminate this option.

(B)In this option we can not able to generate aab which can be generate by given grammar.So we can eliminate this option.
 

(C)In this option we can generate  a. we can't generate anything other than a because there is no terminal present to stop B.So we can eliminate this option.

Answer should be (D)G:S$\rightarrow$aA , A $\rightarrow$ BB/B , B$\rightarrow$ aBb /ab.

Comments

sir please explain it more from basic.i didn't.t get it

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We can yry to genrate string from given grammer.
Example given grammer is genrate string 'a'.
But in option a . String 'a' is not genrayed so we can ellimente option a.
Simmilly we can do

Answer 2

If we remove ∊ production from the grammer

Answer should be

$S \rightarrow a \mid Aa, A \rightarrow BB \mid B, B \rightarrow aBb \mid ab$

Hence none of he options matching here