theory of computation

Question

The language generated by the below grammar is

S --> aSS / b

Answer 1

a w* bb where   wϵ (a+b)* and number of b's= number of a's+1  is the language produced by the grammer

It means any string having prefix a and suffix bb ,

Comments

what about single 'b'

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'b'  is also derived from given grammar which can't be derived from your regular expression.

Answer 2

S --> aSS / b
let us suppose the string is "aaabbbb"
S-->aSS
S-->aaSSS
S-->aaaSSSS
S-->aaabbbb

The language generated by this grammer is a^nb^m such that n<m

Comments

ok..what will be the correct language here? any idea..

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@shubham not getting exact language

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this answer is best answer among these answer

Answer 3

S --> aSS / b

The language generated by this grammer is

$ a^{*}(ab)^{*}a^{n}b^{m}$

where n<m and n>=0 and m>=1.

Comments

abab is also accepted by your regular expression which can't be derived from given grammer.

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m is greater than n , an extra b will always come at end in comparison with second last a.

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aaaababbbbb this string is also derived from given grammar which can't be derived from your regular expression.

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It can be derived now. if Any other please correct me.