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So , 1 is mandatory in Regular expression ; and both of above grammar allows strings without 1 to be genearated.

So ,  I expected None of above to be answer. What Am I missing here ?

Consider for (A)

Production = A0 -> A1

A1-> 0

in Theory of Computation by Junior (611 points)
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+1
I think you are right @vishal. it should be None of these as   0 or 00 or 0* is not in lang unless it conatins one 1 and both grammer are producing it. so it should be none of these.

2 Answers

0 votes
i think the grammar whould be

S -> 0S | A

A -> 1B

B -> 0B | 1B

B  -> epsiloon
by Active (1.8k points)
0 votes
Create NFA for the given expression and from that grammer is as follows:
replace A0 with S and A1 with A
S->0S | 1A
A -> 0A | 1A | epsilon
 the above grammer doesn't match with option A and B

In the options A and B they have production rule A0 -> A1 (As mentioned by @vishal8492) which will produce string "0" which is not accepted by the regular expression.

So the answer would D.
by (113 points)
edited by
0
Smallest string 1 cannot be generate by both so answer option d
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