# Regular Expression to CFG

1 vote
1k views

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

retagged
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.

i think the grammar whould be

S -> 0S | A

A -> 1B

B -> 0B | 1B

B  -> epsiloon
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.

edited
0
Smallest string 1 cannot be generate by both so answer option d

## Related questions

1 vote
1
396 views
1 vote
2
497 views
how $a^nb^nc^n$ n>=1 is not CFL....??
3
307 views
S->AbaC A->BC B->b/epsilon C->D/epsilon D->d I want to know that will A contain epsilon as B and C both are null variables???(In Elimination of epsilon-production)
1 vote