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Consider the regular grammar:

S → Xa | Ya
X → Za
Z → Sa | ϵ
Y → Wa
W → Sa

where S is the starting symbol, the set of terminals is {a} and the set of non-terminals is {S, W, X, Y, Z}.
We wish to construct a deterministic finite automaton (DFA) to recognize the same language. What is the minimum number of states required for the DFA?

1. 2
2. 3
3. 4
4. 5
edited | 680 views
it contains r.e as (aaa)*aa   for that we need  3 state which is min
Would the answer be wrong if I write L= aa(aaa)* ?

S → Xa | Ya
X → Za
Z → Sa | ϵ
Y → Wa
W → Sa

This is left linear grammar having language L. Convert it into right linear using following rule :

S → aX | aY
X → aZ
Z → aS | ϵ
Y → aW
W → aS

is right linear grammar having language LR.

having NFA

Having DFA for language LR

DFA for language L ( reversal)

L = { w : na(w)mod3 =2 , w belong to {a,b}* }   same as Omesh Pandita answered

having 3 states

answered by (53.1k points)
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R: - 1
Sir , here it was quite easy to see that Z will be final state but in general what is the way to determine which non-terminal should be made as final state, e.g. if the grammar given is :

A-->aB/bA/b

B-->aC/bB

C-->aA/bC/a

so here how to determine the final state .

A-->aB/bA/bF

B-->aC/bB

C-->aA/bC/aF

F-> ∊

(B) 3

The string generated by the language is the set of strings with $a$'s such that number of $a$ mod 3 is 2.

So the number of states required should be 3 to maintain the count of number of $a$'s mod 3.
answered by (2.7k points)
S -> Ya productions for Y and W are redundant? The same are done by X and Z productions rt?
Yes, they are redundant.

@Omesh, @Arjun: Set of alphabets contain only symbol  { a } .

I guess you have incorrectly included 'b' in the productions !!

If Z->b is replaced with Z->a , we need minimum 4 states as

L(G)= n(a) =3 *K   ,K>=1

Yes. It was a typo in question, now corrected. It is epsilon and not b.

@Omesh >> It should be (B) 3 and not (D) 3