Skip to content
Related Articles
GATE | GATE-CS-2007 | Question 29
• Last Updated : 24 Feb, 2014

A minimum state deterministic finite automaton accepting the language L={w | w ε {0,1} *, number of 0s and 1s in w are divisible by 3 and 5, respectively} has
(A) 15 states
(B) 11 states
(C) 10 states
(D) 9 states

Answer: (A)

Explanation:

Here a string w of 0’s and 1’s should have the property that, the no of 0’s in the string w should be divisible by 3 ( N(0) % 3 =0 ), and the number of 1’s the string w should be divisible by 5 (N(1) % 5 =0).

Having said that, the Language will contain the strings such as : { ε , 000, 11111, 00011111, 00111101 , 11111000, 10101011 , 00000011111,….and so on }

So, strings accepted by the automaton have to be of length 0, 3, 5, 8, 11, 13, 14, 16….and so on, i.e. equation for length will be 3x + 5y (where x,y>=0 )

Modulo 3 gives remainder as ( 0, 1, 2 ) , and Modulo 5 gives remainder as ( 0, 1, 2, 3, 4 ).  Hence 3 * 5 sates, i.e. there will be 15 states in the automaton to represent this.

Please comment below if you find anything wrong in the above post.

Quiz of this Question

Attention reader! Don’t stop learning now. Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up