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ISRO | ISRO CS 2014 | Question 12

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How many states are there in a minimum state deterministic finite automation accepting the language L = {w | w {0,1} number of 0’s is divisible by 2 and number of 1’s is divisible by 5, respectively} ?
(A) 7
(B) 9
(C) 10
(D) 11


Answer: (C)

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 2 ( N(0) % 2 =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 , 00, 11111, 0011111, 0011111 , 1111100, 1010111 , 000011111,….and so on }

So, strings accepted by the automaton have to be of length 0, 2, 4, 5, 6, 7, 8, 9, 10, 11, 14, 12….and so on, i.e. equation for length will be 2x + 5y (where x,y>=0 )

Modulo 2 gives remainder as ( 0, 1 ) , and Modulo 5 gives remainder as ( 0, 1, 2, 3, 4 ). Hence 2 * 5 sates, i.e. there will be 10 states in the automaton to represent this.

Option (C) is correct.

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Last Updated : 11 May, 2018
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