GATE | GATE-CS-2007 | Question 29
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.
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Last Updated :
28 Jun, 2021
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