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Answer will be (**A**) $15$ states.

We need a separate state for #$0 = 0, 1, 2$ and #$1 = 0, 1, 2, 3, 4$ giving total minimum number of states $= 3 * 5 = 15. $

This is a direct consequence of Myhill-Nerode theorem.

http://courses.cs.washington.edu/courses/cse322/05wi/handouts/MyhillNerode.pdf

0

@Arjun Sir, So this is a cross product between 2 DFAs "number of 0 ...." and "number of 1 ...."

My question is..... Do we always get minimal DFA after cross product or we need to check if it can be minimized?

My question is..... Do we always get minimal DFA after cross product or we need to check if it can be minimized?

0

Yes,

-> divisible by 2 and 4 is clear and direct.. 4 is direct multiple.

-> divisible by 3 and 4 is direct.... they are relatively prime to each other

But if question will be like divisible by 12 and 8... Whats the answer here?

Do we need to check if it can be minimized or answer is 12*8?

-> divisible by 2 and 4 is clear and direct.. 4 is direct multiple.

-> divisible by 3 and 4 is direct.... they are relatively prime to each other

But if question will be like divisible by 12 and 8... Whats the answer here?

Do we need to check if it can be minimized or answer is 12*8?

0

@sid1221 if,L={w∣w∈{0,1}^{∗}, number of 0s and 1s in w are **divisible by 2 and 4**, respectively} Can you show how can we create the finite state machine accepting it in 4 states as per the method of lcm(2,4) ??

0

its lcm only ...24 , not 96 , thats why i wrote lcm(2,4 )

but careful its length of w , which is divsible by x and y number ... for symbol its not case

Can you please explain it in more detail. I am not getting what you are telling about the symbol.

0

I don't think this lcm is going to work here. question is asking about number of a's divisible by 3 and number of b's divisible by 5. here number of states according to myhill nerode will be number of remainders for 3 * number of remainders of 5. Similarly if it would have been 2,4 it has to be 8. for 2 {0,1} * for 4 {0,1,2,3} and our final state is {0,0} and total states = 8.

@Ahwan @Arjun sir please help if approach is correct.

@Ahwan @Arjun sir please help if approach is correct.

+3

for divisble by 2 and 4 , cross product give 8 but it can be made by only 4 states ..(lcm of 2,4)

States in DFA for counting a,b is nothing but all possible combinations of their remainders. I don't think LCM has to do anything with this.

@Shaik Masthan @Ayush Upadhyaya plz correct if wrong.

+1 vote

Given that number of 0's and 1’s are divisible by 3 and 5, it means that the number of 0’s and 1’s must be divisible by 15.

As the LCM of 3 and 5 is 15, so number of 0’s and 1’s are divisible by 3 and 5 is only possible if of 0’s and 1’s are divisible by 15. Also modulo 3 will leave a remainder of 0,1,2 (3 states required) and modulo 5 will leave remainder of 0,1,2,3,4 (5 states required) , so product automata will require (3 × 5=15 states).

As the LCM of 3 and 5 is 15, so number of 0’s and 1’s are divisible by 3 and 5 is only possible if of 0’s and 1’s are divisible by 15. Also modulo 3 will leave a remainder of 0,1,2 (3 states required) and modulo 5 will leave remainder of 0,1,2,3,4 (5 states required) , so product automata will require (3 × 5=15 states).

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