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A minimum state deterministic finite automaton accepting the language

$L=\{w\mid w \in \{0, 1\}^*,$ number of $0$s and $1$s in $w$ are divisible by $3$ and $5$, respectively $\}$   has

  1. $15$ states
  2. $11$ states
  3. $10$ states
  4. $9$ states
asked in Theory of Computation by Veteran (69k points)
edited by | 931 views

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1 Answer

+16 votes
Best answer

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

answered by Veteran (346k points)
edited by
@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?
for divisble by 2 and 4 , cross product give 8 but it can be made by only 4 states ..(lcm of 2,4)
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?
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 

https://gateoverflow.in/723/gate2001-2-5

 



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