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Design a deterministic finite state automaton (using minimum number of states) that recognizes the following language:

$L=\{w \in \{0, 1\}^* \mid w$  interpreted as binary number (ignoring the leading zeros) is divisible by five $\}.$

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I hav forgot from where i read it ...

The minimum DFA that accept all base "M" system which are divisible by "N" contains N state ....

Correct me if i am wrong .....

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yes.|E|=k mod n  or |E| mod n=k where n will be the no of states and k decides which no of state among n  is final  state.
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Language that accepts binary strings whose decimal equivalent is divisible by n.

minimal dfa for this language does not always have n states..

https://gateoverflow.in/263964/gatebook-2019-toc1-7

For binary strings--

$\bullet$  if n is odd then we need n states in mDFA.

$\bullet$   if n is a power of 2 (let's say $2^k$ ) then we need k+1 states.

Suppose we have decimal no $3$ after that we get a symbol $2$ . it becomes $32$ as  $3 x 10 +2 = 32$

In binary if we have $10$ ( i.e $2$ in decimal say old no) and after that we get symbol $1$ it become $101$( i.e $5$ in decimal say new no )

$2$ (old no.) $\times 2$( base)  $+1$( input symbol) $= 5$ (new no.)

Now in the given problem, binary no is divisible by $5$ , i.e $0,101,1010,1111.....$

We need $5$ states in DFA , $0,1,2,3$ and $4$ .Each state represent remainder that comes when we divide no by $5$.

Input symbol $=\{0,1\}$

We get the transition:

[Old state $\times$ base $+$ input symbol] mod $5 =$ New state    , where base is $2$

$[0 \times 2 + 0]mod \ 5 = 0$           $[1 \times 2 + 0]mod \ 5 = 2$         $[2 \times 2 + 0]mod \ 5 = 4$

$[0 \times 2 + 1]mod \ 5 = 1$           $[1 \times 2 + 1]mod \ 5 = 3$         $[2 \times 2 + 1]mod \ 5 = 0$

$[3 \times 2 + 0]mod \ 5 = 1$           $[4 \times 2 + 0]mod \ 5 = 3$

$[3 \times 2 + 1]mod \ 5 = 2$           $[4 \times 2 + 1]mod \ 5 = 4$ by Veteran (56.6k points)
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Then for designing a DFA divisible by n, we require minimum of n states?
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Not in case of even n.
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why not in case of even n??
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some of them get minimized
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@Arjun sir

Ignoring the leading zeros is given in question, So shouldn't it be 6 state from state 0(not final state),1,2,3,4,5(final state).
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then what is the given DFA doing? Or did you interpret the leading zeroes as going to a dead state?
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No, sir

I mean leading zero will loop in the first state...i,e 0th state.

Do i interpret it wrong?
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that's the given DFA rt?
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sir can u tell me.How I can attach a image here?

I am unable to that.
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I.... +1
@ ana , any binary number starting with more than 4 1's will not be accepted by above DFA, say 1111101, i,e 125
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missed these two transitions:

state 4 on 0 goes to state 3

state 5 on 1 goes to state 1
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Then state 5 and state 0 both will become equivalent. And I hope both states 5,0 are finals too
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no only state 5 is the final state.

ignoring the leading zeros, is given in the question....so state 0 is not the final state.

That's what I was saying so long....

M i wrong????
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actually 0 is also divisible by 5 (or by any num). so state 0 should be final too
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yes,  for designing a DFA divisible by n, then total n states required. each state represent remainder ( 0 to n-1 remainder ) that comes when we divide no by n.
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Since initial state is the final state here, so this will accept ⋵  (epsilon) too.. but ⋵
is not a part of the language L.

Ans...   here q0 is the initial state and final state

by Junior (979 points)
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will help in better understanding above described method...........

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