Design DFA accepting binary strings divisible by a number 'n'

Question

I need to learn how to design a DFA such that given any number \'n\', it accepts binary strings {0, 1} whose decimal equivalent number is divisible by \'n\'.

There w

Answer 1

You can build DFA using simple modular arithmetics. We can interpret w which is a string of k-ary numbers using a following rule

V[0] = 0
V[i] = (S[i-1] * k) + to_number(str[i])

V[|w|] is a number that w is representing. If modify this rule to find w mod N, the rule becomes this.

V[0] = 0
V[i] = ((S[i-1] * k) + to_number(str[i])) mod N

and each V[i] is one of a number from 0 to N-1, which corresponds to each state in DFA. We can use this as the state transition.

See an example.

k = 2, N = 5

| V | (V*2 + 0) mod 5     | (V*2 + 1) mod 5     |
+---+---------------------+---------------------+
| 0 | (0*2 + 0) mod 5 = 0 | (0*2 + 1) mod 5 = 1 |
| 1 | (1*2 + 0) mod 5 = 2 | (1*2 + 1) mod 5 = 3 |
| 2 | (2*2 + 0) mod 5 = 4 | (2*2 + 1) mod 5 = 0 |
| 3 | (3*2 + 0) mod 5 = 1 | (3*2 + 1) mod 5 = 2 |
| 4 | (4*2 + 0) mod 5 = 3 | (4*2 + 1) mod 5 = 4 |

k = 3, N = 5

| V | 0 | 1 | 2 |
+---+---+---+---+
| 0 | 0 | 1 | 2 |
| 1 | 3 | 4 | 0 |
| 2 | 1 | 2 | 3 |
| 3 | 4 | 0 | 1 |
| 4 | 2 | 3 | 4 |

Now you can see a very simple pattern. You can actually build a DFA transition just write repeating numbers from left to right, from top to bottom, from 0 to N-1.