Dfa and counter difference?

Question

#digitalelectronics and #deterministicFA
please clarify this..... 
In constructing dfa ,say (num) divisible by 8 or just mod 8 we need minimum of 4 states , 

1. So for constructing mod 8 counter we need minimum of states 4(we have to count only 8 not by shifting so i decided to use 4 states rather 8),

 to represent 2 state - 1 ff needed so for representing 4 states(mod 8 counter) - 2 ff needed .....whether using 2ff can we construct mod 8 counter? 

2. I tried by constructing states representing 0 as 00 ; 1,3,5,7 as 01; 2,6 as 10 and 4 as 11 (see diag below) and I constructed the mod 8 counter using present state and next state (i took this as random not inorder but according to diag), making last transition as 11 to 00; remaining can be anything ,though we have 2 variable and 8 transitions so I used normal algebraic reduction for constructing combi.circuit.... whether my approach is correct ?refer diag 2

3. If there are any other suggestions.. welcomed

Answer 1

actually counter and num mod 8 dfa is 2 different thing. mod 8 counter is a ckt which generate a sequence of 8 different state without taking any input wherease dfa goes to next state by consuming input . mod 8 counter needs only 3 states so 3 fflops. ur method doesnt seems correct . check it https://www.cs.tau.ac.il/~nin/Courses/mivne98/counters/