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In previous two year's a question has been asked for finding number of flip flop's for counting sequence $\color{navy}{0-1-0-2-0-3}$ in 2016 and $\color{navy}{0-0-1-1-2-2-3-3-0-0}$ in 2015. But still, the approach discussed in these questions hasn't arrived at Common answer (or) solution.

For $0-1-0-2-0-3$, there are $6$ different states, and thus it acts as a $mod-6$, counter So, $3$ FF's are suffice. In another approach it is said that $2$ bits are used to distinguish three $0's$ and $2$ for $1,2,3$. So number of FF's required $= 2+ 2 = 4$.

Similar is the case in $0-0-1-1-2-2-3-3-0-0$, one approach says setting clock such that output of FF will be sampled at twice the input frequency makes me think $2$ FF's would suffice. But on the other-hand using $2$ FF's for $1,2,3,4$ and $1$ FF for identifying which $1$ out of two $1's$ makes me think $3$ is also right.

So, I want to be sure in answering these type of questions as they have been asked in recent past. Please also comment about minimum number of $JK$ FF's required for following counting sequence.

$1). 0-0-0-1-1-2-2-3-3-...$
$2). 0-1-0-2-0-2-0-3-0-2-...$
$3). 1-2-3-0-0-1-0-2-2-...$

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1 Answer

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5 votes

For GATE 2015 question

The counting sequence is $0\rightarrow 0\rightarrow 1\rightarrow 1\rightarrow 2\rightarrow 2\rightarrow 3\rightarrow 3\rightarrow 0\rightarrow 0$...................................

The distinct states are 0,1,2,3 and for distinguishing between then 2 bits are sufficient and neccessary.

0 $\rightarrow$ 00; 1 $\rightarrow$ 01; 2 $\rightarrow$ 10; 3 $\rightarrow$ 11

But here repeating sequence are also encountered.

Like 0 is represented by 00 and differentiating between two 0's, a bit is sufficient and neccessary i.e 0 and 1

Hence, 2FF + 1FF = 3FF is minimum which we need here. (Gate key says 3 and not 2)

By same approach for 2016 question, we need 4 FF's.

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