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The finite state machine described by the following state diagram with $A$ as starting state, where an arc label is and $x$ stands for $1$-bit input and $y$ stands for $2$-bit output

1. outputs the sum of the present and the previous bits of the input
2. outputs $01$ whenever the input sequence contains $11$
3. outputs $00$ whenever the input sequence contains $10$
4. none of the above
edited | 1.6k views
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I am unable to understand ( or rather Interpret ) the question, what is arc label? what is the significance of x and y here?? Appreciate your time, Thanks.
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Its a question of mealy machine. where x are input alphabets and y are output alphabets.

In the above question, 1/10 on an arc or arrow can be interpreted  as; 1=input alphabet, 10=output alphabet.

Option (B) and (C)  are clearly wrong . it says input $11$ then $o/p$ $01$ and $i/p \ 10$ then $o/p \ 00$ but here at single bit $o/p$ is $2$ bit sequence

Now, for option (A) we can trace out. Suppose string is $0111$.

at $A$---$0$---> $A$----$1$---> $B$--$1$-->$C$---$1$-->$C$

$O/P$        $00$          $01$       $10$         $10$

We can see here at $(A,0)$---> $(A,00)$ which sum of $0+0=00$, (previous $i/p$ bit $+$ present $i/p$ bit)

$(A,1)$--->$(B,01)$ which is sum of $0+1= 1=01$,

$(B.1)$--->$(C,10)$ which is sum of $1+1=$ (previous $i/p$ bit $+$ present $i/p$ bit)$=10$,

$(C,1)$---> $(C,10)$ which is sum of $1+1=10$

edited
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Then why output of 101 in not 10???

And as per statement A it says sum of present and "previous bits" (not bit).

So according to me answer should be D)
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it says sum of last two bits, not the sum whole string. So A should be correct.

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