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GATE CSE 2021 Set 2 Q38: Two-state Mealy machine that clears the first 1 of each run

Question

Suppose we want to design a synchronous circuit that processes a string of 0's and 1's. Given a string, it produces another string by replacing the first 1 in any subsequence of consecutive 1's by a 0. Consider the following example.

Input sequence: 00100011000011100

Output sequence: 00000001000001100

A Mealy Machine is a state machine where both the next state and the output are functions of the present state and the current input.

The above mentioned circuit can be designed as a two-state Mealy machine. The states in the Mealy machine can be represented using Boolean values 0 and 1. We denote the current state, the next state, the next incoming bit, and the output bit of the Mealy machine by the variables ss, tt, bb and yy respectively.

Assume the initial state of the Mealy machine is 0.

What are the Boolean expressions corresponding to tt and yy in terms of ss and bb?

Try it: multiple choice: exactly one correct option

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Wrong answer costs 2/3 mark in GATE.
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Answer: B (official GATE 2021 final answer key)

Concept used: The state only needs to remember whether the previous input bit was 1.

First step: Let the state be the previous bit, so t=bt = b; the output is 1 only when the current bit is 1 and the previous bit was also 1.

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Source: official GATE 2021 Set 2 CS question paper and final answer key, organised by IIT Bombay. Question paper · Answer key