GATE | GATE-CS-2009 | Question 27
Given the following state table of an FSM with two states A and B, one input and one output:
Present State A |
Present State B |
Input |
Next State A |
Next State B |
Output |
0
|
0
|
0
|
0
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
1
|
1
|
0
|
1
|
0
|
0
|
0
|
0
|
1
|
0
|
1
|
0
|
0
|
1
|
1
|
0
|
0
|
1
|
1
|
0
|
1
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
1
|
If the initial state is A=0, B=0, what is the minimum length of an input string which will take the machine to the state A=0, B=1 with Output = 1?
(A)
3
(B)
4
(C)
5
(D)
6
Answer: (A)
Explanation:
//(0, 0) –1–> (0, 1) –0–>(1, 0) –1–> (0, 1) and output 1
According to the question we have to reach the states A=0, B=1 and output=1.
This state is shown by green color. Thus to reach final states as A=0, B=1 and output=1 we have to reach previous states of A=1, B=0.
Since initial states are A=0,B=0 (red); we provide input=1 (to reach A=0,B=1) Now this will give present states as A=0, B=1 and output=0.
Now we provide (blue) input=0 (to reach A=1,B=0) with present states as A=0, B=1. The present states will become A=1, B=0 and output=0.
This is what is required. On providing input=1 we get final states as A=0, B=1 and output=1.
Hence an input string of 3 i.e. 101 leads to the desired output and states.
Quiz of this Question Please comment below if you find anything wrong in the above post
Last Updated :
11 Oct, 2021
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