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Consider the following Deterministic Finite Automaton $M$.

Let $S$ denote the set of eight bit strings whose second, third, sixth and seventh bits are 1. The number of strings in $S$ that are accepted by $M$ is

1. 0
2. 1
3. 2
4. 3
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There are only Two cases a) and b) in which we get $2^{nd},3^{rd},6^{th},\text{and},\;7^{th}$ bits as $1$

1.

2.

In case 1) we are not reaching at final state with $8$ bits

From Case 2)

The number of $8$ bit strings in S that are accepted by M =2

1. 01110111

2.01110110

Hence,(c)2 is the correct choice.

by Boss (41.2k points)
edited
0
Explanation is good. But please increase the size of the image because numbers are not properly visible.
0
I think Now images are ok :)
Just looking at the finite Automata, there will be two string:

01100111

01100110
by Boss (42.3k points)
+1 vote
by Loyal (9.9k points)