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The following finite state machine accepts all those binary strings in which the number of $1$’s and $0$’s are respectively: 1. divisible by $3$ and $2$

2. odd and even

3. even and odd

4. divisible by $2$ and $3$

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Option (C) and option (D) are cancelled out clearly because with $3$ $1$s we can reach the final state. There is an string where we can reach the final state by $6$ $1$'s now $6$ is nt odd but it is divisivble by $3$. Hence, option (A) is correct.

by Boss (14.4k points)
edited
0
why not D?
0

option d is 1 divisible by 2 and 0 is divisible  3 which is wrong.

because

No. of 0's % 2 = 0

no. of 1's % 3 =0

ε (Empty string) is accepted here. So option (B) & (C) are canclelled out !

No of 0's % 2 = 0 . So answer => A

by Boss (41.9k points)
No. of 1s is represented horizontally and 0s vertically.

There are 3 states each for number of 1s. So, it is understood that states are for 0, 1 and 2 as remainders when no. of 1s is divided by 3.

Similarly, there are 2 states each for number of 0s. So, it is understood that states are for 0 and 1 as remainders when no. of 0s is divided by 2.

Hence option A is correct.
by (369 points)
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What is difference between option A and D?
+7
in option (A) number of 1's divisible by 3 and number of 0's divisible  by 2 and

option (D) number of 1's divisible by 2 and number of 0's divisible  by 3