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Consider the following deterministic finite state automata M:
Let S denotes the set of seven bit binary strings in which the first, the fourth and the last bits are 1. The number of strings in S that are accepted by M is
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- 1
- 5
- 7
- 8
- 1
Correct Option: C
The strings accepted by the given automata are of type.
Option
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 1 | ― | ― | 1 | ― | ― | 1 |
The four blank spaces can have a probability of having 0 or 1, so total 2(pow,4)= 16 strings are possible, but the given automata does not accept all of those.
1. 1 1 1 1 0 0 1
2. 1 1 0 1 0 0 1
3. 1 0 1 1 0 0 1
4. 1 0 0 1 0 0 1
5. 1 0 0 1 0 0 1
6. 1 0 0 1 1 0 1
7. 1 0 0 1 1 1 1
Hence (c) is correct option
