## UGC NET-July-2018-II-32

The finite state machine given in the figure below recognizes:

1. Any string of odd number of a's
2. Any string of odd number of b's
3. Any string of even number of a's and odd number of b's
4. Any string of odd number of a's and odd number of b's

Here ${\mathbf{r}}_{\mathbf{a}}$ identified remainder of a’s and ${\mathbf{r}}_{\mathbf{b}}$, identified remainder of b’s

In the above machine there are four grades

1. Represent even number of  b’s
2. Represent odd number of  b’s
3. Represent even number of a’s
4. Represent odd number of a’s

From the diagram, it is clearly showing that the given final state is an intersection (common state) of odd number of a’s and odd number of b’s.

So, Option D is correct.