Consider the sequential circuit shown in the figure, where both flip-flops used are positive edge-triggered D flip-flops. The number of states in the state transition diagram of this circuit that have a transition back to the same state on some value of “in” is ________.

Let the state of the circuit be represented as $(Q_A, Q_B)$, where:

• $Q_A$ is the state/output of the first (left) flip-flop.

• $Q_B$ is the state/output of the second (right) flip-flop.

From the circuit diagram, the inputs to the D flip-flops are:

• $D_A = \text{in}$

• $D_B = Q_A$

At the next active clock edge, the next state $(Q_A^+, Q_B^+)$ is given by: $$\begin{aligned} Q_A^+ &= D_A = \text{in} Q_B^+ &= D_B = Q_A \end{aligned}$$

For a state $(Q_A, Q_B)$ to transition back to itself (i.e., a self-loop), we require:

1. $Q_A^+ = Q_A \implies \text{in} = Q_A$

2. $Q_B^+ = Q_B \implies Q_A = Q_B$

Thus, a self-loop exists if and only if $Q_A = Q_B$ and the input applied is $\text{in} = Q_A$:

• For state 00 ($Q_A = 0, Q_B = 0$): With $\text{in} = 0$, next state is $(0, 0) = 00$ (Self-loop exists).

• For state 11 ($Q_A = 1, Q_B = 1$): With $\text{in} = 1$, next state is $(1, 1) = 11$ (Self-loop exists).

• For state 01 ($Q_A = 0, Q_B = 1$): Next state is $(\text{in}, 0)$, which can never be $01$.

• For state 10 ($Q_A = 1, Q_B = 0$): Next state is $(\text{in}, 1)$, which can never be $10$.

Therefore, exactly 2 states ($00$ and $11$) have a transition back to themselves.