سوال ۸۱ کنکور ارشد مهندسی کامپیوتر ۱۴۰۰

Each stage is a T flip-flop, for which the next state is

$Q^{+}=Q\oplus T.$

Reading the wiring of the three flip-flops (with the single inverter on the path that drives the right-hand stage), the next-state equations of the circuit are

$Q_2^{+}=Q_2\oplus Q_0,\qquad Q_1^{+}=\overline{Q_2},\qquad Q_0^{+}=\overline{Q_2\oplus Q_0}.$

Equivalently the T-inputs are $T_2=Q_0$, $T_1=Q_1\oplus Q_2\oplus 1$, and $T_0=\overline{Q_2}$ (the lone inverter feeds $T_0$).

Now trace the state $Q_2Q_1Q_0$ starting from $111$:

So the generated sequence is

$111 \to 001 \to 110 \to 100 \to 100 \to \cdots,$

which settles into the fixed state $100$. This matches option 3 exactly.

The other options fail at the very first transition or in the second step: a deterministic FSM must map each state to a single successor, and only option 3's first three transitions are simultaneously consistent with one fixed next-state function.

Exam trick: For T flip-flops just remember $Q^{+}=Q\oplus T$: the bit toggles exactly when its T-input is $1$. Build the next-state table one column at a time and step through the states — you rarely need the full Boolean simplification, only the transition from each reached state.