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Use a three-state Moore machine for a rearmable one-cycle pulse generator. Call the states Ready, Pulse, and Wait. Their outputs are respectively $z=0,1,0$, and initialize in Ready. Labels below are the sampled input pair ab; because simultaneous ones are unspecified, choose reset priority for 11.

The complete chosen state transition table is

A sampled trigger 10 causes a pulse only in Ready. Leaving Pulse unconditionally on the next edge keeps its output high for exactly one clock period, even if a remains high. In Wait, further triggers are ignored until a sampled reset rearms the finite-state machine.

The crucial transition is Pulse --01--> Ready. If reset occurs in the cycle immediately following a trigger, this transition both ends the current pulse and rearms the circuit. The next trigger then produces a pulse after just one intervening low-output cycle. Sending every exit from Pulse to Wait would lose that reset and fail the alternate-cycle requirement. This behavior is explicitly visible in the last three pulses of the original timing diagram.^[ref]

For the sixteen sampled pairs in that diagram, numbering the first edge as 1, z is high just after edges $2,10,13,15$ and low after every other edge. Equivalently,The two low-output states cannot be merged: the same future trigger must produce a pulse from Ready and be ignored from Wait. A separate high-output state is needed to hold the one-cycle pulse, explaining the three-state design.

Use three D flip-flops whose stored bits $A,B,C$ represent $4A+2B+C$. Initialize them to $000$. The Moore machine's output is this stored binary representation, and its state transition table is:

Treat the two unused state encodings $110,111$ as don't-care conditions when minimizing the sum of products. With juxtaposition for AND, $+$ for OR and a bar for NOT, one minimal choice isFor $D_A$, the required ones at $011$ and $100$ cannot share one valid implicant: their differing bits would force inclusion of required zeros. Pair $011$ with the unused $111$ to get $BC$, and $100$ with unused $110$ to get $A\overline{C}$. Neither term can lose a Boolean literal without covering a required zero.

For $D_B$, state $001$ is isolated from every other required one and from the unused states by required zeros, so it forces the three-Boolean literal term $\overline{A}\overline{B}C$. State $010$ pairs with unused $110$ to give $B\overline{C}$. Finally $D_C=\overline{C}$ is a single Boolean literal. Thus the three expressions are minimal sum of products expressions under the stated unused-state convention. In fact these choices also recover from $110\to 111\to 100$, so initialization into an unused state eventually joins the intended modulo-six counter cycle.

Use three positive-edge-triggered D flip-flops, all driven by the same clock signal $CLK$. Write the current binary representation as $B2B1B0$, with $B0$ the least significant bit. The next state is the current value plus one when $A=1$ and minus one when $A=0$, in each case modulo eight. A bit toggles under exclusive or with a one.

For upward counting, $B1$ toggles when $B0=1$, while $B2$ toggles when $B1=B0=1$, the respective carry conditions. For downward counting the corresponding borrow conditions are $B0=0$ and $B1=B0=0$. Define the shared inverted direction $N=\neg A$ and the two direction-match signalsEach $T_i$ is one exactly when $Bi=A$: it equals $Bi$ for upward counting and $\neg Bi$ for downward counting. Thus the Boolean functions for the three D inputs areImplement this with two NOT gates for $N,D_0$, four two-input exclusive or gates for $T_0,T_1,D_1,D_2$, and one two-input AND gate. Connect $D_i$ to the D flip-flop whose output is $Bi$; feed the three outputs back into the indicated gates. All three flip-flops capture simultaneously, so each equation uses the old state.

For completeness, the state transition table is

Every three-bit state is valid. A reset may initialize the binary counter to zero if required; counting itself needs no extra state. The direction input must meet the setup time and hold time requirements at the active clock edge.

Use a four-state Moore machine. In state $A$ wait for an outer-sensor activation; $B$ combines the outer-only and both-sensors phases; $C$ records that the visitor has reached the inner-only phase; $I$ emits the one-cycle count pulse. Only $I$ has INC = 1.

The complete state transition table, with input order $x_1{x}_0$, isFor a normal entry the states are $A\to B\to B\to C\to I$; retreat after the first sensor instead gives $A\to B\to A$ with no pulse. Repeated samples at one sensor remain in the corresponding phase. State $I$ always exits on the next edge, so its output lasts exactly one clock period; an outer activation on that next edge goes directly to $B$, allowing the next visitor to start immediately.

This follows the question's single-visitor entry model and merges the two outer-active phases as its hint requests. Inputs are sampled synchronously; the table also specifies deterministic recovery for other sampled combinations. It is not a specification of a bidirectional occupancy counter for arbitrary overlapping entry and exit traffic.

Five distinct states require three D flip-flops, since $2^2<5\le 2^3$. Choose the state encoding $S_i$ equal to the three-bit binary expansion of $i$. Write the present bits as $A,B,C$, in decreasing significance. The state transition table isThe dashes are don't-care conditions for minimizing the logic. They do not prescribe what the actual circuit will do in those unused states; part (e) determines that from the chosen equations.