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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.

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.

A hardware finite-state machine whose implemented transitions eventually leave every unused state encoding and enter the intended valid state set. If recovery must hold for arbitrary inputs, it is not enough that some input sequence can escape: there must be no indefinitely surviving unused-state path under any input sequence. Since the unused-state graph is finite, absence of a directed cycle among unused states gives a uniform finite recovery bound. A reset or initialization rule may still be required to give the valid state its intended meaning.