= Solution
Evaluate the implemented equations from part (d), rather than the don't-care entries of the original table:
$$
\begin{array}{c|cc}
\text{unused state}&x=0&x=1\\\hline
101&000&010\\110&011&010\\111&000&110
\end{array}.
$$
States $101$ and $110$ enter a valid state after one edge. State $111$ does so after one edge for zero, or moves to $110$ and enters a valid state on the following edge for either next input. Thus \b[every unused state recovers within at most two clock edges], independently of the incoming bits. The machine cannot become trapped there. This <self-starting finite-state machine> property does not guarantee meaningful recognizer output during recovery; initialization to $S_0$ still establishes the intended stream history.
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