= Solution
\b[<Breadth-first tree search> is complete and is still not bonus-optimal.] Fix any deterministic rule for choosing among equally short fox routes; the completeness argument again uses only the finite <branching factor>. The unobstructed corridor of part (g) has unique <shortest paths>, so its fox moves are exactly the same under the new strategy. The one-round bonus is $1$ and the two-round bonus is $2$, proving nonoptimality without any tie-breaking ambiguity.
If there is no route between fox and rabbit, the printed shortest-path rule needs a convention; letting the fox stay supplies a natural extension. The completeness argument is unchanged, and the counterexample above uses only connected squares. When the fox and rabbit initially belong to the same <connected component of a graph>, the <shortest-path pursuit distance bound> also makes every surviving branch finite. That additional property does not make the first escape maximize its bonus.
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