Past exam of the computer science course of the University of Oxford / 2019 / Artificial intelligence / 2 / b

Use the knight-variable representation and build a placement one knight at a time. The initial state is the empty placement. An action places the next knight on an empty square attacking no previously placed knight; the search transition model appends that square. The goal test is that $k$ knights have been placed. Each action adds exactly one, so any solution has depth $k$.

Put $N=n^2$. At depth $j$, there are at most $N-j$ choices, hence branching factor at most $N$. A coarse search tree size bound is $1+N+\cdots +N^k$. Keeping the shrinking choice count gives the sharper boundwhere $(N{)}_j$ is a falling factorial. This counts ordered placements before attack pruning.

For a full expansion down to depth $k$, the number of depth-$k$ leaves is at most $(N{)}_k$, or approximately $n^{2k}$ when $k$ is small relative to $N$. If $I_k$ denotes the number of legal unlabelled $k$-knight placements, the exact number of legal depth-$k$ leaves is $k!I_k$. There can also be earlier dead-end leaves: a legal placement may leave no further nonattacking square. Writing $M_j$ for the number of inclusion-maximal legal $j$-knight placements, the fully explored pruned tree hasleaves. Its leaf count is still at most the unpruned $(N{)}_k$, since each early leaf replaces a disjoint nonempty subtree of the unpruned tree. Breadth-first search that stops at its first goal can construct less of the last layer; if no solution exists, it exhausts the reachable partial placements.

Identifying different insertion orders changes the representation. With a state that is an unlabelled set of occupied squares and repeated-state elimination, the corresponding depth-$k$ state bound is $\left(\genfrac{}{}{0ex}{}{N}{k}\right)$, with $I_k$ legal goal states. The factorial and binomial bounds refer to these different representations, not competing counts of the same tree.