Past exam of the computer science course of the University of Oxford / 2021 / Artificial intelligence / 1 / c

Maintain a certified solution-cost ceiling and prune only search nodes whose lower bound strictly exceeds it. Initially let $B=\infty$, or set $B=u(r)$ at the root $r$ when that value is finite. Whenever a search node $n$ is generated, updateRun ordinary A-star search with $f(n)=g(n)+h(n)$ and a goal test when removing a search node from the search frontier. Discard a generated search node, or a queued search node encountered after a later bound improvement, ifThis is branch and bound using a lower bound $f$ and an upper bound $B$. A finite $u(n)$ certifies that some goal is reachable from $n$ within that remaining-cost ceiling; for a dead end, $h^{\ast }(n)=\infty$ forces $u(n)=\infty$ too. Every finite quantity $g(n)+u(n)$ is consequently at least the global optimum $C^{\ast }$.

On an optimal solution path, admissible heuristic estimates ensure $f(n)\le C^{\ast }\le B$. Hence no search node of that path is pruned. The finite-branching factor and positive step-cost hypotheses from (a) still ensure completeness, and the usual A-star search argument makes the first selected goal optimal. A smaller bound can remove branches earlier; it need not improve every instance.

The strict inequality matters: $u$ is only a numerical bound and need not provide an actual stored solution path. If $B=C^{\ast }$, pruning at $f(n)\ge B$ could remove every optimal continuation. Nor may the algorithm return a path merely because its lower bound meets $B$ when no witnessing path has been constructed. With an actual incumbent solution, the familiar equality-pruning variant can instead retain and return that incumbent.