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

None of the three variants is an optimal search algorithm in general. The following single search tree supplies counterexamples for all three. Its heuristic function is nonnegative, admissible and even a consistent heuristic.

The least-cost goal is $G_4$, reached at cost $1+3=4$; the other goal $G_5$ costs $5$. Assign $h(S)=0$, $h(P)=3$, and zero to both goals. For the first variant, expanding $S$ generates $G_5$ and immediately returns it, before $P$ can be expanded. A goal test at generation therefore returns cost $5$.

For priority $g(n)+h(n^{\prime })$, the root's children have priorities $1$ for $P$ and $5$ for $G_5$. Expanding $P$ generates $G_4$ with priority $4+3=7$. The algorithm next selects $G_5$, again returning cost $5$.

For priority $g(n^{\prime })+h(n)$, the root's children instead have priorities $3$ for $P$ and $0$ for $G_5$. The suboptimal goal is selected immediately. The example uses no tie-breaking assumption. In contrast, ordinary A-star search with its goal test on selection compares the genuine estimated total costs and returns $G_4$.