 Solution

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 $G_4$.