 Solution

Yes. Greedy best-first search needs only the ordering of remaining-cost estimates. Store generated but unexpanded search nodes in a search frontier. To choose the next search node, compare candidates with $\kappa$ and retain the one it returns; a linear scan suffices, or a priority queue can use the comparison rule. Remove that search node, apply the goal test, and otherwise generate its children.

This implements best-first search ordered by $h^{\ast }$ without ever evaluating $h^{\ast }$. If stable tie handling is wanted, comparing both argument orders distinguishes equality, after which insertion order can break ties. No estimate of accumulated-plus-remaining cost is required for this greedy ordering.