The A∗ algorithm is optimal when,

Answer: It always finds the solution with the lowest total cost if the heuristic ' h ' is admissible.

Why this is correct:

• Admissible heuristic: it never overestimates the true minimal cost to reach a goal from any node. In symbols, h(n) ≤ true cost from n to nearest goal.

• A* uses f(n) = g(n) + h(n) to prioritize nodes. With an admissible h, A* will not discard a path that could lead to an optimal goal, so the first goal expanded (under standard A* conditions) has optimal cost.

• Implementation note: for tree-search A* admissibility is sufficient for optimality. For graph-search, having a consistent (monotone) heuristic (h(n) ≤ c(n, n') + h(n')) simplifies guarantees because nodes do not need to be reopened; admissibility is still required for optimality.

Why the other statements are wrong:

1. Statements claiming A* finds the highest-cost solution are incorrect because A* is designed to find minimal-cost paths, not maximal ones.

2. Claims that A* is guaranteed optimal when the heuristic is not admissible are false: if the heuristic overestimates, A* can be misled and may fail to find the lowest-cost solution.

Conclusion: With an admissible heuristic A* is guaranteed to find an optimal (lowest-cost) solution; non-admissible heuristics remove that guarantee, and statements about finding highest-cost solutions are incorrect.