Take as heuristic the numbers in brackets. What would the solution path be if Hill Climbing approach was used?

Question

BlackTom AI · Accepted Answer

Hill climbing selects the neighbor with the best (here, lowest) heuristic value and moves greedily until no neighbor improves the heuristic. 1) Relevant knowledge points and concepts - Hill Climbing is a greedy local search: from the current node it examines neighbors and moves to the neighbor with the best heuristic improvement (here lower numbers are better). - It stops when none of the current node’s neighbors has a better (lower) heuristic — a local minimum. 2) Step-by-step reasoning - Start at A with h(A) = 10. Neighbors: B (h=5) and C (h=8). - Compare: 5 < 8, so choose B (best immediate improvement). - At B (h=5). Neighbors: D (h=3) and E (h=4). - Compare: 3 < 4, so choose D. - At D (h=3). Neighbors: G (h=6) and H (h=7). - Both neighbors have higher heuristic values (6 and 7 > 3), so no neighbor improves on D. - Hill Climbing terminates at D (a local minimum) because no adjacent node has a lower heuristic. 3) Intermediate comparisons (calculations shown) - From A: choose min{h(B)=5, h(C)=8} → 5 → B. - From B: choose min{h(D)=3, h(E)=4} → 3 → D. - From D: min{h(G)=6, h(H)=7} = 6 > h(D)=3 → stop. 4) Why the correct answer is correct - The greedy selection at each step leads A → B → D, and halts at D because no neighbor improves the heuristic. This exactly matches A – B – D. 5) Why other options are incorrect - Option A lists a long path that ignores greedy choices (A would pick B not C, and stops earlier). - Option C (A–C–F–J): A would not choose C because B has better heuristic (5 < 8). - Option D (A–B–E–I): At B hill climbing prefers D (3) over E (4), so it will not go to E→I. 6) Conclusion - The hill climbing solution path is A – B – D, confirming option B.

Take as heuristic the numbers in brackets. What would the solution path be if Hill Climbing approach was used?单项选择题

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