Math List-I with List-II: \(\begin{array}{c} {} & \text{List-I} & {} &…
2019
Math List-I with List-II:
\(\begin{array}{c} {} & \text{List-I} & {} & \text{List-II} \\ (a) & \text{Greedy best-first} & (i) & \text{Minimal cost } (p)+h(p) \\ (b) & \text{Lowest cost-first} & (ii) & \text{Minimal } h(p) \\ (c) & A^* \text{ algorithm} & (iii) & \text{Minimal cost } (p) \\ \end{array}\)
Choose the correct option from those given below:
- A.
\((a) – (i) ; (b) – (ii); (c) – (iii) \) - B.
\((a) – (iii) ; (b) – (ii); (c) – (i) \) - C.
\((a) – (i) ; (b) – (iii); (c) – (ii) \) - D.
\((a) – (ii) ; (b) – (iii); (c) – (i)\)
Attempted by 93 students.
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Correct answer: D
Correct matchings and rationale:
Greedy best-first → minimal h(p).
Rationale: Greedy best-first selects the node that appears closest to the goal according to the heuristic alone, so it orders nodes by the heuristic value h(p).
Lowest cost-first → minimal cost(p).
Rationale: Lowest cost-first (uniform-cost search) expands the node with the smallest path cost from the start, i.e., the smallest cost(p).
A* algorithm → minimal cost(p) + h(p).
Rationale: A* uses the evaluation function f(p) = g(p) + h(p) (here listed as cost(p) + h(p)) to balance path cost and heuristic; with an admissible heuristic this yields an optimal search.
Therefore, the correct overall pairing is: Greedy best-first with minimal h(p); Lowest cost-first with minimal cost(p); and A* with minimal cost(p) + h(p).
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