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NTA | UGC NET 2019 June – II | Question 68

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Match List-I with List-II:

List-I                                                   List-II  

(a) Greedy best-first                     (i) Minimal cost (p)+h(p)

(b) Lowest cost-first                     (ii) Minimal h(p)

(c) A* algorithm                            (iii) Minimal cost (p)

 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)


Answer: (D)

Explanation:

Greedy best first:  Minimal h(p)

Lowest cost: Minimal cost(p)

A* search: Minimal cost(p)+h(p) where, 

=> g = the movement cost to move from the starting point to a given square on the grid, following the path generated to get there. 
=> h = the estimated movement cost to move from that given square on the grid to the final destination. This is often referred to as the heuristic,

So,  option D is correct.


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Last Updated : 10 Mar, 2023
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