UGC NET CSE | June 2019 | Part 2 | Question: 92

Question

Math List-I with List-II:

$$\begin{array}{|c|l|c|l|} \hline {} & \text{List-I} & {}  & \text{List-II} \\ \hline (a) & \text{Greedy best-first} & (i) & \text{Minimal cost } (p)+h(p) \\ \hline (b) & \text{Lowest cost-first} & (ii) & \text{Minimal } h(p) \\ \hline (c) & A^* \text{ algorithm} & (iii) & \text{Minimal cost } (p) \\ \hline \end{array}$$

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)

Answer 1

answer is 4

a)greedy best first ---> ii) Minimal h(p)

b)lowest cost  ----->iii) Minimal cost(p)

c)A* search --->i  Minimal cost(p)+h(p)

 h(p)=heuristic cost from current node to goal node

cost(p)=cost to reach that node

 

https://www.ics.uci.edu/~welling/teaching/271fall09/InfSearch271f09.pdf

Answer 2

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.