Question

Let h1(s) be an admissible A* heuristic. Let h2(s)=2*h1(s)

1. Is the solution found by A* tree search with h2 guaranteed to be an optimal solution? Justify your answer
2. Is the solution found by A* tree search with h2 guaranteed to have a cost at most twice as much as the optimal path? Justify your answer
3. Consider the state space graph shown below in which some of the states are missing a heuristic value. Determine the possible range for each missing heuristic value so that the heuristic is admissible and consistent. If this isn’t possible write so

QUESTION 2 Let h(s) be an admissible A* heuristic. Let h2(s) = 2 h;(s) optimal solution? Justify (a) Is the solution found by A* tree search with h2 guaranteed to be an your answer (2 marks (b) Is the solution found by A* tree search with h2 guaranteed to have a cost at most twice as much as the optimal path? Justify your answer. (2 marks) (c) Consider the state space graph shown below in which some of the states are missing a heuristic value. Determine the possible range for each missing heuristic value so that the heuristic is admissible and consistent. If this isn't pOssible, write so (6 marks) 2 h-2 5 A h ? h ? 3 3 F Goal Start h-5 h 0 h 9 1 6 2 2 h = ? State Range for h(s) h(A) <h(D) h(E)S E

Answer 1

Question 2 а) Solution False Explanation In h2 is not a guaranteed process to an optimal solution because only one side ofthe inequality admissible has to be doubled process that the h2(s) 2h1(s) is false So b) Solution True Explanation In A* tree has always long as the optimal path to the goal state it could not be found prefix state of this optimal path has process of fringe path Hence non optimal is found For the cost of g' is a non optimal value when it is popped such as f-cost

Such that g h0 g 2h1 <2(gh1)2C For C is the optimal cost Therefore g can be found path is twice then that of optimal path 2C for the consistent heuristic part contain h(s)c(s, s0) h(s 0) For s A h(Start state)c(Start state, A) + heuristic (A) ww 6heuristic (A) heuristic (A) c(A, B)heuristic (B) -heuristic (A) S7 heuristic (A) <c(A, G)heuristic (F) heuristic (A)6 to be consistent

For s D Heuristic (D) c(D, G) heuristic (G) heuristic (D) <3 heuristic (B)c(B, D) heuristic (D) - 0heuristic (D) heuristic (F) (F, D)heuristic (D) -1heuristic (D) to be consistent For s E Heuristic (E) < c(E, G) heuristic (G) = heuristic (E) 6 heuristic (C) heuristic (E) c(C, E) =5heuristic (E) heuristic (E) s c(E, F) heuristic (F) =heuristic (E)<6 to be consistent Range for h(s) 6h(A)6 1h(D)3 5h(E)6 state A D E