Suppose we have a heuristic h that over-estimates h* by at most epsilon (i.e., for all n, 0<= h(n) <= h*(n)+epsilon). Show that A* search using h will get a goal whose cost is guaranteed to be at most epsilon more than that of the optimal goal.

Consider the following version of Knapsack. Given are two weight limits W 1 and W 2 , whereW 1 ≤ W 2 . Given are also n items (w 1 , c 1 ), (w 2 , c 2 ), . . . , (w n , c n ), where w i is the weight and c ithe cost of the i-th item. We want to find a subset of these items of the highest cost, wherethe subset weights at least W 1 and at most W 2 . Give an O(nW 2 ) algorithm for this problem.(Recall that the cost (respectively weight) of a subset is the sum of the costs (respectivelyweights) of the items in the subset.)

8. A school is creating class schedules for its students. The students submit their requested courses and then a program will be designed to find the optimal schedule for all students. The school has determined that finding the absolute best schedule cannot be solved in a reasonablo time. Instead they have decided to use a simpler algorithm that produces a good but non-optimal schedule in a more reasonable amount of time. Which principle does this decision best demonstrate? O A Unreasonable algorithms may sometimes also be undecidable O B. Heuristics can be used to solve some problems for which no reasonable algorithm exsts O C. efficiency Two algorithms that solve the same problem must also have the same O B. Approximate solutions are often identical to optimal solutions

I only need part B in the image, I have already completed part A. How do I formally prove that by using consecutive powers for the values of coins, that it will give me the optimal solution? (using induction preferably unless an easier formal proof is available)