Shopping

Give a greedy algorithm to minimize the total cost to purchase all n items.
1. The greedy algorithm to minimize total cost would be to buy the most expensive item first. Since prices double each month, you would save more by buying the expensive items at the current price than the cheaper ones. The algorithm would sort the items by price in descending order and purchase them in that order, one per month.

Prove that your greedy algorithm always outputs the optimal solution.
Greedy Algorithm:
1. Start by listing all the items with their current prices.
1. Sort the items by price in descending order, so the most expensive item is first.
1. Purchase the items in this order, one per month.
Proof of Optimality:
Now, let's prove that this greedy algorithm always gives us the optimal solution.
Proof by contradiction:
Assume that the greedy solution is not optimal, which means there is a cheaper way to buy all items. This cheaper solution would require us to buy a cheaper item before a more expensive one at some point.
Let's consider two items: Item A with a price of a and Item B with a price of b, where a < b. According to our assumption, the cheaper solution buys Item A in the first month and Item B in the second month. The total cost for these two items in the cheaper solution would be a + 2b because Item B's price doubles by the second month.
However, the greedy algorithm would buy Item B first and Item A second, yielding a cost of b + 2a. Since a < b, it follows that 2a < 2b. Thus, b + 2a (greedy approach) is less than a + 2b (assumed cheaper solution), which is a contradiction.
Since buying the more expensive item first leads to a lower total cost, delaying the purchase of more expensive items is never beneficial. Therefore, there can be no cheaper solution than the greedy one, which contradicts our initial assumption that the greedy solution is not optimal.
Consequently, the greedy algorithm must always yield the optimal solution.