The efficient algorithm by which he can determine which water stops he should make; <em>has been determined below and the strategy which is the greedy strategy has been proven to yield an optimal solution.</em>
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To solve this question optimally, we will make use of the greedy solution.
In this method, we will maximize the distance that can be covered from a particular point in such a way that there must be a place where water can be gotten before a run out is experienced.
Now, the first point at which there will be a stop should be located at a point that is farthest from the starting position and is also made to be ≤ m miles from the starting position.
Now, this this situation is one that shows optimal substructure and since our first stopping point will be made to be at p, it means that we are solving the sub-question with the assumption that our starting point is at p.
When we combine the two stated plans above, we will arrive at an optimal solution for the normal reasons via cut and paste.
B) Now we need to show that the greedy approach earlier used produce a first stopping point which is contained in the optimal solution.
Let O represent any optimal solution whereby the professor stops at positions o₁, o₂, o₃....oₙ.
Let h₁ represent the farthest stopping point that is reachable from the starting point. Then we can replace o₁ by h₂ to generate a modified solution H since o₁ - o₂ < o₂ - h₁.
Finally, we can really make it to the positions in H without having to run out of water and since H has the same number of stops, we can conclude that h₁ is contained in one of optimal solution.
Therefore our strategy which is the greedy strategy has been proven to work.
Read more about algorithms at; brainly.com/question/24793921
