Graph traversal algorithm with a twist - minimum number of stops

I am stumped on this homework problem. I think I have the right answer but am unsure how to prove it. I am also unsure of how to approach the proof. Here is the problem:

Professor Gekko has always dreamed of inline skating across North Dakota. He plans to cross the state on highway U.S. 2, which runs from Grand Forks, on the eastern border with Minnesota, to Williston, near the western border with Montana. The professor can carry two liters of water, and he can skate m miles before runni开发者_如何转开发ng out of water. (Because North Dakota is relatively flat, the professor does not have to worry about drinking water at a greater rate on uphill sections than on flat or downhill sections) The professor will start in Grand Forks with two full liters of water. His official North Dakota state map shows all the places along U.S. 2 at which he can refill his water and the distances between these locations.

The professor’s goal is to minimize the number of water stops along his route across the state. Give an efficient method by which he can determine which water stops he should make. Prove that your strategy yields an optimal solution, and give its running time.

I think he should choose his stops so that he goes the maximum distance before running out of water. That is, at every stop, if he were to keep going, he would run out of water before the next stop.

I am not sure how to proceed. I would bet that this has been done before, but I am rather new to this field of CS and could use some guidance.

Your algorithm is correct.

Try to prove the following by induction on the number of stops passed. After passing each water location, no other strategy can have made fewer stops, and of those which made the same number of stops, no other strategy can leave you with more water.

At 0 stops all strategies are equal, so it is trivial to prove the statement.

If you don't drink at a stop under this strategy, the result is easy to prove.

If you do drink at a stop, from the fact that the statement was true at the previous stop, you can prove that either the other strategy made more stops last time (hence they are not ahead on stops, and can't be ahead on water since you just got water), or else the other strategy must have likewise just got water also (else they will run out of water before the next stop).

That is enough to fill out the induction proof.

If you're struggling about the concept of what is required for a formal proof, and how to do them, see How I Do Proofs. I also blogged about my experience using that handout here.

I hope my first attempt is truly deleted. It was wrong.

Proof sketch: If greedy failed, it must have been optimal to take a city closer to the starting point (further from the finish line) than the one greedy chose at some point. Ignore the problem up to that choice, and look at the two subproblems: one starting at the city chosen by greedy and the other one--which includes the greedy point in the subproblem. To avoid a contradiction, the city starting further away from the finish line must have a fewer-hop optimal solution than the optimal solution starting at the greedy point. Ignore how you arrive at this optimal solution for both subproblems, only that it exists. Since the non-greedy starting point includes the greedy subproblem, it must either have the same optimal sub-path hops or more (Prove this statement. It can be done by induction, but I believe there is a simpler proof that I am just too tired to think of. Perhaps you can just say it is obvious?). If they were equal, then greedy worked fine. If they are greater, contradiction.

The trick with these sorts of questions is to state the problem as another problem for which you can prove something.

For example, for this one you could form an unweighted graph where the stops are nodes and you have an edge between two nodes if it's possible to travel between them without stopping. Then all you have to do is find the shortest path in the graph and there you go. This is fine if the distance to be traveled between stops is relatively small, otherwise the graph becomes very dense. I suspect there is a better problem to reduce to since your path is in a line.