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I was looking at the solution for the Code Jam 2014 qualification question but the proof of correctness seems to be incomplete and I was wondering if anyone could help me with it. The full question can be found here but to summarize:

You are given cookies at a rate of 2 cookies/second initially. Your goal is to get to X cookies in the shortest time possible. You can build farms that produce cookies at a rate of F cookies/second but each farm costs C cookies to build. How do you find the shortest time needed to get X cookies given C and F?

Let T(n) be the time it takes get X cookies using n farms. The solution states that to find the shortest time, you can calculate T(n) for n = 0,1,2,.. and once you find T(n+1) > T(n), then you know T(n) is the answer.

Unfortunately the solution omits a proof that T(n) is a global minimum. So far the approach I've been trying to use is that since we know T(n) < T(n+1) and also that T(n) < T(n-1) (because otherwise we would have terminated the algorithm), T(n)i s a local minimum. However I don't know how to show that there is only one minimum and thus the local is the maximum. I also have

$$T(n) = \sum_{i=0}^{n-1} \frac{C}{2 + i*f} + \frac x{2+f*n}$$

Any help would be appreciated! Thanks!

Synderesis
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1 Answers1

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Let's look at $T(n+1)-T(n)$. If we can show that it changes sign exactly once, we're done. (You switched from uppercase letters to lowercase letters in your formula; I'll use the uppercase letters used in your text and in the problem statement.)

\begin{eqnarray} T(n+1)-T(n) &=&\left(\sum_{i=0}^n\frac C{2+iF}+\frac X{2+(n+1)F}\right)-\left(\sum_{i=0}^{n-1}\frac C{2+iF}+\frac X{2+nF}\right)\\ &=&\frac C{2+nF}+X\left(\frac 1{2+(n+1)F}-\frac 1{2+nF}\right)\\ &=&\frac{C(2+(n+1)F)-XF}{(2+(n+1)F)(2+nF)}\;. \end{eqnarray}

The denominator is always positive, and the numerator is linear in $n$ and changes sign exactly once.

joriki
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