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The following question was asked and not (yet) answered at Math Stack Exchange.

There are $n$ horses. At a time only $k$ horses can run in the single race. What is the minimum number of races required to find the top $m$ fastest horses? Please explain your answer.

The $n = 25, k = m = 5$ case was a Google interview question and there are various answers on the web. But I am not sure what the right answer should be for this. Any ideas?

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    $\begingroup$ @Misha: I have already posted the MSE link. $\endgroup$
    – Debanjan Chanda
    Commented Nov 30, 2012 at 16:44
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    $\begingroup$ This seems like a nontrivial algorithms question to me. It may be a standard result for the right people, but I don't think it is obvious and I would be interested in learning the answer. Voting to reopen. $\endgroup$
    – David E Speyer
    Commented Nov 30, 2012 at 20:43
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    $\begingroup$ If $k=2$, this is a well studied but not solved problem en.wikipedia.org/wiki/Partial_sorting . I have to assume that someone knows something about larger $k$. $\endgroup$
    – David E Speyer
    Commented Dec 1, 2012 at 0:32
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    $\begingroup$ I just wanted to say that this is the same as Generalization of a horse-racing puzzle. $\endgroup$
    – Dylan Pizzo
    Commented Dec 24, 2013 at 23:02
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    $\begingroup$ It isn't clearly stated whether we have to plan all the races in advance or we can plan each race after knowing the results of the previous races. The answer could change. $\endgroup$
    – Brendan McKay
    Commented Dec 25, 2013 at 1:36

1 Answer 1

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A trivial lower bound is $n/k$, since clearly every horse must race to obtain the answer.

I think we can get $O(n/k)$ upper bound, by adapting the median-of-medians selection algorithm.

First, note that, up to a constant factor, this problem is equivalent to finding the $m$th best horse. For the reduction one way, simply find the $m$ best horses and the $m-1$ best horses and see which horse mysteriously disappeared. For the reduction the other way, find the $m$th best horse, then race every other against it to find which are better and which are worse. (In fact, you don't need to do this - it's not hard to check that if you've found the $m$th best horse through repeated racing, you already know which are better and which are worse.)

For $k=2$, the median-of-medians algorithm is known to give an $O(n)$ time bound, as was pointed out by Ralph Furmaniak. We will also use this to handle $k\leq 4$.

So assume $k\geq 5$ and is odd. Let $T(n)$ be the time it takes to find the $m$th horse among $n$ horses. Then divide the horses into groups of $k$, race them in time $n/k$, and take the median of each. Then select the median of those medians in time $T(n/k)$, and pivot on it in time $n/(k-1)$. (to pivot, the pivot horse must race each other horse only once. This allows us to remove at least $n(k+1)/4k$ of the horses, so we can find the $m$th horse in time $T(n (3k-1)/4k)$. So we get:

$$T(n) \leq T\left(\frac{n}{k}\right) + T\left( \frac{n(3k-1)}{4k}\right) + \frac{n}{k}+\frac{n}{k-1} $$

By induction,

$$T(n) \leq \frac{4k}{k-3} \cdot \left( \frac{n}{k}+\frac{n}{k-1} \right) = O \left(\frac{n}{k}\right)$$

In fact we obtain an explicit constant of $8+o(1)$.

This ignores non-unique divisibility, which should only lead to a small error term unless $k$ is very large as a fraction of $n$.

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