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A well-known puzzle goes:

"Suppose that you have 25 horses and a racetrack on which you can race up to 5 horses. If the outcome of each race only tells you the relative speeds of the horses in the race, how many races do you need to determine the fastest 3 horses (and what is the strategy)?"

The solution (look away now if you don't want a spoiler) is to arrange the horses into groups of five and race them, labeling the horses $a_1,\dots,a_5$, ..., $e_1,\dots,e_5$ -- for example, the horse in position 3 in the second race gets the label $b_3$.

Then race horses $a_1, b_1, c_1, d_1, e_1$, and relabel the horses so that all those in the same group as the winner of this race get the label $a_j, j=1,\dots,5$ and so on. Finally, race horses $a_2, a_3, b_1, b_2, c_1$ -- the three fastest horses are now $a_1$ and the two fastest from the final race.

The question: Does this strategy generalize to $m$ horses and $n$ tracks where you want to find the fastest $k$ horses?

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Well, this particular strategy generalizes for finding the k best horses when the track size is $n = (k-1)(k+2)/2$ and the number of horses is $n^2$, and it takes n+2 races as in your example:

Split them into n groups of size n, race them in those sets, and label as $a_{11}, a_{12}, \dots, a_{1n}, a_{21}, a_{22}, \dots, a_{2n}, \dots, a_{nn}$ as before (so the horse who came in $j$th place in the $i$th race has label $a_{ij}$. Then race $a_{11}, a_{21},\dots,a_{n1}$, and relabel the first subscripts of all horses using the results of this race. The winner of that race is the best horse. To determine the other k-1 best horses, race the n other horses who have fewer than k horses that are better than them (directly or by transitivity): $a_{12}, a_{13}, \dots, a_{1k}, a_{21}, a_{22}, \dots, a_{2(k-1)}, a_{31}, a_{32}, \dots, a_{3(k-2)},\dots, a_{k1}$. (Note here that conveniently $n = (k-1) + (k-1) + (k-2) + (k-3) + \dots + 2 + 1 = (k-1)(k+2)/2$.)

But this still leaves open the question of what to do for other cases.

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  • $\begingroup$ And by induction, for $n^r$ horses, $n^{r-1}+n^{r-2}+\cdots+n+r=(n^r-1)/(n-1)+(r-1)$ races are sufficient. $\endgroup$ – Thomas Kalinowski Jul 1 '11 at 20:14

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