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

PS: There is no timer.

P.K.
  • 7,742
Quixotic
  • 22,431
  • If $k=2$, I think a lower bound for the number of races is $\log_2\binom {n}{m}$. – Thomas Andrews Nov 30 '12 at 15:55
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    I suppose we have no timer? I.e. any race can only give an ordering of the participating horses, and not information about exactly how fast they actually are? – malin Nov 30 '12 at 15:59
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    This was asked and not answered at http://math.stackexchange.com/questions/209790/how-to-find-a-total-order-with-constrained-comparisons The $n=25, k=m=5$ case was a Google interview question and there are various answers on the web. – Ross Millikan Nov 30 '12 at 16:10
  • More generally, you certainly require at minimum $\log_{k!}\binom{n}{m}$ races. That's because a race of $k$ horses yields the same result as approximately $\log_2{k!}$ pairs of horses. – Thomas Andrews Nov 30 '12 at 16:12
  • Stan Wagon (in his PoW e-mail) said he had written a paper on this subject with Stephen Morris and Richard Stong which was to be published in Math Horizons, but I don't find it online. – Ross Millikan Nov 30 '12 at 16:50
  • @RossMillikan Actually, it looks like the older question wants a complete ranking, not just a list of the top $5$ horses. (Note, the OP doesn't even ask for a ranking of the top $m$, just a list of them. If $m=n$, then it takes $0$ races, and if $n=m-1$ you can do it in $\frac{n-1}{k-1}$ races. – Thomas Andrews Nov 30 '12 at 16:53
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    It's interesting that no-one has bothered to mention the rather unrealistic assumption that the horses always run the race in exactly the same time :-) – joriki Jan 19 '13 at 13:21
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    @RossMillikan Now the Math Horizons article of Morris, Stong and Wagon has been published, but is not available for free. – pepan Sep 11 '13 at 16:53

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