如何决定ñ马与m钢轨的排名？ [英] How to decide the rank of N horses with M tracks?
问题描述
，提供N马和并购（M＆LT; = N）的轨道，但没有计时器，你可以得到一个回合是M马的顺序。这些问题有多少轮至少，如果你想获得的所有马匹的等级？
Providing N horses and M(M <= N) tracks but no timer, all you could get from one round is the order of M horses. The questions how many rounds at least, if you want to get the rank of all horses?
例如。
N = 3，M = 3，圆= 1;
N = 3，M = 2，圆= 3;
N = 4，M = 3，圆= 3;
e.g.
N=3, M=3, Round=1;
N=3, M=2, Round=3;
N=4, M=3, Round=3;
什么是圆的，当N = 1000，M = 3？
what is Round, when N=1000, M=3?
推荐答案
您可以得到一个下界与信息理论。
You can get a lower bound with information theory.
每场比赛给你登录（M！）比特的信息，你需要的log（n！）位。因此，在比赛的数量自然下界，然后登录（N！）/日志（M！）。
Each race gives you log(m!) bits of information, and you need log(n!) bits. So a natural lowerbound on the number of races is then log(n!) / log(m!).
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