Mathematical horse racing problem
So far: 37 games. Game 37: First, randomly arrange the horses into an 8* 8 formation: 0102030405060709101314151. 24 25 26 27 28 29 303132 33 34 35 36 37 38 39 404142 43 44 45 47 49 505152 53 54 55 56 57 59 6162 63 64/kloc-. That is, 0 1 09, 17, 25, 33, 4 1, 49, 57 is the champion of eight games. As mentioned below, after that, you can always decide 8 places every 4 games. (1) each group has a champion, (2) each group has a competition from the bottom, and * * * has two competitions, and the first and sixty-fourth places are decided. Let's assume that every horse in the first train slows down from top to bottom. That is 0 1 is the champion. (3) Now, there are two candidates for the second place, 02,09. Let 02, 09, 10, 17 take part in the third game. Where are the other four horses in the third inning? In similar situations, they are the slowest horses. For example, 64 is the slowest, and the eighth train is 08, 16, 24, 32, 40, 48, 56, 64, so let 56, 63, 55, 48 take part in the third race. From the result of the third game, you can always know that you are second and 63 rd. Let's explain that no matter what the results of 02, 09, 10 and 17 are, there will be no more than four candidates for the third and fourth place. If 02 wins, then the candidates for the third and fourth place are only 03, 04 and 09, and the faster horses 10 and 17 (these two horses have already competed). If 09 wins, the third place is actually known, that is, the one with 02, 10 or 17 is faster. If 02, the fourth candidate horse is 03, 10, 17. If it is 10, the fourth candidate horse is 02, 1 1, 17. If it is 17, the fourth candidate horse is 02, 10,18,25. Therefore, the total number of candidates for the third and fourth places does not exceed four. Similarly, there will be no more than four candidates for the 62nd and 6th1positions. (4) The above-mentioned third and fourth candidates and the sixty-second and sixty-first candidates will not exceed eight people to participate in a competition, so the top four and the last four have decided on eight places. Repeat the above process until 56 places are decided in 7 games and 4 games. 3. Finally, there are 8 places left, and one game is settled. Total: 8+4*7+ 1=37 games.