生成一个单淘汰赛 [英] Generate a single elimination tournament

问题描述

首先，很抱歉，我的英语，我会尽我所能来解释我的问题！

first , sorry for my english I'll do my best to explain my problem !

所以，我想产生一个单淘汰赛与无限数量的球员。

So , i'm trying to generate a single elimination tournament with an unlimited number of players.

现在我只是在想这件事，我什么都没有在纸面上，我想我不会有问题的比赛有两个电源（2 4 8 16 32的球员。），我的大脑挂在球员直接进入到第二轮，我不知道如何确定这个号码在哪里放置它们。

for now i'm just thinking about it , i have nothing on paper , i think i will not have problem for tournament with power of two ( 2 4 8 16 32 players..) , my brain hangs on players going directly to round 2 , i don't know how to determine this number and where to place them.

如（59玩家）

eg (with 59 players)

我觉得有一个公式，但我找不到它，我有一些想法，但我想得特别的情况下，不知道是否会工作的另一回事。

I think there is a formula but I can't find it, I have some ideas but i think too specifically on a case, without knowing if it would work for another .

感谢你，如果你能帮助我！

Thank you if you can help me !

推荐答案

对于给定的数N，发现它和2的最小功率至少大N. 59之间的区别，那将是5（ 64 - 59）。这些5名球员将被添加到赛程的第二轮。

For a given number N, find the difference between it and the smallest power of 2 at least as large as N. For 59, that'll be 5 (64 - 59). Those 5 players will be added to the tournament schedule at the second round.

该算法允许所有玩家成为游戏的一部分，第二轮开始的时候 - 也就是，尽可能早地。它的解释很简单：想象一下，原本有2 ** N的球员 - 但有些只是没有来到他们的比赛，所以他们的对手走得更远束手就擒。 ）

This algorithm allows for all the players to be part of the game when the second round begins - i.e., as early as possible. Its explanation is very simple: imagine that originally there were 2**N players - but some just didn't come to their games, so their opponents went further without a fight. )

一点题外话，你的公式应该考虑到，那就是应该从第二轮，而不是最弱的人进入游戏最强的选手。 ）

As a sidenote, your formula should take into account that it's strongest players that should enter the game from the second round, not the weakest ones. )

第一步显然是计算的播放机将参加在第一轮的数量。现在，让我们继续认为缺少球员的隐喻 - 假设有64名选手原本，所以第一轮应该有32场比赛。但5播放器（64 - 59）没有来为那些游戏 - 这样的真实游戏的数目是27（64/2 - 5），和在第一轮的实参与者的数目是54（27 * 2）。

The first step apparently is calculating the number of players that will participate in the first round. Now, let's continue that 'missing players' metaphor - let's say there were 64 players originally, so the first round should have 32 games played. But 5 players (64 - 59) didn't come for those games - so the number of real games is 27 ( 64/2 - 5 ), and the number of real participants of the first round is 54 (27 * 2).

在第一轮结束后，就会有27人留在了比赛 - 这些人将与其他5人来参加，所以第二轮的球员总人数为32，其余是微不足道的，我想。 ）

After the first round, there'll be 27 people left in the tournament - those people will be joined by those other 5 guys, so the total number of the 2nd round players is 32. The rest is trivial, I suppose. )

其实，这是很容易commonize。比方说，我们有 N 的球员，和2，至少大的最小功率 N 是 P 。现在...

Actually, this is easy to commonize. Let's say we have N players, and the smallest power of 2 at least as large as N is P. Now...

• 在第一轮应该有（N - （P - N））（或只是（2 * N - P））的球员。
• 在第一轮比赛的总人数为（N - P / 2）。
• 显然，同样是球员进入第二轮的数量。
• 这些将被加盟（P - N）玩家留下没有发挥在第一轮， 所以玩家在第二轮总人数将...

• The first round should have (N - (P - N)) (or just (2*N - P)) players.
• The total number of the games in the first round is (N - P/2).
• Apparently, the same is the number of players going into the 2nd round.
• These will be joined by (P - N) players left without a play in the 1st round, so the total number of players in the 2nd round will be...

N - P / 2 + P - N =＆GT;对 - P / 2 =＆GT; P / 2

• ...从现在开始，你只去了2 ^ N球员直接时间表（如P / 2，以及磷，是2的幂）。

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