如何确定的一轮元素树（比赛括号内）？ [英] How to determinate round by element in tree (Tournament brackets)?

问题描述

假设我们有以下三种：

  1
    9
2
        13
3
    10
4
            15
五
    11
6
        14
7
    12
8
 

元素在哪里（匹配）：
1-8是圆的1
9-12是圆的2
13-14是第3轮
15轮4

我怎么能确定的在shuch树轮的元素N？

我目前的公式是：

  total_rounds =地板（日志（totalTeams，2））;

matches_per_round =（totalTeams / POW（2 current_round））

next_match_id =（totalTeams / 2）+ CEIL（match_id / 2）

total_matches = total_teams  -  1
 

解决方案

想象一下，树也被列在相反的。

  15
     7
14
         3
13
     6
12
             1
11
     五
10
         2
9
     4
8
 

在这种情况下，它会简单地将数的对数，四舍五入。现在，我们只需要减去此数轮数，我们就大功告成了。

  reverse_number = total_matches  -  match_number + 1;
reverse_match_round =地板（日志（reverse_number，2））;
match_round = total_rounds  -  match_round;
 

（注意， reverse_match_round 实际上是0索引，不像match_round。但是，由于我们从 total_rounds 减去它，它更容易保持这种方式比1索引它。如果你preFER这1索引，只需添加 +1 到每个最后两行。）

Assume we have following tree:

1
    9
2
        13
3
    10
4 
            15
5
    11
6 
        14
7   
    12
8

Where elements(matches):
1-8 is round 1
9-12 is round 2
13-14 is round 3
15 is round 4

How I can determinate round of element "n" in shuch tree?

My current formulas are:

total_rounds = floor(log(totalTeams,2));

matches_per_round = (totalTeams / pow(2, current_round))

next_match_id = (totalTeams/2) + ceil(match_id/2)

total_matches = total_teams - 1

解决方案

Imagine the tree was numbered in reverse.

15
     7
14
         3
13 
     6
12 
             1
11
     5 
10 
         2
9   
     4
8

In that case, it'd simply be the logarithm of the number, rounded down. Now we simply subtract this number from the number of rounds, and we're done.

reverse_number = total_matches - match_number + 1;
reverse_match_round = floor(log(reverse_number, 2));
match_round = total_rounds - match_round;

(Note, reverse_match_round is actually 0-indexed, unlike match_round. However, since we subtract it from total_rounds, it's easier to keep it that way than to 1-index it. If you prefer it 1-indexed, simply add +1 to each of the last two lines.)

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