如何确定树中的逐个元素（比赛括号）？ [英] How to determinate round by element in tree (Tournament brackets)?

问题描述

假设我们有以下树：

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

元素（匹配）：

1-8第1轮

9-12为第2轮

13-14为第3轮

15为第4轮

我如何确定shuch树中的元素n？

我现在的公式是： >

  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

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

在这种情况下，它只是数字的对数，向下舍入。

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

（注意， reverse_match_round 但是，因为我们从 total_rounds 中减去它，所以比起1-index它更容易保持它，如果你喜欢它1-indexed，添加 +1 到最后两行。）

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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