blocks|key|1607414|text|团队的可能组合/匹配的数量可以在数学上描述为Triangular+Number。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1607415|例如，当有9支球队时，比赛的数量是36场。|1607416|📷|atomic|1607417|​|1607418|请注意，只有当k或k-1可被3整除时，这个数字才能被3整除。对于5个团队，您将最终得到10个可能的游戏。你的最后一周将只有一场比赛，或者你可以改变它的结构。|1607419|如果你想写出匹配的组合，你可以通过迭代两次团队的数量来列出它们。以下是一些示例Java代码。You+may+run+it+in+an+online+Java+compiler.|1607420|++++public+class+MyClass+{

++++++++public+static+void+main(String+args[])+{
++++++++++++int+TEAMS+=+10;+//Number+of+teams
++++++++++++int+combos+=+0;

++++++++++++for(int+i+=+1;+i+<=+TEAMS-1;+i%2B%2B){
++++++++++++++++for(int+j+=+i%2B1;+j+<=+TEAMS;+j%2B%2B){
++++++++++++++++++++System.out.println("Team+"+%2B+i+%2B+"+plays+Team+"+%2B+j);
++++++++++++++++++++combos+%2B%2B;
++++++++++++++++}
++++++++++++}

++++++++++++System.out.println("There+is+"+%2B+combos+%2B+"+possible+matches");
++++++++}
++++}|code-block|syntax|javascript|1607421|我们不只是想要两支球队的每一个组合。我们想看看3个团队的组合。从数学上讲，我们需要一个Combination。|1607422|我们可以将我们的三角数重写为n+rewrite+k。我们前面的例子变成了：|1607423|1607424|1607425|1607426|我们选择的每周有3支球队比赛。总的可能的日期组合是n选择3。在我们的例子中有9个团队。|1607427|1607428|1607429|1607430|我们有84种可能的日期组合。现在很多时候都有重叠的游戏。例如，如果我们让1,2和3队有一天比赛，那么我们不想再有1，2和4队的比赛，因为然后1和2互相比赛2场比赛。解决这个问题的办法可能是忽略重复的游戏。|1607431|我想指出，完美的解决方案是不存在的。对于大多数球队来说，没有一种解决方案可以让每天3支球队一起比赛，而这些球队还没有比赛过。例如，当我们有4支球队时，我们的游戏是:+1-2，1-3，1-4，2-3，2-4，3-4。如果我们在第一天拿到3支球队(1-2，1-3，2-3)，那么第二天我们就得不到完美的组合(1-4，2-4，3-4)。|1607432|无论你如何打破它，你都可以排序出最好的组合，但最终你会得到许多随机的游戏。|1607433|我创建了下面的代码来查看每种可能的日期组合，并打印出不重复的日期。|1607434|++++public+class+MyClass+{

++++++++public+static+void+main(String+args[])+{
++++++++++++int+TEAMS+=+9;+//Number+of+teams

++++++++++++//Keep+track+of+each+game+combination+used
++++++++++++boolean+gamesPlayed[][]+=+new+boolean[TEAMS%2B1][TEAMS%2B1];

++++++++++++int+day+=+1;
++++++++++++for(int+i+=+1;+i+<=+TEAMS-2;+i%2B%2B){
++++++++++++++++for(int+j+=+i%2B1;+j+<=+TEAMS-1;+j%2B%2B){
++++++++++++++++++++for(int+k+=+TEAMS;+k+>=+j%2B1;+k--){

++++++++++++++++++++++++if(!gamesPlayed[i][j]+&&+!gamesPlayed[i][k]+&&+!gamesPlayed[j][k]+)
++++++++++++++++++++++++{
++++++++++++++++++++++++++++System.out.println("Day+"%2B+day%2B%2B+%2B+"+Teams+"+%2B+i+%2B+",+"+%2B+j+%2B+"+&+"+%2B+k+%2B+"+Play");
++++++++++++++++++++++++++++gamesPlayed[i][j]+=+true;
++++++++++++++++++++++++++++gamesPlayed[i][k]+=+true;
++++++++++++++++++++++++++++gamesPlayed[j][k]+=+true;
++++++++++++++++++++++++}
++++++++++++++++++++}
++++++++++++++++}
++++++++++++}


++++++++++++System.out.println("\nLeftover+games");
++++++++++++for(int+i+=+1;+i+<=+TEAMS-1;+i%2B%2B){
++++++++++++++++for(int+j+=+i%2B1;+j+<=+TEAMS;+j%2B%2B){
++++++++++++++++++++if(!+gamesPlayed[i][j])
++++++++++++++++++++++++System.out.println("+Team+"+%2B+i+%2B+"+plays+Team+"+%2B+j);
++++++++++++++++}
++++++++++++}



++++++++}
++++}|1607435|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Triangular_number|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/1711ca4b22e3e3993122215fbaf6e765.gif|imageAlt|2|https://www.jdoodle.com/online-java-compiler|3|https://en.wikipedia.org/wiki/Combination|4|https://ask.qcloudimg.com/http-save/yehe-900000/09cc3bf05aacd383d43b7624f37ee3da.gif|5|https://ask.qcloudimg.com/http-save/yehe-900000/ac0ecf2c51592cdfc41472a005ba620e.gif^0|M|H|0|0|0|0|1|1|0|0|0|1A|16|2|0|0|17|B|3|0|0|0|0|1|4|0|0|0|0|0|1|5|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|21|8|@]|9|@$A|22|B|23|1|24]]|C|$]]|$1|D|3|E|5|6|7|25|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|26|8|@]|9|@$A|27|B|28|1|29]]|C|$]]|$1|I|3|J|5|6|7|2A|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|2B|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|2C|8|@]|9|@$A|2D|B|2E|1|2F]]|C|$]]|$1|O|3|P|5|Q|7|2G|8|@]|9|@]|C|$R|S]]|$1|T|3|U|5|6|7|2H|8|@]|9|@$A|2I|B|2J|1|2K]]|C|$]]|$1|V|3|W|5|6|7|2L|8|@]|9|@]|C|$]]|$1|X|3|J|5|6|7|2M|8|@]|9|@]|C|$]]|$1|Y|3|G|5|H|7|2N|8|@]|9|@$A|2O|B|2P|1|2Q]]|C|$]]|$1|Z|3|J|5|6|7|2R|8|@]|9|@]|C|$]]|$1|10|3|11|5|6|7|2S|8|@]|9|@]|C|$]]|$1|12|3|J|5|6|7|2T|8|@]|9|@]|C|$]]|$1|13|3|G|5|H|7|2U|8|@]|9|@$A|2V|B|2W|1|2X]]|C|$]]|$1|14|3|J|5|6|7|2Y|8|@]|9|@]|C|$]]|$1|15|3|16|5|6|7|2Z|8|@]|9|@]|C|$]]|$1|17|3|18|5|6|7|30|8|@]|9|@]|C|$]]|$1|19|3|1A|5|6|7|31|8|@]|9|@]|C|$]]|$1|1B|3|1C|5|6|7|32|8|@]|9|@]|C|$]]|$1|1D|3|1E|5|Q|7|33|8|@]|9|@]|C|$R|S]]|$1|1F|3|-4|5|6|7|34|8|@]|9|@]|C|$]]]|1G|$1H|$5|1I|1J|1K|C|$1L|1M]]|1N|$5|1O|1J|1P|C|$1Q|1R|1S|-4]]|1T|$5|1I|1J|1K|C|$1L|1U]]|1V|$5|1I|1J|1K|C|$1L|1W]]|1X|$5|1O|1J|1P|C|$1Q|1Y|1S|-4]]|1Z|$5|1O|1J|1P|C|$1Q|20|1S|-4]]]]

The number of possible combinations/matches of teams can mathematically be described as a <a href="https://en.wikipedia.org/wiki/Triangular_number" rel="nofollow noreferrer">Triangular Number</a>.

For example, when there is 9 teams, the number of matches is 36.
<a href="https://i.stack.imgur.com/2uQmk.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2uQmk.gif" alt="Sum k-1 from 1 to 9"></a>

Notice, that this number is only divisible by 3 when k or k-1 is divisible by 3. With 5 teams, you will end up with 10 possible games. Your last week will only have 1 game or you can structure it differently.

If you want to write out the combinations of matches, you can list them by iterating through the number of teams twice. Here is some example Java code. <a href="https://www.jdoodle.com/online-java-compiler" rel="nofollow noreferrer">You may run it in an online Java compiler.</a>

<pre class="lang-java prettyprint-override"><code> public class MyClass {

 public static void main(String args[]) {
 int TEAMS = 10; //Number of teams
 int combos = 0;

 for(int i = 1; i &lt;= TEAMS-1; i++){
 for(int j = i+1; j &lt;= TEAMS; j++){
 System.out.println("Team " + i + " plays Team " + j);
 combos ++;
 }
 }

 System.out.println("There is " + combos + " possible matches");
 }
 }
</code></pre>

We don't just want every combination of 2 teams. We want to look at combinations of 3 teams. Mathematically, we want a <a href="https://en.wikipedia.org/wiki/Combination" rel="nofollow noreferrer">Combination</a>.

We can rewrite our Triangular number as n choose k. Our previous example becomes:

<a href="https://i.stack.imgur.com/LCJgL.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LCJgL.gif" alt="9 choose 2"></a>

Every week that we choose has 3 teams playing. The total possible day combinations is n choose 3. In our example with 9 teams.

<a href="https://i.stack.imgur.com/yFTVI.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yFTVI.gif" alt="9 choose 3"></a>

We have 84 possible day combinations. Many of these days have overlapping games. For example, if we have teams 1, 2, and 3 play one day, then we don't want another day with teams 1,2 and 4 because then 1 and 2 play 2 games against each other. The solution for this could be to ignore duplicated games.

I want to point out that a perfect solution does not exist. For most number of teams, there is not a solution where every day 3 teams can play together that haven't already played. For example, when we have 4 teams, our games are:
1-2, 1-3, 1-4, 2-3, 2-4, 3-4. If we took 3 of those teams the first day (1-2, 1-3, 2-3) then the second day we don't get a perfect combination (1-4, 2-4, 3-4).

No matter how you break it, you can sort for the best combinations but you will end up with many random games at the end.

I created code below to look at every possible day combination and print out days that are not duplicated.

<pre class="lang-java prettyprint-override"><code> public class MyClass {

 public static void main(String args[]) {
 int TEAMS = 9; //Number of teams

 //Keep track of each game combination used
 boolean gamesPlayed[][] = new boolean[TEAMS+1][TEAMS+1];

 int day = 1;
 for(int i = 1; i &lt;= TEAMS-2; i++){
 for(int j = i+1; j &lt;= TEAMS-1; j++){
 for(int k = TEAMS; k &gt;= j+1; k--){

 if(!gamesPlayed[i][j] &amp;&amp; !gamesPlayed[i][k] &amp;&amp; !gamesPlayed[j][k] )
 {
 System.out.println("Day "+ day++ + " Teams " + i + ", " + j + " &amp; " + k + " Play");
 gamesPlayed[i][j] = true;
 gamesPlayed[i][k] = true;
 gamesPlayed[j][k] = true;
 }
 }
 }
 }


 System.out.println("\nLeftover games");
 for(int i = 1; i &lt;= TEAMS-1; i++){
 for(int j = i+1; j &lt;= TEAMS; j++){
 if(! gamesPlayed[i][j])
 System.out.println(" Team " + i + " plays Team " + j);
 }
 }



 }
 }
</code></pre>

I am trying to split X number of Teams into "play days" which consist of 3 teams per day

There is more than one solution to solve this for 15 teams.

What is the best approach to find all possible fixtures/match plans for team count 9-21?
Team count of 11, 14, 17 and 20 also cause problems, because total_matches/3="must be even/integer"
<a href="https://i.stack.imgur.com/reJbu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/reJbu.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/HTbjX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HTbjX.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/MGTue.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MGTue.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/WsJto.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WsJto.png" alt="enter image description here"></a>

<pre><code>15 Teams // 105 Total Matches // 35 Total Days
D1 = 1:2 1:3 2:3
D2 = 2:4 2:5 4:5
D3 = 3:4 3:6 4:6
D4 = 4:1 4:7 1:7
D5 = 5:1 5:6 1:6
D6 = 6:2 6:7 2:7
7 = 7:3 7:5 3:5
8 = 8:1 8:9 1:9
9 = 9:2 9:10 2:10
10 = 10:1 10:11 1:11
11 = 11:2 11:8 2:8
12 = 12:1 12:13 1:13
13 = 13:2 13:14 2:14
14 = 14:1 14:15 1:15
15 = 2:12 2:15 12:15
16 = 3:8 3:10 8:10
17 = 3:9 3:11 9:11
18 = 3:12 3:14 12:14
19 = 4:8 4:12 8:12
20 = 5:8 5:13 8:13
21 = 6:8 6:14 8:14
22 = 7:8 7:15 8:15
23 = 9:4 9:13 4:13
24 = 9:5 9:12 5:12
25 = 10:4 10:14 4:14
26 = 11:4 11:15 4:15
27 = 12:6 12:10 6:10
28 = 13:3 13:15 3:15
29 = 14:5 14:11 5:11
30 = 5:10 5:15 10:15
D31 = 6:9 6:15 9:15
D32 = 6:11 6:13 11:13
D33 = 7:9 7:14 9:14
D34 = 7:10 7:13 10:13
D35 = 7:11 7:12 11:12
</code></pre>

Mathematical solution to split X teams into 3 Teams per day

Java

我试着把X个团队分成“游戏日”，每天由3个团队组成对于15个团队来说，有不止一个解决方案可以解决这个问题。找到9-21球队所有可能的赛程/比赛计划的最佳方法是什么？组数为11、14、17和20也会导致问题，因为total_matches/3=“必须是偶数/整数”​15 Teams // 105 Total Matche...

问每天将X个团队分成3个团队的数学解决方案
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