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我試圖弄清楚什麼是數學公式,將代表以下情況的所有更多鈔票組合:如何找到與2個屬性3個的對象,將總結一共有3種不同性質
- 我有X個顏色,Y球,每個都有2種顏色。
- 我通過以下方法計算Y球的數量: Y = X! /(X - 2)! * 2!
現在,我的任務是找到,將在下面的表格3個球的組合總金額:
- 第一和第二球有一種顏色是類似的。
- 第一個和第二個球的第二個顏色不能相同。
- 的三分球有2個其他球的顏色(不是它們之間共享)
我寫了一個代碼,這將有助於我模擬一下我在尋找,但未能扭轉將其設計爲可幫助我計算數字結果而不模擬它的公式。
的代碼:
var Colors = ['red', 'blue', 'green', 'yellow', 'white' , 'black'];
C = Colors.length;
document.write("Balls: " + Colors.length + '<br>');
document.write(Colors.join() + '<br>');
BallsCount = (sFact(C)/(sFact(C-2) * 2));
document.write("BallsCount: " + BallsCount + '<br>');
var Balls = new Array();
for (i=0, c=1; i<Colors.length;i++)
for (x=i+1; x<Colors.length; c++, x++)
{
document.write(c + ": " + Colors[i] + '/' + Colors[x] + '<br>');
Balls[c-1] = Colors[i] + '/' + Colors[x];
}
//Triangles = BallsCount/3;
TrianglesCount = (sFact(BallsCount)/(sFact(BallsCount-3) * sFact(3)));
document.write("Triangles: " + TrianglesCount + '<br>');
var Triangles = new Array();
for (i=0, c=1; i<Balls.length;i++)
for (y=i+1; y<Balls.length; y++)
for (x=y+1; x<Balls.length; x++)
{
if (Balls[i].split('/')[0] == Balls[y].split('/')[0] && (Balls[i].split('/')[1] + '/' + Balls[y].split('/')[1] == Balls[x] || Balls[y].split('/')[1] + '/' + Balls[i].split('/')[1] == Balls[x]))
{
document.write(c + ": " + Balls[i] + " - " + Balls[y] + " - " + Balls[x] + '<br>');
Triangles[c] = Balls[i] + " - " + Balls[y] + " - " + Balls[x];
c++;
}
}
function sFact(num)
{
var rval=1;
for (var i = 2; i <= num; i++)
rval = rval * i;
return rval;
}
結果:
Colors: 6
red,blue,green,yellow,white,black
Balls: 15
1. red/blue
2. red/green
3. red/yellow
4. red/white
5. red/black
6. blue/green
7. blue/yellow
8. blue/white
9. blue/black
10. green/yellow
11. green/white
12. green/black
13. yellow/white
14. yellow/black
15. white/black
Triangles:
1. red/blue - red/green - blue/green
2. red/blue - red/yellow - blue/yellow
3. red/blue - red/white - blue/white
4. red/blue - red/black - blue/black
5. red/green - red/yellow - green/yellow
6. red/green - red/white - green/white
7. red/green - red/black - green/black
8. red/yellow - red/white - yellow/white
9. red/yellow - red/black - yellow/black
10. red/white - red/black - white/black
11. blue/green - blue/yellow - green/yellow
12. blue/green - blue/white - green/white
13. blue/green - blue/black - green/black
14. blue/yellow - blue/white - yellow/white
15. blue/yellow - blue/black - yellow/black
16. blue/white - blue/black - white/black
17. green/yellow - green/white - yellow/white
18. green/yellow - green/black - yellow/black
19. green/white - green/black - white/black
20. yellow/white - yellow/black - white/black
什麼。將計算的組合的總量的式?
另一個問題是,例如,如果Alice有一組如上所述的球,並且Bob有一組類似球具有Alice所具有的一些顏色以及她所不具備的一些顏色(反之亦然),所以對於N個參與者,我如何計算來自所有參與者的3個組合的最大數量? (每位參賽者允許超過1球)
謝謝。
三角形總數是顏色三元組的數量= X! /(3!*(X-3)!) – 2014-08-27 18:02:16