将集合的分区枚举为大小相等的子集 [英] Enumerate partitions of a set into subsets of equal size

问题描述

我有一个集合，我想将其划分为包含相等数量元素的子集合。

I have a set and I want to partition it into sub-set containing equal number of element.

我正在寻找一种快速算法，最好不要使用启发式算法。

I am looking for a fast algorithm and preferably not heuristic one.

提示：

if  n= number of elements in the main set
    l= number of elements in each subset 

蛮力算法是：

1-x<-一次取n物的所有组合没有
重复。 | x | = nCl = n！/（l！*（n-l）！）

2-y< -x的所有
组合，一次取n次，没有重复。 | y | = xCn

1-x <-All combinations of n things taken l  at a time without repetition. |x|=nCl=n!/(l!*(n-l)!)
2-y <-All combinations of x things taken n at a time without repetition. |y|=xCn

3-选择y中的子集，例如在其
元素内没有任何重叠。

3-Select subsets in y such as there is no any overlap within their elements.

答案的数量为：

n!/(l!^(n/l)*(n/l)!)

例如，如果 S = {a，b，c，d} ，如果需要具有2个元素的子集来划分集合S：

For instance if S={a,b,c,d} and if subsets with 2 elements to partition set S is desired:

集合x是：

   (a,b),(a,c),(a,d),(b,c),(b,d),(c,d)

集合y（潜在答案）是：

The set y (potential answers)is :

{(a,b),(a,c)}
{(a,b),(a,d)}
{(a,b),(b,c)}
{(a,b),(b,d)}
{(a,b),(c,d)}       
{(a,c),(a,d)}
{(a,c),(b,c)}
{(a,c),(b,d)}
{(a,c),(c,d)}
{(a,d),(b,c)}   
{(a,d),(b,d)} 
{(a,d),(c,d)}   
{(b,c),(d,d)}   
{(b,c),(c,d)}   
{(b,d),(c,d)}   

，正确的答案是：

S1={(a,b),(c,d)}
S2={(a,c),(b,d)}
S3={(a,d),(b,c)}

仅当n小时才有用。
例如：

The mentioned algorithm is useful only when n is small. For instance when:

 n=90, l=3 =>
|x|=117480 
|y|=1.28827732e+318
and the number of correct answers is `2.533601e+82`.

因此，由于时间性能和内存问题，该算法在大多数情况下不实用。

So the algorithm is not practical for the most case due to time performance and memory issues.

由于结果数量很多，即使拥有并运行有效的算法也很耗时。例如，在上述问题中，答案的数量= 2.533601e + 82

Even having and running an efficient algorithm would be time consuming as the number of results is a lot. For instance in the above problem the number of answers = 2.533601e+82

我不是该领域的专家理论，所以也许这是一个众所周知的问题。

I am not expertise in the set theory so maybe it is a well known problem.

感谢您的帮助。

推荐答案

我想这可能很近...希望您使用的是Java

I think this might be close...hope you're using java

public static void main(String[] args) {
   ArrayList<String> tokens = new ArrayList(new TreeSet<>(Arrays.asList(new String[]{"a","b","c","d"})));

    for (int i = 0; i < tokens.size(); i++) {
      String tokenA = tokens.get(i);
      for (int x = (i + 1); x < tokens.size(); x++) {
        String tokenB = tokens.get(x);
        //String tokenC = tokens.get(x + 1);

        System.out.println(tokenA + "-" + tokenB);
      }
    }
  }

我的输出是

run:
a-b
a-c
a-d
b-c
b-d
c-d
BUILD SUCCESSFUL (total time: 0 seconds)

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