计算n元笛卡儿积 [英] Calculate n-ary Cartesian Product

问题描述

给定两个列表，我可以生成这两个列表的笛卡尔乘积的所有排列列表

：

  permute :: [a]  - > [a]  - > [[a]] 
 permute xs ys = [[x，y] | x< -xs，y<  -  ys] 
 
例子> permute [1,2] [3,4] == [[1,3]，[1,4]，[2,3]，[2,4]] 
  

如何扩展排列，以便不使用两个列表，而是使用列表的列表（长度为n）并返回列表的列表（长度为n）

  permute :: [[a]]  - > [[a]] 
 
示例> permute [[1,2]，[3,4]，[5,6]] 
 == [[1,3,5]，[1,3,6]，[1,4,5] ，[1,4,6]] --etc 
  

Hoogle上找不到任何相关内容..唯一匹配签名的函数是 transpose ，它不会产生所需的输出。

编辑：我认为这个2列表版本实质上是笛卡尔产品，但我无法换行我的头脑是执行 n-ary Cartesian产品。任何指针？

解决方案

  Prelude>序列[[1,2]，[3,4]，[5,6]] 
 [[1,3,5]，[1,3,6]，[1,4,5]，[ 1,4,6]，[2,3,5]，[2,3,6]，[2,4,5]，[2,4,6]] 
  

Given two lists, I can produce a list of all permutations the Cartesian Product of these two lists:

permute :: [a] -> [a] -> [[a]]
permute xs ys = [ [x, y] | x <- xs, y <- ys ]

Example> permute [1,2] [3,4] == [ [1,3], [1,4], [2,3], [2,4] ]

How do I extend permute so that instead of taking two lists, it takes a list (length n) of lists and returns a list of lists (length n)

permute :: [[a]] -> [[a]]

Example> permute [ [1,2], [3,4], [5,6] ]
            == [ [1,3,5], [1,3,6], [1,4,5], [1,4,6] ] --etc

I couldn't find anything relevant on Hoogle.. the only function matching the signature was transpose, which doesn't produce the desired output.

Edit: I think the 2-list version of this is essentially the Cartesian Product, but I can't wrap my head around implementing the n-ary Cartesian Product. Any pointers?

解决方案

Prelude> sequence [[1,2],[3,4],[5,6]]
[[1,3,5],[1,3,6],[1,4,5],[1,4,6],[2,3,5],[2,3,6],[2,4,5],[2,4,6]]

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