Haskell生成子集 [英] Haskell generate subsets

问题描述

我有一个函数'子集'，它可以生成给定集合的所有子集：

 子集:: [Int] - > [[Int]] 
子集[] = [[]] 
子集（x：xs）=子集xs ++映射（x :)（子集xs）
  
 
 
如何在另一个函数中结合map，foldl和filter来返回所有子元素总和为0的元素？
 
 
 
 **例如：**  
 
 
  set = [1，-1,5 ，2，-2,3] 
 result = [[1，-1]，[2，-2]，[ -  1，-2,3]] 
  
 
 
解决方案
您已有子集。所以我们需要一个函数 
 
 
  filterSubs :: [[Int]]  - > [[Int]] 
 filterSubs =  - 删除所有不加和为0的子集
  
 
 
 <所以接下来我们需要一个谓词 
 
 
  sumZero :: [Int]  - > Bool 
 sumZero xs = sum xs == 0 
  
现在，使用这个和 filter 很容易构造 filterSubs 。我会把这个留给你来弄清楚它是如何工作的。然后我们的解决方案是微不足道的 
 
 
  zeroSubs = filterSubs。子集
  
 
I have a function 'subsets' which generate all the subsets of a given set:
subsets :: [Int] -> [[Int]]
subsets []  = [[]]
subsets (x:xs) = subsets xs ++ map (x:) (subsets xs)
How can I combine map, foldl and filter in another function to return me all the subsets with elements that sum up to 0?

**Example: **
set    = [1,-1,5,2,-2,3]
result = [[1,-1],[2,-2],[-1,-2,3]]

解决方案
You have subsets already. So we need a function
filterSubs :: [[Int]] -> [[Int]]
filterSubs = --remove all subsets which don't sum to 0
So next we'd need a predicate
sumZero :: [Int] -> Bool
sumZero xs = sum xs == 0
Now, using this and filter it's easy to construct filterSubs. I'll leave this to you to figure out exactly how that works. And then our solution is trivial
zeroSubs = filterSubs . subsets


                        
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