Haskell递归子集 [英] Haskell Recursion Subsets

问题描述

我是Haskell的新手.我发现很难用递归来理解下面的子程序.它的评估方式是什么?

i am new to Haskell. i am find it hard to understand the below subset program with recursion.what is the way it evaluates?

subset :: [a] -> [[a]]
subset [] = [[]]
subset (x:xs) = [zs | ys <- subset xs, zs <- [ys,(x:ys)]]

---

输出:[[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

当我手动评估时，其总结如下

When i manually evaluate it is summing up as below

手动输出:[],[3],[2],[1,2]

我在这里缺少一些逻辑，您能帮我理解上面的递归概念吗?请先评估保护声明的哪一部分，评估顺序?

I am missing some logic here can you please help me understand the above recursion concept and which part of guard statement will be evaluated first ,order of evaluation?

推荐答案

最后一行说，x:xs的子集集是xs的子集集，而x都添加到了这些集中

The last line says that the set of subsets of x:xs is the set of subsets of xs along with x added to each of those sets.

另一种表达方式是

zs <- [ys,(x:ys)]

为每个xs的子集之一精确地生成两个集ys和x:ys.

generates exactly two sets, ys and x:ys for each ys which is one of the subsets of xs.

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