在Haskell的嵌套列表的每个排列中添加一个元素

Question

Let's say I have a nested list

[1,2,3]

When I use the Data.List module to permutate this list, the result is the following:

> permutations [1,2,3]
=> [[1,2,3],[2,1,3],[3,2,1],[2,3,1],[3,1,2],[1,3,2]]

But I want it such that, I'm concatenating a 0 at the end of each tail. Example:

=> [[1,2,3,0],[2,1,3,0],[3,2,1,0],[2,3,1,0],[3,1,2,0],[1,3,2,0]]

I'm thinking that the only way of doing this is modifying the source code of permutations as a different function but I'm not sure how to incorporate the concatenation into it. This would be the source:

import Data.List

permutations            :: [a] -> [[a]]
permutations xs0        =  xs0 : perms xs0 []
  where
    perms []     _  = []
    perms (t:ts) is = foldr interleave (perms ts (t:is)) (permutations is)
      where interleave    xs     r = let (_,zs) = interleave' id xs r in zs
            interleave' _ []     r = (ts, r)
            interleave' f (y:ys) r = let (us,zs) = interleave' (f . (y:)) ys r
                                     in  (y:us, f (t:y:us) : zs)

If there's a better way to approach this by using the permutation function without permutating the zero, it would be much appreciated.

Answer 1

Although it might be more efficient to do this at the permutation level, we can simply first do all the permutations, and then post process this, with a map , so:

import Data.List(permutations)

our_perm :: Num a => [a] -> [[a]]
our_perm = map (++[0]) . permutations