Haskell 排列与输出列表的长度

Question

I have such code for creating permutations:

--unique permutation
perm :: [t] -> [[t]]
perm [] = [[]]
perm (x:xs) = [(y:zs) | (y,ys) <- views (x:xs), zs <- perm ys]

--create all possible views
views :: [t] -> [(t,[t])]
views [] = []
views (x:xs) = ((x,xs) : [ (y,(x:ys)) | (y,ys) <- views xs ])

However I want to limit the output to a certain length. For example, it should take a parameter specifying the number of elements in the ouput of the permutation. I want to create k distinct permutations drawn from a list [0..9]. So, for example, if k will be equal to 3, then the output should be something like this:[1,2,3], [9,8,7], [0,6,8] etc..

Right now if i pass [0..9] to the function it will generate a list with permutations of length 10. I am struggling to come up with a solution. In the end, the function should look like perm k list Thank you!

It is connected with this question: Verbal Arithmetics in Haskell (SEND + MORE = MONEY)

Answer 1

Do you mean something like this?

import Data.List (permutations)

choose n list = concatMap permutations $ choose' list [] where
  choose' []     r = if length r == n then [r] else []
  choose' (x:xs) r | length r == n = [r]
                   | otherwise     = choose' xs (x:r) 
                                  ++ choose' xs r

Output:

*Main> choose 2 [0..5]
[[1,0],[0,1],[2,0],[0,2],[3,0],[0,3],[4,0],[0,4],[5,0],[0,5],[2,1]
,[1,2],[3,1],[1,3],[4,1],[1,4],[5,1],[1,5],[3,2],[2,3],[4,2],[2,4]
,[5,2],[2,5],[4,3],[3,4],[5,3],[3,5],[5,4],[4,5]]

Answer 2

Will replicateM do what you need?

Prelude Control.Monad> take 10 $ replicateM 3 [0..9]
[[0,0,0],[0,0,1],[0,0,2],[0,0,3],[0,0,4],[0,0,5],[0,0,6],[0,0,7],[0,0,8],[0,0,9]]

Prelude Control.Monad> take 10 $ replicateM 4 [1,3,3,7]
[[1,1,1,1],[1,1,1,3],[1,1,1,3],[1,1,1,7],[1,1,3,1],[1,1,3,3],[1,1,3,3],[1,1,3,7],[1,1,3,1],[1,1,3,3]]

Prelude Control.Monad> take 10 $ replicateM 2 [4,2]
[[4,4],[4,2],[2,4],[2,2]]