Set permutations

There is a part of my project that calls for a permuted set, that is given the original set of elements create a set of sets where each set contains a permutation of the original.  It turns out that this is more or less simple once you break the problem down into smaller parts.

There are a few algorithms to generate permutations,  I chose interleaving for it being relatively simple and straight forward.
For a set of [x,y, x] permute the first two elements
[[x, y], [y, x]]
and then interleave each set with z
[[z, x, y], [x, z, y], [x, y, z], [z, y, x], [y, z, x], [y, x, z]]

First we create a function that would take a list an a value and create permutations of that list with a value:

interleaveSet :: [a] -> a -> [[a]]
interleaveSet [] value = [[value]]
interleaveSet xs value = [ take pos xs ++ [value] ++ drop pos xs | pos <- [0..length(xs)] ]

Here are some samples:

*Main> interleaveSet [] 1
[[1]]
*Main> interleaveSet [1] 2
[[2,1],[1,2]]
*Main> interleaveSet [1,2,3,4] 5
[[5,1,2,3,4],[1,5,2,3,4],[1,2,5,3,4],[1,2,3,5,4],[1,2,3,4,5]]

So we now have a way of interleaving a set with a value, but remember that what we need to do is build this from an existing set.  What we need is something that will take the output of interleaveSet and interleave the sets created by it with the next value.  I was at first tempted to use a list comprehension once again

interleaveSets sets value = [ interleaveSet set value | set <- sets ]

but since interleaveSet outputs a [[a]] and interleaveSets will create a list of those, our return value will be [[[a]]], which is not what we want at all.


*Main> interleaveSets (interleaveSet [1,2,3,4] 5) 6
[[[6,5,1,2,3,4],[5,6,1,2,3,4],[5,1,6,2,3,4],[5,1,2,6,3,4],[5,1,2,3,6,4],[5,1,2,3,4,6]],[[6,1,5,2,3,4],[1,6,5,2,3,4],[1,5,6,2,3,4],[1,5,2,6,3,4],[1,5,2,3,6,4],[1,5,2,3,4,6]],[[6,1,2,5,3,4],[1,6,2,5,3,4],[1,2,6,5,3,4],[1,2,5,6,3,4],[1,2,5,3,6,4],[1,2,5,3,4,6]],[[6,1,2,3,5,4],[1,6,2,3,5,4],[1,2,6,3,5,4],[1,2,3,6,5,4],[1,2,3,5,6,4],[1,2,3,5,4,6]],[[6,1,2,3,4,5],[1,6,2,3,4,5],[1,2,6,3,4,5],[1,2,3,6,4,5],[1,2,3,4,6,5],[1,2,3,4,5,6]]]

Since we want to be able to feed it back to interleaveSets, this is not what we want.

This will do:

interleaveSets :: [[a]] -> a -> [[a]]
interleaveSets [] value = [[]]
interleaveSets (s:sets) value = interleaveSet s value ++ interleaveSets sets value

*Main> interleaveSets (interleaveSet [1,2,3,4] 5) 6
[[6,5,1,2,3,4],[5,6,1,2,3,4],[5,1,6,2,3,4],[5,1,2,6,3,4],[5,1,2,3,6,4],[5,1,2,3,4,6],[6,1,5,2,3,4],[1,6,5,2,3,4],[1,5,6,2,3,4],[1,5,2,6,3,4],[1,5,2,3,6,4],[1,5,2,3,4,6],[6,1,2,5,3,4],[1,6,2,5,3,4],[1,2,6,5,3,4],[1,2,5,6,3,4],[1,2,5,3,6,4],[1,2,5,3,4,6],[6,1,2,3,5,4],[1,6,2,3,5,4],[1,2,6,3,5,4],[1,2,3,6,5,4],[1,2,3,5,6,4],[1,2,3,5,4,6],[6,1,2,3,4,5],[1,6,2,3,4,5],[1,2,6,3,4,5],[1,2,3,6,4,5],[1,2,3,4,6,5],[1,2,3,4,5,6],[]]

I am not sure what that empty element at the end is, will have to deal with it later (I am sure it is a side effect of my base case)

Ah the solution is that the base case return a [] because any set ++ [] is just set

*Main> [1,2,3] ++ []
[1,2,3]

And here is the new result:
*Main>  interleaveSets (interleaveSet [1,2,3,4] 5) 6
[[6,5,1,2,3,4],[5,6,1,2,3,4],[5,1,6,2,3,4],[5,1,2,6,3,4],[5,1,2,3,6,4],[5,1,2,3,4,6],[6,1,5,2,3,4],[1,6,5,2,3,4],[1,5,6,2,3,4],[1,5,2,6,3,4],[1,5,2,3,6,4],[1,5,2,3,4,6],[6,1,2,5,3,4],[1,6,2,5,3,4],[1,2,6,5,3,4],[1,2,5,6,3,4],[1,2,5,3,6,4],[1,2,5,3,4,6],[6,1,2,3,5,4],[1,6,2,3,5,4],[1,2,6,3,5,4],[1,2,3,6,5,4],[1,2,3,5,6,4],[1,2,3,5,4,6],[6,1,2,3,4,5],[1,6,2,3,4,5],[1,2,6,3,4,5],[1,2,3,6,4,5],[1,2,3,4,6,5],[1,2,3,4,5,6]]

And finally to put it all together:

permuteSet :: [a] -> [[a]]
permuteSet [] = []
permuteSet set = foldl(\acc e -> interleaveSets acc e) seed (tail set)
  where seed = interleaveSet [] (head set)

Lets test

*Main > permuteSet [1,2,3,4]
[[4,3,2,1],[3,4,2,1],[3,2,4,1],[3,2,1,4],[4,2,3,1],[2,4,3,1],[2,3,4,1],[2,3,1,4],[4,2,1,3],[2,4,1,3],[2,1,4,3],[2,1,3,4],[4,3,1,2],[3,4,1,2],[3,1,4,2],[3,1,2,4],[4,1,3,2],[1,4,3,2],[1,3,4,2],[1,3,2,4],[4,1,2,3],[1,4,2,3],[1,2,4,3],[1,2,3,4]]

Looks about right

*Main> length $ permuteSet [1,2,3,4]

24

Number of permutations is equal to n! / (n -k)!

Where k is the number of elements from n per permutation, in our case k = n so that turns into 0! which we know is one

and the factorial of 4 is 4 * 3 * 2 *1 = 24