use recursive function to generate combination in Swift - swift

Question

I have five arrays. I can use five for-loop to generate the result, but I need a recursive function so that I can generate the combinations from more arrays. How do I use a recursive function to generate the combinations? Each combination includes one element from the five arrays.

Answer 1

I'm not sure why you need it to be recursive, but there are a few ways to do it, anyway. You can use higher-order functions to make an Array extension that does what you want:
extension Array {
func everyOf<S: SequenceType where S.Generator.Element == T>(ar: S...) -> [[T]] {
return ar.reduce(self.map{[$0]}){
perms, items in perms.flatMap {
perm in Swift.map(items){ perm + [$0] }
}
}
}
}
This would work like:
[1, 2].everyOf([3, 4]) //[[1, 3], [1, 4], [2, 3], [2, 4]]
or:
[1, 2].everyOf([3, 4], [5, 6])
//[
// [1, 3, 5],
// [1, 3, 6],
// [1, 4, 5],
// [1, 4, 6],
// [2, 3, 5],
// [2, 3, 6],
// [2, 4, 5],
// [2, 4, 6]
//]
But the different arrays don't have to be the same length:
[1, 2].everyOf([3], [5, 6]) // [[1, 3, 5], [1, 3, 6], [2, 3, 5], [2, 3, 6]]
And you can do it as a non-method as well:
func everyOf<T>(ar: [[T]]) -> [[T]] {
return dropLast(ar).reduce(ar.last!.map{[$0]}){
perms, items in perms.flatMap{
perm in Swift.map(items){ perm + [$0] }
}
}
}
everyOf([[1, 2], [3, 4]]) // [[3, 1], [3, 2], [4, 1], [4, 2]]
If you're really into the 'aul recursion, though, here you go:
func everyOf<T>(seqs: [[T]]) -> [[T]] {
return last(seqs).map {
fSeq in everyOf(Array(dropLast(seqs))).flatMap {
seq in fSeq.map{ seq + [$0] }
}
} ?? [[]]
}
everyOf([[1, 2], [3, 4]]) // [[3, 1], [3, 2], [4, 1], [4, 2]]