I was given a puzzle as a present. It consists of 4 cubes, arranged side by side. The faces of each cube are one of four colours.
To solve the puzzle, the cubes must
Here's a first try at a list version. I think it could be cleaned up a bit.
let rec cart nll =
let f0 n nll =
match nll with
| [] -> [[n]]
| _ -> List.map (fun nl->n::nl) nll
match nll with
| [] -> []
| h::t -> List.collect (fun n->f0 n (cart t)) h
Use recursion: the cartesian product of n lists {L1..LN} is the collection of lists you get when you add each element in L1 to each sublist in the cartesian product of lists {L2..LN}.
let rec cart1 LL =
match LL with
| [] -> Seq.singleton []
| L::Ls -> seq {for x in L do for xs in cart1 Ls -> x::xs}
Example:
> cart1 [[1;2];[3;4;5];[6;7]] |> Seq.toList;;
val it : int list list =
[[1; 3; 6]; [1; 3; 7]; [1; 4; 6]; [1; 4; 7]; [1; 5; 6]; [1; 5; 7]; [2; 3; 6];
[2; 3; 7]; [2; 4; 6]; [2; 4; 7]; [2; 5; 6]; [2; 5; 7]]
The cartesian product of [1;2] [3;4;5] and [6;7] is the union of {1 appended to each list in cart [[3;4;5];[6;7]]} and {2 appended to each list in cart [[3;4;5];[6;7]]}. This is the second clause in the match statement.
Here's a solution 'a list list -> Seq<'a list>
to calculate the Cartesian product of n lists, with lazy evaluation. I wrote it to be an F# analogue of Python's itertools.product
let product lists =
let folder list state =
state |> Seq.allPairs list |> Seq.map List.Cons
Seq.singleton List.empty |> List.foldBack folder lists
It's based on List.allPairs
which was introduced in F# 4.0.