Permutations of two lists

问题

I have a char list in OCAML. I would like to create a ( (char * bool) list) list of every combination of the chars with true and false.

What I have guess I have to do is something like a List.fold_left, but I am not quite sure how to pull it off.

This is the outline that I tried (OCAML syntax, but not run-able):

let rec var_perm var_list options = 
    match var_list with
        | [] -> options
        | x :: v' ->
            ((x, true) :: (var_perm_intern v')) :: ((x, false) :: (var_perm_intern v'))
;;

let all_options = var_perm ['a';'b'] [];;

should return

[
    [('a',true);('b',true)];
    [('a',true);('b',false)];
    [('a',false);('b',true)];
    [('a',false);('b'false)];
]

Edit: Another example:

let all_options = var_perm ['u';'w';'y'] [];;

should return (order is not important)

[
    [('u',false);('w',false);('y',false)];
    [('u',false);('w',false);('y',true )];
    [('u',false);('w',true );('y',false)];
    [('u',false);('w',true );('y',true )];
    [('u',true );('w',false);('y',false)];
    [('u',true );('w',false);('y',true )];
    [('u',true );('w',true );('y',false)];
    [('u',true );('w',true );('y',true )];
]

回答1:

You're close to a correct solution. Specifically:

you must remove the _intern suffix in your recursive call
the "options" parameter is useless (look at how you do your recursive call, passing only one parameter v'), so you must find out what to return in the [] case
the concatenation of "the results of v', plus true for the head var" and " the results of v', plus false for the head var" should be written foo @ bar rather than foo :: bar, because those are two lists you're concatenating, not one element added to a list.

来源：https://stackoverflow.com/questions/14740866/permutations-of-two-lists

标签

ocaml