Question

The resulting tree of this recursion is not what I what, which probably proves that I don’t fully understand the behaviour of lists/tuples in recursion. If someone could explain what I did wrong in this example and also explain the right way to think I’d be very grateful.

move([],{Main, One, Two}) ->
    {Main, One, Two};
move([X|Xr], {Main, One, Two}) ->
    [{Main, One, Two}, move(Xr, single(X, {Main, One, Two}))].

Desired result (one list containing 3 tuples):

[{[a,b],[],[]}, {[a],[b],[]}, {[],[b],[a]}, {[b],[],[a]}]

Actual Result(a list containing a tuple and a list, containing a tuple and a list…):

[{[a,b],[],[]},[{[a],[b],[]},[{[],[b],[a]},[{[b],[],[a]}]]]]

Answer 1

You have two problems:

1. The | instead of , as @nmichaels mentioned.
2. The function move/2 returns a list so the terminating clause must also return a list. This is not seen in your example as the first problem hides it.

So the resulting code would be:

move([X|Xr], {Main, One, Two}) ->
    [{Main, One, Two} | move(Xr, single(X, {Main, One, Two}))];
move([], {Main, One, Two}) ->
    [{Main, One, Two}].

I flipped the order of the clauses as I personally prefer writing them this way. No fundamental difference in this case. I am assuming that single/2 returns a tuple.

You can actually optimise this code by removing all knowledge of the tuple from move/2 as it never actually uses the internal structure. So:

move([X|Xr], Tuple) ->
    [Tuple | move(Xr, single(X, Tuple))];
move([], Tuple) ->
    [Tuple].