Prolog: find all numbers of unique digits that can be formed from a list of digits

Question

The best thing I could come up with so far is this function:

 numberFromList([X], X) :-  
    digit(X), !.  
 numberFromList(List, N) :-  
    member(X, List),     
    delete(List, X, LX),  
    numberFromList(LX, NX),  
    N is NX * 10 + X.

where digit/1 is a function verifying if an atom is a decimal digit.

The numberFromList(List, N) finds all the numbers that can be formed with all digits from List.
E.g. [2, 3] -> 23, 32. but I want to get this result: [2, 3] -> 2, 3, 23, 32

I spent a lot of hours thinking about this and I suspect you might use something like append(L, _, List) at some point to get lists of lesser length.

I would appreciate any contribution.

Answer 1

You are missing case when you skip digit from list.

 numberFromList([X], X) :-  
    digit(X), !.  
 numberFromList(List, N) :-
    member(X, List),     
    delete(List, X, LX),
    numberFromList(LX, NX),  
    ( % use X
        N is NX * 10 + X
    ; % skip X
        N = NX
    ).

BTW, as @Roland Illig mentioned there is select(X, List, LX) to replace member(X, List), delete(List, X, LX)