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Home/ Questions/Q 814765
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T01:34:26+00:00 2026-05-15T01:34:26+00:00

The best thing I could come up with so far is this function: numberFromList([X],

  • 0

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.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T01:34:27+00:00Added an answer on May 15, 2026 at 1:34 am

    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)

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