[--I，++ J]的Prolog递归构建列表，而J> 0

Question

I'm trying to put together a Prolog program that will take in a number and an output list, and essentially make a list of the form [I, J], where I+J = N. Here's the code I have, with the output I'm getting:

loop(I, J, [[X,Y]|Lout]):-
  I > 0,
  I1 is I-1,
  J1 is J+1,
  X is I1,
  Y is J1,
  loop(I1, J1, Lout).
loop(I, J, [[I,J]]).

listFromLoop(N, W):-
  loop(N,0,W).

Output:

?- listFromLoop(4,W).
W = [[3, 1], [2, 2], [1, 3], [0, 4], [0, 4]] ;
W = [[3, 1], [2, 2], [1, 3], [1, 3]] ;
W = [[3, 1], [2, 2], [2, 2]] ;
W = [[3, 1], [3, 1]] ;
W = [[4, 0]].

I'm getting two extra elements at the end ([0,4], [0,4]) that I don't need, as well as having to step through it, rather than it all being displayed as a single list.

Here is what it is supposed to look like:

?- listFromLoop(4,W).
W = [[3, 1], [2, 2], [1, 3]].

I'm close, but I think my main problem is that I'm trying to code it like I would a C++ or Java while-loop. However, that doesn't seem to be the proper way to go about it in Prolog.

Any help would be much appreciated!

EDIT

I figured it out! After a little bit more reading, it seemed I was missing my base case for the recursive version. Here is what I ended up with that fixed it. Although, I'm not certain this is necessarily the correct/most efficient way to do this in Prolog.

loop(1,_,[]):- !.
loop(I, J, [[X,Y]|Lout]):-
  I > 0,
  I1 is I-1,
  J1 is J+1,
  X is I1,
  Y is J1,
loop(I1, J1, Lout).

listFromLoop(N, W):-
  loop(N,0,W).

Answer 1

Your code is fine, but I would do this kind of thing to clean it up a bit:

build(1, _, []).
build(I, J, [[X, Y]|Zs]) :-
    I > 1,
    X is I - 1,
    Y is J + 1,
    build(X, Y, Zs).

build(N, W) :- build(N, 0, W).

You're not really using a loop - it's recursion that builds the list. So just call it build . It's also perfectly fine to define predicates with the same name but a different number of parameters.

I've also named Lout as Zs as this helps indicate that it's a list - prolog doesn't usually care about inputs or outputs anyway. You could run the query ?- build(6, [[5, 1], [4, 2], [3, 3], [2, 4], [1, 5]]). and you'll get success even though you've sent the output list in as an input parameter.

Answer 2

a more idiomatic way:

listFromLoop(U,W) :- U_ is U-1,
    findall([I,J],(between(1,U_,J),I is U_-J+1),W).