queens(N):-
length(Qs,N),
Qs :: 1..N,
foreach(I in 1..N-1, J in I+1..N,
(Qs[I] #\= Qs[J],
abs(Qs[I]-Qs[J]) #\= J-I)),
labeling([ff],Qs),
writeln(Qs).
The array notation on lists helps shorten the description. Without it, the foreach loop in the program would have to be written as follows:
foreach(I in 1..N-1, J in I+1..N,[Qi,Qj],
(nth(Qs,I,Qi),
nth(Qs,J,Qj),
Qi #\= Qj,
abs(Qi-Qj) #\= J-I)),
where Qi and Qj are declared local to each iteration. The following gives a program for the N-queens problem, which uses a Boolean variable for each square on the board.
bool_queens(N):-
new_array(Qs,[N,N]),
Vars @= [Qs[I,J] : I in 1..N, J in 1..N],
Vars :: 0..1,
foreach(I in 1..N, % one queen in each row
sum([Qs[I,J] : J in 1..N]) #= 1),
foreach(J in 1..N, % one queen in each column
sum([Qs[I,J] : I in 1..N]) #= 1),
foreach(K in 1-N..N-1, % at most one queen in each diag
sum([Qs[I,J] : I in 1..N, J in 1..N, I-J=:=K]) #=< 1),
foreach(K in 2..2*N,
sum([Qs[I,J] : I in 1..N, J in 1..N, I+J=:=K]) #=< 1),
labeling(Vars),
foreach(I in 1..N,[Row],
(Row @= [Qs[I,J] : J in 1..N], writeln(Row))).