% there are n teams - the number of teams is odd 
% so, we assume that it is of the form 2k+1
% and will receive the input k 
% and define a constant n 

const n = 2*k + 1.

team(1..n).

#domain team(T1;T2;T3).

% each pair of teams (X,Y) plays one game, at the location 
% of the first team or the second team 

1 { game(X,Y,Z) : good_location(X,Y,Z)} 1 :- team(X), team(Y), X > Y.

good_location(T1,T2,T1).
good_location(T1,T2,T2).

% this is to count the number of games plays at a location 
% X,Y plays at one location is the same as Y,X plays at the same
% location 

play_at(T1,T2,T3) :- game(T1,T2,T3).
play_at(T2,T1,T3) :- game(T1,T2,T3).

% there should be only k games at each location -- we count each 
% game as play twice so there should be only 2k games at each 
% location -- if there are more than that (2k+1 is n) it is 
% not a good schedule 

:- n { play_at(X,Y,Z) : team(X) : team(Y) }, team(Z).

hide team(_).
hide good_location(_,_,_).
hide play_at(_,_,_).