% 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.
const ngames = k*(2*k+1).

team(1..n).
location(1..n).

#domain team(T1;T2;T3;T4).

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

2*ngames { game(X,Y,Z) : team(X) : team(Y) : location(Z) } 2*ngames.

% a team does not play itself 

:- game(T1,T1,T3).

% teams do not play at neutral place

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

% each pair plays only one 

% one game is scheduled only one 

:- game(T1,T2,T3), game(T1,T2,T4), neq(T3,T4).

% if A plays B means that B plays A 

game(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 { game(X,Y,Z) : team(X) : team(Y) }, location(Z).

hide team(_).
hide location(_).