我正在尝试使用答案集编程，Clingo和MKAtoms解决二进制数独难题。我有一套有效的代码，但是没有返回任何答案。
我已经检查过代码，但是不确定问题所在。
% Binary Sudoku solver: Binary Sudoku is a similar puzzle to sudoku.
% The rules are as follows
% 1. Each row and column must have the same number of 1's and 0's
% 2. Each row and column must be unique
% 3. Each number may not appear in consecutive triples
%
% The program works in conjuction with python 3.6, Clingo, and Mkatoms to create the required files and solve the user input puzzle.
%
% Required Files:
% BinaryTemp.sm
% BinaryOut.sm
% BinaryPy.sm - This File
%
% Using the defined rules Clingo will attempt to find an answer set that will statisfy all conditions
num(0..1).
range(0..6).
% There may only be one number in each X,Y position
1{pos(N,X,Y):num(N)}1:-num(X), num(Y).
% Each row and column must have the same number of 1’s and 0’s
% Row check - check each rows sum to see if they hold the same number of 1's
total_row(S) :- S = #sum{ N : pos(N,X,Y1), pos(N,X+1,Y1), pos(N,X+2,Y1), pos(N,X+3,Y1), pos(N,X+4,Y1), pos(N,X+5,Y1)}.
total_row1(S) :- S = #sum{ N : pos(N,X,Y2), pos(N,X+1,Y2), pos(N,X+2,Y2), pos(N,X+3,Y2), pos(N,X+4,Y2), pos(N,X+5,Y2)}.
:- total_row(S) != total_row1(S), range(S).
% Column Check
total_column(S) :- S = #sum{ N : pos(N,X1,Y), pos(N,X,Y+1), pos(N,X1,Y+2), pos(N,X1,Y+3), pos(N,X1,Y+4), pos(N,X1,Y+5)}.
total_column1(S) :- S = #sum{ N : pos(N,X2,Y), pos(N,X2,Y+1), pos(N,X2,Y+2), pos(N,X2,Y+3), pos(N,X2,Y+4), pos(N,X2,Y+5)}.
:- total_column(S) != total_column1(S), range(S).
% Each row and column must be unique
% Row check - check if 2 values in seperate rows are different
different(X1,X2) :- pos(N1,X1,Y), pos(N2,X2,Y), N1!=N2.
:-pos(N,X1,Y), pos(N,X2,Y), X1!=X2, not different(X1,X2).
% Column check
different(Y1,Y2) :- pos(N1,X,Y1), pos(N2,X,Y2), N1!=N2.
:-pos(N,X,Y1), pos(N,X,Y2), Y1!=Y2, not different(Y1,Y2).
% Each number may not appear in consecutive triples
:-pos(N,X,Y), pos(N,X+1,Y), pos(N,X+2,Y), num(N).
:-pos(N,X,Y), pos(N,X,Y+2), pos(N,X,Y+2), num(N).
#show pos/3.
输入文件：
pos（0,1,1）。
pos（1,1,4）。
pos（1,2,2）。
pos（0,2,5）。
pos（1,3,6）。
pos（1,4,3）。
pos（0,4,4）。
pos（1,5,1）。
pos（0,5,4）。
pos（0,6,2）。
pos（1,6,5）。
没有错误，但应返回与上述规则匹配的答案的位置。