function [ x_num, x ] = r8mat_rref_solve_binary ( m, n, a, b )
%*****************************************************************************80
%
%% r8mat_rref_solve_binary seeks binary solutions of an RREF system.
%
% Discussion:
%
% An MxN linear system Ax = b is considered, in which only binary
% solutions are allowed; that is, each entry of x is either 0 or 1.
%
% The matrix A and right hand side B are assumed to have been converted
% to row-reduced echelon form. Therefore, the matrix A satisfies:
%
% * The first nonzero entry in each row is 1.
%
% * The leading 1 in a given row occurs in a column to
% the right of the leading 1 in the previous row.
%
% * Rows which are entirely zero must occur last.
%
% The matrix is in reduced row echelon form if, in addition to
% the first three conditions, it also satisfies:
%
% * Each column containing a leading 1 has no other nonzero entries.
%
% Licensing:
%
% This code is distributed under the GNU LGPL license.
%
% Modified:
%
% 13 May 2018
%
% Author:
%
% John Burkardt
%
% Input:
%
% integer M, N, the number of rows and columns of
% the RREF matrix A.
%
% real A(M,N), the RREF matrix to be analyzed.
%
% real B(M), the RREF right hand side.
%
% Output:
%
% integer X_NUM, the number of binary solutions.
% Note that there may be no binary solutions at all.
%
% real X(N,X_NUM), the solutions.
%
%
% Remove the zero rows of A.
%
while ( 0 < m )
if ( sum ( abs ( a(m,1:n) ) ) == 0.0 )
a = a(1:m-1,1:n);
m = m - 1;
else
break
end
end
%
% Number of dimensions of freedom is n - m.
%
klog = n - m;
%
% Determine the indeterminate variables.
% Insert corresponding data in A and B.
%
freedom = [];
if ( 0 < klog )
for j = 1 : n
if ( j <= m )
if ( a(j,j) ~= 1 )
a = [ a(1:j-1,1:n); r8vec_identity_row(n,j); a(j:m,1:n) ];
b = [ b(1:j-1); 0; b(j:m) ];
freedom = [ freedom; j ];
m = m + 1;
end
else
a = [ a; r8vec_identity_row(n,j) ];
b = [ b; 0 ];
freedom = [ freedom; j ];
m = m + 1;
end
end
end
%
% The indeterminate variables have a simple equation
% x(i) = b(i) = 0 or 1
% Set up and solve every variation of this system.
% If a solution is binary, accept it.
%
x_num = 0;
x = [];
binary = zeros ( klog, 1 );
while ( true )
b2 = b;
for k = 1 : klog
i = freedom(k);
b2(i) = binary(k);
end
y = r8mat_u_solve ( n, a, b2 );
if ( r8vec_is_binary ( n, y ) )
x_num = x_num + 1;
x(1:n,x_num) = y(1:n,1);
end
binary = r8vec_binary_next ( klog, binary );
if ( sum ( binary ) == 0 )
break
end
end
return
end