function [ a, done, active, dir, change ] = vec_gray_next ( n, base, ...
a, done, active, dir )
%*****************************************************************************80
%
%% vec_gray_next() computes the elements of a product space.
%
% Discussion:
%
% The elements are produced one at a time.
%
% This routine handles the case where the number of degrees of freedom may
% differ from one component to the next.
%
% A method similar to the Gray code is used, so that successive
% elements returned by this routine differ by only a single element.
%
% A previous version of this routine used internal static memory.
%
% Example:
%
% N = 2, BASE = ( 2, 3 ), DONE = TRUE
%
% A DONE CHANGE
% --- ----- ------
% 0 0 FALSE 1
% 0 1 FALSE 2
% 0 2 FALSE 2
% 1 2 FALSE 1
% 1 1 FALSE 2
% 1 0 FALSE 2
% 1 0 TRUE -1
%
% Licensing:
%
% This code is distributed under the MIT license.
%
% Modified:
%
% 16 May 2015
%
% Author:
%
% John Burkardt
%
% Reference:
%
% Dennis Stanton, Dennis White,
% Constructive Combinatorics,
% Springer, 1986,
% ISBN: 0387963472,
% LC: QA164.S79.
%
% Input:
%
% integer N, the number of components.
%
% integer BASE(N), contains the number of degrees of
% freedom of each component. The output values of A will
% satisfy 0 <= A(I) < BASE(I).
%
% integer A(N). On the first call, the input value of A doesn't
% matter. Thereafter, it should be the same as
% its output value from the previous call.
%
% logical DONE. On the first call, the user must
% set DONE to TRUE. Thereafter, the input value DONE should simply
% be the output on the previous call.
%
% integer ACTIVE(N), DIR(N), work arrays needed by the function.
% The user should create them before the first call, but
% thereafter should not change their values, passing in the output values
% of the previous call for the next call.
%
% Output:
%
% integer A(N), the next vector.
%
% logical DONE. If DONE is FALSE, then the program has computed another
% entry in A. If the output value of DONE is TRUE, then there are no
% more entries.
%
% integer ACTIVE(N), DIR(N), work arrays needed by the function.
%
% integer CHANGE, the index of the element whose
% value was changed. On return from the first call, CHANGE
% is 1, even though all the elements have been "changed". On
% return with DONE equal to TRUE, CHANGE is -1.
%
%
% The user is calling for the first time.
%
if ( done )
done = false;
a(1:n) = 0;
dir(1:n) = 1;
active(1:n) = 1;
for i = 1 : n
if ( base(i) < 1 )
fprintf ( 1, '\n' );
fprintf ( 1, 'vec_gray_next(): Warning!\n' );
fprintf ( 1, ' For index I = %d\n', i );
fprintf ( 1, ' the nonpositive value of BASE(I) = %d\n', base(i) );
fprintf ( 1, ' which was reset to 1!\n' );
base(i) = 1;
active(i) = 0;
elseif ( base(i) == 1 )
active(i) = 0;
end
end
change = 1;
return
end
%
% Find the maximum active index.
%
change = -1;
for i = 1 : n
if ( active(i) ~= 0 )
change = i;
end
end
%
% If there are NO active indices, we have generated all vectors.
%
if ( change == -1 )
done = true;
return
end
%
% Increment the element with maximum active index.
%
a(change) = a(change) + dir(change);
%
% If we attained a minimum or maximum value, reverse the direction
% vector, and deactivate the index.
%
if ( a(change) == 0 || a(change) == base(change) - 1 )
dir(change) = - dir(change);
active(change) = 0;
end
%
% Activate all subsequent indices.
%
for i = change + 1 : n
if ( 1 < base(i) )
active(i) = 1;
end
end
return
end