% 《工程線性代數（MATLAB版）》第一章化為對角行階梯形式子程序ref2
% 
function [A,jb] = ref2(A,tol)
%REF2   求矩陣的對角元素未歸一化的最簡行階梯形式
%   R = REF2(A)  生成A的對角未歸一化的行階梯形式.
%
%   [R,jb] = REF2(A) 同時返回數組 jb:
%       r = length(jb) 是A的秩。
%       x(jb) 為線性系統Ax = b中約束變量, ,
%       A(:,jb) 是A空間中的一組基,
%       R(1:r,jb) 是上三角矩陣.
%
%   [R,jb] = REF2(A,TOL) 按給定的容差判斷秩.
%
%   舍入誤差可能使此程序算出的秩與用 RANK, ORTH and NULL等函數算出的不同.
%
%   參閱 RREFMOVIE, RANK, ORTH, NULL, QR, SVD. RREF

%   CBM, 11/24/85, 1/29/90, 7/12/92.
%   Copyright 1984-2001 The MathWorks, Inc. 
%   $ 5/19/2006 陳懷琛修改

[m,n] = size(A);

% Does it appear that elements of A are ratios of small integers?
[num, den] = rat(A);
rats = isequal(A,num./den);

% Compute the default tolerance if none was provided.
if (nargin < 2), tol = max(m,n)*eps*norm(A,'inf'); end

% Loop over the entire matrix.
i = 1;
j = 1;
jb = [];
while (i <= m) & (j <= n)
   % Find value and index of largest element in the remainder of column j.
   [p,k] = max(abs(A(i:m,j))); k = k+i-1;
   if (p <= tol)
      % The column is negligible, zero it out.
      A(i:m,j) = zeros(m-i+1,1);
      j = j + 1;
   else
      % Remember column index
      jb = [jb j];
      % Swap i-th and k-th rows.
      A([i k],j:n) = A([k i],j:n);
      % Divide the pivot row by the pivot element.
      %% A(i,j:n) = A(i,j:n)/A(i,j);
      % Subtract multiples of the pivot row from all the other rows.
      for k = [1:i-1 i+1:m]
         A(k,j:n) = A(k,j:n) - A(k,j)*A(i,j:n)/A(i,j);
      end
      i = i + 1;
      j = j + 1;
   end
end

% Return "rational" numbers if appropriate.
if rats
    [num,den] = rat(A);
    A=num./den;
end