## Copyright (C) 2001, 2007 Paul Kienzle
##
## This file is part of Octave.
##
## Octave is free software; you can redistribute it and/or modify it
## under the terms of the GNU General Public License as published by
## the Free Software Foundation; either version 3 of the License, or (at
## your option) any later version.
##
## Octave is distributed in the hope that it will be useful, but
## WITHOUT ANY WARRANTY; without even the implied warranty of
## MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
## General Public License for more details.
##
## You should have received a copy of the GNU General Public License
## along with Octave; see the file COPYING.  If not, see
## <http://www.gnu.org/licenses/>.

## -*- texinfo -*-
## @deftypefn {Function File} {@var{s} =} rat (@var{x}, @var{tol})
## @deftypefnx {Function File} {[@var{n}, @var{d}] =} rat (@var{x}, @var{tol})
##
## Find a rational approximation to @var{x} within tolerance defined
## by @var{tol} using a continued fraction expansion. E.g,
##
## @example
## rat(pi) = 3 + 1/(7 + 1/16) = 355/113
## rat(e) = 3 + 1/(-4 + 1/(2 + 1/(5 + 1/(-2 + 1/(-7))))) 
##        = 1457/536
## @end example
##
## Called with two arguments returns the numerator and denominator separately
## as two matrices.
## @end deftypefn
## @seealso{rats}

function [n,d] = rat(x,tol)

  if (nargin != [1,2] || nargout > 2)
    print_usage ();
  endif

  y = x(:);

  ## replace inf with 0 while calculating ratios
  y(isinf(y)) = 0;

  ## default norm
  if (nargin < 2)
    tol = 1e-6 * norm(y,1);
  endif

  ## First step in the approximation is the integer portion
  n = round(y);  # first element in the continued fraction
  d = ones(size(y));
  frac = y-n;
  lastn = ones(size(y));
  lastd = zeros(size(y));

  nd = ndims(y);
  nsz = numel (y);
  steps = zeros([nsz, 0]);

  ## grab new factors until all continued fractions converge
  while (1)
    ## determine which fractions have not yet converged
    idx = find(abs (y-n./d) >= tol);
    if (isempty(idx))
      if (isempty (steps))
	steps = NaN .* ones (nsz, 1);
      endif
      break;
    endif

    ## grab the next step in the continued fraction
    flip = 1./frac(idx);
    step = round(flip); # next element in the continued fraction

    if (nargout < 2)
      tsteps = NaN .* ones (nsz, 1);
      tsteps (idx) = step;
      steps = [steps, tsteps];
    endif

    frac(idx) = flip-step;

    ## update the numerator/denominator
    nextn = n;
    nextd = d;
    n(idx) = n(idx).*step + lastn(idx);
    d(idx) = d(idx).*step + lastd(idx);
    lastn = nextn;
    lastd = nextd;
  endwhile

  if (nargout == 2)
    ## move the minus sign to the top
    n = n.*sign(d);
    d = abs(d);

    ## return the same shape as you receive
    n = reshape(n, size(x));
    d = reshape(d, size(x));

    ## use 1/0 for Inf
    n(isinf(x)) = sign(x(isinf(x)));
    d(isinf(x)) = 0;

    ## reshape the output
    n = reshape (n, size (x));
    d = reshape (d, size (x));
  else
    n = "";
    nsteps = size(steps, 2);
    for i = 1: nsz
      s = [int2str(y(i))," "];
      j = 1;

      while (true)
	step = steps(i, j++);
	if (isnan (step))
	  break;
	endif
	if (j > nsteps || isnan (steps(i, j)))
	  if (step < 0)
	    s = [s(1:end-1), " + 1/(", int2str(step), ")"];
	  else
	    s = [s(1:end-1), " + 1/", int2str(step)];
	  endif
	  break;
	else
	  s = [s(1:end-1), " + 1/(", int2str(step), ")"];
        endif
      endwhile
      s = [s, repmat(")", 1, j-2)];
      n_nc = columns (n);
      s_nc = columns (s);
      if (n_nc > s_nc)
	s(:,s_nc+1:n_nc) = " "
      elseif (s_nc > n_nc)
	n(:,n_nc+1:s_nc) = " ";
      endif
      n = cat (1, n, s);
    endfor
  endif

endfunction
