From bug-octave-request at bevo dot che dot wisc dot edu Sat Apr 20 13:26:48 2002
Subject: rgb2hsv patch
From: Paul Kienzle <pkienzle at jazz dot ncnr dot nist dot gov>
To: bug-octave at bevo dot che dot wisc dot edu
Date: Sat, 20 Apr 2002 14:26:19 -0400

octave:3> rgb2hsv([0 0 0])
warning: division by zero
warning: in /home/pkienzle/cvs/octave/scripts/image/rgb2hsv.m near line 48, column 5:

>>> s = (v' > 0) .* (1 - min (rgb') ./ v)';


ans =

  0  0  0



The following patch removes this.  Note that the code would be
cleaner if rgb(condition,idx) would work even if condition is
always false.


Paul Kienzle <pkienzle at users dot sf dot net>
	* image/rgb2hsv.m: faster, more accurate, remove the
	divide by zero warning.

Index: rgb2hsv.m
===================================================================
RCS file: /cvs/octave/scripts/image/rgb2hsv.m,v
retrieving revision 1.1
diff -c -r1.1 rgb2hsv.m
*** rgb2hsv.m	2001/02/28 08:24:43	1.1
--- rgb2hsv.m	2002/04/20 18:18:31
***************
*** 43,90 ****
      error ("rgb2hsv: argument must be a matrix of size n x 3");
    endif
  
!   # get saturation and value
!   v = max (rgb');
!   s = (v' > 0) .* (1 .- min (rgb') ./ v)';
  
!   # if v==0 set s to 0 too
!   s(isnan (s)) = 0;
! 
!   # subtract minimum and divide trough maximum
!   # to get the bright and saturated colors
!   sc = (rgb - kron ([1, 1, 1], min (rgb')'));
!   sv = sc ./ kron([1, 1, 1], max (sc')');
! 
!   # if r=g=b (gray value) set hue to 0
!   sv(isnan (sv)) = 0;
! 
!   # hue=f(color) must be splitted into 6 parts 
!   # 2 for each color
! 
!   # h1(green)
!   tmp = (sv(:, 1) == 1 & sv(:,3) == 0) .* (1/6 * sv(:,2) + eps);
!   # avoid problems with h2(red) since hue(0)==hue(1)
!   h = (tmp < 1/6) .* tmp; 
!   # h2(green)
!   h = h + ((h == 0) & sv(:,1) == 0 & sv(:,3) == 1) \
!       .* (-1/6 * sv(:,2) + 2/3 + eps);
! 
!   # h1(red)
!   h = h + ((h == 0) & sv(:,2) == 1 & sv(:,3) == 0) \
!       .* (-1/6 * sv(:,1) + 1/3 + eps);
! 
!   # h2(red)
!   h = h + ((h == 0) & sv(:,2) == 0 & sv(:,3) == 1) \
!       .* (1/6 * sv(:,1) + 2/3 + eps);
! 
!   # h1(blue)
!   h = h + ((h == 0) & sv(:,1) == 1 & sv(:,2) == 0) \
!       .* (-1/6 * sv(:,3) + 1 + eps);
! 
!   # h2(blue)
!   h = h + ((h == 0) & sv(:,1) == 0 & sv(:,2) == 1) \
!       .* (1/6 * sv(:,3) + 1/3);
! 
!   hsval = [h, s, v'];
  
  endfunction
--- 43,82 ----
      error ("rgb2hsv: argument must be a matrix of size n x 3");
    endif
  
!   ## get the max and min
!   s = min (rgb')';
!   v = max (rgb')';
! 
!   ## set hue to zero for undefined values (gray has no hue)
!   h = zeros(size(v));
!   notgray = (s != v);
!     
!   ## blue hue
!   idx = (v == rgb(:,3) & notgray);
!   if any(idx)
!     h(idx) = 2/3 + 1/6 * (rgb(idx,1)-rgb(idx,2)) ./ (v(idx) - s(idx));
!   endif
! 	      
!   ## green hue
!   idx = (v == rgb(:,2) & notgray);
!   if any(idx)
!     h(idx) = 1/3 + 1/6 * (rgb(idx,3)-rgb(idx,1)) ./ (v(idx) - s(idx));
!   endif
! 
!   ## red hue
!   idx = (v == rgb(:,1) & notgray); 
!   if any(idx)
!     h(idx) =       1/6 * (rgb(idx,2)-rgb(idx,3)) ./ (v(idx) - s(idx));
!   endif
! 
!   ## correct for negative red
!   idx = (h < 0);
!   h(idx) = 1+h(idx);
! 
!   ## set the saturation
!   s(!notgray) = 0;
!   s(notgray) = 1 - s(notgray) ./ v(notgray);
  
!   hsval = [h, s, v];
  
  endfunction
----- End forwarded message -----



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