Google| Subject | Re: Is RGB to Lab lossy? - was(Re: Lenses and sharpening) |
| From | Martin Brown |
| Date | 10/06/2014 17:33 (10/06/2014 16:33) |
| Message-ID | <IDyYv.521499$PG2.150632@fx12.am4> |
| Client | |
| Newsgroups | rec.photo.digital |
| Follows | PeterN |
| Followups | Alan Browne (1h & 29m) |
On 06/10/2014 14:19, PeterN wrote:
PeterNThe problem is not one of interpolation but that there are unavoidable minor rounding errors in the nonlinear transform from RGB to CIELAB and also on the way back due to the finite representation of the results. See:
On 10/5/2014 10:37 PM, Alan Browne wrote:Alan BrownePeterN
On 2014.10.05, 20:55 , PeterN wrote:PeterNAlan Browne
On 10/5/2014 6:57 PM, Alan Browne wrote:Alan BrownePeterN
On 2014.10.05, 14:42 , PeterN wrote:We went through all this some many months ago. I demonstrated clearly that the amount of 'loss' was negligible in practical terms.PeterNAlan Browne
I would use the terem "color change." anstead of loss.
Any change is a quality loss. Whether that is colour difference, tone, brightness, sharpness ... whatever, it's a loss.
Then you are using a different definition of quality.
Not at all. A non lossy process would have:
RGB-A -->X-format -->RGB-B
with RGB-A identical to RGB-B
But - the fact is that with Lab
RGB-A -->Lab -->RGB-B
RGB-A =/= RGB-B, therefore there was quality loss.
It seems to me that the assumption in that logic is: the quality of RGB-A >quality of RGB-B. LAB has a larger color gamut than RGB. If there is no processing in LAB I would think that there would be no need for interpolation on the return trip.
http://en.wikipedia.org/wiki/Lab_color_space
There is no assumption here beyond the very definition of a lossless transform. Applying a lossless transform and then its inverse to the result you must be able to get to back your original image *EXACTLY*.
Do the round trip x + 10 times without processing and one might see a difference. It is doubtful that there will be a noticable difference from 10 round trips. Meanwhile there are color modification processes that are easier to perform in LAB than RGB. I would think that if the changes made in LAB created color outside the RGB gamut there would have to be some interpolation. The interpolation coud mae a better image, or it could make the changed image horrific.It isn't interpolation - if anything it is a quantisation effect with very different steps in CIELAB than in RGB. See for example:
http://en.wikipedia.org/wiki/Lab_color_space
Some pixel values may converge to an attractor on the first or second pass around the loop and after that do not change at all. A few unlucky ones may drift away from the original colour (but probably not by much).
Out of gamut values pretty much end up clipped against the nearest RGB boundary on a minimise the absolute perceived colour error basis.
In another area, I have found images to be fine with a color cast, but when I remove the cast, to my eye the image looks horrific.CIELAB does a better job of managing just noticeable visual differences and allows better adjustment of visual saturation and lightness.
-- Regards, Martin Brown
09/13/2014-
