Re: [Algorithms] RGB to XYZ conversion
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Re: [Algorithms] RGB to XYZ conversion
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From: Adrian B. <ad...@gm...> - 2008-04-26 01:51:00
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Or you could have been referring to the linearity of the color space
affine transformation (Rx + t). That would work too.
On Fri, Apr 25, 2008 at 6:46 PM, Adrian Bentley <ad...@gm...> wrote:
> "anything that's a linear op will work."
>
> Your assertion about a+b=1 implies you meant affine op. Now _that's_
> a technicality :-).
>
>
>
> On Fri, Apr 25, 2008 at 8:57 AM, Sam Martin <sam...@ge...> wrote:
> > I thought for a moment there could be something to do with the range of
> > the output. I don't think there is, but I've done it now and thought I'd
> > point this out anyway in case it's helpful:
> >
> > If your RBG colour components are in the range 0-1 then the YCoCg output
> > is actually signed:
> > Y = [0 .. 1]
> > Co = Cg = [-1/2 .. 1/2]
> >
> > So if you want to store this in an unsigned 0-1 range texture you need
> > to shift the CoCg values up, which you can do by adding a 'translation'
> > colour vector to the YCoCg colours: t = {0,1/2,1/2}. So your transform
> > is actually (Rx + t) and your inverse is R'(c - t).
> >
> > Will this work? Well:
> >
> > (Setting the rbg interp equal to what happens when you are interpolating
> > YCoCg with the translation vector)
> > a*x + b*x = R'(a(Rx + t) + b(Ry + t) - t)
> >
> > (expanding out the brackets)
> > = R'(a*Rx + a*t + b*Ry + b*t - t)
> >
> > (regrouping to separate the t's)
> > = R'(a*Rx + b*Ry) + R'(a*t + b*t - t)
> > = R'(a*Rx + b*Ry) + R'((a+b)*t - t)
> > = a*R'Rx + b*R'Ry + R'((a+b)*t - t)
> > = a*x + b*y + R'((a+b)*t - t)
> >
> > So we have the first bit and this extra term involving t's. Now, as long
> > as a+b = 1 (which they do when you are doing an interpolation), or you
> > subtract a t appropriately scaled, then the t's will cancel and the
> > equality will be valid.
> >
> > In a nutshell I guess I'm really just demonstrating that anything that's
> > a linear op will work. Although again, I could have ballsed this up
> > somewhere :)
> >
> >
> > Cheers,
> > Sam
> >
> >
> >
> > From: gda...@li...
> > [mailto:gda...@li...] On Behalf Of
> > Jon Olick
> > Sent: 25 April 2008 15:39
> >
> > To: Game Development Algorithms
> > Subject: Re: [Algorithms] RGB to XYZ conversion
> >
> >
> >
> > Hi Jon,
> >
> > So, thats interesting then that the bilinear interpolation of RGB colors
> > have artifacts. As soon as I get a chance I'll take a closer look at our
> > assets and both in straight up RGB and DXT5 to verify. After reading
> > your e-mail, I went over to Jan Pal (our local YCoCg expert) to ask him
> > about this very interpolation problem and he swears that the
> > interpolation in YCoCg is not the same as interpolation in RGB. We were
> > looking at this about 6 months ago now, but we very well could have made
> > a mistake.
> >
> > Best,
> > Jon Olick
> > On Thu, Apr 24, 2008 at 4:54 PM, Jon Watte <hp...@mi...>
> > wrote:
> > Willem H. de Boer wrote:
> > > But, this doesn't take away from the fact that the interpolation
> > > can choose colours in between (i.e. for t in [0,1]) that are
> > > in some ways 'wrong' or at least counterintuitive to what
> > > we would expect - which is what Jon points out. This is
> > > a result of the way YCoCg maps out in colour space.
> > >
> > Note: that "Jon" was a different Jon than me :-)
> >
> > It actually does mean that all the interpolated colors will be the same,
> > because for _any_ "a" and "b" (interpolation values, equal to t and
> > 1-t), a*x + b*y equals R'(aRx +bRy), through mathematic equality. So,
> > no, there would not be interpolation artifacts based on the math being
> > in a different space. This is because the two spaces are linearly
> > equivalent. You don't get interpolation differences until you go to a
> > non-equivalent space, such as HSV, or a space with a different gamma.
> >
> > The interpolation of DXT compression, though, is somewhat quantized, and
> > will generally result in a slight green/purple banding in RGB. That
> > quantization may cause a different color banding effect in the Y Co Cg
> > space, because quantization is not a linearly independent operation.
> >
> > Sincerely,
> >
> > jw
> >
> >
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