Re: [Algorithms] Filtering

[Stefan S. <kef...@gm...>]

Assuming you're after precision, what's wrong with doing it manually? :)
If performance is what you're after, and you're working on textures as they
were intended(ie, game textures or video or something like that, not
'data'), you could separate contrast & color separately, keeping high
contrast resolution, and downsampled color, and
you'd save both bandwidth and instr.
If you simply want to know 'why', I'm guessing loss of precision in the tex
units?
You've already ruled out shader precision from your own manual filtering,
so doesn't leave much else, imo..
Other than manipulating the data you're working on, which is the only thing
you -can- change I guess, I cant really see a solution,
but far greater minds linger here than mine, so hold on for what I assume
will be a lengthy description of floating point math as
it is implemented in modern gpu's :)



On Fri, Oct 1, 2010 at 9:57 AM, Andreas Brinck <and...@gm...>wrote:

> Hi,
>
> I have a texture in which I use the R, G and B channel to store a
> value in the [0, 1] range with very high precision. The value is
> extracted like this in the (Cg) shader:
>
> float
> extractValue(float2 pos) {
>    float4 temp = tex2D(buffer, pos);
>    return (temp.x * 16711680.0 + temp.y * 65280.0 + temp.z * 255.0) *
> (1.0 / 16777215.0);
> }
>
> I now want to sample this value with bilinear filtering but when I do
> this I don't get a correct result. If I do the filtering manually like
> this:
>
> float
> sampleValue(float2 pos) {
>        float2 ipos = floor(pos);
>        float2 fracs = pos - ipos;
>        float d0 = extractValue(ipos);
>        float d1 = extractValue(ipos + float2(1, 0));
>        float d2 = extractValue(ipos + float2(0, 1));
>        float d3 = extractValue(ipos + float2(1, 1));
>        return lerp(lerp(d0, d1, fracs.x), lerp(d2, d3, fracs.x), fracs.y);
> }
>
> everything works as expected. The values in the buffer can be seen as
> a linear combination of three constants:
>
> value = (C0 * r + C1 * g + C2 * b)
>
> If we use the built in texture filtering we should get the following
> if we sample somewhere between two texels: {r0, g0, b0} and {r1, g1,
> b1}. For simplicity we just look at filtering along one axis:
>
> filtered value = lerp(r0, r1, t) * C0 + lerp(g0, g1, t) * C1 +
> lerp(b0, b1, t) * C2;
>
> Doing the filtering manually:
>
> filtered value = lerp(r0 * C0 + b0 * C1 + g0 * C2, r1 * C0 + g1 * C1 +
> b1 * C2, t)  =
>                   = (r0 * C0 + b0 * C1 + g0 * C2) * (1 - t) + (r1 *
> C0 + g1 * C1 + b1 * C2) * t =
>                   = (r0 * C0) * (1 - t) + (r1 * C0) * t + ... =
>                   = lerp(r0, r1, t) * C0 + ...
>
> So in the world of non floating point numbers these two should be
> equivalent right?
>
> My theory is that the error is caused by an unfortunate order of
> floating point operations. I've tried variations like:
>
> (temp.x * (16711680.0 / 16777215.0) + temp.y * (65280.0/16777215.0) +
> temp.z * (255.0/16777215.0))
>
> and
>
> (((temp.x * 256.0 + temp.y) * 256.0 + temp.z) * 255.0) * (1.0 / 16777215.0)
>
> but all exhibit the same problem. What do you think; is it possible to
> solve this problem?
>
> Regards Andreas
>
>
> ------------------------------------------------------------------------------
> Start uncovering the many advantages of virtual appliances
> and start using them to simplify application deployment and
> accelerate your shift to cloud computing.
> http://p.sf.net/sfu/novell-sfdev2dev
> _______________________________________________
> GDAlgorithms-list mailing list
> GDA...@li...
> https://lists.sourceforge.net/lists/listinfo/gdalgorithms-list
> Archives:
> http://sourceforge.net/mailarchive/forum.php?forum_name=gdalgorithms-list
>