I'm thinking maybe older gpu's did even courser filtering than this - 6-bit or something instead of 8. Maybe this was the recent improvement?

On Fri, Oct 1, 2010 at 12:59 PM, Nathaniel Hoffman <naty@io.com> wrote:
Didn't the newer NVIDIA GPUs fix this?

> You guessed right. The loss of precision is in the texture units.
> Unfortunately, 8 bit components are filtered to 8 bit results (even though
> they show up as floating point values in the shader). This is true for
> nvidia gpus for sure and probably all other gpus.
>
> -mike
>   ----- Original Message -----
>   From: Stefan Sandberg
>   To: Game Development Algorithms
>   Sent: Friday, October 01, 2010 1:45 AM
>   Subject: Re: [Algorithms] Filtering
>
>
>   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
> <andreas.brinck@gmail.com> 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
>
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