## There can be curious gaps, in what some people understand.

One of the concepts which once dominated CGI was, that textures assigned to 3D models needed to include a “Normal-Map”, so that even early in the days of 3D gaming, textured surfaces would seem to have ‘bumps’, and these normal-maps were more significant, than displacement-maps – i.e., height- or depth-maps – because shaders were actually able to compute lighting subtleties more easily, using the normal-maps. But additionally, it was always quite common that ordinary 8x8x8 (R,G,B) texel-formats needed to store the normal-maps, just because images could more-easily be prepared and loaded with that pixel-format. (:1)

The old-fashioned way to code that was, that the 8-bit integer (128) was taken to symbolize (0.0), that (255) was taken to symbolize a maximally positive value, and that the integer (0) was decoded to (-1.0). The reason for this, AFAIK, was the use by the old graphics cards, of the 8-bit integer, as a binary fraction.

In the spirit of recreating that, and, because it’s sometimes still necessary to store an approximation of a normal-vector, using only 32 bits, the code has been offered as follows:


Out.Pos_Normal.w = dot(floor(normal * 127.5 + 127.5), float3(1 / 256.0, 1.0, 256.0));

float3 normal = frac(Pos_Normal.w * float3(1.0, 1 / 256.0, 1 / 65536.0)) * 2.0 - 1.0;



There’s an obvious problem with this backwards-emulation: It can’t seem to reproduce the value (0.0) for any of the elements of the normal-vector. And then, what some people do is, to throw their arms in the air, and to say: ‘This problem just can’t be solved!’ Well, what about:


//  Assumed:
normal = normalize(normal);

Out.Pos_Normal.w = dot(floor(normal * 127.0 + 128.5), float3(1 / 256.0, 1.0, 256.0));



A side effect of this will definitely be, that no uncompressed value belonging to the interval [-1.0 .. +1.0] will lead to a compressed series of 8 zeros.

Mind you, because of the way the resulting value was now decoded again, the question of whether zero can actually result, is not as easy to address. And one reason is the fact that, for all the elements except the first, additional bits after the first 8 fractional bits, have not been removed. But that’s just a problem owing to the one-line decoding that was suggested. That could be changed to:


float3 normal = floor(Pos_Normal.w * float3(256.0, 1.0, 1 / 256.0));
normal = frac(normal * (1 / 256.0)) * (256.0 / 127.0) - (128.0 / 127.0);



Suddenly, the impossible has become possible.

N.B.  I would not use the customized decoder, unless I was also sure, that the input floating-point value, came from my customized encoder. It can easily happen that the shader needs to work with texture images prepared by an external program, and then, because of the way their channel-values get normalized today, I might use this as the decoder:


float3 normal = texel.rgb * (255.0 / 128.0) - 1.0;



However, if I did, a texel-value of (128) would still be required, to result in a floating-point value of (0.0)

(Updated 5/10/2020, 19h00… )

## Wavelet Decomposition of Images

One type of wavelet which exists, and which has continued to be of some interest to computational signal processing, is the Haar Wavelet. It’s thought to have a low-pass and a high-pass version complementing each other. This would be the low-pass Haar Wavelet:

[ +1 +1 ]

And this would be the high-pass version:

[ +1 -1 ]

These wavelets are intrinsically flawed, in that if they are applied to audio signals, they will produce poor frequency response each. But they do have as an important Mathematical property, that from its low-pass and its high-pass component, the original signal can be reconstructed fully.

Now, there is also something called a wavelet transform, but I seldom see it used.

If we wanted to extend the Haar Wavelet to the 2D domain, then a first approach might be, to apply it twice, once, one-dimensionally, along each axis of an image. But in reality, this would give the following low-frequency component:

[ +1 +1 ]

[ +1 +1 ]

And only, the following high-frequency component:

[ +1 -1 ]

[ -1 +1 ]

This creates an issue with common sense, because in order to be able to reconstruct the original signal – in this case an image – we’d need to arrive at 4 reduced values, not 2, because the original signal had 4 distinct values.

And so closer inspection should reveal, that the wavelet reduction of images has 3 distinct high-frequency components: ( :1 )