glsl sqrt result is negative - glsl

I have a blur function in my shader which takes a float as the blur radius.
I calculate the distance from the center of the screen like this:
vec2 nPos = texCoord.xy - vec2(0.5, 0.5);
float d = sqrt(nPos.x * nPos.x + nPos.y * nPos.y)/sqrt2; //ranged from 0 to 1
Now I call my blur function like this:
gl_FragColor = chromaticBlur(pr, pg, pb, d, size);
The weird thing is, if I set d to 0, there is no blur, just as expected (values above 0 work as expected too). But if I use the function shown above, a huge part of the screen (in fact an oval) will be black. That means that the blur value for this area is negative, which SHOULD be impossible. How can that happen?

Related

What is this technique, if not bilinear filtering?

I'm trying to replicate the automatic bilinear filtering algorithm of Unity3D using the next code:
fixed4 GetBilinearFilteredColor(float2 texcoord)
{
fixed4 s1 = SampleSpriteTexture(texcoord + float2(0.0, _MainTex_TexelSize.y));
fixed4 s2 = SampleSpriteTexture(texcoord + float2(_MainTex_TexelSize.x, 0.0));
fixed4 s3 = SampleSpriteTexture(texcoord + float2(_MainTex_TexelSize.x, _MainTex_TexelSize.y));
fixed4 s4 = SampleSpriteTexture(texcoord);
float2 TexturePosition = float2(texcoord)* _MainTex_TexelSize.z;
float fu = frac(TexturePosition.x);
float fv = frac(TexturePosition.y);
float4 tmp1 = lerp(s4, s2, fu);
float4 tmp2 = lerp(s1, s3, fu);
return lerp(tmp1, tmp2, fv);
}
fixed4 frag(v2f IN) : SV_Target
{
fixed4 c = GetBilinearFilteredColor(IN.texcoord) * IN.color;
c.rgb *= c.a;
return c;
}
I thought I was using the correct algoritm because is the only one I have seen out there for bilinear. But I tried it using unity with the same texture duplicated:
1º texture: is in Point filtering and using the custom bilinear shader (maded from the default sprite shader).
2º texture: is in Bilinear filter with the default sprite shader
And this is the result:
You can see that they are different and also there is some displacement in my custom shader that makes the sprite being off-center when rotating in the Z axis.
Any idea of what I'm doing wrong?
Any idea of what is doing Unity3D different?
Are there another algorithm's who fits in the Unity3D default filtering?
Solution
Updated with the complete code solution with Nico's code for other people who search for it here:
fixed4 GetBilinearFilteredColor(float2 texcoord)
{
fixed4 s1 = SampleSpriteTexture(texcoord + float2(0.0, _MainTex_TexelSize.y));
fixed4 s2 = SampleSpriteTexture(texcoord + float2(_MainTex_TexelSize.x, 0.0));
fixed4 s3 = SampleSpriteTexture(texcoord + float2(_MainTex_TexelSize.x, _MainTex_TexelSize.y));
fixed4 s4 = SampleSpriteTexture(texcoord);
float2 TexturePosition = float2(texcoord)* _MainTex_TexelSize.z;
float fu = frac(TexturePosition.x);
float fv = frac(TexturePosition.y);
float4 tmp1 = lerp(s4, s2, fu);
float4 tmp2 = lerp(s1, s3, fu);
return lerp(tmp1, tmp2, fv);
}
fixed4 frag(v2f IN) : SV_Target
{
fixed4 c = GetBilinearFilteredColor(IN.texcoord - 0.498 * _MainTex_TexelSize.xy) * IN.color;
c.rgb *= c.a;
return c;
}
And the image test with the result:
Why don't substract 0.5 exactly?
If you test it you will see some edge cases where it jumps to (pixel - 1).
Let's take a closer look at what you are actually doing. I will stick to the 1D case because it is easier to visualize.
You have an array of pixels and a texture position. I assume, _MainTex_TexelSize.z is set in a way, such that it gives pixel coordinates. This is what you get (the boxes represent pixels, numbers in boxes the pixel number and numbers below the pixel space coordinates):
With your sampling (assuming nearest point sampling), you will get pixels 2 and 3. However, you see that the interpolation coordinate for lerp is actually wrong. You will pass the fractional part of the texture position (i.e. 0.8) but it should be 0.3 (= 0.8 - 0.5). The reasoning behind this is quite simple: If you land at the center of a pixel, you want to use the pixel value. If you land right in the middle between two pixels, you want to use the average of both pixel values (i.e. an interpolation value of 0.5). Right now, you have basically an offset by a half pixel to the left.
When you solve the first problem, there is a second one:
In this case, you actually want to blend between pixel 1 and 2. But because you always go to the right in your sampling, you will blend between 2 and 3. Again, with a wrong interpolation value.
The solution should be quite simple: Subtract half of the pixel width from the texture coordinate before doing anything with it, which is probably just the following (assuming that your variables hold the things I think):
fixed4 c = GetBilinearFilteredColor(IN.texcoord - 0.5 * _MainTex_TexelSize.xy) * IN.color;
Another reason why the results are different could be that Unity actually uses a different filter, e.g. bicubic (but I don't know). Also, the usage of mipmaps could influence the result.

DirectX Converting Pixel World Position to Shadow Map Position Gives Weird, Tiled Results

I've been trying for some time now to get a screen-space pixel (provided by a deferred HLSL shader) to convert to light space. The results have been surprising to me as my light rendering seems to be tiling the depth buffer.
Importantly, the scene camera (or eye) and the light being rendered from start in the same position.
First, I extract the world position of the pixel using the code below:
float3 eye = Eye;
float4 position = {
IN.texCoord.x * 2 - 1,
(1 - IN.texCoord.y) * 2 - 1,
zbuffer.r,
1
};
float4 hposition = mul(position, EyeViewProjectionInverse);
position = float4(hposition.xyz / hposition.w, hposition.w);
float3 eyeDirection = normalize(eye - position.xyz);
The result seems to be correct as rendering the XYZ position as RGB respectively yields this (apparently correct) result:
The red component seems to be correctly outputting X as it moves to the right, and blue shows Z moving forward. The Y factor also looks correct as the ground is slightly below the Y axis.
Next (and to be sure I'm not going crazy), I decided to output the original depth buffer. Normally I keep the depth buffer in a Texture2D called DepthMap passed to the shader as input. In this case, however, I try to undo the pixel transformation by offsetting it back into the proper position and multiplying it by the eye's view-projection matrix:
float4 cpos = mul(position, EyeViewProjection);
cpos.xyz = cpos.xyz / cpos.w;
cpos.x = cpos.x * 0.5f + 0.5f;
cpos.y = 1 - (cpos.y * 0.5f + 0.5f);
float camera_depth = pow(DepthMap.Sample(Sampler, cpos.xy).r, 100); // Power 100 just to visualize the map since scales are really tiny
return float4(camera_depth, camera_depth, camera_depth, 1);
This yields a correct looking result as well (though I'm not 100% sure about the Z value). Also note that I've made the results exponential to better visualize the depth information (this is not done when attempting live comparisons):
So theoretically, I can use the same code to convert that pixel world position to light space by multiplying by the light's view-projection matrix. Correct? Here's what I tried:
float4 lpos = mul(position, ShadowLightViewProjection[0]);
lpos.xyz = lpos.xyz / lpos.w;
lpos.x = lpos.x * 0.5f + 0.5f;
lpos.y = 1 - (lpos.y * 0.5f + 0.5f);
float shadow_map_depth = pow(ShadowLightMap[0].Sample(Sampler, lpos.xy).r, 100); // Power 100 just to visualize the map since scales are really tiny
return float4(shadow_map_depth, shadow_map_depth, shadow_map_depth, 1);
And here's the result:
And another to show better how it's mapping to the world:
I don't understand what is going on here. It seems it might have something to do with the projection matrix, but I'm not that good with math to know for sure what is happening. It's definitely not the width/height of the light map as I've tried multiple map sizes and the projection matrix is calculated using FOV and aspect ratios never inputing width/height ever.
Finally, here's some C++ code showing how my perspective matrix (used for both eye and light) is calculated:
const auto ys = std::tan((T)1.57079632679f - (fov / (T)2.0));
const auto xs = ys / aspect;
const auto& zf = view_far;
const auto& zn = view_near;
const auto zfn = zf - zn;
row1(xs, 0, 0, 0);
row2(0, ys, 0, 0);
row3(0, 0, zf / zfn, 1);
row4(0, 0, -zn * zf / zfn, 0);
return *this;
I'm completely at a loss here. Any guidance or recommendations would be greatly appreciated!
EDIT - I also forgot to mention that the tiled image is upside down as if the y flip broke it. That's strange to me as it's required to get it back to eye texture space correctly.
I did some tweaking and fixed things here and there. Ultimately, my biggest issue was an unexpectedly transposed matrix. It's a bit complicated as to how the matrix got transposed, but that's why things were flipped. I also changed to D32 depth buffers (though I'm not sure that helped any) and made sure that any positions divided by their W affected all component (including W).
So code like this: hposition.xyz = hposition.xyz / hposition.w
became this: hposition = hposition / hposition.w
After all this tweaking, it's starting to look more like a shadow map.
Oh and the transposed matrix was the ViewProjection of the light.

Advanced moiré a pattern reduction in HLSL / GLSL procedural textures shader - antialiasing

I am working on a procedural texture, it looks fine, except very far away, the small texture pixels disintegrate into noise and moiré patterns.
I have set out to find a solution to average and quantise the scale of the pattern far away and close up, so that close by it is in full detail, and far away it is rounded off so that one pixel of a distant mountain only represents one colour found there, and not 10 or 20 colours at that point.
It is easy to do it by rounding the World_Position that the volumetric texture is based on using an if statement i.e.:
if( camera-pixel_distance > 1200 meters ) {wpos = round(wpos/3)*3;}//---round far away pixels
return texturefucntion(wpos);
the result of rounding far away textures is that they will look like this, except very far away:
the trouble with this is i have to make about 5 if conditions for the various distances, and i have to estimate a random good rounding value
I tried to make a function that cuts the distance of the pixel into distance steps, and applies a LOD devider to the pixel_worldposition value to make it progressively rounder at distance but i got nonsense results, actually the HLSL was totally flipping out. here is the attempt:
float cmra= floor(_WorldSpaceCameraPos/500)*500; //round camera distance by steps of 500m
float dst= (1-distance(cmra,pos)/4500)*1000 ; //maximum faraway view is 4500 meters
pos= floor(pos/dst)*dst;//close pixels are rounded by 1000, far ones rounded by 20,30 etc
it returned nonsense patterns that i could not understand.
Are there good documented algorithms for smoothing and rounding distance texture artifacts? can i use the scren pixel resolution, combined with the distance of the pixel, to round each pixel to one color that stays a stable color?
Are you familiar with the GLSL (and I would assume HLSL) functions dFdx() and dFdy() or fwidth()? They were made specifically to solve this problem. From the GLSL Spec:
genType dFdy (genType p)
Returns the derivative in y using local differencing for the input argument p.
These two functions are commonly used to estimate the filter width used to anti-alias procedural textures.
and
genType fwidth (genType p)
Returns the sum of the absolute derivative in x and y using local differencing for the input argument p, i.e.: abs (dFdx (p)) + abs (dFdy (p));
OK i found some great code and a tutorial for the solution, it's a simple code that can be tweaked by distance and many parameters.
from this tutorial:
http://www.yaldex.com/open-gl/ch17lev1sec4.html#ch17fig04
half4 frag (v2f i) : COLOR
{
float Frequency = 0.020;
float3 pos = mul (_Object2World, i.uv).xyz;
float V = pos.z;
float sawtooth = frac(V * Frequency);
float triangle = (abs(2.0 * sawtooth - 1.0));
//return triangle;
float dp = length(float2(ddx(V), ddy(V)));
float edge = dp * Frequency * 8.0;
float square = smoothstep(0.5 - edge, 0.5 + edge, triangle);
// gl_FragColor = vec4(vec3(square), 1.0);
if (pos.x>0.){return float4(float3(square), 1.0);}
if (pos.x<0.){return float4(float3(triangle), 1.0);}
}

Calculate clipspace.w from clipspace.xyz and (inv) projection matrix

I'm using a logarithmic depth algorithmic which results in someFunc(clipspace.z) being written to the depth buffer and no implicit perspective divide.
I'm doing RTT / postprocessing so later on in a fragment shader I want to recompute eyespace.xyz, given ndc.xy (from the fragment coordinates) and clipspace.z (from someFuncInv() on the value stored in the depth buffer).
Note that I do not have clipspace.w, and my stored value is not clipspace.z / clipspace.w (as it would be when using fixed function depth) - so something along the lines of ...
float clip_z = ...; /* [-1 .. +1] */
vec2 ndc = vec2(FragCoord.xy / viewport * 2.0 - 1.0);
vec4 clipspace = InvProjMatrix * vec4(ndc, clip_z, 1.0));
clipspace /= clipspace.w;
... does not work here.
So is there a way to calculate clipspace.w out of clipspace.xyz, given the projection matrix or it's inverse?
clipspace.xy = FragCoord.xy / viewport * 2.0 - 1.0;
This is wrong in terms of nomenclature. "Clip space" is the space that the vertex shader (or whatever the last Vertex Processing stage is) outputs. Between clip space and window space is normalized device coordinate (NDC) space. NDC space is clip space divided by the clip space W coordinate:
vec3 ndcspace = clipspace.xyz / clipspace.w;
So the first step is to take our window space coordinates and get NDC space coordinates. Which is easy:
vec3 ndcspace = vec3(FragCoord.xy / viewport * 2.0 - 1.0, depth);
Now, I'm going to assume that your depth value is the proper NDC-space depth. I'm assuming that you fetch the value from a depth texture, then used the depth range near/far values it was rendered with to map it into a [-1, 1] range. If you didn't, you should.
So, now that we have ndcspace, how do we compute clipspace? Well, that's obvious:
vec4 clipspace = vec4(ndcspace * clipspace.w, clipspace.w);
Obvious and... not helpful, since we don't have clipspace.w. So how do we get it?
To get this, we need to look at how clipspace was computed the first time:
vec4 clipspace = Proj * cameraspace;
This means that clipspace.w is computed by taking cameraspace and dot-producting it by the fourth row of Proj.
Well, that's not very helpful. It gets more helpful if we actually look at the fourth row of Proj. Granted, you could be using any projection matrix, and if you're not using the typical projection matrix, this computation becomes more difficult (potentially impossible).
The fourth row of Proj, using the typical projection matrix, is really just this:
[0, 0, -1, 0]
This means that the clipspace.w is really just -cameraspace.z. How does that help us?
It helps by remembering this:
ndcspace.z = clipspace.z / clipspace.w;
ndcspace.z = clipspace.z / -cameraspace.z;
Well, that's nice, but it just trades one unknown for another; we still have an equation with two unknowns (clipspace.z and cameraspace.z). However, we do know something else: clipspace.z comes from dot-producting cameraspace with the third row of our projection matrix. The traditional projection matrix's third row looks like this:
[0, 0, T1, T2]
Where T1 and T2 are non-zero numbers. We'll ignore what these numbers are for the time being. Therefore, clipspace.z is really just T1 * cameraspace.z + T2 * cameraspace.w. And if we know cameraspace.w is 1.0 (as it usually is), then we can remove it:
ndcspace.z = (T1 * cameraspace.z + T2) / -cameraspace.z;
So, we still have a problem. Actually, we don't. Why? Because there is only one unknown in this euqation. Remember: we already know ndcspace.z. We can therefore use ndcspace.z to compute cameraspace.z:
ndcspace.z = -T1 + (-T2 / cameraspace.z);
ndcspace.z + T1 = -T2 / cameraspace.z;
cameraspace.z = -T2 / (ndcspace.z + T1);
T1 and T2 come right out of our projection matrix (the one the scene was originally rendered with). And we already have ndcspace.z. So we can compute cameraspace.z. And we know that:
clispace.w = -cameraspace.z;
Therefore, we can do this:
vec4 clipspace = vec4(ndcspace * clipspace.w, clipspace.w);
Obviously you'll need a float for clipspace.w rather than the literal code, but you get my point. Once you have clipspace, to get camera space, you multiply by the inverse projection matrix:
vec4 cameraspace = InvProj * clipspace;

Generating a normal map from a height map?

I'm working on procedurally generating patches of dirt using randomized fractals for a video game. I've already generated a height map using the midpoint displacement algorithm and saved it to a texture. I have some ideas for how to turn that into a texture of normals, but some feedback would be much appreciated.
My height texture is currently a 257 x 257 gray-scale image (height values are scaled for visibility purposes):
My thinking is that each pixel of the image represents a lattice coordinate in a 256 x 256 grid (hence, why there are 257 x 257 heights). That would mean that the normal at coordinate (i, j) is determined by the heights at (i, j), (i, j + 1), (i + 1, j), and (i + 1, j + 1) (call those A, B, C, and D, respectively).
So given the 3D coordinates of A, B, C, and D, would it make sense to:
split the four into two triangles: ABC and BCD
calculate the normals of those two faces via cross product
split into two triangles: ACD and ABD
calculate the normals of those two faces
average the four normals
...or is there a much easier method that I'm missing?
Example GLSL code from my water surface rendering shader:
#version 130
uniform sampler2D unit_wave
noperspective in vec2 tex_coord;
const vec2 size = vec2(2.0,0.0);
const ivec3 off = ivec3(-1,0,1);
vec4 wave = texture(unit_wave, tex_coord);
float s11 = wave.x;
float s01 = textureOffset(unit_wave, tex_coord, off.xy).x;
float s21 = textureOffset(unit_wave, tex_coord, off.zy).x;
float s10 = textureOffset(unit_wave, tex_coord, off.yx).x;
float s12 = textureOffset(unit_wave, tex_coord, off.yz).x;
vec3 va = normalize(vec3(size.xy,s21-s01));
vec3 vb = normalize(vec3(size.yx,s12-s10));
vec4 bump = vec4( cross(va,vb), s11 );
The result is a bump vector: xyz=normal, a=height
My thinking is that each pixel of the image represents a lattice coordinate in a 256 x 256 grid (hence, why there are 257 x 257 heights). That would mean that the normal at coordinate (i, j) is determined by the heights at (i, j), (i, j + 1), (i + 1, j), and (i + 1, j + 1) (call those A, B, C, and D, respectively).
No. Each pixel of the image represents a vertex of the grid, so intuitively, from symmetry, its normal is determined by heights of neighboring pixels (i-1,j), (i+1,j), (i,j-1), (i,j+1).
Given a function f : ℝ2 → ℝ that describes a surface in ℝ3, a unit normal at (x,y) is given by
v = (−∂f/∂x, −∂f/∂y, 1) and n = v/|v|.
It can be proven that the best approximation to ∂f/∂x by two samples is archived by:
∂f/∂x(x,y) = (f(x+ε,y) − f(x−ε,y))/(2ε)
To get a better approximation you need to use at least four points, thus adding a third point (i.e. (x,y)) doesn't improve the result.
Your hightmap is a sampling of some function f on a regular grid. Taking ε=1 you get:
2v = (f(x−1,y) − f(x+1,y), f(x,y−1) − f(x,y+1), 2)
Putting it into code would look like:
// sample the height map:
float fx0 = f(x-1,y), fx1 = f(x+1,y);
float fy0 = f(x,y-1), fy1 = f(x,y+1);
// the spacing of the grid in same units as the height map
float eps = ... ;
// plug into the formulae above:
vec3 n = normalize(vec3((fx0 - fx1)/(2*eps), (fy0 - fy1)/(2*eps), 1));
A common method is using a Sobel filter for a weighted/smooth derivative in each direction.
Start by sampling a 3x3 area of heights around each texel (here, [4] is the pixel we want the normal for).
[6][7][8]
[3][4][5]
[0][1][2]
Then,
//float s[9] contains above samples
vec3 n;
n.x = scale * -(s[2]-s[0]+2*(s[5]-s[3])+s[8]-s[6]);
n.y = scale * -(s[6]-s[0]+2*(s[7]-s[1])+s[8]-s[2]);
n.z = 1.0;
n = normalize(n);
Where scale can be adjusted to match the heightmap real world depth relative to its size.
If you think of each pixel as a vertex rather than a face, you can generate a simple triangular mesh.
+--+--+
|\ |\ |
| \| \|
+--+--+
|\ |\ |
| \| \|
+--+--+
Each vertex has an x and y coordinate corresponding to the x and y of the pixel in the map. The z coordinate is based on the value in the map at that location. Triangles can be generated explicitly or implicitly by their position in the grid.
What you need is the normal at each vertex.
A vertex normal can be computed by taking an area-weighted average of the surface normals for each of the triangles that meet at that point.
If you have a triangle with vertices v0, v1, v2, then you can use a vector cross product (of two vectors that lie on two of the sides of the triangle) to compute a vector in the direction of the normal and scaled proportionally to the area of the triangle.
Vector3 contribution = Cross(v1 - v0, v2 - v1);
Each of your vertices that aren't on the edge will be shared by six triangles. You can loop through those triangles, summing up the contributions, and then normalize the vector sum.
Note: You have to compute the cross products in a consistent way to make sure the normals are all pointing in the same direction. Always pick two sides in the same order (clockwise or counterclockwise). If you mix some of them up, those contributions will be pointing in the opposite direction.
For vertices on the edge, you end up with a shorter loop and a lot of special cases. It's probably easier to create a border around your grid of fake vertices and then compute the normals for the interior ones and discard the fake borders.
for each interior vertex V {
Vector3 sum(0.0, 0.0, 0.0);
for each of the six triangles T that share V {
const Vector3 side1 = T.v1 - T.v0;
const Vector3 side2 = T.v2 - T.v1;
const Vector3 contribution = Cross(side1, side2);
sum += contribution;
}
sum.Normalize();
V.normal = sum;
}
If you need the normal at a particular point on a triangle (other than one of the vertices), you can interpolate by weighing the normals of the three vertices by the barycentric coordinates of your point. This is how graphics rasterizers treat the normal for shading. It allows a triangle mesh to appear like smooth, curved surface rather than a bunch of adjacent flat triangles.
Tip: For your first test, use a perfectly flat grid and make sure all of the computed normals are pointing straight up.