Processing 3 - how to set default opacity - colors

This is driving me crazy. Here is my setup:
background(0);
stroke(255,255,255,255);
When I put a pixel out, then query the same pixel, the colors does not match:
point(2,2);
int x = get(2,2);
print(red(x));
print(green(x));
print(blue(x));
What I get is: 223 223 223
Why do I get 223 instead of 255?

This is happening because by default, Processing enables anti-aliasing. This adds blurriness and transparency to shapes, including points. Usually this is a good thing, as it makes your drawings look nicer.
But if you want to disable anti-aliasing because you need exact colors, then just call the noSmooth() function before you do your drawing:
noSmooth();
background(0);
stroke(255,255,255,255);
point(2,2);
int x = get(2,2);
print(red(x));
print(green(x));
print(blue(x));
This prints out 255 like you're expecting. More info can be found in the reference.

Related

Transformed colors when painting semi-transparent in p5.js

A transformation seems to be applied when painting colors in p5.js with an alpha value lower than 255:
for (const color of [[1,2,3,255],[1,2,3,4],[10,11,12,13],[10,20,30,40],[50,100,200,40],[50,100,200,0],[50,100,200,1]]) {
clear();
background(color);
loadPixels();
print(pixels.slice(0, 4).join(','));
}
Input/Expected Output Actual Output (Firefox)
1,2,3,255 1,2,3,255 ✅
1,2,3,4 0,0,0,4
10,11,12,13 0,0,0,13
10,20,30,40 6,19,25,40
50,100,200,40 51,102,204,40
50,100,200,0 0,0,0,0
50,100,200,1 0,0,255,1
The alpha value is preserved, but the RGB information is lost, especially on low alpha values.
This makes visualizations impossible where, for example, 2D shapes are first drawn and then the visibility in certain areas is animated by changing the alpha values.
Can these transformations be turned off or are they predictable in any way?
Update: The behavior is not specific to p5.js:
const ctx = new OffscreenCanvas(1, 1).getContext('2d');
for (const [r,g,b,a] of [[1,2,3,255],[1,2,3,4],[10,11,12,13],[10,20,30,40],[50,100,200,40],[50,100,200,0],[50,100,200,1]]) {
ctx.clearRect(0, 0, 1, 1);
ctx.fillStyle = `rgba(${r},${g},${b},${a/255})`;
ctx.fillRect(0, 0, 1, 1);
console.log(ctx.getImageData(0, 0, 1, 1).data.join(','));
}
I could be way off here...but it looks like internally that in the background method if _isErasing is true then blendMode is called. By default this will apply a linear interpolation of colours.
See https://github.com/processing/p5.js/blob/9cd186349cdb55c5faf28befff9c0d4a390e02ed/src/core/p5.Renderer2D.js#L45
See https://p5js.org/reference/#/p5/blendMode
BLEND - linear interpolation of colours: C = A*factor + B. This is the
default blending mode.
So, if you set the blend mode to REPLACE I think it should work.
REPLACE - the pixels entirely replace the others and don't utilize
alpha (transparency) values.
i.e.
blendMode(REPLACE);
for (const color of [[1,2,3,255],[1,2,3,4],[10,11,12,13],[10,20,30,40],[50,100,200,40],[50,100,200,0],[50,100,200,1]]) {
clear();
background(color);
loadPixels();
print(pixels.slice(0, 4).join(','));
}
Internally, the HTML Canvas stores colors in a different way that cannot preserve RGB values when fully transparent. When writing and reading pixel data, conversions take place that are lossy due to the representation by 8-bit numbers.
Take for example this row from the test above:
Input/Expected Output Actual Output
10,20,30,40 6,19,25,40
IN (conventional alpha)
R
G
B
A
values
10
20
30
40 (= 15.6%)
Interpretation: When painting, add 15.6% of (10,20,30) to the 15.6% darkened (r,g,b) background.
Canvas-internal (premultiplied alpha)
R
G
B
A
R
G
B
A
calculation
10 * 0.156
20 * 0.156
30 * 0.156
40 (= 15.6%)
values
1.56
3.12
4.7
40
values (8-bit)
1
3
4
40
Interpretation: When painting, add (1,3,4) to the 15.6% darkened (r,g,b) background.
Premultiplied alpha allows faster painting and supports additive colors, that is, adding color values without darkening the background.
OUT (conventional alpha)
R
G
B
A
calculation
1 / 0.156
3 / 0.156
4 / 0.156
40
values
6.41
19.23
25.64
40
values (8-bit)
6
19
25
40
So the results are predictable, but due to the different internal representation, the transformation cannot be turned off.
The HTML specification explicitly mentions this in section 4.12.5.1.15 Pixel manipulation:
Due to the lossy nature of converting between color spaces and converting to and from premultiplied alpha color values, pixels that have just been set using putImageData(), and are not completely opaque, might be returned to an equivalent getImageData() as different values.
see also 4.12.5.7 Premultiplied alpha and the 2D rendering context

Web Assembly drawing gray canvas

I'm using Go and compiling it to web assembly.
I'm trying to render a bunch of rectangles next to eachother with a random colour, but they keep rendering as just gray.
My render function looks something like this:
for row,_ := range rows {
for col,_ := range row {
ctx.Set("fillStyle", fmt.Sprintf("#%06x", rand.Int()))
ctx.Call("fillRect", 20, 20 + (col * width), maxHeight - (row*height))
}
}
With which it renders a big block (all rectangles are next to eachother) but just all in gray, instead of doing them in different colours.
Is this enough code in the example to help further? If not I can post it to a gist, as I'm new to WASM I'm unsure which parts could really be relevant - but those 2 functions are the only ones doing something with rendering as far as I can tell.
The problem is that you use this expression to construct the fill style:
fmt.Sprintf("#%06x", rand.Int())
rand.Int() returns a non-negative pseudo-random int. Size of int is 64 bits if GOOS=js and GOARCH=wasm. What this means is that the random int number will be random 8 bytes (first bit being always 0 due to being non-negative).
If you format such a number with the %06x verb, like almost all the time it will be more than just 6 hex digits. The width 6 means to be at least 6, and the flag 0 means to pad with zeros if less. But if it's longer, it is not truncated.
And if you set an invalid color to canvas.fillStyle, it will disregard it and the last set valid fill style will remain active. And I'm guessing it was a gray color you used before the loop.
Fix is easy, just make sure the random number has no more than 3 bytes, or in other words, 6 hex digits. Use a simple bitmask:
ctx.Set("fillStyle", fmt.Sprintf("#%06x", rand.Int()&0xffffff))
Or use rand.Intn() instead of rand.Int():
ctx.Set("fillStyle", fmt.Sprintf("#%06x", rand.Int(0x1000000)))
Also context.fillRect() expects 4 arguments: x, y, width and height, so it should be something like this:
ctx.Call("fillRect", 20+(col*width), maxHeight-(row*height), width, height)

Merging overlapping transparent shapes in directx

This is the problem I am facing simplified:
Using directx I need to draw two(or more) exactly (in the same 2d plane) overlapping triangles. The triangles are semi transparent but the effect I want to release is that they clip to transparency of a single triangle. The picture below might depict the problem better.
Is there a way to do this?
I use this to get overlapping transparent triangles to not "accumulate". You need to create a blendstate and set it on output merge.
blendStateDescription.AlphaToCoverageEnable = false;
blendStateDescription.RenderTarget[0].IsBlendEnabled = true;
blendStateDescription.RenderTarget[0].SourceBlend = D3D11.BlendOption.SourceAlpha;
blendStateDescription.RenderTarget[0].DestinationBlend = D3D11.BlendOption.One; //
blendStateDescription.RenderTarget[0].BlendOperation = D3D11.BlendOperation.Maximum;
blendStateDescription.RenderTarget[0].SourceAlphaBlend = D3D11.BlendOption.SourceAlpha; //Zero
blendStateDescription.RenderTarget[0].DestinationAlphaBlend = D3D11.BlendOption.DestinationAlpha;
blendStateDescription.RenderTarget[0].AlphaBlendOperation = D3D11.BlendOperation.Maximum;
blendStateDescription.RenderTarget[0].RenderTargetWriteMask = D3D11.ColorWriteMaskFlags.All;
Hope this helps. Code is in C# but it works the same in C++ etc. Basically, takes the alpha of both source and destination, compares and takes the max. Which will always be the same (as long as you use the same alpha on both triangles) otherwise it will render the one with the most alpha.
edit: I've added a sample of what the blending does in my project. The roads here overlap. Overlap Sample
My pixel shader is as:
I pass the UV co-ords in a float4.
xy = uv coords.
w is the alpha value.
Pixel shader code
float4 pixelColourBlend;
pixelColourBlend = primaryTexture.Sample(textureSamplerStandard, input.uv.xy, 0);
pixelColourBlend.w = input.uv.w;
clip(pixelColourBlend.w - 0.05f);
return pixelColourBlend;
Ignore my responses, couldn't edit them...grrrr.
Enabling the depth stencil prevents this problem

Fade through more more natural rainbow spectrum in HSV/HSB

I'm trying to control some RGB LEDs and fade from red to violet. I'm using an HSV to RGB conversion so that I can just sweep from hue 0 to hue 300 (beyond that it moves back towards red). The problem I noticed though is that it seems to spend far to much time in the cyan and blue section of the spectrum. So I looked up what the HSV spectrum is supposed to look like, and found thisL
I didn't realize that more than half the spectrum was spent between green and blue.
But I'd really like it to look much more like this:
With a nice even blend of that "standard" rainbow colors.
I'd imagine that this would end up being some sort of s-curve of the normal hue values, but am not really sure how to calculate that curve.
An actual HSV to RGB algorithm that handles this internally would be great (any code really, though it's for an Arduino) but even just an explanation of how I could calculate that hue curve would be greatly appreciated.
http://www.fourmilab.ch/documents/specrend/ has a fairly detailed description of how to convert a wavelength to CIE components (which roughly correspond to the outputs of the three kinds of cone sensors in your eyes) and then how to convert those to RGB values (with a warning that some wavelengths don't have RGB equivalents in a typical RGB gamut).
Or: there are various "perceptually uniform colour spaces" like CIE L*a*b* (see e.g. http://en.wikipedia.org/wiki/Lab_color_space); you could pick one of those, take equal steps along a straight line joining your starting and ending colours in that space, and convert to RGB.
Either of those is likely to be overkill for your application, though, and there's no obvious reason why they should be much -- or any -- better than something simpler and purely empirical. So why not do the following:
Choose your starting and ending colours. For simplicity, let's suppose they have S=1 and V=1 in HSV space. Note them down.
Look along the hue "spectrum" that you posted and find a colour that looks to you about halfway between your starting and ending points. Note this down.
Now bisect again: find colours halfway between start and mid, and halfway between mid and end.
Repeat once or twice more, so that you've divided the hue scale into 8 or 16 "perceptually equal" parts.
Convert to RGB, stick them in a lookup table, and interpolate linearly in between.
Tweak the RGB values a bit until you have something that looks good.
This is totally ad hoc and has nothing principled about it at all, but it'll probably work pretty well and the final code will be basically trivial:
void compute_rgb(int * rp, int * gp, int * bp, int t) {
// t in the range 0..255 (for convenience)
int segment = t>>5; // 0..7
int delta = t&31;
int a=rgb_table[segment].r, b=rgb_table[segment+1].r;
*rp = a + ((delta*(b-a))>>5);
a=rgb_table[segment].g; b=rgb_table[segment+1].g;
*gp = a + ((delta*(b-a))>>5);
a=rgb_table[segment].b; b=rgb_table[segment+1].b;
*bp = a + ((delta*(b-a))>>5);
}
(you can make the code somewhat clearer if you don't care about saving every available cycle).
For what it's worth, my eyes put division points at hue values of about (0), 40, 60, 90, 150, 180, 240, 270, (300). Your mileage may vary.
FastLED does a a version of this: https://github.com/FastLED/FastLED/wiki/FastLED-HSV-Colors
HSLUV is another option: http://www.hsluv.org/. They have libraries in a bunch of different languages.
Also, this is an interesting technique: https://www.shadertoy.com/view/4l2cDm
const float tau = acos(-1.)*2.;
void mainImage( out vec4 fragColor, in vec2 fragCoord )
{
vec2 uv = fragCoord.xy / iResolution.xy;
vec3 rainbow = sqrt( //gamma
sin( (uv.x+vec3(0,2,1)/3.)*tau ) * .5 + .5
);
fragColor.rgb = rainbow;
}
Also see:
https://en.wikipedia.org/wiki/Rainbow#Number_of_colours_in_spectrum_or_rainbow for more info.

HLSL tex2d sampler seemingly using inconsistent rounding; why?

I have code that needs to render regions of my object differently depending on their location. I am trying to use a colour map to define these regions.
The problem is when I sample from my colour map, I get collisions. Ie, two regions with different colours in the colourmap get the same value returned from the sampler.
I've tried various formats of my colour map. I set the colours for each region to be "5" apart in each case;
Indexed colour
RGB, RGBA: region 1 will have RGB 5%,5%,5%. region 2 will have RGB 10%,10%,10% and so on.
HSV Greyscale: region 1 will have HSV 0,0,5%. region 2 will have HSV 0,0,10% and so on.
(Values selected in The Gimp)
The tex2D sampler returns a value [0..1].
[ I then intend to derive an int array index from region. Code to do with that is unrelated, so has been removed from the question ]
float region = tex2D(gColourmapSampler,In.UV).x;
Sampling the "5%" colour gave a "region" of 0.05098 in hlsl.
From this I assume the 5% represents 5/100*255, or 12.75, which is rounded to 13 when stored in the texture. (Reasoning: 0.05098 * 255 ~= 13)
By this logic, the 50% should be stored as 127.5.
Sampled, I get 0.50196 which implies it was stored as 128.
the 70% should be stored as 178.5.
Sampled, I get 0.698039, which implies it was stored as 178.
What rounding is going on here?
(127.5 becomes 128, 178.5 becomes 178 ?!)
Edit: OK,
http://en.wikipedia.org/wiki/Bankers_rounding#Round_half_to_even
Apparently this is "banker's rounding". I have no idea why this is being used, but it solves my problem. Apparently, it's a Gimp issue.
I am using Shader Model 2 and FX Composer. This is my sampler declaration;
//Colour map
texture gColourmapTexture <
string ResourceName = "Globe_Colourmap_Regions_Greyscale.png";
string ResourceType = "2D";
>;
sampler2D gColourmapSampler : register(s1) = sampler_state {
Texture = <gColourmapTexture>;
#if DIRECT3D_VERSION >= 0xa00
Filter = MIN_MAG_MIP_LINEAR;
#else /* DIRECT3D_VERSION < 0xa00 */
MinFilter = Linear;
MipFilter = Linear;
MagFilter = Linear;
#endif /* DIRECT3D_VERSION */
AddressU = Clamp;
AddressV = Clamp;
};
I never used HLSL, but I did use GLSL a while back (and I must admit it's terribly far in my head).
One issue I had with textures is that 0 is not the first pixel. 1 is not the second one. 0 is the edge of the texture and 1 is the right edge of the first pixel. The values get interpolated automatically and that can cause serious trouble if what you need is precision like when applying a lookup table rather than applying a normal texture. You need to aim for the middle of the pixel, so asking for [0.5,0.5], [1.5,0.5] rather than [0,0], [1, 0] and so on.
At least, that's the way it was in GLSL.
Beware: region in levels[region] is rounded down. When you see 5 % in your image editor, the actual value in the texture 8b representation is 5/100*255 = 12.75, which may be either 12 or 13. If it is 12, the rounding down will hit you. If you want rounding to nearest, you need to change this to levels[region+0.5].
Another similar thing (already written by Louis-Philippe) which might hit you is texture coordinates rounding rules. You always need to hit a spot in the texel so that you are not in between of two texels, otherwise the result is ill-defined (you may get any of two randomly) and some of your source texels may disapper while other duplicate. Those rules are different for bilinar and point sampling, you may need to add half of texel size when sampling to compensate for this.
GIMP uses banker's rounding. Apparently.
This threw out my code to derive region indicies.

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