Should you normalize RGB values by 255 or 256?

You don’t divide a float by 256 by shifting it right eight bits; that would yield complete garbage. You subtract 8 from the exponent, then check if you got an underflow.

groundzeros2015• 2 months ago

I’m dumb. Doesn’t 0 start at the beginning?

dylan6044• 2 months ago

It's right up there with the confusion if 2000 was the new year of the 21st century or the last year of the 19th century.

Timwi• 2 months ago

I'm confused by that analogy. Is the “ruler” a 255-inch ruler with 256 points labeled 0–255, or is it a 256-inch ruler with 256 1-inch segments, making L = 256×1?

zephen• 2 months ago

The analogy is pretty straightforward.

When you have a 12 inch ruler, you effectively have 13 numbers on the ruler. The fact that zero isn't marked is neither here nor there -- the numeral one is not at the far end of the ruler.

So if you extend the ruler to be as long as you can hold in eight bits, it will range from 0 to 255, and the total length will be 255.

The ruler analogy may seem overly simplistic, but then the real world is likewise fairly simplistic.

At the end of the day, the numbers presumably come from a sensor, or go to a display, and, often, in either case, zero represents as dark as you can get and 255 represents as light as you can get, so the physics dictate that the intervals associated with the 0 and 255 are half the size of the rest of the intervals.

Audio is more interesting than video, because in audio, you care deeply about not having an offset, and about having a balanced signal, so the question of whether the midpoint is actually on a number or not is pertinent.

In audio, it is often useful to simply discard a code so that 0 is the midpoint (e.g -65535 to +65535, discarding 0xFFFF). But this still gives you smaller intervals at both ends.

layer8• 2 months ago

But who says that the numbers are representing the points, rather than representing the intervals between the points?

wky• 2 months ago

It doesn't even need to represent intervals. A 13 inch ruler with 13 markings at 0.5, 1.5, etc inches is still a valid ruler, albeit an odd construction.

m463• 2 months ago

the correct way is to use a slide rule

knappa• 2 months ago

Fencepost errors aren't errors if you are actually trying to count fenceposts.

herf2• 2 months ago

I'll argue for the +0.5 solution. First, I don't like half-sized intervals at the edges, and second, a 255-based representation is typically a SDR (not HDR) image.

RGB values represent luminances against some adapted state, and a "zero" in a daylit scene is not "zero luminance" - it's just about 0.001x as bright as the brightest point - it's millions of photons, way more than zero. In a sense our eyes experience contrast on a sliding scale, and there is no absolute zero in the system. For example, broadcast systems historically used 16-235 as their luminance range for SDR. I think any argument that says "we must have zero" is going to have a bias, but I don't think zero is needed for most things.

yxhuvud• 2 months ago

Both solutions add 0.5, the difference is where in the process it happens.

account426• 2 months ago

They don't - trunc(result * 255 + 0.5) is just round(result * 255), i.e. error is on average 0 whereas trunc(result * 256) has an average bias of -0.5.

amavect2• 2 months ago

I agree. Additionally, both 0.0 and 1.0 don't really exist for dithered signals, so a byte should map to [0.5, 255.5] before division by 256. This also solves the signed integer asymmetry, as a signed byte maps to [-127.5, 127.5] before division by 128. I wonder if audio DSP folks have done this already.

amavect2• 2 months ago

Thinking about this more, dithering requires negative values to cancel out when adding. Works for audio, but color doesn't have negative numbers.

themafia• 2 months ago

In a sense our eyes experience contrast on a sliding scale

There's a whole visual center to check the amount of incoming light and adjust your pupils for you. It's intentionally reactive.

and there is no absolute zero in the system.

There maybe is. I think we call that "blind."

broadcast systems historically used 16-235 as their luminance range for SDR

Mostly because it was a fully analog system and these all translate down to signal voltage. Jokingly NTSC used to be referred to as "Never Twice the Same Color" due to being a compromise bolted onto the side of an already compromised system.

a_conservative• 2 months ago

> and there is no absolute zero in the system.
There maybe is. I think we call that "blind."

If you go looking into that, you'll see that the reality is far far more complex [0]

"The number of people with no light perception is unknown, but it is estimated to be less than 10 percent of totally blind individuals."

[0] https://chicagolighthouse.org/sandys-view/what-blind-people-...

infinet• 2 months ago

Interesting idea, but somehow I feel the world is shaking. For the processing program, what used to black(0.0) and white(1.0) has became very dark gray and very bright gray.

pixelesque3• 2 months ago

As someone with a lot of experience in this area doing image processing and rendering for VFX (including writing image readers and writers for my own software and commercial VFX software), I think you might be forgetting that colourspace conversion (to sRGB 'linear' rec709 for old-school SDR, but other more wider gamuts for newer formats) would happen after this, so the 'squish' of the dynamic range would happen after loading.

Also, a lot of workflows for image processing and compositing do assume that 0 means zero, whether correctly or not (often incorrectly). So there are often assumptions that for 8-bit, 0u maps to 0.0f and 255 maps to 1.0f for things like masking or alpha: as soon as you have 0 values which become just over 0.0, you then have artifacts because some code somewhere is using a hard threshold of 0.0 to mask some other operation, and vice-versa for 1.0 with alpha, where suddenly because the 255 values are no longer 1.0f, you have very slightly see-through objects (often only visible in certain situations or when pixel-peeping) after pre-multiplication.

(Same thing can happen when 254 becomes 1.0f after +0.5 with masking).

herf2• 2 months ago

good point - alpha is a notable exception, it is not luminance

yuriks• 2 months ago

I think more to the point, if 0 doesn't represent 0.0, and 255 doesn't represent 1.0, congratulations you've just lost your additive and multiplicative identities and most of the math used in colors falls apart.

The argument for 0-256 feels compelling when thinking about the physical display, but it seems like a very poor fit for any digital image processing or rendering.

HelloNurse• 2 months ago

Remember how the 0 and 255 bins poked slightly beyond the [0,1][0,1] range’s edges? In the standard approach, the range of representable values is actually [−0.5/255,255.5/255][−0.5/255,255.5/255], meaning the bins are spaced further apart than strictly needed for [0,1][0,1] inputs

This is of course silly: the "range of representable values" of floating point colour components is [0,1] independent of quantization and how an invalid input would be quantized is irrelevant.

Looking at the actual "big picture" there are 256 representable values and (taking into account gamma correction, arbitrary ranges other than [0,1], deliberately nonuniform quantization bins, and other plausible complications) their correspondence to 256 floating point values should be regarded as a generic lookup table, abandoning all hope of using elegant and cheap formulas and making it obvious than encoding and decoding differently is not an option.

dylan6044• 2 months ago

broadcast systems historically used 16-235

For 8-bit, 16 maps to 7.5IRE which is the well understood legal black. Mapping 235 means they mapped peak to 110IRE. This is based on a 0-120IRE scale. This gets weird as the broadcast limit for video was 100IRE allowing for the chroma to reach 110IRE. So if you're trying to limit your white values to 235, that'll be higher than is broadcast safe. Of course, nobody cares about NTSC broadcast limits any more. However, to this day, I still see out of spec tapes marked as "broadcast master" that have been ingested for streaming use. It drives me crazy to this day, and it's only getting worse as people don't even have scopes to adjust the VTR's TBC properly.

variaga• 2 months ago

Ugh. Sudden flashbacks to having to switch analog output between Japanese NTSC (no pedestal) and US NTSC (with pedestal) without getting weird noise in the black regions.

But IIRC the MPEG-2 standard had luma==235 -> 100IRE for all of the analog formats (pal/ntsc-j/ntsc/secam) so I'm not sure why you say that would violate the broadcast limits?

keithwinstein• 2 months ago

For 8-bit, 16 maps to 7.5IRE which is the well understood legal black. Mapping 235 means they mapped peak to 110IRE.

Generally no -- in an 8-bit NTSC-M Rec. 601 system, 16 maps to E'Y = 0 at 7.5 IRE, and 235 maps to E'Y = 1 at 100 IRE. See https://www.poynton.ca/pdf/Poynton-1996-TechIntrDigiVide.pdf

The "16" digital black level is independent of the "7.5 IRE" analog setup. E.g. in Japan with an 8-bit "NTSC-J" Rec. 601 system, my understanding is that 16 still maps to E'Y = 0 which is now at 0 IRE, and 235 is still E'Y = 1 at 100 IRE.

pornel2• 2 months ago

Although the post focuses on RGB, the same quantization issue exists for any type of signal being mapped between discrete and continuous representations.

The issue isn't in having a representation for 0 photons, but about maximizing information stored in a byte. Ideally you shouldn't be underutilizing the byte value 0, nor add bias to data that should have been assigned to the 0th bucket, regardless of what it represents (you could have a color space that goes from bright to super bright, and still want to ensure that every byte represents equal chunk of your brightness range).

PaulDavisThe1st• 2 months ago

Yep, the exact same problem arises in digital audio, mapping between integer sample formats and the floating point representation that is generally used internally.

kazinator2• 2 months ago

They are not half sized at the edges, unless negative black bothers you.

account426• 2 months ago

For example, broadcast systems historically used 16-235 as their luminance range for SDR.

Unfortunately "modern" HDMI is still plagued by this insanity so if your display and source don't agree you can either get washed out or crushed blacks.

theyeenzbeanz• 2 months ago

Should always be 0-255 as that fits an unsigned byte.

Retr0id2• 2 months ago

assume that in both cases the output values are clamped before the final typecast

crazygringo3• 2 months ago

That's not what the article is about.

Retr0id2• 2 months ago

Both of these assume a linear transfer function, which is rarely the case.

leni536• 2 months ago

Basically never for 8-bit color channels.

crazygringo3• 2 months ago

Advice for anyone on mobile: read in landscape mode if you want to be able to see the division by 256 version code example at the start.

The HTML/CSS is bad that lets it completely overflow the right edge of the page instead of wrapping.

I re-read this post three times in total confusion before I figured out the most important piece was off-screen entirely.

dist-epoch3• 2 months ago

A similar issue exists in the audio world, for example 16-bit integer audio is between [-32768, 32767] (non-symmetric), but floating point audio is [-1.0, 1.0].

adzm• 2 months ago

note that floating point audio very often exceeds [-1.0, 1.0] within the pipeline, just to be tamed at the very end of the mix to fit within those bounds. this is pretty much why every modern DAW uses floating point these days.

ack_complete• 2 months ago

There is an analogous situation in graphics with signed normalized formats. The solution there is that the R16_SNORM format maps -1 to +1 as [-32767, 32767] with -32768 being a special value (not normally emitted, and mostly but not always interpreted as -32767). Some audio storage formats seem to use this mapping too.

Sesse__2• 2 months ago

You should multiply by 255.0, optionally add a dither (triangular is okay), and then let the FPU round using its default IEEE 754 round-to-nearest-ties-to-nearest-even mode. None of this crazy 0.5 stuff. :-)

ctdinjeu8• 2 months ago

Both. 255 for each color and the last 1 as the alpha for each channel.

Why not??? Fight me

atilimcetin2• 2 months ago

Interesting article. I tend to use

- i = min(floor(f * 256), 255) (from float to uint8)

- f = i / 255 (from uint8 to float)

Basically a mix of the 2 approaches mentioned in the article.

For all integers between [0,255], if I do uint8 -> float -> uint8 conversion, I will get the same result.

edit: I wondered what's the maximum jitter amount that I can introduce to the float and get the same uint8 value. And also these 0->0.0 and 255->1.0 should map properly.

With my approach at the top, the jitter margin that I can introduce is 1/65280.

But with the article's approach

- i = floor(f * 255 + 0.5)

- f = i / 255

maximum jitter margin is 1/510 (which is better).

vitorsr• 2 months ago

This is what I do for the former:

    floor( nextafter( 256, 255 ) * value )

atilimcetin2• 2 months ago

Oh very nice idea to get rid of the min operator.

AgentME2• 2 months ago

It's worth pointing out that the article explicitly calls out your first mixed technique:

Finally, one should never mix the encode and decode steps of the two quantizers. That’s just broken code. It’s an easy mistake to make, though.

dgently72• 2 months ago

"Let’s say you’re writing an image processing program. The program takes in an image, converts it to floating point, does some processing and finally saves the modified pixels to disk as 8-bit colors. "

excuse to argue about the best way aside, if this is the goal you should not be rolling your own image file reading. you should use openimageio. idk what approach it takes in its internal conversion to float, but that library is more likely to have the right answer than you trying to roll it yourself given its the library used internally by tons of professional image manipulation software...

pixelesque3• 2 months ago

If you're a beginner, or just want something which works quickly, sure.

However OIIO is far from perfect in all situations (having had to debug and fix issues with its mip-map generation filtering code in the past), so don't always assume that just because there's a mature open source library out there doing something that it's always perfect.

dgently72• 2 months ago

sure of course nothing is perfect and oiio has a lot of surface area / is still oss. thats good advice.

ive just seen a lot of "ai researchers" who are getting into professional image processing and are both beginners and want things quickly and so could do much worse than just starting from what they get out of oiio. especially for a lot of the non-obvious stuff (more of that in color handling than just the io stuff though)

AgentME2• 2 months ago

OpenImageIO uses the standard division by 255 technique: https://openimageio.readthedocs.io/en/latest/imageoutput.htm...

Nuthen• 2 months ago

That was a fun article to read of something I haven't had to think about in a while. It brought to mind moments in game development of having pixel art needing to be drawn on an integer value despite the game logic using floating point math. I tried something similar to the +0.5 in places so that it wouldn't look as bad (especially when there's a moving camera, which also needed to be truncated..).

I also enjoyed the 2002 article by Jonathan Blow [1] that's linked at the bottom. The visualization from the first article helped a lot once this started to go more in-depth.

[1] https://web.archive.org/web/20240706043551/https://number-no...

jessetemp6• 2 months ago

The author is confusing bins with bin edges. In their first plot, the standard approach looks strange because 0-7 should be the bin edges, not the center points as shown in the plot.

You can see this confusion again in the histogram example. There are only 255 bins, not 256. If you fix that mistake and remove the 0.5 offset, then the histogram is distributed correctly at both ends.

bjourne• 2 months ago

How do you fit 256 distinct values into 255 bins?

jessetemp6• 2 months ago

By counting the edges

nomel• 2 months ago

2*8 = 256. You can represent 256 distinct values, bins, with an 8 bit number. If you stick a 0 in that first one, it takes a bin. If you fill the rest with by-one increasing integers, then the max value will be 255, thus the 2*bits - 1, which is the max value you can store.

pornel2• 2 months ago

No, the author understands the problem way deeper than you do.

You haven't grasped the fact that the choice isn't obvious, and has subtle trade-offs.

If you don't believe the author, check the other posts he references.

jessetemp6• 2 months ago

Judging by your other comment in this thread, you might agree with my rational [1] more than you realize

[1] https://news.ycombinator.com/item?id=48365800

alterom3• 2 months ago

Sorry, you seem to be confused.

There are only 255 bins, not 256

There are 256 bins because there are 256 values.

The questions are:

1. What are the boundaries of these bins?

2. Which sample represents a particular bin?

With 1-bit color, we have sample values {0, 1}. What bins do they represent?

Here's one choice:

     [0, 1), [1, 2)

Two equally sized bins, spanning the interval [0, 2] of length 2, each defined by its sample at lower bound.

Alternatively, we could consider these bins:

    [-0.5, 0.5), [0.5, 1.5)

These are also equally sized bins, spanning the interval [-0.5, 1.5] of length 2, defined by samples at the center.

We could also define bins like this:

    [0, 0.5), [0.5, 1]

Two equally sized bins spanning the interval [0, 1] of length 1, where we sample the first bin at the lower bound, and the last bin at the upper bound.

This, in a nutshell is what the author is trying to explain.

Let's look at this again, with 2 bits.

With 2-bit color, we have sample values {0, 1, 2, 3}.

Which bins do they come from?

The three options above yield:

    [0, 1), [1, 2), [2, 3), [3, 4)

    [-.5, 0.5), [0.5, 1.5), [1.5, 2.5), [2.5, 3.5)

    [0, 0.5), [0.5, 1.5), [1.5, 2.5), [2.5, 3]

The first two span an interval of length 4, the third spans an interval of length 3.

In the third case, the tail bins are short (have size ½), and the rest have size 1.

The last bin must be a closed interval in the third case, so that it includes the value we picked to represent it.

None of these choices is inherently invalid or better than the others; and none stems from "confusing bins with edges".

The third option does have the distinction that the first and last bins are smaller than the rest. But it's not necessarily a drawback. Especially when we're talking about color, hardware interpretation, and human perception.

When you remap these bins into the [0, 1] interval, you're "dividing by 4" in the first two cases, and by 3 in the third case.

The maps are:

     x → x/4
     x → (x + ½)/4
     x → x/3

The inverse maps (that yield a sample in {0, 1, 2, 3} given a floating point value in interval [0, 1]) are:

    x → trunc(4x)
    x → round(4x - ½) = trunc(4x)
    x → trunc(3x + ½)

In the first two options, the domain is [0, 1). It might be necessary to apply clipping because the exact value 1.0 falls outside the range of the forward transform.

The 2nd option is the most symmetric, of course, but the 3rd one is the most straightforward (and cheapest) to implement, so that's the default.

The choice amounts to making the highest and lowest bins slightly smaller to make the rest sightly larger.

That's to say, if you generate uniform noise between 0 and 1, you'll get the following samples from your function with equal probability:

    0 or 3
    1
    2

As the author points out, this hardly matters when you are talking about having 256 bins.

That, and with color specifically, the "good" histograms aren't uniform anyway (and any photographer wants to avoid getting much at either extreme).

TL;DR: The author is not confusing anything — but their diagram and explanation are, indeed, a bit confusing.

jessetemp6• 2 months ago

Thank you for the thoughtful reply. Maybe bins is the wrong word to use, so I'll try with intervals. Starting with 1 bit data, there are two numbers and one interval. I think where bins makes it confusing is that inside the interval there are two big rounding errors mapping everything to either 0 or 1 and many people seem to be considering those the bins.

Taking a step back, remember we're ultimately mapping these discrete numbers to some real world continuous variable like the saturation of red, frequency, mass on a scale, whatever. And our digital device can only represent a finite amount of numbers. For 2 bit data, we can represent 0-3, and for 3 bit data we can represent 0-7.

The important part is that 0 represents the minimum and 1,3, and 7 all represent the same maximum real value, and everything that can be measured by the device will fall within those ranges. So comparing 1, 2 and 3 bit data on a linear number line looks like this:

  0                    1
  0      1      2      3
  0  1  2  3  4  5  6  7

You could assume that everything gets assigned to whatever number is nearest in the number scale or come up with another scheme, but that is ultimately defined by the ADC and likely nonlinear. All we know is that those are the numbers we have available to represent the real values we're measuring.

The question is about how to normalize the data. 1 bit data is already normalized. If you normalize 2 bit data by 3 you get [0, 1/3, 2/3, 1]. LGTM. If you normalize it by 4, you get [0, 1/4, 2/4, 3/4] and you're effectively throwing away some of the range of the ADC. You can try to get it back by offsetting by 0.5 then normalizing but now you get [1/8, 3/8, 5/8, 7/8]. And you could stretch that with some clever formula to fill from 0 to 1, but if you do it right then it's the equivalent to normalizing by 3, so why not normalize by 3?

So the answer is, if you have N bit data, you normalize by 2^N-1.

RobRivera• 2 months ago

Are we talking 0 or 1 based values? HONKHONK*

AlienRobot• 2 months ago

Case against 255: it looks wrong in the graph :(

Case against 256: no 0 or 1 values :(

Considering how important having a 0 and 1 value is for arithmetic in general, I think 255 is better.

moefh• 2 months ago

This problem of what exactly a color value means is mostly inconsequential when you have 8 bits per component, the difference in the denominator being either 255 or 256 makes the errors tiny, you must have really good color perception and get really close to the screen to see any difference at all, and your monitor/phone screen is probably not calibrated anyway, so who cares.

It becomes a pain in the ass when you're generating a VGA signal with a microcontroller with 8 color output pins (3 red, 3 green, 2 blue). The meaning of a color value is very real in this setup: it corresponds to a voltage level you must send to the VGA monitor, 0V-0.7V.

So the blue channel will map (0->0V, 1->0.23V, 2->0.47V, 3->0.7V), and the red/green will map (0->0V, 1->0.1V, ..., 7->0.7V). Notice how none of the blue voltages match any of the red/green ones (other than the extremes)? That means you don't get to see any pure grays -- the closest ones will have bit of blue or yellow tint, depending on the direction of the difference.

Not only that, any gradients at all (other than the ones not mixing blue with the other channels) will be noticeable off: for example, the closest colors in the line between pure red to pure white will all be slightly orange or purple.

Code for VGA output in 8-bit color with double-buffered 320x240 framebuffer for the Raspberry Pi Pico 2 here, if anyone cares: https://github.com/moefh/pico-vga-8bit-demo

chongli• 2 months ago

You forgot about gamma correction. Before converting a value in the range of 0-255 into a voltage, PCs typically raise that value to the power of 2.2. This makes the difference between small values and large values far more apparent:

2^2.2 = 4.595, 255^2.2 = 196,964.699

spider-mario2• 2 months ago

Differences between small and large values are irrelevant to the point being made here, though. Much more relevant is the difference between nearby values, and the gamma just gets that closer to logarithmic perception, instead of perceptual steps being disproportionately large for small values.

(This may be more apparent when you frame gamma as being applied in the 0-1 range, so it doesn’t really turn 2 into 4.595 and 255 into ~200k; it turns (2/255)≈0.00784 into (2/255)^2.2 ≈ 0.0000233, and leaves (255/255)=1 as is.)

flufluflufluffy• 2 months ago

That means you don't get to see any pure grays -- the closest ones will have bit of blue or yellow tint, depending on the direction of the difference.

OMG I remember as a kid staring at static-y CRT displays, and seeing these faint blue and yellow lines at the borders of them. I’d always wondered why they appeared and why they were specifically blue and yellow. I finally know! (at least, assuming those specific artifacts are due to the same thing)

adastra22• 2 months ago

specifically on the edges? I would guess that is the phosphor layout, that the 'gray' beam is hitting a blue phosphor but not the red & green, or vice-versa.

tdeck• 2 months ago

Notice how none of the blue voltages match any of the red/green ones (other than the extremes)? That means you don't get to see any pure grays

I assume this is why RGBI color was so common in the 80s.

londons_explore• 2 months ago

Dithering in time seems like the best solution to this problem. Delta sigma modulation per pixel can be done reasonably easily.

Changing at 30Hz I doubt a human can tell the difference between slightly blue and slightly yellow.

account426• 2 months ago

This is called frame rate control (FRC) in display terminology, at least when used for LCD displays where the panel has fewer distinct levels than the manufacturer wants to advertise.

kazinator2• 2 months ago

No, the "alternative" approach looks strange in the 7 bit example.

1.0 lies on the right side of the bin 7. But 0.0 lies on the left of bin 0.

The standard approach assumes that we have centered samples: that zero is dead black, plus (and minus!) some uncertainty and so is bin 7.

If the sampling of the intensity is distortion-free (no clipping took place due to overexposure) then bin 7 represents a range of possible values centered around 1.0.

It is not a half-sized interval.

This means that when converting floating-point values in the [0,1] range back to integers, the extreme bins have effectively half the width of other bins.

Under any interpretation whatsoever of the image samples, there is latitude for interpreting the maximum value 255 as being distortion: clipping from an arbitrarily higher value. Shifting things by 0.5 doesn't fix this issue of not knowing whether 255 means that an intensity close to 1.0 is being represented (no distortion), or an outlier intensity of 37.49 (severely clamped). That could go the other way too.

In other words, there is a possible bias in the extreme bin. The signal could be limited such that the bin's full sampling range is not in effect, or the signal could be overwhelming, so that values far outside of the range are clipped and included.

The only way around this is to make the highest value a canary which represents "clipped value". That is to say, 255 means "clipped datum", so that only 254 and below is sampling of unclipped signal. Machine-generated image (e.g. 3D rendering) then avoid the 255 value, and camera sensors are calibrated so that it doesn't occur when technical images are being shot.

wyager• 2 months ago

You don't need to make this judgement; it's fixed by the colorspace you're working in.

First, figure out what colorspace the processing needs to happen in. Usually this is linear RGB.

Then, figure out what OETF and EOTF your input/output format use. This will be something like PQ or HLG. This will exactly specify the meaning of each integer value.

This fixes the choice of representation and conversion.

dpark3• 2 months ago

The entire issue arises from the use of truncation, right? It guarantees that only an exact 1.0 could land in the 255 bin so the net effect is a reduction of 256 bins to 255 bins. (Using random numbers as shown also guarantees no 1.0.)

Why not scale to fill the available bins, though? i.e. trunc(result * 255.999)?

klodolph3• 2 months ago

Why not scale to fill the available bins, though? i.e. trunc(result * 255.999)?

That’s half of the mid-riser staircase quantizer discussed in the article. (The other half is coming up with the reverse.)

(I would implement it as min(floor(x * 256), 255).)

dpark3• 2 months ago

Kind of. According to the article, the mid-riser vs mid-tread distinction correlates to where the 0.5 is applied in the transform. I’m proposing that there is no 0.5 applied at all. Instead of counteracting the compression of the truncation operation by adding a fixed offset, it multiplies by a scale factor.

Possibly my proposal doesn’t hold up to repeated transforms and operations. It might skew toward 255 in real operations.

BearOso• 2 months ago

There is a fallacy here in assuming there's 256 steps from 0 to 255. That's not true, there's 256 values that can be represented in 8 bits, and 255 steps (spaces between those values) from 0 (black) to (255) pure white. Thus, the division by 255 isn't problematic. Of course, 128 isn't half grey, it isn't in 0-255 and quantized 8-bit values are almost always in sRGB, not linear perceptive space.

This is the same kind confusion that happens with sampling positions in modern APIs, where the location is specified in coordinates and not in pixel centers.

smallstepforman• 2 months ago

BeOS API is based on pixel centers, not that anyone cares anymore …

1xn• 2 months ago

I care! :-)

2001zhaozhao• 2 months ago

There are only two real solutions after factoring in the need to preserve black as zero.

They are "rgb / 255.0" vs. "rgb / 256.0". Both have different tradeoffs. Pick your poison. (If you're using a 8 bit display signal then you better match whatever value the OS picked for the mapping back to the display, so your RGB values pass through unchanged)

fps-hero2• 2 months ago

This was a genuinely thought provoking article. I had to challenge some personal assumptions.

Coming from an electrical engineering background, I disagree with how the author presented "Two types of quantizers". Mathematically rigorous, but not grounded in practical systems.

In ADCs, there is always an inherent +-1/2 LSB of quantisation uncertainty. The transfer characteristic is always mid-tread sampling, or at least I haven't come across any counter examples. This is true for bipolar or unipolar ADCs.

The lowest code is negative voltage reference, and the highest code the positive reference. The transfer characteristic plot will show what the author has demonstrated, that the highest and lowest bins will effectively be 1/2LSB in width.

In a unipolar system, this has the consequence of not being able to represent the midpoint voltage precisely, or in other words, the gray problem. In a bipolar system, 0V will be mid-tread N/2 value, but that doesn't mean it has "256 ranges".

So, I'll be sticking with (VREF+ - VREV-) * k / (2^N - 1). Or in other words I agree with the normalisation by 255. It's the fence post error all over again, you have N values, but N-1 ranges. If you have less ranges than you do values, you need to distribute 1 of those ranges between two values, hence the 1/2LSB range endpoints.

kroeckx• 2 months ago

All ADCs I have looked at document that they can't represent the positive full scale. For instance, for an 8 bit ±1 V ADC, -128 represents -1 V, +127 represents 127/128=0.99219 V. The transition from 126 to 127 happens at 1.5 LSB from the positive full range. 1 LSB difference represents 1/128 = 0.00781 V difference, and not 2 / 255 = 0.00784 V.

But if you actually care about what the voltage (and uncertainty) is, most of this is difference is mostly pointless, you're reference will have a bias, there are linearity errors and so on. 1 LSB will not match either the 1/128 or 2/255, you will need parameters to compensate for it.

fps-hero2• 2 months ago

You've made me do even more digging and now I'm even more confused.

ATMega328P, data sheet specifically calls out using that it cannot represent full range, ADC = VIN * 1024 / VREF. The STM32F4 datasheet shows an idealised transfer function which is mid-tread, but the "actual transfer function" is sort of a shifted mid-tread (lowest code is larger than a single LSB, highest code is VREF and 1/4 LSB). Other STM32 references I found show that it should follow an ideal mid-tread, unless you are using VADC as your voltage reference. High resolution differential input ADCs are, as best as I can tell, always mid-tread with both end codes representing the positive and negative voltage reference.

The best data sheet I've found was for a PIC32, very detailed transfer characteristic diagram. This shows that each LSB is VREF/1024, but also why! The transfer function shows that is is still a mid-tread transfer, but the highest code is assigned to (1023/1024) * VREF, hence the divide by 1024, but it doesn't solve the 1/2LSB end bin problem.

This is different to the ideas presented the article, there is no reason a single ended ADC couldn't use mid-tread quantisation with VREF as the end code. What it does show is that by using divide by 256, you are truncating the range and arbitrarily deciding where the end points should be. This doesn't fix any inherent 1/2LSB quantisation uncertainty, which was the main argument against using divide by 255.

As you say, in ADCs it's a moot point with all of the other sources of error.

jason_s3• 2 months ago

Largely those datasheets are poorly written... but in practice it doesn't matter much because the manufacturing variations of offset and gain error are larger than the quantization level.

jason_s3• 2 months ago

The datasheet specifications are largely standardized across ADC manufacturers and do not represent the exact mathematical norm but rather the range of ADC performances. So it's not that "the transfer characteristic is always mid-tread sampling" but rather that "the ideal model of the transfer characteristic is always X".

I do have some historical examples where mid-riser was used, but in practice it doesn't matter when you consider ADC variations and the variation of gain and offset are multiple counts. (8-bit or more; this exact definition would have been important for ADCs less than 8 bits.)

MyMemoryfails• 2 months ago

As game dev, i never understood why floats are used to present colors? Isn't integers better? The issues which this article mentioned wouldn't exist.

I can only think its due integers having undefined behavior what happens on overflow, usually its wrapping but not always.

sheept• 2 months ago

Floats are better for calculating lighting. I would think some GPUs are probably also more optimized for float processing than ints.

somat2• 2 months ago

It is because light(colors) is a fundamentally exponential process and floats are also a fundamentally exponential process and as such the two are a good match for one another.

account426• 2 months ago

The problem would still exist, for example when you want to multiply colors which is common in lighting calculations - with integers you would need to divide by either 255 or 256 after such an operation. Which floats you only need to make that decision when converting to/from float representation, which is part of what makes them convenient.

jason_s3• 2 months ago

Dammit, I have an 80%-written article covering the same issue but for ADCs, and had to put it aside for the past few months. There's historical precedent here from the 1960s and 1970s, and in large part it involves testing and definitions of gain and offset error in ADCs.

Someday I'll finish... :-(

orlp• 2 months ago

When going from float to u8 you should add a triangular dither. It makes a world of difference for grayscale gradients, even in 24bit truecolor.

JamesTRexx• 2 months ago

Why not (uint8_t) ( float * (256/255) ) * 255?

mwkaufma• 2 months ago

See: cbloom's rant on quantization for deeper investigation: https://cbloomrants.blogspot.com/2020/09/topics-in-quantizat...

"While in theory there are cases where you might want to use either type of quantization, if you are in games don't do that!

The reason is that the GPU standard for UNORM colors has chosen "centered" quantization, so you should do that too."

Zardoz84• 2 months ago

0-255

Using 1-256 i find it weird

glkindlmann2• 2 months ago

This is like a 1D version of what a scientific computing person might describe as the distinction between node-centered ("standard" or "mid-tread" in this post) vs cell-centered ("alternative", "mid-riser") samples: do consider values to be at the middles of bins (or middles of triangles, or middle of tetrahedra), or the boundaries between intervals (or vertices of triangles, or of tets).

In a scientific computing setting it would be insane to start doing data processing without knowing how to interpret the values. In the context of audio signal processing, if you just get a stream of integers, you'd have to know the representational intent of those integers (mu-law encoding or linear?) if you're going to compute anything about the underlying signal. The meta-data accompanying the values would hopefully provide the answer.

But with 8-bit pixel values, absent any meta-data from a competent file format that can communicate the representational intent, we're adrift and there's no right answer (like the author says). Certainly no one can fault you for picking whichever one seems to be give better results for your application, but you can raise awareness that bits without context have had their meaning undermined.

davidkwast• 2 months ago

Your comment remembered me of ESA normalization value for the Sentinel-2 level-2 satellite image quantization.

Something like this:

Digital Number DN=0 remains the “NO_DATA” value

For a given DN in [1; 1;215-1], the L2A SR reflectance value will be:

L2A_SRi = (L2A_DNi + BOA_ADD_OFFSETi) / QUANTIFICATION_VALUE

https://sentiwiki.copernicus.eu/web/s2-products

mark-r• 2 months ago

In 8-bit (per channel) color there is a standard for when the context is unspecified: sRGB. It was developed to match the characteristics of a "typical" display at the time of its creation. It won't be 100% accurate, but it represents the best you can assume. You're not forced to throw up your hands and give up.

glkindlmann2• 2 months ago

I see that https://en.wikipedia.org/wiki/SRGB#Translation_to_integers implies what I would call node-centered sampling ("standard" or "mid-tread" in the post), but are you aware of sRGB documentation that precisely specifies the quantization semantics?

Most of what I've read about sRGB focuses on the semantics of continuous (floating-point) R,G,B channels.

virtualritz2• 2 months ago

255.0. Everything else makes no sense and is actually dangerous when working with colors. Trust me. :)

And when you go from float to 8bit you should dither to avoid banding.

If in doubt, error diffusion with a random number between -0.5..=0.5 is fine. 0.5 here is dither_amplitude:

round(255 * input_value + dither_amplitude * random(-1, 1))

See e.g. my dithereens crate: https://crates.io/crates/dithereens

jyounker2• 2 months ago

From an algebraic standpoint, the answer is clearly f(x) -> [0, 255].

If you don't have f(n * 0) == n * f(0), then all sorts of weird stuff happens, like:

For f(x) -> [0, 255] then f(0) + f(0) + f(0) = 0 + 0 + 0 = 0 = f(0)

For f(x) -> [0.5/8,7.5/8] then f(0) + f(0) + f(0) = 0.5/8 + 0.5/8 + 0.5/8 = 1.5/8 != f(0)

Choosing the latter means that if you do a calculation on the x side, then you can't expect it to match the calculation on the f(x) side.

nasitsony• 2 months ago

No, I do not think so

jyounker2• 2 months ago

From an algebraic standpoint, the answer is clearly f(x) -> [0, 255].

If you don't have f(n * 0) == n * f(0), then all sorts of weird stuff happens. Compare these two:

  For f(x) -> [0, 255] then f(0) + f(0) + f(0) = 3 * 0 = 0

  For f(x) -> [0.5/8,7.5/8] then f(0) + f(0) + f(0) = 3 * 0.5/8 = 1.5/8

Choosing f(x) -> [0, 255] means that if you do a calculation on the x side and you do the same calculation on the f(x) side, then you'll get the same result when you convert from one to the other.

Choosing f(x) -> [0.5/8,7.5/8] breaks the algebraic correspondence.

zajio1am• 2 months ago

When converting between different depths, the 255-like approach has a property that minimum maps to minimum and maximum maps to maximum.

GuB-42• 2 months ago

As an argument for 255, take the extreme case of a black and white image, single bit, 0 is black and 1 is white.

It seems pretty obvious that you want 0 to map to 0.0 and 1 to 1.0. It is black and white, not light gray (0.25) and dark gray (0.75). It means that you normalize a black and white image by 1, not 2.

For 2 bits, you would normally have 0 = black, 1 = light grey, 2 = dark grey, 3 = black. So it is natural to map it to 0.0, 0.33, 0.66, 1.0, again, you want your black to be black and your white to be white, you also want equal spacing, so normalize by 3.

Follow the logic to 8 bits and you will normalize by 255, because you want black to be 0.0 and white to be 1.0, even though the difference starts to get really small with 8 bits.

Another way to see things is that with the alterative normalization (256 for 8 bits), the output range depends on the number of bits: [0.25, 0.75] for 1 bit, [0.125, 0.875] for 2 bits, etc... You typically don't want that, more bits should mean more nuances, but the contrast should stay the same.

gwerbin• 2 months ago

I appreciate that the author concluded you should use 255, because this approach here (making sure the minimum and maximum of the range actually map to 0 and 1) makes so much sense I never considered you could or ever would want to do it any other way.

davidladdsource• 2 months ago

Great Breakdown!

Should you normalize RGB values by 255 or 256?

Related Stories