Should you normalize RGB values by 255 or 256?

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.

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

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)

I see what you’re saying - index 0 holds values from 0-1, index 2 from 1-2 etc, but then you have index 255 holding values between 255 and 256. So you’re sort of arguing that the 0-255 8-bit quantization is actually representing ‘real’ values of 0-256?…

Edit: somehow missed alterom’s reply - they explain it much better than my question above does.

> 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)