Toy code can be found here.

This is an extract of a blog post that I haven’t been able to motive myself to complete. The topic of which is some high level brushstrokes of rank-1 lattices and how they are related to irrational generating vectors (n-dimensional Weyl sequences or additive recurrence) . Examples of irrational generating vectors are various flavors of Fibonacci/Golden ratio sets and Martin Robert’s $R_n$-sequences presented in the blog post: The Unreasonable Effectiveness of Quasirandom Sequences. Here I’m going to ignore n-dimensional and focus on computing a 1D sequence. This is all gonna be quick and dirty…let’s go:

The $i^\text{th}$ element of the sequence generated by an irrational constant $\alpha$ can be expressed in closed form:

$\begin{equation*} \begin{array}{l} s_i & = i~\alpha \bmod 1 \\ & = \left\{i~\alpha\right\} \end{array} \end{equation*}$

where $\left\{ \cdot \right\}$ is just shorthand for the modulus (or the fractional part) and we can progressively walk elements with the recurrence:

$s_{i+1} = \left\{s_i + \alpha\right\}$

A couple quick comments on $\alpha$. First we can require it to be on $\left(0,1\right)$ because:

$\left\{k + \alpha\right\} = \alpha$

for all integers $k$. Also 0 and 1 are obviously useless. Like the previous we also have:

$\left\{1 - \alpha\right\} = -\alpha$

So negating $\alpha$ produces the same sequence “mirrored” about $1/2$ and we can choose to limit ourselves to either $\left(0,1/2\right)$ or $\left(1/2,1\right)$. (Really interesting values are never near 0 or 1..but that doesn’t matter here)

Now let’s spitball how addition/products of computer primitive types work:

1. float point gives us the $p$ most significant bits of the result (and there’s that whole rounding thing)
2. integers (are modulo width) and gives us the $p$ least significant bits of the result.

So integers automatically perform the operations we’re really interested in. Onward. As an example irrational let’s use the perennial favorite, the golden ratio (well, really its inverse..or the fractional part of it…they’re the same):

$\alpha = \frac{1}{\varphi} = \frac{\sqrt{5}-1}{2} \approx 0.618034$

To work with integers we need to convert $\alpha$ into a $0.b$ fixed point value $A$:

$A = \text{RO}\left( 2^b~\alpha \right)$

where $\text{RO}\left(\cdot\right)$ denotes round-to-odd and $b$ is the bit-width of the integer. The rounding to odd is so $A$ is coprime to $2^b$ which ensures our sequence has $2^b$ elements (the best we can do with $b$ bits). A code sketch:

Now let’s consider what happens if we follow the same recipe for floating point:

$\text{alpha} = \text{RO}\left(\alpha \right)$

where $\text{RO}\left(\cdot\right)$ will give us a significand with the smallest bit set. If we compute fmodf(i*alpha,1) (or an equivalent) what’s going to happen is we’re going to lose information as a function of $\log_2\left(i\right)$. Think of how long multiplication works (the number of digits needed to represent the result). If we compute v += alpha things will be a little better but it’s still a function of $\log_2$ the number of elements we’ve walked. The toy code computes these values and there’s a table at the end with the output. These numbers probably don’t look very impressive, but remember we care about a collection of values and some measurement between this point set (and most likely in a dimension greater than one). Actually, just forget about that and let’s fix it instead.

Since truncated fixed-point does exactly what we want the question becomes can we mimic that in floating-point? The good news is yeah and the bad news is it’s in all ways inferior to using integers if we try to use all available bits of precision. However with a little TLC we can make it work. If we have a $b$ bit unsigned integer then the maximum value we can represent is $\left(2^b-1\right)$ so the largest value we can get by adding two of them is:

$\left(2^b-1\right) + \left(2^b-1\right) = 2^{b+1} - 2$

bringing us the to well known fact that we need one extra bit to be exact. So for additive recurrence in floating point if we can ensure that the smallest bit is always clear then we can prevent rounding and we’ll be exact. Using single precision we have 24-bits and if we limit ourselves to 23 then the only time our sum will have the full 24 will be when it has become greater than one which we will then subtract away taking us back to 23. The previous statement is only true if our additive constant is on $\left(1/2,1\right)$ which places the “decimal” point at the required fixed position (left as exercise to reader..I’ve always wanted to type that).

Let’s step through this in painful detail. Say we want to use the plastic number $(\rho)$:

\begin{align*} \rho & =\sqrt{\frac{9+\sqrt{69}}{18}}+\sqrt{\frac{9-\sqrt{69}}{18}} \approx 1.324717957244746 \\ \left\{\rho\right\} & \approx 0.324717957244746 \\ 1-\left\{\rho\right\} & \approx 0.675282042755254 \\ 2^{23}\left(1-\left\{\rho\right\}\right) & \approx 5664676.346113066 \\ \text{RO}\left(2^{23}\left(1-\left\{\rho\right\}\right)\right) & = 5664675 \end{align*}

The value has an integer part, kill it, on wrong range so we mirror it about $1/2$, scale it by the number of bits and round to odd. Here we have no particular reason to choose one odd over the other..so pick one.

Here’s a toy sketch. To get the initial value it simply uses the integer method and then walks the sequence using floating point.

## Summary

A quick recap.

The pure integer method:

• no special rules to remember other than “make it odd”. This isn’t a hard rule: for every trailing zero we divide the period by two.
• it’s period is $2^b$
• downside is it requires: int-to-float and scale

The float-point method:

• if constant doesn’t fit the listed requirements then things go abit wrong
• it’s period is $2^{p-1}$, where $p$ is the number of normal significand bits of the format.
• downside is it requires: manually handling the “wrap” (subtract one when greater than)

## Toy code output

This is section is simply the output of the toy code formatted into tables.

First computing si = fmodf(i*alpha,1) with odd alpha compared to the exact result. Each line of the table is when we’ve lost a bit of precision.

$i$ $\log_2\left(i\right)$ $s_i$ exact $s_i$ float exact float
3 1.584962 0.854102 0.854102 daa66b00 daa66c00
6 2.584963 0.708204 0.708204 b54cd600 b54cd800
12 3.584963 0.416407 0.416408 6a99ac00 6a99b000
24 4.584962 0.832815 0.832815 d5335800 d5336000
48 5.584962 0.665629 0.665630 aa66b000 aa66c000
96 6.584962 0.331259 0.331261 54cd6000 54cd8000
192 7.584962 0.662518 0.662521 a99ac000 a99b0000
384 8.584963 0.325035 0.325043 53358000 53360000
768 9.584963 0.650070 0.650085 a66b0000 a66c0000
1536 10.584963 0.300140 0.300171 4cd60000 4cd80000
3072 11.584963 0.600281 0.600342 99ac0000 99b00000
6144 12.584963 0.200562 0.200684 33580000 33600000
12288 13.584963 0.401123 0.401367 66b00000 66c00000
24576 14.584963 0.802246 0.802734 cd600000 cd800000
49152 15.584963 0.604492 0.605469 9ac00000 9b000000
98304 16.584963 0.208984 0.210938 35800000 36000000
196608 17.584963 0.417969 0.421875 6b000000 6c000000
393216 18.584963 0.835938 0.843750 d6000000 d8000000
786432 19.584963 0.671875 0.687500 ac000000 b0000000
1572864 20.584963 0.343750 0.375000 58000000 60000000
3145728 21.584963 0.687500 0.750000 b0000000 c0000000
6291456 22.584963 0.375000 0.500000 60000000 80000000
12582912 23.584963 0.750000 0.000000 c0000000 00000000
25165824 24.584963 0.500000 0.000000 80000000 00000000

and the same for computing si += alpha; if (si >= 1.f) si -= 1.f; with odd alpha compared to the exact result.

$i$ $\log_2\left(i\right)$ $s_i$ exact $s_i$ float exact float
4 2.000000 0.090170 0.090170 17155d00 17155c00
9 3.169925 0.180339 0.180339 2e2aba00 2e2ab800
43 5.426265 0.193494 0.193493 3188cc00 3188c800
111 6.794416 0.219802 0.219801 3844f000 3844e800
255 7.994353 0.216690 0.216689 37790000 3778f000
543 9.084808 0.210466 0.210464 35e12000 35e10000
1119 10.127995 0.198019 0.198015 32b16000 32b12000
2263 11.144021 0.228853 0.228845 3a961800 3a959800
4559 12.154502 0.234792 0.234776 3c1b5000 3c1a5000
9159 13.160975 0.190941 0.190911 30e18800 30df8800
18343 14.162942 0.214697 0.214636 36f66800 36f26800
36719 15.164239 0.206481 0.206359 34dbf000 34d3f000
73471 16.164886 0.190048 0.189804 30a70000 30970000
146967 17.165133 0.212911 0.212423 36815800 36615800
293967 18.165295 0.202909 0.201932 33f1d000 33b1d000
587967 19.165375 0.182903 0.180950 2ed2c000 2e52c000
1175959 20.165407 0.198621 0.194715 32d8d800 31d8d800
2351943 21.165422 0.230057 0.222245 3ae50800 38e50800
4703919 22.165432 0.237201 0.221576 3cb93000 38b93000
9407871 23.165438 0.251488 0.220238 40618000 38618000
18815775 24.165440 0.280062 0.217562 47b22000 37b22000
37631583 25.165442 0.337210 0.212210 56536000 36536000
75263199 26.165442 0.451506 0.201506 7395e000 3395e000
150526423 27.165442 0.735826 0.235826 bc5f1800 3c5f1800