2026-08-21 21:14:28
A few weeks ago I wrote about the number system used in labeling spreadsheet columns. Labels run from A through Z, then AA through AZ, etc. This looks a lot like base 26, but it’s not quite the same. It has no analog of zero. If Z were like zero, Y would be followed by AZ. The Excel labeling system is not base 26, but what’s called bijective base 26.
If you found fragments of writing from an ancient culture and inferred that five symbols were used as digits, how could you distinguish base 5 from bijective base 5? Suppose you believe these five symbols were digits
★ 
☗ 
but you don’t know in what order. You just see sequences like and ★★ and believe they’re numbers.
If you noticed that numbers often contain , but never appears at the beginning of a number, you might infer that is a zero. But this would take a fairly large sample. If you found only 20 numbers, for example, you could hardly conclude never appears at the beginning of a number just because it doesn’t come at the beginning of any number you’ve seen.
Now suppose you’ve found writing with more number symbols. Say you’ve found 17 numeric symbols. You might infer that the writing used a base 20 system, because it would be hard to imagine a human culture using base 17. Now imagine you find more fragments and confirmed that indeed there are 20 numeric symbols. Approached as a purely statistical problem, you’d need a very large sample to infer what the digits correspond to and whether they use a base 20 or bijective base 20 system (or some other system).
You’re best hope is to find numbers in some context where you know what number is being represented. If you knew somehow that some symbol corresponds to 20, then you’d know they didn’t use base 20 because base b doesn’t have a single symbol for b.
If you had a huge collection of numbers but no context, which is highly unlikely, you could use Benford’s law to infer the meaning of the number symbols: the most common leading digit is probably 1, the next most common is probably 2, etc. This is interesting to think about, but it seems much more realistic that a number system would be decoded by finding context, such as a list of consecutive numbers or numbers with known meaning.
The post How would you know whether an ancient culture had zero? first appeared on John D. Cook.
2026-08-20 21:48:43
I like AI-generated ASCII diagrams. Because nobody would ask AI to generate ASCII diagrams, and so, it’s congruous. I like incongruity [1].
Aside from the incongruity of using a gazillion-parameter neural network to make 1970’s style ASCII art, ASCII diagrams have some uses. They’re absolutely tiny compared to image files. But more importantly they can be inserted into plain text files, such as source code or markdown. A diagram embedded directly into a source file cannot become separated from the code.
ASCII diagrams are tedious to create, though there are tools to mitigate the tedium. But if an AI can generate the diagram, the tedium goes away.
I was curious how well Claude could create ASCII diagrams, so I tried a few examples. I hope these render well in whatever format you’re reading this post. They look fine for me previewing the post in a browser. I expect they might not turn out so well in an RSS reader.
I asked it to reproduce the graphs from my recently post on the graph imbalance theorem and the first diagram turned out nicely.
+-------+ +-------+
| A |--------------------------------| B |
+-------+ +-------+
| |
| |
------------------|------------------ ---------|---------
| | | | | | | |
| | | | | | | |
+-----+ +-----+ +-----+ +-----+ +-----+ +-----+ +-----+ +-----+
| a0 | | a1 | | a2 | | a3 | | a4 | | b0 | | b1 | | b2 |
+-----+ +-----+ +-----+ +-----+ +-----+ +-----+ +-----+ +-----+
The second network is more complicated and so the corresponding ASCII diagram is hard to read.
+----------------------------------------------------------------+
| |
|+-----------------------------------------------+ |
|| | |
||+-------------------------------+ | |
+-------+ +-------+ +-------+ +-------+ +-------+
| R1 |-------| R2 |-------| R3 |-------| R4 |-------| R5 |
+-------+ +-------+ +-------+ +-------+ +-------+
| | | | | | | | | | |
++ | | | | | | | | | |
| +---------------------------+ | | | | | |
| | ++| | | | | | | |
| | || +--------------+ | | | | |
| | || | +---------------------------------------+ | |
| | |+-|--|-------------+ | | | |
| | | | | | +------+ | | |
| | | | | | | +----------------------+ |
| | | +--|-------------|--|--|-------------+ |
| | | | | | | | +-----|--+ |
| | | | | | | | | | +-----+
| | | | | | | | | | |
+-------+ +-------+ +-------+ +-------+
| G1 | | B1 | | B2 | | B3 |
+-------+ +-------+ +-------+ +-------+
For a third example, here is a fairly complicated diagram that nevertheless lends itself to a readable ASCII diagram. It’s a Feistel network diagram for DES encryption.
+-------------+ +-------------+
| L(i-1) | | R(i-1) |--------
+-------------+ +-------------+ |
| | |
| | |
| +-----------------------+ |
| | E (expand 32->48) | |
| +-----------------------+ |
| | |
| +-----------------------+ |
| | XOR with K(i) | |
| +-----------------------+ |
| | |
| +-----------------------+ |
| | S-boxes S1..S8 | |
| +-----------------------+ |
| | |
| +-----------------------+ |
| | P (permutation) | |
| +-----------------------+ |
| | |
| | |
| +-------+ | |
+------------| XOR |-------------+ |
+-------+ |
| |
+----------------|--------------------------------+
| +------------------+
| |
+-------------+ +-------------+
| L(i) | | R(i) |
+-------------+ +-------------+
[1] See Christian Wolff’s discussion of dogs playing poker in The Accountant (2016).
The post AI-generated ASCII diagrams first appeared on John D. Cook.2026-08-19 08:22:36
A new paper just came out, The Maximum-Area Small Polygon Problem. The paper solves the problem of finding, for each n, the n-gon with diameter 1 and maximum area.
For odd n, the solution is what you might expect: a regular n-gon. I would expect this to be the solution for even n as well, but it’s not.
In 1974 [1] Ron Graham found a solution for n = 6, a hexagon with unit diameter and area larger than a regular hexagon with unit diameter. Polygons with diameter ≤ 1 are called “small”, and he found the “largest” (i.e. maximum area) small hexagon.
The vertices of Graham’s hexagon are given below.
A = (0.0000000000, 0.0000000000) C = (0.4023506913, -0.5000000000) F = (0.9390533483, -0.3437714489) B = (1.0000000000, 0.0000000000) E = (0.9390533483, 0.3437714489) D = (0.4023506913, 0.5000000000)
You can verify that the distance between any pair of vertices is no more than 1 and that the area of Graham’s hexagon is 0.674981.
The area of a regular hexagon of diameter 1 is (3/8)√3 = 0.649519, and the area of Graham’s hexagon is about 3.9% larger.
[1] R. L. Graham. The Largest Small Hexagon. Journal of Combinatorial Theory (A) 18, 165–170 (1975). The paper was submitted February 22, 1974 and published in 1975.
The post Big little hexagon first appeared on John D. Cook.2026-08-18 23:54:57
The imbalance conjecture is now a theorem. James Alexander Schreib and Yousof Yavari posted a proof last week.
What does the conjecture theorem say? Start with a graph G with no edge between two nodes of the same degree. Then for every edge, calculate the absolute value of the difference of the degree of each end. The imbalance theorem says there exists another graph H whose vertices have degrees corresponding to the differences of degrees in G.
For example, let G be the graph below.
The edges from the top red vertex A to each of the blue vertices around it all have degree difference 5 because A has degree 6 and the vertices a0 to a4 have degree 1. The edge between the two red vertices, A and B, has degree difference 2. The remaining vertices have degree difference 3.
So the multiset of degree differences is
{5, 5, 5, 5, 5, 2, 3, 3, 3}
The imbalance theorem says there exists a graph H whose nodes have these degrees. Here is an example of such an H.
Note that in H, the 5 red nodes have degree 5, the single green node has degree 2, and the three blue nodes have degree 3.
2026-08-18 21:42:06
Suppose you have a planet in an elliptical orbit around a star. The math is identical for any light object orbiting a heavy object, such as a moon or satellite orbiting a planet, but we’ll call the heavy object a star and the light object a planet.
The center of the star is not quite the center of the orbit. The planet moves along an ellipse with the star at one focus of that ellipse.
Let a be the semi-major axis of planet’s orbit, the maximum distance from the center of the ellipse to a point on the ellipse. Then the distance of a focus to the center of the ellipse is ae where e is the eccentricity of the ellipse. This defines eccentricity. The center of earth’s orbit is between three and four solar radii away from the center of the sun [1].
The planet is farthest from the star when it is along the major axis of the ellipse on the opposite side as the star. The distance is then a + ae, the distance to the center plus the distance from the center to the star. On the opposite side of its orbit, the planet is closest to the star. There the distance is a − ae. In summary the maximum distance to the star is
a(1 + e)
and the minimum distance is
a(1 − e).
If you had to guess the average distance between the planet and its star, a would be a good guess since it’s the average of the maximum and minimum distance. And that’s a good approximation, provided e is small. The mean distance over time is
a(1 + ½e²).
See derivation. The average distance is greater than a because the planet moves faster when nearest the star and slower when further from the star.
The relative error in approximating the mean distance by a is then ½e². When e is small, ½e² is very small. For the earth’s orbit, e = 0.01671, and so the approximation is off by around 0.014%.
The eccentricity of Pluto’s orbit is 0.2488, and so in that case the approximation is off by about 3.1%. The eccentricity of a Molniya orbit, used by some Russian satellites, is 0.74 [2]. For such satellites the error in approximating the mean distance to earth as the semimajor axis is around 27%.
[1] For earth’s orbit, e = 0.01671, a = 1.496×1011 m, and the sun’s radius is r = 6.957×108 m. And so ea / r = 3.59.
[2] An object in such a highly elliptical orbit will spend a long time at the far side of its orbit, i.e. over Russia. Sort of a poor man’s geostationary orbit.
The post Mean distance to the sun first appeared on John D. Cook.2026-08-17 09:16:39
The first post in the recent series of posts on Hadamard matrices describes a way of constructing new Hadamard matrices from two other Hadamard matrices by taking their Kronecker product.
Starting with a Hadamard matrix H0 and a Hadamard matrix G, you can construct a sequence of Hadamard matrices by
Hn+1 = G ⊗ Hn
for positive integers n. This is known as the generalized Sylvester method.
Let pn be the proportion of 1s in Hn and let q be the proportion of 1s in G. Then you can show that the recurrence holds
pn+1 = q pn + (1 − q)(1 − pn).
You can solve the recurrence to show that
limn → ∞pn = ½
and so as the iterations proceed, the ratio of number of 1s to the number of −1s approaches 1.
This doesn’t say anything Hadamard matrices in general, but it does apply to all Hadamard matrices created by repeatedly applying the generalized Sylvester method.
If you set G and H equal to the matrix
then p0 = q = ¾. Then for n = 1, 2, 3, …, 8 the values of pn are
0.625
0.5625
0.53125
0.515625
0.5078125
0.50390625
0.501953125
0.5009765625.
The post Proportion of 1s in a Hadamard matrix first appeared on John D. Cook.
