Geometry · Aska Digital’s own research
From a doodle to the densest packing of balls
A square, its two diagonals, and one dot in each triangle. Followed with AI models, the dot leads to a number that turns up in every dimension, and to an exact way to fold a loose crystal of balls into the tightest stacking there is.
Aska Digital’s own work. Founding member Ahraz Arifuddin had the idea, chose every question and decided every claim; AI models from Anthropic and OpenAI did most of the proofs, code and checking, and the project records which model did what.
Live sitePreprint v2 · DOI 10.5281/zenodo.23340326RepositoryOn the Ledger
The doodle, one step at a time
How to read: pick a step. The dots are the triangles’ incentres; each circle is the largest that fits in its triangle. Steps 6 and 7 carry the rule into a cube: drag the picture, or focus it and use the arrow keys, to turn it.
- 22.3% → 74.05%space filled by the balls along the exact fold, from loose scaffold to the densest possible stacking, π/√18fold-continuous-fcc/certificate.json
- 16,240exact certificate numbers, every one ≥ 0, ruling out a collision anywhere along the fold (3,248 pair tests)PROOF.md
- 1/(n2−n−1)how much the framework thins sideways when stretched, in every dimension n ≥ 3 (written proof)rigidity-nd/PROOF.md
- ~11 minfor an AI search loop to climb from 46.9% to the 74.05% ceiling, in 5 logged runsautoresearch-fold/results.tsv
00//The doodle
First principlesOne dot, exactly where it has to be.
For years Ahraz Arifuddin, who founded Aska Digital, drew the same figure on the backs of tests and on napkins: a square, its two diagonals, and a dot in each of the four triangles.
Which dot? In any triangle there is exactly one point that is the same distance from all three sides. Draw a circle around it with that distance as its radius and you get the biggest circle that fits inside the triangle.
In a square of side 2 that distance is √2 − 1 ≈ 0.4142. Put a ball of that radius on each of the four dots and something tidy happens: neighbouring balls just touch, across the diagonals and across the square’s edges into the next square. Tile the plane with the doodle and the touching balls form a pattern of octagons and squares, a tiling known for centuries as 4.8.8.
Carry the same rule into a cube, cut from its centre into six pyramids, and the six dots form an octahedron. Stack the cubes and the dots land exactly where the boron atoms sit in a real crystal, calcium hexaboride (CaB6), whose structure Linus Pauling and Sidney Weinbaum worked out in 1934.
Most of that is old, and the project says so: the tiling, the boron skeleton and the size limits chemists use to predict how atoms pack were all known, and the site credits the people who got there first.
The new questions start when you treat the balls and the rods between touching balls as a hinged toy. A rod (the papers call it a bar) joins the centres of two touching balls and never changes length. How floppy is the toy? How far can it fold? Could it fold all the way into the tightest stacking of balls without stretching a rod, and without two balls ever passing through each other?
Those questions were worked on in October 2026 with AI models doing most of the proofs, code and checking, under Ahraz Arifuddin’s direction. The sections below take them in order: where the number comes from, what it does in every dimension, what two sizes of ball can do, the fold, and how an AI search found the fold’s end point.
01//One number
The number is not chosen. Mirrors force it.
Treat every line of the tiled doodle as a mirror. A mirror puts a copy of a point on the far side of the wall, exactly as far away as the original, so the gap between a point and its copy is twice its distance to the wall.
The mirrors cut each square into eight identical small triangles. Pick one of them, the shaded chamber, and put a dot anywhere inside it. Reflect the dot in all the mirrors, again and again, and its copies fill the plane. Join each copy to the copies one reflection away. Every such link crosses exactly one wall, so its length is twice the dot’s distance to that wall.
The chamber has three walls: the square’s edge (the floor), the diagonal, and the square’s centre line. In the doodle the dot sits on the centre line, so that wall gives no link at all: the dot is its own reflection there. Two kinds of link remain, one across the floor and one across the diagonal.
For a pattern of balls that all touch, every link must have the same length, so the dot must be equally far from the floor and the diagonal. Points equally far from two walls lie on the line that splits the angle between them. That angle is 45°, so the dot sits on the 22.5° line, and on the centre line its height is tan 22.5° = √2 − 1. Nothing else works.
Move the dot off the centre line and three kinds of link appear. They are all equal only at one other point, the chamber’s own incentre, which also sits on the 22.5° line. The rule “place a point on some walls of a mirror chamber, equally far from the rest, and reflect” is the Wythoff construction (Willem Wythoff, 1918), later developed by H. S. M. Coxeter. The project credits it.
In a cube the same argument puts the dot at 2 − √2 from the centre with ball radius √2 − 1, and the same numbers come out in every number of dimensions.
The mirror room
How to read: drag the white dot inside the shaded chamber, or focus the picture and use the arrow keys. Link colours show which wall each link crosses; the bars below compare their lengths.
All links equal: the dot is at height √2 − 1 = 0.4142 on the centre line.
Show the numbers
| Dot | x | y | Floor link | Diagonal link | Centre-line link |
|---|---|---|---|---|---|
| The doodle’s dot (on the centre line) | 1 | 0.414214 | 0.828427 | 0.828427 | none |
| The chamber’s incentre | 0.707107 | 0.292893 | 0.585786 | 0.585786 | 0.585786 |
| Your dot now | 1 | 0.414214 | 0.828427 | 0.828427 | none |
02//Every dimension
One formula for how the scaffold squeezes, in any number of dimensions.
The cube rule works in a cube of any dimension n: 2n pyramids, 2n dots, the same ball radius √2 − 1, and each ball touching 2n − 1 others. The project asked how stiff that scaffold is.
Imagine every rod is a perfect spring and you stretch the whole scaffold a little along one cube axis. Most materials get thinner sideways when stretched. The fraction is called the Poisson ratio: a ratio of 1/5 means that stretching by 1% makes it 0.2% thinner in the other directions.
The same work counts the scaffold’s floppy motions: n(n − 1)/2 independent ways to wiggle per cube that no rod resists, and n(n − 3)/2 ways to load the rods that cancel out.
How sure: a written proof covers every n ≥ 3, and a finite computation checks n = 2 to 8. Only AI models have reviewed the proof. When an AI-led literature search found that the same scaffold appears in work by Borcea and Streinu (2021), the project publicly withdrew its broader novelty claim and now presents the counts as consequences of existing theory. The spring formulas above remain its strongest candidate for something new, and that is still unsettled.
Every dimension at once
How to read: choose a dimension. Left: the 2n balls around one cube centre, turned in n-dimensional space and projected onto the page (balls drawn small so the rods show; drag it, or press Turn). Each ball is joined to every other ball except its opposite partner. Right: the Poisson ratio for each dimension; the one you chose is marked.
The balls around one cube centre in the chosen dimension, projected onto the page: each ball is joined to every other ball except its opposite partner.
Bar chart of the Poisson ratio 1/(n squared minus n minus 1) for n from 3 to 8: 0.2, 0.091, 0.053, 0.034, 0.024, 0.018.
- Balls per cell
- 6
- Each ball touches
- 5
- Rods per cell
- 15
- Floppy motions
- 3
- Poisson ratio
- 1/5 = 0.2000
- Stiffness ratio
- 4
Show the numbers
| n | Balls per cell 2n | Touches 2n−1 | Rods 2n²−n | Floppy n(n−1)/2 | Self-stresses n(n−3)/2 | Poisson ratio |
|---|
03//Two sizes
Two sizes of ball beat one, in a sliver of size ratios.
Equal balls can never fill more than π/√18 ≈ 74.05% of space. Put big balls where calcium sits in calcium hexaboride and small balls on the doodle’s six dots, where the boron sits, and the best this pattern can do is (π/6)(30√2 − 41) ≈ 74.69%, at a size ratio of exactly √2 − 1.
Take a cube of side 1. A big ball sits at every corner, and six small balls sit inside, one on each line from the cube’s centre to a face, as an octahedron. The size ratio γ is the small radius divided by the big one. Two things limit how big the balls can be: neighbouring big balls along a cube edge must not overlap, and the small balls must not overlap each other, inside one octahedron or across to the next cube.
When the small balls are relatively small, the big ones touch first: the big radius is fixed at ½, and the fraction filled is (π/6)(1 + 6γ3). When they are relatively large, the small ones touch first, at radius (√2 − 1)/2. The best is where both touch at once, and that happens at γ = √2 − 1: the doodle’s number again.
It beats equal balls only between γ = 0.41023 and 0.41591, and at its best by 0.64 percentage points. That is a geometric fact about this one pattern. Other patterns do better: the salt (NaCl) pattern reaches 79.31% at the same ratio.
Two sizes beat one
How to read: move the size ratio. Left: one cube of the pattern, big balls at the corners (drawn see-through) and the six small balls inside; drag it to turn it. Right: the most space this pattern can fill at each ratio, against the 74.05% limit for equal balls; the gold sliver is where it wins.
Line chart of the most space the pattern can fill against the size ratio from 0.30 to 0.60. It rises to a peak of 74.69% at the square root of 2 minus 1, about 0.4142, then falls; it is above the 74.05% limit for equal balls only between 0.41023 and 0.41591.
- Size ratio γ
- 0.4142
- Space filled
- 74.6865%
- Against equal balls
- beats 74.05% by 0.64 points
- What touches
- both: big balls along the edges, small balls with each other
Show the numbers
| γ | Big radius | Small radius | Space filled | Against 74.048% |
|---|
04//The fold
Fold the loose scaffold into the densest packing, without stretching a single rod.
In three dimensions the balls fill only 22.33% of space. The densest any stacking of equal balls can be is π/√18 ≈ 74.05%: Johannes Kepler guessed it in 1611, and Thomas Hales proved it in 2005.
The question: can the scaffold fold, like a hinged toy, from 22.33% all the way to 74.05%? Every rod must keep its length the whole time, and no two balls may ever pass through each other, at any of the infinitely many in-between moments, anywhere in the endlessly repeating crystal.
The answer is yes, and the motion has a short formula. Keep the six balls of one cell fixed and let the box they repeat in lean and shrink. One number, s, runs from 0 to 1; the box’s edges are L(s) = (2 + a) I + t C with t = 2s/(1+s²) and a = √2(1−s²)/(1+s²). Because a² + 2t² = 2 for every s, the rods that join neighbouring cells keep length √2 exactly.
Why no collision, for infinitely many moments? Snapshots cannot prove that. Instead, each pair of balls that could ever meet gives a distance that is a polynomial in s. Written in a particular form (the Bernstein form), a polynomial whose coefficients are all zero or positive can never dip below zero anywhere between 0 and 1. The project checked 3,248 such pairs: 16,240 coefficients, computed exactly with √2 kept as a symbol, all ≥ 0. Copies of the cell farther away are ruled out by one short inequality.
Who did it: as the repository records, OpenAI’s GPT-6 (through Codex) derived the curve, its first proof and the exact certificate; GPT-6.1 Sol checked it independently with a separate replay; Claude wrote the symmetry-reduced hand proof for version 2 of the preprint. Ahraz Arifuddin directed the work and decided what to claim. No person other than the author has checked the proofs.
Fold without stretching
How to read: move the slider from the open scaffold (s = 0) to the densest packing (s = 1). Drag the picture to turn it. Green rods join neighbouring cells, grey rods join balls within a cell; all of them stay the same length. The curve on the right is the space filled.
Line chart of the space filled against s: it rises steadily from 22.3% at s = 0 to 74.05% at s = 1, which is the Kepler bound.
- Space filled
- 22.3266%
- Volume per cell
- 39.799
- Every rod
- √2 · error 2×10−16
- Nearest unjoined pair
- 2.0000 (touch: 1.4142)
Show the numbers
| s | Volume per cell | Space filled | Touching neighbours per ball |
|---|---|---|---|
| 0 | 39.79899 | 22.3266% | 5 |
| 0.25 | 35.18725 | 25.2528% | 5 |
| 0.5 | 25.96034 | 34.2282% | 5 |
| 0.75 | 17.73225 | 50.1108% | 5 |
| 1 | 12 | 74.0480% = π/√18 | 12 |
Certificate: 171 nearby cell offsets × 19 differences between balls, less the ball itself, gives 3,248 distance polynomials of degree 4, with 5 Bernstein coefficients each; 114 coefficients are exactly zero and the rest positive. The volume shrinks 3.3 times.
05//The search
Propose freely, verify strictly.
Before anyone had the exact fold, an AI agent was asked to find a folding path by trial. It could change one file, its own search script. A separate checker, which it could not touch, decided what counted.
The checker’s file is locked by its fingerprint: a SHA-256 hash, a 64-character code that changes completely if one character of the file changes. The run script refuses to start if the fingerprint is off. So the agent could try any idea, but it could not make a bad path pass by editing the judge.
Early on 12 October 2026 the loop ran five times in about eleven minutes. The first run reproduced a known fold that stops at 46.86%. The second, a free continuation after that fold, reached 71.09%. The third skipped the fold and got stuck. The fourth tried a bigger block of cells and moved too far in one step; the checker rejected it. The fifth folded only part of the way, then let the box shear freely, and reached 74.05% with every ball touching 12 others: the densest packing, in its hexagonal form.
The limit matters too. The checker tests a long list of snapshots, each at most 0.01 apart, plus the midpoints between them. That is strong evidence, not a proof that nothing goes wrong in between, and the project labels it VERIFIED rather than PROVEN. The exact fold in section 04, a different motion that ends at the cubic form of the densest packing, was found the next day.
The paper’s disclosure says Claude models ran the numerical searches for fold paths. The loop follows Andrej Karpathy’s open “autoresearch” pattern.
The search ladder
How to read: each mark is one logged run, by clock time, at the highest density its path reached. Step through the runs with the buttons or the arrow keys; the card shows what the agent tried, in its own words.
Five runs on 12 October 2026: 07:56:10, 46.86%, kept as the baseline; 07:56:30, 71.09%, kept; 07:59:03, 57.29%, discarded; 08:03:36, rejected by the checker; 08:06:56, 74.05%, kept as best, at the Kepler ceiling.
- Kept · valid and better
- Discarded · valid, not better
- Rejected · failed the checker
verify.py · locked · sha256 05d2385d…35cd
- Every rod keeps its length (to one part in a billion)
- No two balls overlap
- No ball moves more than 0.01 between snapshots, and the box changes by no more than 0.01
- The midpoint between snapshots is checked for overlaps too
Show the log
| Time | Run hash | Valid | Max density | Outcome | What the agent tried |
|---|
06//Who did what
As recorded in the repositoryA person with a question, and models that did most of the work.
The repository says plainly that most of the computation, proofs, checking and writing was done by AI models under direction, and it names which model did what.
Human
Ahraz Arifuddin · Aska Digital
- Drew the doodle for years and decided to follow it into three and more dimensions.
- Chose the questions, directed every model, and decided what is claimed and what is published.
- Sole author of both preprints; set up the research checks (locked checkers, certificates, a public hash posted to a dated gist, evidence labels on every claim).
- Withdrew the broad novelty claim in public when prior work surfaced, and issued a corrected version 2.
Two early results, the exact end states of the sampled fold and a one-cell version of it, came from pull requests #1 and #2 by another contributor.
Digital
Models named by the project
- Claude Opus 5.5 · Anthropic
- Proofs, derivations, literature and novelty review, the fold-path searches, the site. Co-author line on 33 of the 35 commits.
- Claude Sonnet · Claude Haiku
- Checkers, tests, certificates, automated checks and documentation; running scripts and housekeeping.
- GPT-6 via Codex · OpenAI
- The explicit fold curve, its first proof and exact certificate; the proof that the two-size CaB6 packing reaches its maximum.
- GPT-6.1 Sol · GPT-6 Luna
- An independent exact replay of the certificate and review; repository publishing and link checks.
Model names as recorded in CONTRIBUTORS.md and the preprint’s disclosure; Foundation0 has not verified them independently.
07//How sure
Written down, replayable, and not yet checked by an outside mathematician.
The project labels every claim. PROVEN means a written proof exists, so far reviewed only by AI models. VERIFIED means a finite computation that anyone can replay, which is not a proof. KNOWN means published by others and cited.
| Result | Exact value | Label | What backs it |
|---|---|---|---|
| Exact fold from the open scaffold to the densest packing (cubic form) | 22.3266% → 74.0480%, every rod exact, density rising throughout | Proven | Written proof, exact certificate, two separate exact implementations, a hand proof; fold-continuous-fcc |
| Spring stiffness in every dimension n ≥ 3 | Poisson ratio 1/(n²−n−1); stiffness ratio (n−1)²; no shear stiffness | Proven | Written proof; finite check for n = 2 to 8; rigidity-nd |
| Floppy-motion counts | n(n−1)/2 per cell | Lemma from known theory | Same scaffold appears in Borcea and Streinu 2021, section 7; the novelty claim was withdrawn (PR #27) |
| Best packing of the CaB6 pattern with two ball sizes | (π/6)(30√2 − 41) = 74.6865% at size ratio √2−1 | Proven | Upper bound and attainment proofs; the attainment proof was prepared with GPT-6; not found elsewhere, not confirmed new; cab6-packing |
| Sampled fold of 48 balls to close packing | 793 snapshots, 22.33% → 74.05% (hexagonal form) | Verified | Sampled steps of at most 0.01, two in-house checkers; not a proof of continuity |
08//Sources
Checked 13–14 Oct 2026Read it, replay it, argue with it.
Every result ships with code that re-runs it. Running make verify on a copy of the repository regenerated every certificate byte for byte when Foundation0 replayed it.
The project
Preprints
Proofs, certificates, logs
Prior work cited
The figures on this page adapt fold-continuous.js and mirrors.html from the project, which are MIT-licensed (text and figures CC BY 4.0), © Ahraz Arifuddin, Aska Digital, and contributors. Status lines here follow the repository on 13 October 2026, not older notes on the live site.