A simple program, a forty-year question
Start with a single black square. Apply one tiny rule, row after row. What grows is part orderly, part wild, and nobody can yet prove whether the column running straight down its middle ever repeats.
1 · The rule
Picture a row of squares, each black or white. To make the next row, every square looks at three squares above it: the one directly above, and that one's left and right neighbours. Those three colours can come in eight combinations, and the rule simply says, for each combination, what colour the new square is.
Write the eight answers as a binary number and you get 00011110, which is 30. That is the whole of Rule 30. Below, the rule is applied one square at a time: the three boxed squares decide the new one, and the matching entry of the rule lights up.
2 · Why it matters
In the early 1980s Stephen Wolfram set out to look systematically at the simplest possible programs of this kind. There are only 256 of them. Most do something dull: they die out, repeat, or make neat nested patterns. Rule 30 does not. Its left side settles into regular stripes, but its right side never settles at all, and its centre column passes the standard tests of randomness.
That observation, that a very simple rule can produce behaviour which looks random, became a cornerstone of Wolfram's A New Kind of Science (2002). For many years Rule 30 generated Mathematica's random integers. Its pattern clads the walls of Cambridge North railway station. And it was the subject of a challenge.
It's been nearly 40 years since I first saw rule 30—but it still amazes me. Stephen Wolfram, announcing the Rule 30 Prizes, 1 October 2019
On that day Wolfram offered $30,000 in prizes, $10,000 for each of three questions about the centre column. Seven years on, all three are open. Read the column as a coin, black for heads and white for tails:
The coin is not random: every toss is fixed by the rule. So each answer is simply yes or no, and only a proof can settle it. Wolfram's team has computed a billion cells of the column. They look perfectly random, and that proves nothing.
3 · Order at the edges
Look again at the pyramid. Its left side is striped and its right side is wild, but neither is quite what it seems. Both edges carry order with an exact shape, and both shapes were found on 9 October 2026 by looking hard at the picture.
Read the pyramid in lines parallel to its left edge, its diagonals. The first few diagonals settle into short repeating patterns, and once a run of them repeats, it repeats for ever, because those diagonals depend only on one another. So at every row there is an exact count of diagonals that have reached their eternal stripes. Below, they are shaded, and the magenta line is the edge of the shaded band: the exact place where order ends.
It does not run down the side of the pyramid. It starts near the centre at the very top and leans out at about a quarter of the speed of the pyramid's own edge, 0.244 squares per row on average. That is close to the speed, about 0.246, at which a change made on the right spreads leftwards through a random row; whether the two are exactly equal is not known. The chart follows the edge to row 524,288: about that straight line it wanders like a random walk.
Gareth noticed this by eye: along the wild right edge, the white triangles that touch it arrive at perfectly even spacing, and only their sizes differ. It is exactly true. At every second row a white triangle starts one square inside the edge, and its width depends only on how many times 2 divides the row number, like the marks on a ruler. Rows 2, 6, 10, 14 and so on (twice an odd number) get the smallest triangle, rows 4, 12, 20 the next size up, rows 8, 24, 40 the next, and so on.
The reason is that every diagonal parallel to the right edge repeats with a period that is a power of 2. At a row that is a multiple of a diagonal's period, the diagonal is white, as it was at the start; the triangle ends at the first diagonal whose period does not divide the row. This is proved for all time.
Neither kind of order reaches the prize column. At row t the centre column sits on the t-th diagonal from either edge, ever deeper, where the periods have grown too long to see: the 54th diagonal from the right edge repeats only after 4,194,304 rows. Order at the edges says nothing yet about the middle, and that is why the problems are hard.
4 · The whole picture
Now to the prize column itself. Problem 1 asks whether it can ever repeat. So assume it does, in the simplest way it could: white, black, white, black, for ever. Hold the centre to that beat, like a wall, and run Rule 30 on the squares to its right from a random start. The column right next to the wall does something remarkable. It falls into a fixed loop of 56 steps, a wheel, and stays on it until something from further right knocks it off, after which it settles back at a new position on the wheel.
The plate below puts the pieces together, after Leonardo's Vitruvian Man. At the top is the triangle Rule 30 draws from one cell, with the prize column down its middle and the off-centre region where the chaos lives. It is the top half of a square standing on its corner. Below is the wheel, which is really two clocks at once, of 8 and of 7 positions. Beside it runs the wall experiment that drives it. The dotted lines follow the wheel's current position to both clocks and to the cell it stands for.
5 · Assume the opposite
Rule 30 has a useful property: if you know the centre column and the column beside it, every square to the left is forced, one after another. So a repeating centre forces the whole left side. The prize asks about one black square; we ask the same of every finite start, a pattern with only finitely many black squares, which includes it. For any of those the forced left side would have to turn white for good beyond some depth. A counterexample to Problem 1 is exactly that: a white edge that never breaks.
The same property works the other way round. Choose anything for the centre to say, the primes, the Fibonacci numbers, the digits of π in binary, your own name, and build the start one square at a time, leftwards: the square t places left of the centre decides the centre at step t outright, without disturbing the steps before. So every sequence can be made. The catch is that the start never ends; each step needs one more square. Cut it to a finite start and the sequence breaks, just as the first square it lacks reaches the centre. Even the best of every start of 18 squares follows the primes for only 18 steps, and π for 19. A finite start buys about one step per square, whatever it is asked to say, unless the sequence is one Rule 30 makes by itself. Problem 1 asks whether any repeating pattern is one of those.
A start word can also be animated. Flip one of its bits and the picture below changes, but only inside a cone that opens from the flipped square, and the change arrives one step at a time, so it sweeps down the page. Its right edge is a straight line, exactly one square per step, for the same reason the start could be built square by square. Its left edge moves only when the colours let it through: about a quarter of a square per step across the random-looking inside, a full square once it reaches the white outside. A flip left of the centre reaches the centre column after exactly as many steps as it is squares away. Flip the bits in Gray-code order and the frames walk through every possible start, each one flip from the last.
Try it. Below, the centre is held to its beat (teal) and the column beside it (amber) is given a repeating pattern; pick one from the menu or ask for a random one. Rule 30 then fixes the columns to the left one at a time, so the picture grows leftwards from the centre. Whatever the pattern, after a short stretch (the dashed line) the left side freezes into a crystal: a wall built from one brick, repeating across as well as down. The brick is never blank, so the left side can never turn white. That is Jen's theorem of 1990. Across every pattern of even length up to 20, 1,398,100 of them, only 20 different bricks ever appear.
So a counterexample's second column can never repeat, and its left side has no crystal to settle into. The picture below shows how that left side is built. It is turned on its side: time runs left to right, the top row is the centre's beat, the next is the column beside it, driven by a random right side, and each row below is the next column to the left. Built this way, each row records where the row above it changes, plus scattered dots, the edge events, which the pattern makes for itself.
That makes the left side a pile of Sierpinski triangles. Start with one black square, and fill each row below it by a far simpler rule than Rule 30: a square is black when exactly one of the two squares above it, straight up and up-right, is black (this comparison is the Gray code, which chapter 7 comes back to). The single dot grows into a triangle made of three half-size copies of itself, each made of three smaller copies, and so on for ever: the Sierpinski triangle, which is also the pattern of the odd numbers in Pascal's triangle. Give the same rule several dots and each grows its own triangle, but where two triangles overlap they cancel, so a square ends up black exactly when an odd number of triangles cover it.
This is not special to the left side, nor to running sideways. Read the ordinary way, down the page, Rule 30 is the same kind of rule: do what Rule 60 does, black when the square up-left and the square above differ, then flip the square under every kick, a white square with a black one to its right. Rule 60 alone grows one black square into a Sierpinski triangle, and each kick starts another, so every square of the whole pyramid, the striped side and the wild side alike, is an odd overlap of Sierpinski triangles. About a quarter of all squares are kicks on both sides; what differs is only how they are arranged, in regular stripes on the left and scattered on the right. All the triangles lean the same way, down and to the right, because Rule 30 uses its left neighbour exactly and its right neighbour only through the kick, which is also why only the left side can be built sideways.
The demonstration below shows both readings of that one equation. Down the page is the familiar pyramid: point at any square, on either side, to see the triangle of kicks that decides it. Sideways, beside the wall is the left half built outwards from a centre held to the beat, the picture this chapter has been describing. There, click or drag on the green line, time 0, to ask what a counterexample would need: if its white edge started at that depth, everything in the shaded wedge below would have to stay white, one depth deeper at every step. The darker squares are where this random start is black instead. With every row shown, the wedge closes into a whole triangle where the picture itself ends; in a real counterexample it would go on for ever.
We looked at what such an edge would cost. It drags a disturbance along with it, and that disturbance pulls on the squares that must stay white with a beat: odd, odd, even, odd, odd, even, the pattern of the Fibonacci numbers 1, 1, 2, 3, 5, 8, 13. Something inside has to answer that beat for ever. The four short animations below show the beat, the one pattern that could pay it and why Rule 30 forbids it, why the payment always needs older and older events, and how real Rule 30 keeps trying regardless.
6 · What we found
We have not solved any of the three problems. Here is what has been settled along the way, in plain terms. Every proof below was checked by a second reader inside the project; none has yet been reviewed outside it. Computations were announced with their expected outcomes before they were run, and the few exploratory ones are marked as such in the record.
Apart from the first point, which is Jen's theorem of 1990, none of this was in our record on 4 October, and a limited search did not find it in print. Together it makes a counterexample implausible, but it is not a proof. Everything comes down to one statement nobody has a method for. Holding the centre to its beat has to cost real information: the number of starting patterns that keep the beat should fall off exponentially with time. We can measure that fall-off, and it is close to one bit per step. The part of the cost paid by the start's left side is proved. The part paid by its right side is the open problem.
One construction shows how close the edge of possibility is. A row of 84 squares, repeated end to end for ever, comes back to itself every six steps. Its centre ticks white, black, white, black for good, and the column beside it plays the short loop over and over. It is no counterexample, because it has infinitely many black squares, but it is a real Rule 30 pattern doing exactly what a counterexample would have to do.
Watched as it runs, the ring is simpler than it looks. Each step of Rule 30 turns the whole ring by 14 of its 84 squares, one sixth of a turn, which is why it comes back every six steps. Press Turning with the ring to redraw it as if the paper turned too: every column becomes a single colour for ever. Its whole infinite history is one brick, 14 squares wide and 6 rows tall with 43 black squares, laid like bricks in a wall, each column of bricks one row lower than the one before. It is one of the 20 crystals above.
Behind it is a general fact we proved. Rule 30 can be cut into eight jigsaw pieces, one for each case of the rule, and its histories are exactly the ways of fitting them together. A row that comes back moved further than a change could have travelled in the time, as this ring does, must repeat across as well, so its history is a wall of one brick. The same pieces make walls like this one, so Problem 1 is not a question about the pieces: it asks whether the one picture grown from a single black piece can ever repeat down its middle.
7 · A twin in arithmetic
Take any whole number. If it is even, halve it; if it is odd, triple it, add one and halve it. Repeat. Lothar Collatz asked in 1937 whether every start eventually falls to 1. Computers have checked every start below 271, about 2.4 × 1021; nobody can prove it; and a prize of ¥120 million (Bakuage, 2021) waits for a proof or a counterexample.
It looks like another world, but in binary it is the same machine as Rule 30. A number written in binary is a row of black and white squares, and multiplying it by 3/2, the heart of Collatz's odd step (which is 3n/2 plus a half), works on that row much as Rule 30 does. Both start from one simple step, the Gray code, and add a correction.
The Gray code is a way of counting in binary in which each number differs from the one before in a single digit: 000, 001, 011, 010, 110, 111, 101, 100. It is named after Frank Gray of Bell Labs, and it is used wherever a reading must never jump, as in a rotary encoder. Its recipe is one comparison: write the number in binary and compare each digit with the one on its left, black where they differ. Repeat that step row after row from a single black square and the Sierpinski triangle of chapter 5 appears. Counting in Gray code changes digit number 0, 1, 0, 2, 0, 1, 0, 3, ... at steps 1, 2, 3, 4, ...: the digit is how many times 2 divides the step, which is the edge ruler of chapter 3 again, so the ruler's rhythm, and its sound, is the Gray code counting. Below, one number goes through three rules side by side.
So the two problems are two ways of getting the same simple rule slightly wrong, and both unroll into overlapping Sierpinski triangles, one hanging from each correction. Here both run from a single black square: Rule 30, and the powers of 3, which are what repeated multiplication by 3/2 does to a single 1 (the halving only shifts the row).
The likeness reaches the edges too. Every second row of the powers of 3 has a run of white squares just inside its black edge, and its length depends only on how many times 2 divides the row number: the same kind of ruler that Gareth spotted on Rule 30's edge. For the powers of 3 it is a classical theorem, the "lifting the exponent" lemma, and the ruler is perfectly regular, one square longer for each doubling. Rule 30's grows by about two and a half squares for each doubling, irregularly; that rate is measured, not proved. In both pictures every diagonal parallel to the right edge repeats with a period that is a power of 2, and in both the middle is where nobody can prove anything.
The questions are the same shape as well. In both, a finite start, a number's finitely many digits or a row's finitely many black squares, is fed to a rule that can be run backwards, and the question is whether what comes out beyond the part the start controls behaves like coin tosses. One exact statement holds word for word in both: a stretch of the output can repeat only if it is no longer than the current state is large, counted in digits or in squares (on the Collatz side this was already known). And the method that settles the easy cases stops at the same place in both, and in a third famous problem, Mahler's 3/2 problem of 1968. It shows that no counterexample exists when the constrained part can behave in only finitely many ways, and says nothing once it can behave in infinitely many, which is exactly where Rule 30's white-black case and Mahler's question sit.
So each side holds a tool the other lacks. Collatz has arithmetic: remainders modulo powers of 3, and linear forms in logarithms, which bound how close a power of 3 can come to a power of 2. Rule 30 has locality: its corrections are single squares, which is what made the exact rules about edge events possible. A proof on either side would show what a proof on the other has to replace. Neither side has one yet. The details, with every claim's source, are in the Collatz part of the record.
8 · How this was done
This work was done between 4 and 9 October 2026 by Gareth, working with AI collaborators: Claude, by Anthropic, in two roles, and GPT, by OpenAI. Each proof was proposed by one party and checked by another before it was filed as proved. Predictions were written down before experiments, and every prediction that failed was kept. The full record, with its proofs, refutations and computations, is public.