The same problem, three times
A free side and a constrained side that must agree for ever
Three famous open questions turn out to have one skeleton. In each, something chaotic runs freely, something narrow is required of it, and the question is whether the two can stay in step for ever. Where the narrow side can do only finitely many things, all three are solved by one argument. Where it can do infinitely many, all three are open.
The three, in plain words
Rule 30
Can a finite pattern keep an alternating centre column for ever? The free side is the left half of the pattern, which can be anything. The constrained side is the column beside the centre, which must keep the centre alternating, and that forces it into a thin language of allowed words (the wheel). The prize is Wolfram's, ten thousand dollars.
Collatz
Take any whole number; halve it if even, triple it and add one if odd; repeat. Does every start reach 1? The free side is the number's low bits, which choose the first steps outright. The constrained side is what remains after those free steps, a remainder modulo a power of three, which must keep the orbit from climbing for ever. Bakuage's prize is 120 million yen. Multiplying by three is a Gray step plus carries, and Rule 30 is the same Gray step plus a carry of its own: the carry of Rule 30.
Mahler's 3/2 problem (1968)
Is there a positive number ξ such that ξ, ξ·3/2, ξ·(3/2)², … all have fractional part below one half? Such a ξ is called a Z-number, and nobody knows whether one exists. The free side is the integer parts, which form a full shift. The constrained side is the fractional parts, which must keep to half the circle for ever. No prize, and open for 58 years. An exact census rules out every number below 263: no Z-number below 263.
A fourth figure has the same shape. Erdős asked in 1979 whether 2n, written in base 3, always contains a digit 2 once n > 8. The powers of 2 read in base 3 are the Collatz and Mahler arithmetic at its barest. It has been checked for every n up to 2·345, and, like the other three, it is open for one specific orbit (the statement exactly; the one missing method).
Side by side
The shared skeleton
| Rule 30, period two | Collatz | Mahler's 3/2 | |
|---|---|---|---|
| The free side | the left half at the wall: every itinerary occurs once (a full shift) | the first w−1 parities are a bijection with the number mod 2^(w−1) (Terras) | the integer parts: every itinerary occurs once, by residues mod q^k |
| The constrained side | the column beside the centre: a kicked rotation, few admissible words | the state after the free bits: a remainder modulo 3^a (the least-residue lemma) | the fractional parts: a linear mod-one map with few admissible strings |
| The object sought | a finite pattern whose centre alternates for ever | an orbit that never comes down | a Z-number: the two itineraries agree for ever |
| Sparsity | the entropy squeeze (proved): such a pattern's left side is almost frozen; that such patterns are exponentially rare is measured | almost every orbit attains almost bounded values (Tao, 2019, proved); the count follows a coin past the free bits (measured) | fewer than x^γ Z-numbers below x (Flatto, Lagarias, Pollington, 1995) |
| Emptiness (proved when) | the constrained column is eventually periodic (Jen) or monotone (Condrey) | the parity sequence is periodic or Sturmian (the window principle; Dubickas, 2009) | only finitely many fractional orbits are admissible (FLP, Theorem 3.2) |
| Open where | the alternating centre, where the column has positive entropy | every single orbit, as opposed to almost all | Mahler's interval of length one half, where infinitely many orbits are admissible |
The argument that works in the solved rows is the same in all three: if the constrained side has only finitely many behaviours, the pigeonhole principle makes the coupled orbit periodic, and the free side's injectivity forbids a periodic orbit. Where the constrained side has infinitely many behaviours, nothing forces periodicity, and the method stops. Rule 30's period two sits exactly there, as Mahler's conjecture does, and as Collatz does for a single orbit.
Sources. RULE30-PRIZE.md section 8.45 (the side-by-side, from a reading of Flatto, Lagarias and Pollington 1995), 8.47 (the survey), 8.58 (the window principle); COLLATZ-PRIZE.md, honest summary and sections 1, 4 and 5; PRIOR-ART.md (Kari and Kopra's restatement of Mahler's problem as a question about one column of the automaton that multiplies by 3/2; Terras 1976; Lagarias 1985; Dubickas 2009; Tao 2019).
What the comparison says, and what it does not
It is a calibration, not a route. It says precisely where the problem sits: in the regime where the one method that closes such problems has not worked for 58 years on Mahler's. It explains in one frame why every counting result we proved gives sparsity but never emptiness, and why the solved cases are exactly the ones where the constrained side is eventually periodic. It does not say the three problems stand or fall together; a proof for one would show what a proof for the others has to replace.
One difference matters. In Collatz the state after the free bits is an explicit integer, and arithmetic can be done on it. In Rule 30 there is no such number; the constrained side is a word in a thin language, and the tools are combinatorics on words and automata. That is why Rule 30's record holds exact automata, certified forbidden words and a census of records where Collatz's holds residues and exponential sums.
Sources. RULE30-PRIZE.md section 8.45, "What it means"; COLLATZ-PRIZE.md sections 2 and 3.
The same correspondence with definitions and theorems: the technical page.