Each fold needs its own two

Twenty-two is a floor adds the alignments the original enumeration left out — a point sent onto a crease being made in the same instant, a fold square to one, one simultaneous crease onto another — and takes the two-fold count from twenty-two to eighty-six. It also says, plainly, that eighty-six is a floor too: the enumeration spends its constraints from one pool without tracking which fold each alignment attaches to, so a condition that only makes sense between two particular creases is counted once instead of once per pair.

Attaching them is the obvious repair and it is not the only one it makes. Tracking which fold an alignment names turns out to expose two further defects in the pooled count, and the second is the awkward one: twenty-two was not the number of two-fold operations determined by the paper alone. That number is twenty-eight, and it needs no enumeration at all.

What a fold line is entitled to #

A fold line in the plane has two degrees of freedom. That is the whole basis of the count, and it is a statement about one line.

Two simultaneous fold lines have four freedoms between them, which is where $2m$ comes from — and where the trouble starts, because four constraints on two lines is not the same thing as two constraints on each. A pooled budget admits three conditions naming the first line and one naming the second: the first is over-determined and the second is a one-parameter family, and the pair as a whole determines nothing. Every such spending is counted as an operation by the pooled enumeration, and none of them is one.

So the budget is per fold, not per operation. Each fold must carry alignments worth exactly two, naming it. That is the first repair and it is the one that moves the paper-only count.

Two things a named crease costs #

The second repair is the one the earlier account asked for, and it is larger than it sounds.

A point onto a crease being made at the same time is not one alignment. At three folds it is two — the crease could be either of the other two — and at $m$ folds it is $m-1$. The same goes for square to a crease being made. Every appearance of a cross alignment therefore multiplies the catalogue by the number of folds it could have meant, which is why the gap between the two counts widens so fast: 1.2 times at two folds, 10.3 at three, 183.6 at four, 5,459 at five.

One simultaneous crease onto another is worse behaved. It names an ordered pair of other folds, so it needs at least three folds to exist at all. At two folds there is no pair to name, and the pooled count admits it anyway — an alignment counted with nothing it could possibly refer to.

The third repair goes the other way and takes operations out. The single-fold enumeration discards “square to two lines” because a fold square to two lines exists only when they are parallel and is then the same condition twice. The pooled count then quietly admits square to a named line together with square to a simultaneous crease, because the two carry different keys — and it is the same condition twice for the same reason. A fold may carry at most one perpendicularity, whatever the perpendicularity is to.

The case with nothing simultaneous in it #

Set the cross alignments aside entirely and the repaired count has an answer anyone can check without a computer, which makes it the right place to calibrate.

Folds that name only the paper do not interact. Each one is determined by two constraints on itself, which is to say each one is an axiom, and an $m$-fold operation of that kind is $m$ axioms chosen independently — a multiset, since the folds are not numbered. The number of multisets of size $m$ from seven is $\binom{m+6}{m}$: 7, 28, 84, 210, 462.

The repaired enumeration returns exactly those, at every size, by construction and not by being told. The pooled enumeration returns 22 where the answer is 28 — and 50 where it is 84, and 161 where it is 462.

That is the finding that matters most, because it is not about the difficult part of the subject at all. The cross alignments are genuinely hard to count and it was reasonable to call the result a floor. Two simultaneous folds each pinned by the paper are not hard to count; they are two axioms, and twenty-two is simply the wrong number for them. The pooled budget was losing operations in the easiest case in the catalogue while the discussion was entirely about the hardest.

Three kinds, where there had been one #

With every alignment attached to a fold, an operation has a shape: a directed graph, with an arrow from each fold to every fold its own alignments name. And the shape decides what kind of thing the operation is.

An operation whose graph has no arrows names nothing simultaneous. It is $m$ single folds that happen to be made at once, and nothing is lost by making them one at a time. There are 28 at two folds.

An operation whose graph has arrows but no cycle can be sorted. Some fold names nothing, so it can be made first; the folds naming only it can be made second; and so on down the order. It is a sequence wearing the costume of a multifold, and everything it constructs is constructible by single folds. There are 49 at two folds.

An operation whose graph has a cycle cannot be sorted, because every fold in the cycle is waiting for another. It has to be solved at one instant, as a simultaneous system, and it is the only kind that makes multifolds more than a notation. There are 28 at two folds — twenty-seven per cent of the hundred and five.

That trichotomy is invisible to a pooled count, which has no folds to draw arrows between. It is also the answer to a question the earlier essays framed in terms of hands: seven, and then twenty-two counts the coincidences a pair of hands has to achieve in the same instant, and this says which operations actually require one. Two of the three kinds require none.

The two counts are not the same shape #

The pooled count grows roughly as a polynomial in the number of folds: it is counting multisets from a fixed alphabet with a budget that grows linearly. The attached count grows as a product over folds of a per-fold catalogue that itself grows with the number of folds, and then divides by $m!$ — so it grows faster than exponentially, and the ratio between the two grows without limit.

Which of those is the right shape for the subject is a question this essay cannot settle, and it is worth saying why not. Both are counts of descriptions. Neither asks whether the resulting system of equations has a solution, how many, or whether the solutions are distinct from those of some smaller operation. A catalogue that grows as fast as this one certainly contains enormous redundancy, and finding it is a different piece of work.

What the earlier count was, and what it was not #

It is worth putting the pooled count back beside this one, because it was not wrong in the way a miscalculation is wrong.

Read as a count of ways to spend $2m$ constraints drawn from a fixed list, both of its columns are correct and its own footer says the rest. Read as a count of operations they are neither an upper nor a lower bound of anything one would want: too small because a cross alignment is many alignments, too large because a pooled budget builds things that determine nothing, and — the awkward one — smaller than the trivial case it contains.

The number 22 was doing a great deal of work across two essays and it was never the count of anything. Seven, and then twenty-two sets it against the coincidences a pair of hands must achieve, which is a comparison between a catalogue and a difficulty; with 28 in place of 22 that comparison changes in size and not in kind, since both numbers are small. What changes in kind is the three-fold row, where 50 becomes 84 in the easy case and 3,042 in the full one.

What it does to the reading about hands #

Two of those essays make an argument about who can perform these operations, and the trichotomy sharpens it in a way neither could.

A pair of hands making two folds at once has to achieve every alignment in the same instant, and the earlier account priced that by counting coincidences. The reference graph says which operations genuinely demand it. An independent operation demands nothing simultaneous: two hands, two folds, and no relation between them. A sequential one demands only that the folds be made together, not that they be solved together — the folder could in principle work out the second fold from the first before starting. Only a cyclic operation is a simultaneous system in the strict sense, where neither fold can be described without the other.

That matters for what a machine is for. The fold a machine can make asks what a device folding one crease at a time reaches; a device holding two adjustable lines and iterating until both conditions hold is a solver, and a solver’s whole advantage is over the cyclic cases. So the machine’s advantage is not the 105 and not the 77 that name a crease being made — it is the 28, which is a quarter of the catalogue and the only quarter a sequence cannot reach.

Two creases at once found the hendecagon reachable by two simultaneous folds, and the largest equilateral triangle a square holds found an optimisation that turned out to need nothing beyond a quadratic — a useful calibration in the other direction. Between them sits which of the largest inscribed polygons are foldable at all, and the question this count makes askable is which of those need a cyclic operation and which are a sequence in disguise.

Two routes to the number #

The count is large enough that a single derivation of it would be worth distrusting, so it is done twice.

The first route builds every numbered diagram and reduces it to a canonical form under all $m!$ renumberings of the folds. It is exact and it is exponential: 17,576 structures at three folds, 3.4 million at four, and it stops being the right tool there.

The second is Burnside’s lemma — the number of orbits is the average number of structures a renumbering leaves alone — which is a sum of $m!$ products and costs nothing at any size. The two agree at one, two and three folds, which is where both can run, and past that only the second can.

Two implementations agreeing is the whole of the evidence for the large numbers, and it is worth being clear about what it is evidence for. It says the orbit counting is right. It says nothing about whether the model of an operation is right, which is the part that had to be chosen.

What the count does not settle #

The model is a set of choices and each one could have gone another way.

It does not ask whether an operation is solvable. A fold carrying two constraints is determined only when the constraints are independent and consistent, and plenty of these are neither — a fold told to pass through two points that are the same point, or to be square to a crease that turns out parallel to it. Why the list stops at seven discards one of eight single-fold combinations for exactly that reason, and the corresponding discarding here has not been done.

It does not ask whether an operation is new. Fifty years in the wrong language is this subject’s standing warning that a result can exist and be unavailable, and a catalogue of descriptions is exactly the kind of object in which a known construction appears many times under different names.

It does not ask whether two operations construct the same thing. An operation is counted by its description, and the catalogue certainly contains many descriptions of one construction — most obviously the sequential ones, every one of which is reachable by single folds and so adds nothing to what can be built.

And it does not price a solution’s multiplicity. An alignment that sends a point onto a line is a quadratic and has two answers; several such in one system multiply. The number of constructions an operation offers is therefore larger than one and is not counted here at all.

What the model assumes #

A fold line has exactly two freedoms and needs exactly two constraints naming it. That is the assumption the whole repair is, and it is the same one the single-fold count rests on.

Alignments cost one or two and the costs add. No alignment is worth a fraction, and no pair of alignments is worth less than the sum of its parts — which is exactly what the perpendicularity rule is an exception to, handled by forbidding the pair rather than by pricing it.

Cross alignments name folds, not creases-so-far. A condition refers to a fold being made in the same instant; there is no notion here of a crease made a moment earlier, because that is a sequence and belongs to the single-fold theory.

And the folds are unnumbered. An operation is a shape, so two descriptions differing only in which fold is called first are one operation. That is the quotient the Burnside count is over, and it is why the answer is not simply a product.

How the numbers were checked #

The single-fold count must still be seven, by the repaired enumeration rather than by fiat. Any repair that disturbed the axioms would be repairing the wrong thing.

The paper-only count is checked against a closed form at every size — $\binom{m+6}{m}$, computed independently of the enumeration — so the calibration case has two answers that must agree.

The exact and averaged orbit counts are required to agree wherever both can run, which is three sizes, at the largest of which the exact route builds 17,576 structures.

And the trichotomy is required to be a partition: the three kinds must sum to the total at every size, which would fail immediately if the cycle test or the reference graph were wrong.

Still open: the count of constructions, not of descriptions #

Everything above counts ways of asking, and what a subject about constructibility wants is the count of things built.

The gap has three parts and the middle one is the largest. Sequential operations construct nothing a sequence of single folds cannot, so by the trichotomy at least 49 of the 105 two-fold operations and 1,253 of the 3,042 three-fold ones are already inside the single-fold theory; the independent ones, 28 and 84, are inside it by definition. That leaves 28 and 1,705 as the operations that could add anything, which is a far smaller and far more interesting catalogue than either of the counts this essay computes.

Reducing that further needs the two things the model does not have: a test for whether a system is consistent and determined, and a notion of when two operations construct the same set of points. Neither is exotic — the first is a rank computation on the constraint system and the second is an equivalence on the resulting field extensions — and together they would turn a count of descriptions into the count the subject actually wants.

Sideways from here, the trichotomy is the right frame for the machine question. The fold a machine can make asks what a device folding one crease at a time can reach; a device that can hold two adjustable fold lines and iterate to a consistent state is a device for the cyclic operations specifically, and the 28 of them at two folds are exactly its advantage over hands. Two creases at once found the hendecagon on the far side of that advantage, which suggests the twenty-eight are not a small catalogue in what they can do.

The habit worth carrying is about enumerations that pool a budget. Ask whether the budget belongs to the object or to its parts. Four constraints on two lines and two constraints on each line are different counts of different things, and the difference is invisible in the arithmetic — $2m$ is $2m$ either way — right up to the moment somebody asks what the operations with three on one side determine.