Seven, and then twenty-two

The restriction to one fold is a rule somebody imposed rather than a property of paper. Allow two creases to be made simultaneously and the reachable degree rises: the hendecagon nobody could fold becomes foldable, and the operation set stops being the seven axioms.

That leaves an obvious question unanswered. It stops being the seven — and becomes what?

Where seven comes from #

Why the list stops at seven is the derivation for one fold and it takes about a minute. A fold line in the plane has two degrees of freedom, so it is pinned down by alignments worth two constraints between them. There are five kinds of alignment and they cost one or two:

Worth one — the fold passes through a named point; the fold sends a named point onto a named line; the fold is square to a named line. Worth two — the fold sends a named point onto a named point; the fold sends a named line onto a named line.

Enumerate the ways to reach two. Two of the one-cost alignments, in any combination, gives six; either of the two-cost alignments alone gives two more; eight in all. One of the eight — square to two lines at once — determines nothing, because a fold square to two lines exists only when the lines are parallel and is then the same condition twice.

Seven.

Nothing in that argument mentions paper. It is a count of ways to spend two numbers, and the seven come out identified with the operations a folder actually performs afterwards rather than being put in.

It is worth noticing what the derivation does not need. It does not need any of the seven to be described, drawn or named. It does not need a sheet of paper. It does not use the fact that a fold is a reflection, or that paper does not stretch, or anything about what happens after the crease is made. The seven fall out of counting how many numbers a line has and how many each kind of coincidence removes, and everything else about folding is downstream of that.

The same count, run again #

m fold lines have 2m degrees of freedom. The alignments are the same five and cost the same. So the count of operations is the number of ways to spend 2m constraints with those five alignments, discarding the perpendicular-to-two-lines degeneracy exactly as before.

At one fold it gives seven, which is the check on the method. At two it gives twenty-two. At three, fifty. At four, ninety-five. At five, a hundred and sixty-one.

The growth is cubic — the differences between consecutive counts are 15, 28, 45 and 66, whose third differences are constant — which is what a count of multisets over a fixed alphabet with a linearly growing budget does.

The count has a closed form #

The growth is reported as cubic and established by taking differences, which is the right way to notice it and stops one step short. The polynomial can be written down, and having it is worth more than five values.

Set up the enumeration exactly as the derivation does. Three alignments cost one constraint each and two cost two, so an operation at $m$ folds is a multiset of them totalling $2m$; and the exclusion is that no operation may use the perpendicular alignment twice, since a fold square to two lines has been given the same condition over again.

Counting those multisets and subtracting the excluded ones gives, at $m$ folds,

$$N(m) = \frac{(m+1)(m+2)(4m+3)}{6}.$$

Put the numbers in. At one fold: two times three times seven over six, which is seven. At two: three times four times eleven over six, which is twenty-two. Then fifty, ninety-five and a hundred and sixty-one — every value in the table, exactly, with nothing fitted.

That the formula reproduces seven is the same check the enumeration itself passes and it is worth twice as much here, because a closed form can be wrong in ways a count cannot: a polynomial fitted to five points will always fit five points. This one is derived from the weights and the exclusion rather than fitted to the values, and it lands on all five.

What the form says #

Two things fall out of the factored shape that the differences do not show.

The leading term is $\tfrac{2}{3}m^3$, so the catalogue grows as two thirds of the cube of the number of simultaneous folds. Doubling the folds multiplies the operations by about eight, and there is no size at which that slows down.

And the factor $(m+1)(m+2)$ says where the cubic comes from. Two of the three factors count how the two-cost alignments can be distributed and the third counts the one-cost ones — which is the enumeration’s own structure, visible in the answer. A count with a factored closed form is a count whose combinatorics has been understood rather than merely performed, and that is the difference between a table and a theorem.

It also sharpens the essay’s own comparison. The catalogue is $(m+1)(m+2)(4m+3)/6$ and the width is $2m$, so operations per coincidence go as $m^2/3$ — quadratically better with every extra fold, for ever. The reachable degree rises too, and the numbers a fold can name grow with it. Every column that measures what is gained improves without limit, and the one column that measures what a hand can do stopped at the first row.

And what a folder has to hold #

The other column grows too, and much more slowly: 2m. Two coincidences at one fold, four at two, six at three.

That looks like the smaller difficulty and it is the binding one, because of where it starts. Two is already at the limit. A single fold — bringing a corner onto a point while the crease passes through another point — is a thing a folder does by holding the paper in a configuration where both are satisfied and then flattening. It is achievable because the two conditions leave a one-parameter family of positions to slide along until both are met, and a pair of hands can slide along one parameter.

At two simultaneous folds there are four conditions to satisfy at one instant, on a sheet that is in a bent, unflattened, three-dimensional intermediate state whose shape is not determined until the folds are complete. No amount of sliding along one parameter reaches it. What a two-fold construction is, in practice, is a sheet held in a shape that nobody can see all of, adjusted until four things happen together.

There is a further asymmetry worth naming. The two coincidences of a single fold are satisfied by the same motion: a folder brings the paper over, and as the flap swings there is a one-parameter family of positions, and somewhere in that family both conditions hold at once. The search is one-dimensional and a hand can perform a one-dimensional search by feel.

At two simultaneous folds the intermediate state has more freedom than a hand can sweep. The sheet is not flat, not folded, and its shape depends on both creases at once; there is no single motion whose parameter can be run until four conditions are met, because the conditions are not all monotone along any motion a person can make. That is why a two-fold construction, where it is published at all, comes with an account of how to approximate it rather than how to do it.

The two columns #

So the table has one column that grows and one that binds, and they are not in competition — they are measuring different subjects.

The operation count is a fact about the algebra. It says how many distinct kinds of construction exist at each level, which is the same sort of statement as the seven, and it is what a theory of multifold constructions has to enumerate before it can say anything.

The alignment count is a fact about hands. It says how many things must be true at one moment, and it goes from two — which is what a fold is — to four, which is what a person does with paper only in a very small number of published constructions and never reliably.

That gap is what the phrase the machine is not the hand names on this site, arriving from the construction side rather than from the folding side. A machine can hold four alignments; a machine that folds paper along two lines at once, adjusting both until four coincidences are met, is a device somebody could build. Nothing about the algebra forbids it. What forbids it is that nobody has, and that the tradition the axioms come from is a tradition of hands.

There is a third column the table does not have and it is worth naming so that nobody assumes it. How many of the twenty-two two-fold operations are useful — determine a fold pair rather than a degenerate one, and reach something the seven do not — is a different question from how many exist. At one fold the answer is known and is all seven: each of them determines a fold and each is used. At two folds nobody has sorted them, and the sorting is not a counting exercise: it means examining each combination for the kind of degeneracy the perpendicular-to-two-lines case is the single-fold instance of.

So twenty-two is the size of the catalogue somebody would have to work through, not the size of the answer they would end up with. The single-fold case suggests the two numbers are close — eight combinations, seven operations — and there is no argument here that the ratio holds.

Which theorem was checked, and how #

The enumeration is over multisets of alignments with a weight budget, and the whole of its content is the weights and the one exclusion. Both are stated in the module and neither is tuned.

The check that licenses the extension is that the same code, run at one fold, returns seven — and returns it as a count of enumerated combinations, not as a constant. If the weights were wrong, or the exclusion were wrong, or an alignment were missing, the single-fold answer would not be seven, and there is nowhere for a mistake to hide because seven is the number the subject has agreed on since 2001.

Three further conditions are asserted: the count must strictly increase with the number of folds; it must grow faster than linearly; and a request for zero folds must be refused rather than answered.

What settles a column is not the drawing but the arithmetic under it. A degree either reaches a given prime or it does not, and the regular polygons an operation set constructs are exactly those whose totient is built from the primes its own degree can reach.

Where the model stops #

The count is a floor, and the omission is the interesting one.

Alignments that refer to the folds being made at the same time are not enumerated. A point landing on the crease that another simultaneous fold is producing is a perfectly good condition, it is not expressible with the five alignments above, and it is precisely what makes multifolds strictly more powerful than a sequence of single folds. Every published two-fold construction uses one. So the true operation count at m folds is larger than the number here, and by an amount this method cannot reach.

The count also treats operations as combinations of conditions, which is the right abstraction for asking how many kinds there are and the wrong one for asking how many are useful. Several of the twenty-two will be degenerate for reasons the perpendicular-to-two-lines case is the only single-fold instance of, and finding them means examining each rather than counting.

And the alignment count assumes the folds are independent lines with two freedoms each. Two folds forced to be parallel, or to meet at a named point, have fewer — which is a restriction a construction might impose deliberately and which would change both columns.

What the picture cannot show #

The table is a table because a count is a number. What would be worth drawing is a two-fold operation — a sheet in an intermediate three-dimensional shape with four coincidences marked — and that picture is genuinely hard to make honestly, because the intermediate shape is not determined until the operation is complete. Every published illustration of a two-fold construction shows the result and not the doing, for exactly that reason.

The generalisation #

There is a shape here that recurs whenever a formal system is asked to be performed rather than merely stated, and folding gives an unusually clean instance of it.

A system has a catalogue — how many kinds of thing it can express — and a width — how much has to be true at once for one of them to be carried out. Enlarging the system usually grows both, and almost always grows the catalogue faster: more operations, each requiring a little more simultaneity. That is a good trade for a theory and a bad one for an agent, because a theory pays for the catalogue and an agent pays for the width.

Folding’s numbers are 7 → 22 → 50 → 95 → 161 against 2 → 4 → 6 → 8 → 10. The first sequence is what a book about origami constructions would have to contain; the second is what a person would have to do. And the second is already, at its first term, the whole of what a pair of hands can manage: a single fold satisfying two coincidences is the operation the subject is built on, and it is at the limit rather than comfortably inside it.

That is the honest reason the axioms stop at one fold, and it is not a mathematical reason at all.

Who found it, and when #

The axioms are Humiaki Huzita’s and Jacques Justin’s, from 1989 and 1986 respectively, with the seventh recovered by Koshiro Hatori in 2001 and the completeness proof settled shortly after. The degrees-of-freedom argument for the count is folklore by now and appears in several places.

Multifolds are Roger Alperin and Robert Lang’s subject, from a 2006 paper that shows two-fold operations solve quintics and that m-fold operations reach further still — one step past the cubic a single fold reaches. That paper is about the algebra. The count of operations at each level is, as far as this site can find, not tabulated anywhere — and it is a natural thing to want, because the reason no one has built a theory of three-fold constructions is not that the algebra is uninteresting but that fifty operations is a lot to characterise and six simultaneous coincidences is not a thing anybody can do.

One number is worth extracting from all of this, because it is the one a reader will want. The reachable degree rises with the number of simultaneous folds and the catalogue rises cubically, but the ratio of what is gained to what must be held is falling: seven operations for two coincidences is three and a half operations per coincidence; a hundred and sixty-one for ten is sixteen. So more simultaneous folds are, by this measure, a better and better deal — right up to the point where the coincidences cannot be achieved at all, which is at the second row.

That is an uncomfortable shape for a subject. The economics improve monotonically and the feasibility stops immediately, so there is no optimum to find and no trade-off to balance: there is a cliff at m = 2, and everything past it belongs to a machine or to a proof.

Where the ladder goes next #

The immediate continuation is the omission. Enumerating the operations that refer to the simultaneous creases needs a different combinatorial model — the conditions are between the unknowns rather than between an unknown and a named object — and it would give the true count, of which twenty-two is a floor.

The other direction is a machine. Nothing in this essay’s argument about hands applies to a device, and the question of what a mechanism that makes two adjustable folds at once could construct is a question with a definite answer that nobody appears to have asked. It is the counterpart of what a machine that folds one crease at a time can reach, run at the construction end instead of the folding end.