Counting operations is not counting power

Each fold needs its own two counts the operations a set of simultaneous folds admits once every alignment is attached to the fold it constrains, and the count is startling: 7, 105, 3,042, 145,211, 9,782,771 at one to five folds. A catalogue growing by a factor of five thousand between two folds and five looks like a subject whose power is exploding.

It is a catalogue of ways to ask. What a construction reaches is decided somewhere else — by how many fold lines an operation actually admits — and that number grows about as slowly as the catalogue grows fast.

An alignment is a curve in the plane of lines #

A fold line in the plane has two numbers in it, so the lines form a plane of their own, and every alignment is a curve in that plane. The degree of the curve is what matters.

Through a named point is the set of lines through a point — a pencil, and a line in the plane of lines. Degree one.

Square to a named line is the set of lines in a fixed direction. That is also a pencil, but the point it passes through is at infinity, which turns out to matter. Degree one.

A named point onto a named line is the set of perpendicular bisectors of the point and each point of the line. Those are exactly the tangents of the parabola with that point as focus and that line as directrix: a conic’s tangents, degree two.

A point onto a point is one line, the perpendicular bisector. A line onto a line is two, the bisectors of the angle between them.

A fold carrying two alignments of cost one therefore admits the product of their degrees — except that the product counts solutions the plane of lines has and the sheet does not.

Where the missing fourth tangent went #

Two parabolas ought by Bézout’s count to have four common tangents, and the sixth axiom has three. The missing one is not missing: it is the line at infinity, and every parabola is tangent to it.

That is the whole of the exception, and it is the same exception twice. The seventh axiom — a point onto a line, square to a line — pairs a parabola’s tangents with a pencil, and the pencil’s point is at infinity, so one of the two tangents from that point is the line at infinity again: two minus one is one, which is the count the seventh axiom has. The fifth axiom pairs a parabola with a pencil through a finite point, nothing is shared at infinity, and the count is two.

So the rule is one sentence. The number of folds is the product of the alignments’ degrees, less one when both of them contain the line at infinity. It gives 1, 1, 2, 1, 2, 3, 1 for the seven axioms — which is what they are known to have — and it explains the pattern rather than recording it. Why the list stops at seven counts the ways to spend two constraints; this counts what each way then delivers, and the two questions have very different answers.

Seven operations, eleven constructions. Four of the axioms offer one fold, two offer two, and the sixth offers three. That single axiom carries more of the subject’s constructive power than any three of the others, and it is the one whose three answers are the three roots of a cubic.

The eleven, one at a time #

The seven counts are worth reading individually, because each has a geometric reason and the reasons are not the same.

Through two named points is one line and needs no argument. A named point onto a named point is the perpendicular bisector, again one line, and it is the only axiom whose alignment is worth two constraints and still admits a single answer. A named line onto a named line is two, the pair of angle bisectors — and that is the first place the subject gets a choice, which is why it is also the first place a folder can go wrong.

Through a point, square to a line is one: a direction and a point determine a line with nothing left over. A point onto a line, through a point is two — the tangents to a parabola from a point outside it, which is the familiar picture of two tangent lines from an external point and the reason that axiom is sometimes drawn with a ruler pivoting.

A point onto a line, square to a line is one, and it is the subtle member. Pairing a conic with a pencil should give two; the pencil’s point is at infinity, the parabola is tangent to the line at infinity, and one of the two is therefore the line at infinity itself. A point onto a line, and another point onto another line is three by the same subtraction applied twice over.

What each axiom is worth prices them by what they add to the reachable field; this prices them by how many lines they offer, and the two orderings are not the same. The third axiom offers two folds and adds nothing to the field — the bisectors of two constructible lines are constructible with a compass. The sixth offers three and adds the cube root. A count of answers and a count of new answers are again different quantities, which is the essay’s own point applied one level down.

Why three and not four is the whole story #

The subtraction is easy to state and easy to dismiss as bookkeeping, and it is not bookkeeping: it is the reason this subject has a cubic in it at all.

Bézout’s theorem counts intersections in the projective plane, where the line at infinity is an ordinary line and a parabola is an ordinary conic tangent to it. Two parabolas meet in four tangents there. A sheet of paper is not projective — it has no line at infinity to fold along — so one of the four is unavailable, and what is left is three.

Three is odd. A polynomial of odd degree over the rationals with no rational root is irreducible over a quadratic tower, which is exactly what makes the sixth axiom construct things a compass cannot: folding beats the compass because the number of its answers is not a power of two. Had the fourth tangent been available, the sixth axiom would offer four answers, four is a power of two, and the whole subject would be a slower way of doing what a compass does.

So the missing tangent is not a defect in the count. It is the source of the one thing folding has that the compass does not, and it goes missing for a reason that has nothing to do with paper: every parabola passes through the same two points at infinity, so any two of them share a tangent there. The axiom that names two folds is where the same axiom’s multiplicity causes trouble of a different kind, and the multiplicity is this one.

A catalogue that multiplies against a reach that multiplies more slowly #

An operation on several folds admits the product of what its folds admit, so the ceiling is three to the power of the number of folds.

The distribution is as lopsided as the single-fold one. Of the 105 two-fold operations, thirty-six admit exactly one fold — a third of the catalogue, offering one answer each — and only six reach the maximum of nine. The 288 solutions across 105 operations average under three, and the average is doing what averages do: most operations are worth one and a handful are worth nine.

Between one fold and five, the catalogue multiplies by about 1.4 million and the ceiling multiplies by 81. There are 2.3 operations per available solution at one fold and 40,258 at five. Almost all of the catalogue’s growth is growth in the number of ways to describe the same small set of answers.

And the reach is smaller than the ceiling #

The ceiling of three to the $m$ is a bound on solutions, and what a constructibility argument wants is the degree of the polynomial those solutions satisfy — which is smaller.

A single fold settles a cubic: three solutions, one irreducible cubic, and that is what buys the regular heptagon and the trisected angle. Two simultaneous folds admit up to nine solutions, and what they settle is a quintic — degree five, not nine. The general statement is that $m$ simultaneous folds settle an irreducible polynomial of degree at most $2m + 1$.

That is where two creases at once gets the hendecagon. The eleven-gon needs a fifth root, $\varphi(11) = 10 = 2\cdot 5$, and five is exactly $2\cdot 2 + 1$. The twenty-three-gon needs an eleventh root and so needs five simultaneous folds, which is a great deal of hands.

So there are three counts of this subject and they grow at three speeds: the catalogue as something faster than exponential, the solution ceiling as $3^m$, and the reach as $2m+1$. At five folds those are 9,782,771, 243 and 11. The gap between the second and the third is the redundancy inside a single operation; the gap between the first and the second is the redundancy across the catalogue. Both are enormous and neither has been measured.

What the earlier count was counting #

Read beside the reach, the whole enumeration argument changes character. It is a count of notations — of distinguishable ways to specify a fold — and a notation is worth having for its own reasons. It tells a folder what may be asked for. It tells a mechanism what it must be able to express. It does not tell anybody what can be built.

Seven, and then twenty-two sets the catalogue against the coincidences a pair of hands must achieve, and that comparison survives everything here — it was never about constructibility. What does not survive is any inference from the catalogue’s size to the subject’s power. A count of descriptions and a count of constructions are different quantities, and this one runs ahead of that one by four orders of magnitude before the fifth fold.

What a folder actually gets #

The gap between an algebraic count and a folder’s experience is worth making explicit, because the numbers above are all algebraic.

Three common tangents is the count over the complex numbers. In the configuration drawn all three are real and all three cross the sheet, which is why the figure can show them; move the second point a little and two of them become a conjugate pair, leaving one fold and a folder with no choice to make. That instability is a property of the configuration rather than of the axiom, and it is the reason a construction using the sixth axiom has to say which of its solutions it wants and how to recognise it.

None of that changes the degree. The polynomial is still a cubic whether its roots are three reals or one; the reachable field is the same; and a construction that specifies a root by some geometric test works wherever that root is real. What changes is whether the fold can be made by aligning two things by eye, which is the sense in which the sixth axiom is harder than the other six in practice as well as in theory.

The same distinction runs through the two-fold catalogue. An operation admitting nine solutions is an operation whose system has nine roots somewhere; how many of them are real, distinct and on the paper is a question per configuration, and the catalogue is silent about every instance of it.

What the count of solutions does not settle #

The rule above prices an operation by Bézout’s theorem with one correction, and Bézout counts solutions over the complex numbers with multiplicity.

It does not say how many solutions are real. The sixth axiom admits three folds in the configuration drawn and admits one in most configurations; the number a folder actually has depends on where the points and lines sit, and the catalogue cannot see that. What the three in the table means is the degree of the polynomial, which is the quantity constructibility cares about, and not a promise of three creases.

It does not say how many are distinct. A degenerate configuration collapses solutions together — two points that coincide, a line parallel to another — and every such coincidence lowers the count for that instance without changing the operation.

And it does not say whether a solution is on the paper. A fold line meeting the sheet nowhere is a solution of the system and not a fold, which is a restriction the whole subject lives with and which the reference closure has to handle case by case.

What the model assumes #

Alignments are algebraically independent unless they share a solution at infinity. That is what lets the degrees be multiplied, and the correction is applied only for the one coincidence that is present in every instance rather than in special ones.

A fold carries exactly two alignments of cost one, or one of cost two. That is the per-fold budget the labelled count rests on, and the solution counts are computed against it.

A cross alignment behaves like its paper counterpart. A point sent onto a crease being made in the same instant is priced as a parabola’s tangents, exactly as a point onto a named line is. That is right when the other crease is treated as known and is the place where the simultaneity is being idealised away.

And the operations multiply across folds. For an operation whose folds refer to nothing, or refer in an order, that is exact. For one whose references form a cycle it is an upper bound, since a cyclic system may have fewer solutions than the product of its parts.

How the numbers were checked #

The rule is checked against the seven axioms, which have known solution counts. A rule that produced anything other than 1, 1, 2, 1, 2, 3, 1 would be the wrong rule, and the check is on the produced list rather than on the three that matter.

The three common tangents are found numerically and then verified by folding. Each solution is used to reflect both points, and each reflected point is required to land on its own line to within a part in a billion — so the solver’s answers are checked against the condition rather than against the solver.

The distribution over two-fold operations is required to have as many entries as there are operations, checked against the independent orbit count, so no operation is priced twice or left out.

And the ratio of catalogue to ceiling is required to grow at every step, which is the claim the essay is about and would fail immediately if either column were wrong.

Still open: how much of the catalogue constructs anything new #

The two gaps named above are both measurable and neither has been measured.

The nearer one is the redundancy across the catalogue. An operation constructs nothing new if its references form no cycle, since it can then be performed one fold at a time, and the earlier count found 49 of the 105 two-fold operations and 1,253 of the 3,042 three-fold ones to be of that kind — with another 28 and 84 needing no simultaneity at all. The catalogue’s constructive part is therefore already known to be a minority of it, and reducing it further needs a test for when two operations reach the same points.

The further one is the gap between $3^m$ and $2m+1$. Nine solutions settling a quintic means the nine split as five and four over the field the references lie in, and which operations produce an irreducible factor of the full available degree is the question that would turn the ceiling into a reach. That is a Galois-theoretic question rather than a combinatorial one, and it is the point where this line of counting stops being able to answer its own questions.

Sideways from here, the three speeds are an argument about what a machine should be built to do. A device that can express the whole catalogue is a device with an enormous specification language and no more reach than one that can express the cyclic operations alone. The fold a machine can make asks what a one-crease device reaches; the useful version of that question at two folds is not what can be asked but what the asking is worth, and that is 288 solutions rather than 105 operations.

The habit worth carrying is about catalogues generally. Before reading growth in a catalogue as growth in power, price one entry. A catalogue whose entries are worth a bounded amount each is a catalogue of names, and its size is a fact about the language rather than about the subject.