Gauss's polygon is the expensive one

Which regular polygons a fold reaches is told as a verdict. A polygon of $n$ sides is constructible by folding when the odd prime factors of $n$ are Pierpont primes appearing once each — which is a condition on a factorisation, and the answer it returns is yes or no. The heptagon is yes and no compass reaches it; the hendecagon is no.

Reachable is not cheap observes that a condition deciding membership by factorising something can usually be read again as a cost, by counting the factors instead of testing them, and that the free second result is almost never taken. This is that result taken.

Every reachable polygon has a height: the number of extension steps the shortest tower to it must climb, which is the sum of the exponents of two and three in the degree of the equation its cosine satisfies. Read the heights instead of the verdicts and the ordering the subject uses turns over.

Where the degree comes from #

A regular $n$-gon is constructible from a circle when the angle $2\pi/n$ can be laid off, and the quantity a construction actually has to reach is $2\cos(2\pi/n)$. That number generates the real subfield of the $n$-th cyclotomic field, and its degree over the rationals is

$$\frac{\varphi(n)}{2}$$

where $\varphi$ is Euler’s totient — the count of numbers below $n$ sharing no factor with it. So the degree of the heptagon is $\varphi(7)/2 = 3$, of the pentagon $\varphi(5)/2 = 2$, of the seventeen-sided polygon $\varphi(17)/2 = 8$.

That is the whole input. Everything else is factorising it.

A fold reaches the polygon when the degree is $2^a 3^b$, which is the condition the reachable degrees satisfy. A compass reaches it when $b = 0$. And the height — the shortest tower — is $a + b$, because each extension step multiplies the degree by two or by three and a tower of $a$ twos and $b$ threes is the shortest that arrives.

Three columns that do not agree #

The heptagon costs one step. Its degree is three, which is one cube root, which is one application of the conic axiom — Beloch’s fold — and nothing else. It is the cheapest non-trivial polygon a folder can make, and it is the standing example of a polygon the compass cannot reach at all.

The seventeen-sided polygon costs three. Its degree is eight, which is $2^3$: three square roots, one after another, and no cube root anywhere. A compass reaches it, which is the fact Gauss proved at nineteen and asked to have carved on his headstone, and a folder reaching it is climbing the tallest tower on the list.

So the two instruments do not order the polygons the same way, and they do not even order them compatibly. The polygon that is impossible for one is cheap for the other, and the polygon that is the other’s most celebrated achievement is the first one’s most expensive construction under forty sides.

Half of the polygons a fold reaches in a single step are polygons a compass cannot draw at all. The one-step column holds eight polygons — five, seven, eight, nine, ten, twelve, fourteen and eighteen — and four of them are out of the compass’s reach entirely. Every one of those four has degree three, needs the conic axiom, and needs it once.

The shape of the four columns #

Counting the columns rather than reading them says how much of the list each height holds, and the distribution is not what a reader of the verdict would guess.

Thirty-one of the thirty-eight polygons up to forty sides are reachable by folding and sixteen of them by a compass — so folding roughly doubles the list, which is the comparison everybody makes. What the height column adds is that the compass’s share is highest at the two ends and lowest in the middle: it has all three of the free polygons, half of the one-step ones, under two-fifths of the two-step ones, and over half of the three-step ones again.

The bottom end is trivial — the triangle, the square and the hexagon have degree one and need no extension at all, so every instrument has them. The top end is the finding. The compass is over-represented among the most expensive polygons because expensive, for a compass, is the only way it has of being anything, and the polygons it reaches at all are the ones whose degree is a tall stack of twos.

The polygons that are not there #

Seven of the thirty-eight are unreachable, and it is worth naming them because the reason is the same in every case and it is not about the polygon.

Eleven, twenty-two, twenty-three, twenty-five, twenty-nine, thirty-one and thirty-three. Their degrees are five, five, eleven, ten, fourteen, fifteen and ten — and every one of those has a prime factor that is neither two nor three. No tower reaches them, at any height, by any route, because a tower’s degree is a product of the steps’ degrees and every step contributes a two or a three.

That is a stronger kind of impossibility than the one the heights describe, and the table prints it as a dash rather than as a large number for exactly that reason. A polygon of height three is expensive; a polygon with a factor of five in its degree is outside the field. The fifth root is the standing example, and the hendecagon is the same fact wearing a polygon.

The twenty-five-sided polygon is the instructive one. Twenty-five is five squared and $\varphi(25)/2 = 10 = 2 \cdot 5$, so it fails on a single factor of five — while the fifteen-sided polygon, whose factors are three and five, has $\varphi(15)/2 = 4$ and is reachable in two steps by a compass. A prime in $n$ is not a prime in the degree, and reading the polygon’s own factorisation instead of its totient’s is the mistake the Pierpont condition exists to prevent.

Why the inversion is not a coincidence #

The mechanism is short and it is the reason to state the result as a rule rather than as a curiosity.

A compass-constructible polygon has degree $2^a$, so its tower is $a$ square roots and its height is $a$. To get a large compass-constructible degree, the only thing available is more twos, and more twos is more steps — there is nothing else the degree can be made of.

A fold-constructible polygon has degree $2^a 3^b$, and a three costs the same one step that a two does while contributing more degree. So for a given height, the degrees a folder can reach are larger, and conversely a given degree is reached in fewer steps the more of it is threes.

Push that to the extreme and the statement is clean. The cheapest tower to degree $d$ has height $\log_3 d$ when $d$ is a power of three and $\log_2 d$ when it is a power of two, and $\log_2 d$ is larger by a factor of $\log 3 / \log 2 \approx 1.585$. A compass-only degree is about sixty per cent more expensive, in steps, than a fold-friendly degree of the same size, and the polygons are one place that ratio becomes visible because their degrees are set by the totient rather than chosen.

The lattice makes it immediate. Towers of a given height are the points of a diagonal; the degrees on that diagonal run from $2^h$ at one end to $3^h$ at the other; and the compass is confined to the single row $b = 0$, which is the cheapest end of every diagonal it touches and the only end it has.

What the totient supplies, and what it does not #

The inversion depends on the degrees being handed out rather than chosen, so it is worth looking at what the totient actually supplies.

$\varphi(n)/2$ is even far more often than it is odd. It is odd only when $n$ is 3, 4, 6 or a product involving a single factor of a Fermat-or-Pierpont kind that leaves a three — which is why the one-step polygons are so few and why they are exactly the ones with degree three. And the degrees that come out are dominated by powers of two, because $\varphi$ of a prime $p$ is $p-1$ and half the primes leave a large power of two behind.

So the polygons are a biased sample of the degrees, and the bias runs against the folder: the totient produces many powers of two and comparatively few clean multiples of three. That is the sense in which the result is stronger than it looks. Even on a sample drawn to favour the compass, the compass’s polygons are the expensive ones.

The bias also explains the gaps. The hendecagon’s degree is five, which is neither a two nor a three, so no tower reaches it and no amount of cleverness will — the same argument that puts a fifth root out of reach at any number of folds. The twenty-three-sided polygon’s degree is eleven, the twenty-five-sided’s is ten, and each of those failures is a prime the lattice does not contain.

What a height is and is not #

A height is not a number of creases. One extension step is one cube root or one square root, and performing either takes several folds — locating the references, making the alignment, transferring the length. What the height counts is the algebraic depth, which bounds the construction from below and does not describe it.

Nor is it an accuracy. What buys the reach costs the accuracy finds the conic axiom bringing a tail of badly conditioned constructions with it, so the cheap column of this table is the column with the error in it. A one-step heptagon and a three-step seventeen-gon are not comparable as things a hand does, and the essay makes no claim that they are.

And the shortest tower is shortest among towers, not among procedures. A construction might reach a polygon by a route whose intermediate quantities have larger degree, and nothing forbids it; the height is a lower bound on the number of extensions any tower to that number must take, achieved by the factorisation. A folder taking a longer route is doing something inefficient, not something the arithmetic forbids.

The heights are heights of towers over the rationals, not over each other. A folder constructing a seventeen-gon does not construct a heptagon on the way, and nothing in the table says one polygon’s tower contains another’s. The column is a depth from the ground in each case.

The list stops at forty sides. Nothing changes in the argument past it — the totient keeps supplying degrees and the factorisation keeps deciding — but the inversion is sharpest where the famous examples are, and past forty the polygons stop having names.

How the table was computed #

The totient is computed by trial division and the factorisation by repeated division, both directly, with no table of values and no special cases. A polygon’s row is four divisions and a comparison.

The verdict is derived from the degree rather than from the Pierpont condition. The two are equivalent — a polygon’s totient halves into twos and threes exactly when its odd prime factors are distinct Pierpont primes — and computing the degree and factorising it is the statement this essay is about, so it is the one the figure runs. The agreement with the Pierpont condition at every $n$ on the list is the check.

And the claim about the heptagon and the seventeen-gon is checked rather than described. The figure stops if the heptagon’s height is not one, if the seventeen-gon’s is not three, or if the compass reaches the first or fails to reach the second.

The same result without the polygons #

Strip the geometry out and the finding is a statement about two numbers that is easy to check and easy to miss.

Nine is larger than eight. A ninth root sits at $(0,2)$ on the lattice — two cube roots — and an eighth root sits at $(3,0)$ — three square roots. The larger number is on the shorter tower, and every instance of the inversion in this essay is that sentence dressed in a polygon.

Reachable is not cheap states it that way and stops, because at that point the essays here had no family of degrees to run it on. The polygons are that family: their degrees are supplied by the totient rather than chosen, they are famous individually, and the two orderings disagree on the two most famous of them.

What the polygons add beyond a worked example is the bias. A ninth root and an eighth root are a pair somebody picked; the polygons are a sample nobody picked, drawn by a function with its own preferences, and those preferences favour powers of two. So the inversion is being observed on a sample selected against it — which is the difference between an example and evidence, and it is why the column is worth printing for every $n$ rather than for the two.

What a step is made of #

The height counts extension steps and says nothing about what one is, which is worth a paragraph because the two ends of the lattice are not the same operation at the paper.

A square-root step is a bisection: fold a marked angle onto itself, or bring a point to a line. It uses alignments every folder performs without thinking and it needs references that are already on the sheet. A cube-root step is an alignment that brings two points onto two lines at once, which has to be found by sliding rather than by matching — the coincidence is not visible until it happens.

So the heptagon’s single step is the harder kind and the seventeen-gon’s three are the easy kind, and a fair account has to say so in the same breath as the count. The inversion is in the algebra and it is not obviously in the hand. What would settle it is a count of creases rather than of extensions, and the operation counts made for simultaneous folds are counts of what an axiom set contains rather than of what a construction spends.

The second qualification is the sheet. Most of what the plane specifies falls off the paper — seventy-two per cent of two rounds’ crossings on a square — and a tall tower needs more references than a short one, so the height is also a rough measure of how much of the sheet a construction has to still have unused when it gets there. That is a cost the algebra cannot see at all.

Still open: what the second column would cost a hand #

The height orders the polygons by algebraic depth and nothing here connects that to a folder.

A construction’s real cost is folds, and nobody has counted them for a polygon here. A heptagon by Beloch’s fold is a known sequence and so is a seventeen-gon by repeated bisection; laying the two side by side as fold counts would say whether the height ordering survives the translation or reverses again. There is a reason to expect it might reverse: a square root is a bisection, which is one fold from references a folder already has, and a cube root is a conic alignment, which needs a sliding coincidence and several references to set up. A step is not a step, and the arithmetic above has been treating them as one.

The other direction is the sample. The polygons are one family of degrees the totient happens to produce, and the inversion was measured on them. Whether it holds on a family drawn some other way — the degrees of the numbers a folder actually constructs, say, or of the roots of the equations this subject’s own theorems produce — is a question about whether the polygons are representative, and the bias noted above says they are not.

The habit worth carrying is about conditions that decide by factorising. A yes-or-no test that works by factorising something is carrying a cost inside it, and the cost is free to read. The Pierpont condition has been stated as a verdict for a century; the same factorisation it performs already contains the height, and nobody had printed the column.