The price is on the equation

This field measures size. Five bars for a line, four hundred for a quintic counted nine machines and found the bar count growing like the fourth power of the degree; a circle costs one term found that what a curve costs is not its degree but how many frequency pairs its own symmetry fails to cancel. Both readings treat the cost as belonging to the curve — a line costs five bars — and the arithmetic behind them is careful about everything except that.

The demand is an equation, and the compiler is handed that equation rather than the curve. It expands $p(x,y)$ into a sum of cosines of whole-number combinations of the arm’s two angles, builds one gadget per term, and sums them head to tail. Nowhere in that process is there a step that asks what set $p$ vanishes on. So the cost it returns is a cost of the polynomial, and a curve does not determine its polynomial.

The demonstration takes one line. The factor $1 + x^2 + y^2$ is at least one at every real point of the plane, so $p \cdot (1 + x^2 + y^2) = 0$ holds at exactly the same real points as $p = 0$. Hand the compiler the second one and it returns a bigger machine for the same curve.

The padded machine is on the curve, not near it #

The first thing to establish is that this is not a trick played on the solver, because the obvious suspicion is that the bigger machine is merely nearly right and the comparison is between a mechanism and a worse mechanism.

It is not. A machine compiled from $pq$ satisfies $pq = 0$ at every position it reaches, exactly, to the solver’s own floor — that is what this field’s whole apparatus is for. And $q > 0$ everywhere real, so $pq = 0$ forces $p = 0$. The padded machine is therefore on the original curve for the same reason the unpadded one is: not approximately, not to a tolerance, but as a consequence of the constraints.

Measured rather than argued: at the points the padded machines actually reached, the original polynomial reads $1.2 \times 10^{-12}$ for the line, $1.0 \times 10^{-13}$ for the circle and $4.9 \times 10^{-14}$ for the hyperbola. The unpadded machines read $7.0 \times 10^{-14}$, $1.3 \times 10^{-11}$ and $1.3 \times 10^{-11}$ at theirs. The padded ones are, if anything, the tidier set, and the difference between the two columns is which position of which march happened to be worst rather than anything structural.

That measurement is worth pausing on, because the padded machine has never heard of $p$. Its terms are different cosines. Its bars are different lengths. The polynomial it was compiled from has a degree two higher and coefficients that share no factor with the original’s. It lands on the original curve because the original curve is where its own equation vanishes, and that is the only reason.

So the situation is exactly the one the section title says. There are two machines, of twenty and seventy-five bars, both exact, both tracing the same hyperbola, and the second is not worse at the job. It is only larger.

There is one asymmetry between them worth naming before it is mistaken for a defect. The padded machine reaches fewer positions — twenty-nine against ninety-six — so the column of numbers it is being judged on is shorter, and a shorter march has fewer chances to find its own worst case. That could in principle flatter it. It does not flatter it enough to matter: the unpadded hyperbola’s worst reading over ninety-six positions is $1.3 \times 10^{-11}$, and the padded one would have to be three orders of magnitude worse on the sixty-seven positions it never reached before the comparison reversed. What the short march is evidence of is something else entirely, and it is the subject of two sections below.

Five curves, three equations each #

Running the comparison across the catalogue gives the size of the effect, and it is not small.

A line goes from five bars to fifty and then to a hundred and fourteen. A circle goes from eleven to twenty-seven to forty-nine. A hyperbola goes twenty, seventy-five, a hundred and forty-four; a lemniscate fifty, a hundred and fourteen, a hundred and ninety-two; a cubic a hundred and one, two hundred and nine, three hundred and fifty. Fifteen machines, five curves, and every curve in the set is the same curve in all three of its rows.

The reading is unavoidable. The field’s headline numbers are upper bounds attached to particular representations. “A line costs five bars” is the claim that this equation of this line compiles to five bars. What a line costs, taken over all its equations, is a minimum nobody here has computed and which the numbers above only bound from above.

That is not a complaint about the measurements. They are the right measurements of the thing they measure, and an upper bound is what a construction gives — the field has been explicit from the start that the compiler is one way of building a machine rather than the cheapest, and what an approximation costs against what exactness costs is a comparison between constructions throughout. What changes here is the attribution. A statement like the cost grows like the fourth power of the degree reads as a fact about polynomials of that degree, and it survives intact; a statement like a lemniscate is expensive reads as a fact about lemniscates, and it does not.

There is a sharper version. Since a polynomial and its multiple have different degrees, any curve here has equations of arbitrarily high degree — so the curve has machines of arbitrarily many bars, and a maximum cost does not exist. Every cost quoted in this field is therefore one end of an unbounded interval, and the interesting end, the minimum, is the one not measured.

A line, at fifty bars #

The case worth looking at rather than tabulating is the cheapest one, because the penalty there is the largest and the picture is legible.

Five bars for a line is the field’s smallest machine and its clearest result: the arm, one reflector, one mean and the closure, and the traced point runs along a straight line exactly. Multiply that line’s equation by the factor and the compiler returns fifty bars — a machine with sixteen translators in it, nearly all of them carrying a direction from one place to another so that a term which sums to nothing useful can be added to a chain.

That is the field’s own transport result arriving from a new direction. A parallelogram carries an angle and only so far established that transport grows quadratically in the number of terms while the arithmetic grows like $d \log d$, so most of a large machine is carrying. What the padding shows is that the carrying can be made to dominate a machine whose curve needs none of it — the line still has two frequency pairs’ worth of content, and the other three are pure overhead introduced by an equation nobody had to write that way.

It also makes the measurement’s independence from the drawing plain. Dragging the padding through its three stops changes the mechanism from five bars to fifty to a hundred and fourteen while the thin line it is drawn against does not move by a pixel, because it is the same line. That is the whole argument in one control.

The penalty is not the factor’s either #

The next question is whether the padding at least has a fixed price — whether multiplying by $1+x^2+y^2$ costs some constant factor in bars whatever it is applied to, which would make the effect a nuisance to be divided out rather than a hole in the cost model.

It does not. One copy of the factor costs a factor of 10.00 on the line, 3.75 on the hyperbola, 2.45 on the circle and 2.07 on the cubic. The cheapest curve in the catalogue pays the most, and it pays five times what the most expensive curve pays.

The reason is the field’s own central mechanism seen from an unfamiliar side. A curve’s cost is the number of frequency pairs its symmetry fails to cancel: the expansion of $p$ produces a great many terms $\cos(m\alpha + n\beta + \varphi)$ and most of them cancel against each other, leaving a few. Every curve is a sum of cosines, and multiplying $p$ by $q$ multiplies out into a new and larger set of frequency pairs, and how many of those cancel depends on how the symmetries of $p$ and $q$ interact. A circle is the most symmetric object in the catalogue and cancels almost everything, before and after; a line is only symmetric under a reflection and cancels very little of what the multiplication produces.

So the penalty is not a property of the multiplier, is not a property of the curve on its own, and is not a function of the degree — it is a property of the pair. A cost model with that character cannot be repaired by normalising the degree or by dividing out a constant, which is what makes it a hole rather than a correction.

The cost that is not counted in bars #

The bar count is the field’s currency and it is not the quantity that decides whether a machine works.

Marching each padded machine and asking how far it can be driven before a position fails to solve gives a different and more brutal picture. The hyperbola runs over 1.68 radians unpadded, 0.51 with one factor, and nothing at all with two: a hundred and forty-four bars, every length computed, and not one position solvable. The cubic does the same — 1.63, then 0.56, then nothing. The circle degrades gently, 1.56 to 1.56 to 1.14. The line does something else again, going up from 2.26 to 2.80 before collapsing to 0.80, which is worth recording as the one place the pattern is not monotone and is not explained here.

The mechanism behind the collapse is already in the field. A gadget stops being the function it was built to be at the configuration where its rhombus flattens: the machine goes on turning, every bar stays the length it was, the closure residual stays at the solver’s floor, and the thing it computes is a different function. Every gadget has such a configuration, so a machine with three times as many gadgets has three times as many chances for one of them to fall inside the arc being marched. The machine itself does not notice: thirty-five equations go on being satisfied and the polynomial at the traced point is the only quantity that knows. Lengthening the equation shortens the run.

That is the sense in which padding is not free even though it buys exactness for nothing. The bars are a cost to a manufacturer. The arc is a cost to a reader: a machine with no arc draws no curve, and a hundred and forty-four bars that cannot be assembled anywhere is not a slower way to trace a hyperbola but a failure to trace one.

It also sharpens what what universality is worth found. The existence theorem says a linkage exists for every algebraic curve; the four-hundred-bar quintic says how many bars one construction needs; and this says that the arc — which the theorem does not mention at all — shrinks as the construction grows, in a way that depends on the equation rather than on the curve. Three facts, and only the first is in the theorem.

Nothing tried is cheaper #

The obvious hope is that the effect runs only one way — that padding always costs and there is no equation of a curve cheaper than the natural one, so the natural one is the minimum and the cost model is saved by a convention.

Over the multiplications tried, that hope holds and proves nothing. Multiplying each catalogue curve by each of five factors — $1+x^2+y^2$, $2+x^2+y^2$, $1+x^2$, $1+y^2$, $1+x^4$ — never produces fewer terms than the equation it multiplies. A circle goes from one term to two at best; a quintic from eighteen to twenty-four.

Five multiplications are not a search. The equations of a curve are every polynomial vanishing on it, which for an irreducible curve is every multiple of its minimal polynomial — so a search over multiples is the right family, and five of them is a vanishingly small sample of it. Worse, for a reducible or non-real-radical situation the minimal polynomial need not be the natural one at all, and the catalogue here is chosen from curves whose natural equation is already the obvious generator. Scaling the coefficients is the one family of alternative equations the field had already measured, and it is the one that changes nothing.

What can be said is narrow and worth saying anyway: nothing observed here suggests that a redundant factor ever cancels more than it creates, and the reason to expect it might is real. The expansion’s cancellation is exactly a coincidence between frequency pairs, and a multiplier chosen to make coincidences is not obviously impossible. Whether some curve has a cheaper equation of higher degree is open, and it is the question a cost model for curves would have to settle first.

What this does not settle #

The factor is harmless only over the reals. $1 + x^2 + y^2 = 0$ is a perfectly good complex conic, so $pq$ is a reducible curve with a component the machine can never reach, and the machine’s branch structure is correspondingly different. Nothing here counts its assembly modes or asks whether the extra component contributes spurious branches of the kind bracing removes.

No minimum is computed. The whole point is that the minimum over equations is the quantity the cost model wants, and it is not measured here for any curve — not even for the circle, where one term looks unimprovable and is not proved to be.

Five curves and two paddings. The table is fifteen machines. The effect is large enough at that sample that its existence is not in doubt; its behaviour at higher paddings and on curves with other symmetries is not charted.

The arc is measured by marching. A machine is driven from its seed and a position counts when its Newton solve converges, so the arc reported is the arc this march found from this start. A machine with a short reported arc may have a longer one elsewhere in its configuration space, and the arc has been predicted as well as found only for the unpadded machines.

Nothing is claimed about compilers in general. A different construction — not this one — might charge for something closer to the curve. What is shown is that this compiler charges for the equation, and that the field’s numbers inherit that.

Still open: the cheapest equation of a curve #

The measurement above turns a statement the field has been making into a question it has not asked. What a curve costs is the minimum of the compiler’s bar count over every polynomial vanishing on it, and no instrument here computes a minimum over an infinite set.

Its distinct argument would be a lower bound rather than a search. The compiler’s cost is set by the number of surviving frequency pairs in the expansion, and that number is a rank: the expansion is linear in the polynomial’s coefficients, so the set of polynomials of degree at most $d$ vanishing on a given curve is a linear subspace, and the cheapest equation of degree at most $d$ is the vector in that subspace with the sparsest expansion. That is a sparsest-vector problem on an explicit subspace, and although the general version is hard, the dimensions here are small enough to settle by exhaustion at low degree. Two things would come out of it. Whether the natural equation of any catalogue curve is beaten at its own degree — which would mean the field has been quoting a cost that is not even locally minimal; and whether a higher degree ever wins, which is the question the five multiples above could only fail to answer.