Two circles for the price of one

The price is on the equation established that a compiled machine’s size belongs to the polynomial it was handed rather than to the curve that polynomial vanishes on, and ended by asking what a curve’s own cost is — the minimum over all its equations. It could not answer, because the evidence there was five hand-chosen multiplications, and five guesses are not a search.

The minimum can be searched for exactly, and the reason is that the quantity is linear in the right variables. Every curve is a sum of cosines, and the map from a polynomial’s coefficients to the amplitudes of those cosines is linear: expanding a sum is summing the expansions, because the whole of it is multiplying and adding Laurent series built monomial by monomial. So fix a curve’s equation $p$ and a multiplier degree, and every equation in the family $\{pq\}$ is a point of a linear space, with each frequency pair’s complex amplitude a linear function of the point.

A term of the compiled machine is a frequency pair whose amplitude is not nought, and what a curve costs is how many of those its own symmetry fails to cancel. The cheapest equation in the family is therefore the vector in that space with the most vanishing amplitudes, which is a sparsest-vector problem — and at these dimensions it can be solved by enumeration rather than approximated. The answer is a minimum, not a best-so-far.

The search, and what makes it exhaustive #

A multiplier of degree at most two has six coefficients, so the family is a six-dimensional space. A multiplier annihilating a set $Z$ of frequency pairs satisfies two homogeneous equations per pair — one for the real part of the amplitude and one for the imaginary — so $2|Z|$ equations in six unknowns.

Take $|Z| = 3$. That is six equations, generically of rank five, leaving a one-dimensional answer; and any sparser vector than the ones found this way would annihilate some three pairs and therefore appear among them. So enumerating the three-element subsets and solving each one exactly finds the sparsest vector, full stop. There are 165 such subsets for the circle, 220 for the line, 1,140 for the hyperbola and 2,300 for the lemniscate, and each is a singular-value decomposition of a six-by-six matrix.

What comes back is the same on every curve tried: the minimum is the curve’s own equation. The line costs two terms and its family’s cheapest is two; the circle one and one; the hyperbola three and three; the lemniscate five and five. A multiplier with nothing special about it costs between eleven and twenty-five on the same curves, so there is a great deal of room below the generic and none below the constant.

That is a stronger statement than the one it replaces, and it is worth being precise about how much stronger. It is not no multiple is cheaper — the family here is multiples by polynomials of degree at most two, and a curve’s equations include multiples of every degree. It is nothing in this family is cheaper, established by exhaustion rather than by sampling. The next degree up is not reachable the same way: a quartic multiplier has fifteen coefficients, the enumeration is over seven-element subsets of some thirty pairs, and two million solves is the wrong shape of computation to put behind a figure.

The one tie #

Three of the four curves have exactly one minimiser, which is the constant multiplier — the curve’s own equation, up to scale. The circle has two.

The second is $q = 1 - 0.391(x^2+y^2)$, and it reaches one term as well. It is not a constant, so $pq$ is a quartic rather than a conic, and it is not positive: it vanishes on the circle of squared radius $2.56$. So $pq = 0$ describes two circles — the original at $r^2 = 1.44$ and a second at $r^2 = 2.56$ — and the two of them together cost exactly what one of them costs.

That is a counterexample to something the field had not stated but had been assuming everywhere. The cost model treats a bigger curve as a bigger job; the whole shape of five bars for a line, four hundred for a quintic is a scale of increasing difficulty. Here is a curve strictly containing another, drawn by a machine of the same term count. Cost is not monotone in the curve, and no amount of care about which equation is used repairs that, because the cheap equation of the pair is the cheap one and there is no cheaper equation of the single circle to compare it against.

It is also the reason the search had to carry a second test. A sparsest-vector search that forgot to ask whether the multiplier vanishes anywhere real would have reported the circle’s minimum as attained by two multipliers and left it there — and one of them is not an equation of the circle at all. Distinguishing them is not a refinement of the search; it is what makes the search’s answer mean what it says.

The law, and it is about the arm #

The two radii are not a coincidence and they are not a property of circles. Their squared radii sum to four, which is $4\ell^2$ for the arm’s link length $\ell = 1$: the two circles are the ones whose squared radii are symmetric about half the arm’s greatest squared reach.

The derivation is three lines. The tip of a two-link arm of equal links sits at $r^2 = 2\ell^2(1 + \cos\theta)$ with $\theta = \alpha - \beta$, so a circle $r^2 = a$ is the linear factor $2\ell^2(1+u) - a$ in the variable $u = \cos\theta$, with its root at $u = a/2\ell^2 - 1$. A product of two circles is a quadratic in $u$, and a quadratic in $\cos\theta$ expands into a constant, a $\cos\theta$ and a $\cos 2\theta$ — two terms. It is an even function of $u$ exactly when its two roots are negatives of each other, which is $a + b = 4\ell^2$; and an even polynomial in $\cos\theta$ has only even harmonics, so the $\cos\theta$ term vanishes and one is left.

Sweeping the second radius confirms it and shows how narrow the condition is: the term count is two at every radius except one, where it is one. And the law tracks the arm rather than the circles — at $\ell = 0.8$ the two squared radii must sum to $2.56$, at $\ell = 1.5$ to $9$, and in each case the pair costs one term and a pair a little off it costs two.

So the free circle is an artefact of the mechanism, not of the geometry. The same two circles handed to any other construction have no reason to be cheap together, and what makes them cheap here is that the arm’s reach supplies a natural centre of symmetry in $\cos\theta$ and a pair placed symmetrically about it cancels an odd harmonic.

Six circles for three terms #

The parity argument does not stop at two, and neither does the measurement.

A stack of $k$ concentric circles is a polynomial of degree $k$ in $\cos\theta$. If its roots pair up under $u \mapsto -u$ — that is, if the circles come in pairs summing to $4\ell^2$ — the polynomial is even, only the even harmonics survive, and the term count is $\lceil k/2 \rceil$. Measured: two circles cost one term, four cost two, six cost three. The same circles with their partners moved off the law cost two, four and six.

The harmonics come out exactly as predicted. The paired stacks carry $\cos 2\theta$; $\cos 2\theta$ and $\cos 4\theta$; $\cos 2\theta$, $\cos 4\theta$ and $\cos 6\theta$. The unpaired ones carry every harmonic up to their degree. Nothing about that is a fit — the prediction is a parity statement about a polynomial, and the amplitudes it says are nought are nought to the expansion’s floor.

There is a limit to the measurement and it is arithmetic rather than mathematical. At eight circles the count comes back sixteen instead of four. The cancellation that removes the odd harmonics is between coefficients of a degree-sixteen polynomial running to several thousand, and in double precision the residues left behind are above the amplitude floor the expansion uses. The parity argument is a proof and the reading at eight circles is the arithmetic failing — which is floating point silently ceasing to be exact arithmetic — the failure this subject meets most often — arriving in a place where the answer is known in advance, and is therefore worth leaving in rather than tuning away.

What the free circles are paid for with #

Terms are one currency and the field has two others, and neither of them is free.

The machine for one circle has eleven bars; the pair has seventeen, four circles thirty-three and six circles sixty-one. So the second circle costs six bars even though it costs no terms, and the reason is in the field’s own arithmetic: the pair’s single surviving term is $\cos 2\theta$ rather than $\cos\theta$, and an angle has to be multiplied before it can be used — one doubling gadget, which is a reflector and its three bars, plus the offsets that go with it.

The working arc does something stranger, and it goes the other way. The single circle’s machine runs over 1.560 radians; the pair’s runs over 5.200, which is more than three times as far and nearly a full turn. That is the opposite of what padding an equation does, where lengthening it shortened the run in every case but one. Here the longer equation lengthens it, and the mechanism is presumably the same one in reverse — the gadget that computes a doubled angle has its flattening configurations somewhere else, and the somewhere else happens to be off the arc being marched — a gadget that flattens goes on turning and computes something else, so where its flattening sits is the whole of what an arc is. Past two circles the arc falls again, to 3.10 at four and 1.15 at six, which is the ordinary behaviour.

The machine traces one circle and not both. Over two hundred and forty solved positions its radius varies by $4 \times 10^{-13}$, and the radius it sits at is the outer one. That is not a failure of the compiler but a fact about mechanisms: the two circles are disjoint sets, a linkage moves continuously, and a continuous path cannot cross from one to the other. It is the same fact the branch census met from the other side, where four of eight closing assemblies put the tracing point somewhere that was not the curve at all and no closure residual could tell them apart. The free component is free and unreachable, which is the honest summary of what was bought.

What this does not settle #

The search is one degree of multiplier. Multipliers of degree at most two, on four curves. Degree four is fifteen coefficients and two million exact solves, which is not attempted; whether a quartic multiplier ever beats the constant is open and is the obvious next thing to ask. Padding by a positive quartic was measured there and is the one point of that space anybody has looked at.

A curve’s equations are not only its multiples. For an irreducible $p$ they are, up to the ideal’s structure, and every curve in the catalogue is irreducible — but the statement the cheapest equation of this curve has content only relative to that, and a reducible curve has equations that are not multiples of any single one of them.

Admissibility is measured, not proved. Whether a multiplier has a real zero is decided here by sampling it on a disc of radius six and asking whether it changes sign. That is enough to separate the two circle minimisers, which differ by a great deal; it is not a proof, and a multiplier that dips just below nought outside the sampled disc would be misclassified.

The free circles are concentric and centred on the pivot. Everything above uses $r^2$, which is the arm’s own natural variable. Whether an off-centre pair of circles, or two ellipses, admits the same cancellation is not measured, and there is no reason from the argument to expect it — the parity is about $\cos\theta$, and only a circle about the pivot is a function of $\cos\theta$ alone.

Nothing here is a lower bound on a curve’s cost. The search finds the minimum over a family. What a curve costs over all its equations remains unmeasured, and the tie on the circle shows that even the question is subtler than it looks, because the cheapest equation of a set need not be an equation of that set alone.

What the search says about the field’s own numbers #

The result that the curve’s own equation is minimal in its family is a relief, and it is worth being clear about how small a relief it is.

It means that the costs this field has published are not wrong in the way the padding measurement raised as possible: nobody has been quoting an inflated figure because an unlucky equation was chosen. On four curves, at one degree of multiplier, the obvious equation is the cheap one, and the field’s table of costs is a table of the cheapest equations available in that family.

It does not mean the costs are properties of the curves. The family searched is a thin slice of the equations a curve has, and the direction it does not reach is exactly the one where a surprise would live — a multiplier of higher degree, contributing enough cancellation to pay for itself. Nothing here rules that out, and the circle’s tie is a demonstration that the space of equations does contain coincidences at higher degree: a quartic that costs one term, where every generic quartic in the same family costs eleven.

There is also a reading of the search’s shape that matters more than its answer. The generic multiplier costs eleven to twenty-five terms on these curves and the best costs one to five, which is a factor of five to eleven between a multiplier chosen carelessly and the best available. So a designer handed an equation that came out of some other computation — an elimination, a resultant, a fit — is holding a polynomial with no reason to be near the minimum, and the machine compiled from it will be several times larger than it needs to be. The cheapest equation is not merely a theoretical minimum; it is the first thing worth computing before compiling anything.

Still open: which curves come free with which #

The circle’s tie is a single instance of a phenomenon that ought to have a shape: a component that can be added to a curve without adding a term. Nothing here says which curves have such a partner, or how many.

Its distinct argument would be that question asked as a search rather than as a discovery. The family of equations of degree at most $d$ is a linear space, the term count is a block-sparsity on it, and the sparsest vectors of that space are computable by the same enumeration used above — but run on the whole space rather than on the multiples of one $p$. Each sparse vector found is a curve that costs less than its degree suggests, and factoring it says whether it is one curve or several. Two things would come out of it. A census of the cheapest curves at each degree, which is the positive form of everything this field has measured negatively; and whether the free partners are always circles — the parity above needs a variable the arm supplies, and $\cos\theta$ is one of exactly two such variables, the other being $\cos(\alpha+\beta)$, whose level sets are not circles at all.