Five marks and the sixth

A general conic has six coefficients and they matter only up to a common scale, so it has five degrees of freedom. Five points impose five linear conditions. The arithmetic closes exactly: five marks on a photograph determine one conic, with nothing left over and no choice made anywhere.

That is a familiar counting argument and it is worth doing something with rather than reciting. What follows is what the count actually buys, what it does not, where it fails, and what changes the instant a sixth mark joins the fit.

The count, and what makes it exact #

Write the conic as

$$A x^2 + Bxy + Cy^2 + Dx + Ey + F = 0.$$

Multiplying every coefficient by the same number changes nothing, so the object is a point in a five-dimensional projective space. Each point $(x_i, y_i)$ that must lie on the conic gives one linear equation in the six coefficients. Five points give five equations, and a five-by-six homogeneous system generically has a one-dimensional null space — one conic, up to the scale that was never meant to matter.

The word “generically” is doing real work there and it is the subject of a later section. Take it at face value for now.

What is worth noticing immediately is that this is not a fit. There is no residual to minimise, no compromise between marks, and no sense in which some marks matter more than others. The conic through five points is as determined as the line through two.

The withheld mark #

A count is not evidence. The test is whether the conic through five marks passes through a sixth that had no say in it.

Six points are taken on a circle in the world, all six are projected, and five of them are handed to the solver. The sixth is measured against the answer afterwards, along the curve’s own normal so the number is a distance in pixels rather than an algebraic residual whose units depend on how the conic was scaled.

It comes back at 1.9e-13 px — the same arithmetic floor the dual construction on five tangents reaches, and the same one two circles in one picture compares against.

That number is arithmetic, not geometry, and saying so is the point. The five marks did not approximately determine the sixth; they determined it, and what is left is the accumulated rounding of a few dozen floating-point operations.

What five marks leave open #

The counting argument is exact and it is also modest, and the modesty is easy to lose.

Five marks determine the conic. They do not determine the circle — the two points a picture hides are what would. They do not say how big it was, how far away it was, or which way its plane was tilted; they do not say whether the thing photographed was a circle at all.

That last one deserves care, because the phrase “the conic through five marks on a photographed circle” quietly contains an assumption that the arithmetic never used. The solver was given five image points and returned the conic through them. It would have returned a conic through five marks on a photographed ellipse just as happily, and the two results are indistinguishable — which is what a projection destroys and why a circle needs its own recovery — which they must be, since an ellipse in the world and a circle in the world can produce the same picture.

So the honest statement of the result has two clauses. Five marks fix the picture of the circle exactly, and fix nothing about the circle. The second clause is not a caveat on the first; it is a different and equally sharp fact, and it has an essay of its own.

Determined is not well determined #

The withheld mark lands at $1.9\times10^{-13}$ px, and that is a statement about arithmetic on exact inputs. A reader with a photograph has marks with a pixel or two in them, and the question that decides whether the count is useful is how a pixel propagates — which turns out to depend almost entirely on one thing the count does not mention.

Where the five marks sit on the curve. Five marks spread around a whole ellipse pin it; five marks on a short arc pin it exactly and uselessly. Fitting a circle — three parameters — to an arc of angular half-extent $\beta$ with marks read to $\varepsilon$ gives a centre and radius uncertain by about $\varepsilon/\beta^{2}$, so the curve’s far side, which no mark touched, is out by that much. A general conic has five parameters instead of three and extrapolates more steeply still.

Put numbers on the circle case, which is the gentler one. A 30° arc gives $\beta = 0.26$ and an amplification of about 15; a 60° arc gives 3.6; a 90° arc gives 1.6. So a pixel of marking on a quarter-arc is fifteen pixels of error on the far side, and on a fully surrounded curve it is a pixel.

The count is right and it is silent about this. Five marks determine the conic at every arrangement that is not degenerate, and the conic they determine ranges from firmly pinned to wildly loose without the count changing at all. That is the same separation this collection keeps meeting between exactness and conditioning, arriving on the one construction where the exactness is a dimension count rather than a theorem.

Which degeneracies the count excludes #

The genericity the count assumes has a name and it is checkable. Five points fail to determine a conic when three of them are collinear: the only conics through such a set are degenerate ones — a pair of lines, with the line through the three as one of them — and the solver returns a degenerate conic rather than nothing. Four collinear leaves a whole pencil of them.

So the condition is no three collinear, which is the conic’s version of the no three collinear that a homography’s four points need, and it is the condition an arc of a genuine curve satisfies automatically. It is also the condition that fails silently: three nearly-collinear marks give a conic that is nearly degenerate, which is a hyperbola with two very close branches or an extremely elongated ellipse, and it will pass every algebraic residual test while being nothing like the curve. The determinant that vanishes is the one written out below, and its near-vanishing is the warning.

And where the sixth mark should go #

The test’s own strength depends on the same geometry, which is worth saying because the natural choice of sixth mark is the worst one.

A sixth mark between two of the five is interpolated, and interpolation between nearby constraints is nearly free — the conic passes close to it whatever the five were. A sixth mark on the far side of the curve is extrapolated, and it is where the $\varepsilon/\beta^{2}$ lives.

So the withheld mark should be placed as far from the five as the curve allows. A test with the sixth mark tucked among the others reports a small number and means very little; one with the sixth mark opposite reports the extrapolation, which is the quantity a reader is actually relying on when they use the fitted conic anywhere the marks are not.

That is the same design as every other withheld-point test in this collection — the four-point shadow prediction is worthless if the fifth point is chosen among the four, and the rolled print is measured on the sixteen anchors did not touch. The principle is one line: a prediction is only tested where the fit had no say, and how much it is tested is how far from the fit’s own reach the test point sits.

Where five marks are not enough #

The generic case is not the only case, and a solver that reports the exceptional one as an answer is worse than one that fails.

Five points determine a conic unless four of them are collinear. If four are on a line $\ell$, then every conic of the form “$\ell$ together with any line through the fifth point” passes through all five — a whole pencil of degenerate conics — and the system’s null space is two-dimensional rather than one.

The failure is not a large residual. The solver returns a conic; it just returns an arbitrary member of a family, chosen by whichever direction the numerics happened to favour. Nothing about the answer looks wrong.

The guard is on the eigenvalue gap: the ratio of the smallest singular value of the design matrix to the next smallest. In the good case that ratio is tiny — the null space is genuinely one-dimensional and its companion is not. As four points approach collinearity the two collapse together, and the ratio is the quantity that actually degrades.

That is the same shape of failure this site has recorded before, in a homography fitted to four nearly-collinear shadow marks: a confident answer with a small residual, from a system that had nothing to say. The tell is never the residual. It is always the conditioning.

The sixth mark, used #

Now change one thing. Instead of holding the sixth mark back, put it into the calculation.

The arithmetic changes character completely. Six equations in six unknowns, homogeneous, and generically the only solution is zero — which is to say there is no conic through six points, in general. The six points a photograph of a circle supplies are not general, so a solution does exist; but the problem is now overdetermined, and the object the solver returns is the smallest-eigenvector solution of a least-squares problem rather than a null vector.

Three things follow, and each is useful.

There is now a residual, and it means something. With five marks the residual is identically zero and carries no information. With six it is the amount by which the marks fail to lie on any one conic, which is a measurement of the marks rather than of the solver.

The residual is a test of the assumption. Six marks on a genuine circle give a residual at arithmetic noise. Six marks on something that is not a conic at all — a rounded rectangle, a hand-drawn oval, a curve with a flat on one side — give a residual that is not, and the size of it says how far from conic the thing was.

And more marks buy accuracy against noise. A mark located to within a pixel makes the five-point answer wrong by roughly a pixel’s worth. Twenty marks around the same curve average that down, and the improvement is the ordinary square-root one.

So five is the number at which the problem is determined, and it is almost never the number to use. This is the same trade the site meets in every recovery it makes: the minimal solution is what proves the theorem, and the overdetermined one is what a reader with a photograph should actually compute.

Why five points, projectively #

There is a second way to see the number five that explains why it is not four or six, and it is worth having because it makes the result look inevitable rather than lucky.

A conic in the projective plane is a symmetric $3\times3$ matrix up to scale: six entries, minus one for the scale, is five. That count is projectively invariant — it does not change under any projectivity — which is why the same number governs a circle, an ellipse, a hyperbola and a pair of lines. They are one object in five dimensions, and the affine names are labels for where the object sits relative to a chosen line.

The five conditions are equally projective. “This point lies on this conic” is an incidence, and incidences survive projection. So the entire statement — five marks, one conic, exactly — is a projective statement, and it can be made on a photograph without knowing anything about the camera. That is the property that makes it usable.

Five marks, and the pencil they can fail to leave #

It is worth drawing the degenerate case rather than only describing it, because what makes it dangerous is that it looks like the good case.

Put five marks on a photograph so that four of them fall on one straight edge — a kerb, a window mullion, the join between two paving slabs — and the fifth anywhere else. Every conic consisting of that straight line together with any line through the fifth mark passes through all five. There is a one-parameter family of answers and the solver returns one of them.

What the returned conic looks like is a pair of crossing lines, which is a perfectly legitimate conic and is exactly what the marks describe. The mistake is not in the arithmetic; it is in having asked five marks a question that needed a different five.

The practical rule follows from the failure rather than from taste: spread the marks around the curve. Four on one side and one on the other is nearly the degenerate case and inherits most of its conditioning, even though no four are exactly collinear. The eigenvalue gap says how nearly, and it is the number to look at before the residual.

Doing it on a real picture #

The construction is worth stating as a procedure, because it is one of the few things in this subject that a reader can carry out with a photograph and a straightedge and get an answer that is exactly right.

Mark five points on the curve. Anywhere on it. There is no need to find its ends, its axes or its centre — none of those survive projection anyway, so a method that needed them would be asking for something the picture does not contain.

Solve the five-by-six system. By hand this is a determinant; the conic through five points is the vanishing of a six-by-six determinant whose first row is $(x^2, xy, y^2, x, y, 1)$ and whose other five rows are the same expressions at the marks.

Then use it as a curve. The conic can be intersected with any line in the picture, tangents can be dropped to it from any point, and its pole–polar relation can be constructed with a straightedge — all of which are things about the world, transported through the picture, and none of which needed the camera.

The determinant, written once #

For a reader who wants the answer rather than the machinery, the conic through five marks is one determinant. Set

$$\begin{vmatrix} x^2 & xy & y^2 & x & y & 1 \\ x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1 \\ \vdots & & & & & \vdots \\ x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1 \end{vmatrix} = 0.$$

Expanding along the top row gives the six coefficients directly, as five-by-five minors of the marks’ own coordinates. It vanishes at each mark because the determinant then has two equal rows, which is the whole proof, and it degenerates to nothing when four marks are collinear because the corresponding rows become linearly dependent — the same failure the eigenvalue gap detects numerically, seen algebraically.

What the number five is really counting #

One last reading, and it is the one that connects this essay to the rest of the field.

Five is the dimension of the space of conics. Every constraint that cuts that space down by one is worth the same amount, and a point on the curve is only the most obvious kind. A tangent line is worth one too — a conic tangent to five given lines is equally determined, by exactly the same count in the dual plane. So is “passes through this point with this tangent direction”, which is worth two.

Which is why the circular points are worth so much. They are two points that every circle passes through, so knowing that the thing photographed was a circle is worth two of the five conditions before any mark is made — and three marks on a photographed circle determine the conic, provided the plane’s horizon is known.

That is not a saving in effort. It is the reason a photographed circle calibrates a camera and a photographed ellipse does not: the two conditions the roundness supplies are the two that carry the metric information, and they are on the horizon.