Two numbers about the paper buy nothing about the places

A mixed picture holds one more, and it is north counted what a flat picture of $n$ places can hold, and the count came out of a single idea. A picture has four freedoms — it can be moved two ways, turned, and scaled — and a constraint holds only what its own blindness leaves:

The pattern has an obvious continuation. Position is reached by neither kind, and for a reason rather than by accident: nothing about how places stand to one another can say where on the paper to put them. But a constraint that is about the paper would reach it.

A pin — this dot goes here — is two numbers and is blind to none of the four freedoms. Two of them ought to exhaust the picture at $2n$, and they do. That part is bookkeeping, and if it were all this measurement had to say it would not be worth making.

What is worth making is the other question the earlier essay asked and could not settle: whether a picture holding $2n$ constraints of three kinds behaves like the ones holding $2n-2$ of two, or is a different problem wearing the same arithmetic.

The count, which is the easy half #

A pin’s row in the constraint Jacobian has a single 1 in it. Every other row here is a difference between two places, which is what makes it blind to a shift of the whole picture; a row with one entry is not.

At four places, over every subset of the eighteen place constraints and the two pins:

• no pin — independent sets stop at 6, which is $2n-2$ and is the earlier result;
• one pin — they stop at 7, because one pin leaves the picture free to slide along a line;
• both pins — they reach 8, which is $2n$, and there are 7,596 of them;
• at 9 there is nothing, of any mixture, which is the statement that a picture of four places has eight numbers in it and they have all been spent.

The finding is in what did not change #

A largest pinned picture holds eight constraints, of which two are pins — so six are about the places. A largest unpinned picture holds six, all about the places.

The same six. Not the same number by coincidence: there are 7,596 largest pinned pictures and 7,596 largest unpinned ones, and each of the first is one of the second with the two pins added. Nothing was gained, exchanged or reshuffled.

The pair-by-pair accounting says the same thing from the direction the pair that cannot be held is not the one the geometry names approached it from. Every pair of the four cities is held by exactly 80.0 per cent of largest pictures, with pins and without, to every digit. The equality across the six pairs as well as across the two columns is a fact about this particular set of cities rather than a theorem — the parameters are not independent is the general warning about reading an equality of shares as an absence of structure — and what the comparison establishes is the second equality, not the first. Pinning a place to the sheet has no opinion about which places are hard to tie together — which is what one would want it to mean for a constraint to be about the paper, and is not something the count could have told anyone.

The constraint system is a direct sum #

The strongest form of the statement is a prediction that could have failed at any of twelve places and does not.

If the pins and the place constraints do not interact at all, then an independent set of size $k$ holding $p$ pins is exactly a choice of $p$ pins from two together with an independent set of $k-p$ place constraints. Its count must therefore be $\binom{2}{p}$ times the number of independent place sets of size $k-p$ — for every size and every mixture, not merely at the top.

The independent place sets of four cities number 1, 18, 147, 720, 2,355, 5,274, 7,596 at sizes 0 to 6, and nothing at 7. Every one of the twelve pinned counts is one of those numbers times one or two:

• 8 constraints, both pins: $1 \times 7{,}596$;
• 7, one pin: $2 \times 7{,}596 = 15{,}192$;
• 7, both: $1 \times 5{,}274$;
• 6, one: $2 \times 5{,}274 = 10{,}548$;
• 5, both: $1 \times 720$.

That is a direct sum. The picture’s eight numbers are two separate accounts — six that the geography can spend and two that only the sheet can — and there is no transaction between them.

What a pin is, on a sheet somebody drew #

The word is doing real work and it is worth saying what it stands for, because a reader may reasonably suspect a pin of being an invention.

It is a control point: a place on the ground whose position on the page is stated, in the page’s own units. Every map has them. A graticule crossing is one, a corner tick is one, a registration mark on a printing plate is one, and the entire method of a map with no graticule is about what can be done when there are none. A surveyor setting out a plan pins it to the sheet by two marks and then fills in the rest from measurements between points; a compiler fitting a new sheet pins it at two corners of the old one.

Two pins is not an arbitrary number either. It is what it takes: one pin fixes the picture against translation and leaves it free to turn and scale about that point, which is why one-pin sets stop at $2n-1$ in the count above. A second fixes the rest.

So the arithmetic above is a statement about ordinary practice and not about a contrived third kind. A sheet with two control points and $2n-2$ measurements between places is a fully determined drawing; a sheet with four control points and $2n-4$ measurements is also fully determined, and it is a worse drawing of the geography, because two of its four control points have gone into constraining the sheet a second time where they could have gone into the places. The direct sum says exactly how much worse: two constraints’ worth, with no partial credit.

Why it had to be a direct sum #

The twelve agreeing counts are a measurement, and there is an argument underneath them that says the measurement could not have come out otherwise. It is worth setting out, because it also predicts the table rather than fitting it.

A budget is $2n$ less the dimension of a group. Write the group of transformations that leave every constraint of a given kind unchanged. For distances it is the isometries of the plane — two translations and a rotation, dimension 3 — so the constraints can never determine more than $2n - 3$ of the $2n$ coordinates. For bearings it is translations and scalings, also dimension 3. For the two together it is the intersection, which is translations alone: dimension 2, budget $2n-2$. For a pin it is the trivial group, dimension 0, budget $2n$.

Every number in the table above is that subtraction, and nothing about the places enters it. That is already the first half of the direct sum: the pins do not raise the budget by being extra constraints, they raise it by removing the invariance the place constraints were subject to.

The second half follows from which invariance. The place constraints all annihilate the two translation directions and nothing else; the pins are supported on those same two directions and nowhere else. Two sets of rows whose row spaces meet only at zero contribute their ranks independently, which is exactly the statement that the counts factorise. So the measurement is a check that the row spaces really are complementary at this configuration — which is not guaranteed by the argument, because a degenerate arrangement could make a place constraint’s row lie partly along a translation direction.

That reading also says what would break it, and it is worth naming. A pin is complementary to the place constraints only because it fixes a place rather than a direction. A constraint of the form this line runs east–west on the sheet would remove the rotation instead, and rotation is not a freedom the distance constraints are blind to — a distance-only picture is already unable to see it. Such a constraint would raise the bearing budget and not the distance budget, and the accounts would share a wall.

The table extends downwards, and the pattern is the group #

Once the budget is read as a subtraction the table has rows nobody asked for, and they check the reading.

A ratio of two distances is blind to position, turning and size — scaling the picture multiplies both distances and leaves the ratio alone — so its group has dimension 4 and its budget is $2n - 4$. An angle at a place, being a difference of two bearings, is blind to the same three the bearings are blind to and also to turning, since turning the picture turns both directions equally: dimension 4 again, budget $2n-4$.

So the four kinds sort into three budgets, and the sorting is by the dimension of a group rather than by anything about what the quantity measures. A ratio and an angle are as different as two geometric statements can be and they cost the same, because they are invariant under groups of the same size.

An angle is a difference, and the difference doubles the error prices what a difference of two directions does to an error, and it is the same object seen from the other end: the invariance that makes an angle cheap in this budget is the invariance that makes it behave badly under a substituted sphere. A quantity blind to a transformation is a quantity that cannot see an error along it either.

What a separate account is worth to somebody building one #

The practical reading is short and it is the reason the distinction matters outside a rank calculation.

Pins cannot make up for missing geography. A picture short of one distance is short of one distance, and no amount of registration to the sheet repairs it: the six place constraints a largest picture holds are six whether or not the picture is pinned. What another common point buys measures the converse for a datum transformation — an extra observation that is about the places buys real information — and the contrast is the point. An observation about the paper is not a weak observation about the places; it is not an observation about the places at all.

And geography cannot make up for a missing registration. A picture holding every one of its $2n-2$ place constraints exactly is still free to sit anywhere on the sheet, and no further distance or bearing will pin it. A grid has an origin that is not there reaches the same fact from the practical side: a grid’s false origin is a decision, not a measurement, precisely because nothing measurable determines it.

The two together give the rule a compiler of maps already works by without stating it. The shape of a sheet and its registration are established by different operations, from different sources, and a shortfall in one is never repaired by a surplus in the other — which is why the sheet moved before it was measured can find paper shrinkage absorbed into the fitted parameters while the shape information survives, and why a map with no graticule is a hard problem rather than a slightly harder one. Take the registration away and the shape is still there; take the shape away and no registration recovers it.

Five places, sampled, and the same shape #

At five places the budget is again reached only with both pins, at exactly $2n = 10$, with nothing at 11. And the three curves are one curve moved along: the share at size $2n$ with both pins is 0.3225 and the share at size $2n-2$ with none is 0.3235.

The worst departure from the diagonal is 0.005 on twenty thousand draws, which is sampling noise rather than structure. So the direct sum is not a four-place coincidence.

What each number was checked against #

Two pins must take the budget to exactly $2n$. At four places the largest independent set is 8. A larger one would mean a pin was reaching something the four freedoms do not contain; a smaller one would mean a pin was blind to something after all.

And $2n$ must be reachable only with both pins. One pin leaves the picture free to slide along a line, so a set of $2n$ holding one pin must be dependent. None is.

No set of $2n-1$ place constraints may be independent, which is the earlier measurement recovered inside this one rather than assumed. Zero are.

A largest picture must hold the same number of place constraints with pins as without. Six against six. This is the control the whole essay turns on: if two pins had bought two more distances, the budget would be a single currency and the third kind would be the first two under another name.

The counts must factorise, at every size and every mixture. Twelve of twelve.

Adding pins must not reorder which pairs are easiest to hold. All fifteen orderings are preserved — trivially here, because all six shares are equal, and the check is kept because on a set where they are not equal it is the one that would catch a pin quietly favouring a place.

And the shares at five places must shift by exactly the number of pins. The worst disagreement is 0.005.

Where the model stops #

The rank is taken at one configuration. Independence here is the rank of the Jacobian at the best flat picture of the four cities, which is the generic answer at that point and not a statement about every configuration. A degenerate arrangement — three places on a line — has a smaller rank for some sets, and five bearings of twelve, and only eight ways to draw them meets that distinction for bearings alone.

Independent is not drawable. A set can be independent at the linearisation and still have no picture that holds it exactly, which is the difference between counting and solving, and it is why the obstruction the pair that cannot be held is not the one the geometry names found does not appear in these counts. Everything above is about the count; nothing above says a pinned picture can be drawn.

A pin is exact. A real registration to a sheet has error, and a pin with a tolerance is a different object: it constrains a region rather than a point, and two of them would leave the picture nearly rather than exactly fixed. That is a question about a weighted fit rather than about a rank, and nothing here prices it.

Four places and five. The exhaustive count is $\binom{20}{8}$ at four and $\binom{32}{10}$ at five; six places would be $\binom{47}{12}$, which is four hundred billion subsets. The direct sum is checked exactly at four, by sampling at five, and claimed nowhere else.

Still open: whether the two accounts are ever really separate #

The measurement settles the arithmetic and sharpens the question the earlier essay was actually asking, which was not about counting at all.

A map’s registration to a sheet is a real operation with real error — it is the operation every projection performs, and what a coordinate refers to sets out what it costs. The direct sum says the registration’s two numbers and the geography’s $2n-2$ do not trade against each other in a rank sense. It does not say they do not trade at all.

They plainly do when the constraints are inexact. A picture fitted to noisy distances and noisy pins has one residual, and the fit will move places to satisfy pins and move the registration to satisfy places; the two accounts share a denominator the moment anything has a tolerance. Whether the direct sum survives that — whether the covariance of a fitted pinned picture blocks into the same two pieces, or whether the pins’ error leaks into the places’ — is a question a rank cannot answer, and it is the one worth asking next.