A mixed picture holds one more, and what it is holding is north

Two counts have been arrived at here by two quite different arguments and they came out the same. Five distances of six, and never more found that a flat picture of $n$ places can be exactly right about $2n-3$ of their distances: the picture has $2n$ coordinates, three of them buy nothing a distance can see, and the remainder is the budget. Five bearings of twelve, and only eight ways to draw them found the same $2n-3$ for bearings, from an argument with different pieces in it — a linear system rather than ruler and compasses, and a count of ways rather than a construction.

The obvious reading is that there is one budget of $2n-3$ and two currencies to spend it in. That reading is wrong, and the way it is wrong is worth a picture.

Three freedoms, and each kind sees two of them #

A picture of $n$ dots has $2n$ numbers in it, and four of them are not about the places at all. Slide the whole picture across the page and nothing about the arrangement changes: that is two. Turn it: one more. Enlarge it: one more again. Four freedoms that any measurement of how the places stand to each other must be blind to.

Except that the two kinds of measurement are not blind to the same ones.

A distance does not change when the picture is slid or turned, and does change when it is enlarged. So a distance-only problem is trying to determine $2n-3$ numbers — the $2n$ less the two of position and the one of turning — and $2n-3$ distances determine them.

A bearing does not change when the picture is slid or enlarged, and does change when it is turned, because a bearing is measured from north and north is fixed on the page. So a bearing-only problem is trying to determine $2n-3$ numbers as well — the $2n$ less the two of position and the one of size.

Both counts are $2n-3$ and the subtractions are different. Put one constraint of each kind into the same picture and the only freedom left invisible is position: two, not three. The budget is $2n-2$.

What independence is being tested #

Two of eighteen constraints can say almost the same thing, and a count of how many a picture can hold is a count of how many of them are different. The test used throughout is the one every such count reduces to: write each constraint as a function of the picture’s coordinates, take its gradient there, and ask how many of those gradient vectors are linearly independent. A set of $k$ constraints can be held exactly, near a picture, precisely when its $k$ gradients are independent — that is the implicit function theorem doing the whole of the work, and it is why a linear test answers a question about distances, which are not linear at all.

The gradients have a simple shape. A distance between two dots pushes them apart or together, so its gradient points along the segment joining them. A bearing turns one dot about the other, so its gradient points across it. That is the same orthogonality the two freedoms rest on, seen one level down: a constraint whose gradient lies along every segment can never resist a rotation, and one whose gradient lies across every segment can never resist a change of size.

It also says where the dependence at six constraints comes from. Three distances round one triangle are three gradients in the plane spanned by that triangle’s own motions, and the third adds nothing the first two had not already fixed once the scale is pinned; two opposite bearings of the same pair — London to New York and New York to London — have gradients that differ only in sign, so the pair contributes one constraint and not two, whatever the asymmetry between the two numbers. Neither is a near-miss. Both are exact dependencies, and they are why 41 per cent rather than all of the six-constraint choices work.

The enumeration says it without any theory. Four cities offer eighteen constraints: six distances and twelve one-way bearings, the two directions of a pair being different numbers on a sphere. Of the 18,564 ways of choosing six, 7,596 can be held exactly and every one of them mixes the kinds. The 924 sets of six bearings are all dependent. The single set of six distances is dependent. And the split that is five distances and one bearing works every time — all seventy-two of them.

The construction, in the order a hand would do it #

The five-distances-and-one-bearing case is the one worth carrying, because it is the only split that works every time and because it can be drawn without solving anything.

Place London. Place New York at its true distance from London, on the bearing London measures to it — one distance and one bearing, two constraints, and the pair is now fixed absolutely rather than up to a turn. Place Tokyo at its true distances from both, by intersecting two circles, which is two more constraints and two choices of side. Place Sydney the same way from any two already placed: two more. Six constraints, no equations, and every one of them exact.

What that construction makes visible is which constraint is doing which job. The bearing is spent entirely on the second dot and never appears again; every dot after the second is placed by distances alone, because once two dots are down the picture cannot turn. So the extra constraint is not distributed over the picture — it is one operation at the start, and it is the operation that turns a shape into a map.

All seventy-two sets of five distances and one bearing being independent is that construction counted: any five distances that pin the shape, plus any of the twelve bearings to fix the turn, and nothing about which five or which bearing can go wrong.

Seven is not rare, it is impossible #

The share of choices that can be held falls as more are asked for — 77 per cent at four, 62 at five, 41 at six — and then it does not taper. At seven constraints it is nought of 31,824, in every mixture from seven bearings to six distances and one bearing.

That is the shape of a counting bound rather than a difficulty. A picture of four places has eight coordinates, two of which no constraint here can reach, so six is what there is. The falling share below six is about dependence — two bearings of the same pair in opposite directions say almost the same thing, three distances round one triangle are not three independent facts about a picture with a fixed scale — and the cliff at seven is about there being nothing left to say.

The gain is one and it stays one. At eight places a picture can hold thirteen distances or thirteen bearings or fourteen of a mixture, and the fourteenth is the only extra there ever is, because it is a single freedom rather than a share of anything. The best flat picture is not a map turns on the ratio between $2n-3$ free numbers and $n(n-1)/2$ demands, and that ratio collapses as the set grows; the extra constraint here does not scale with either, so as a share of what a picture is asked to do it becomes negligible almost at once.

What the extra constraint actually buys #

Which leaves the question of what it is worth, and the honest answer is that it is worth exactly one thing and that thing is not accuracy.

Five distances fix the shape of a picture of four cities. They leave it free to turn: every rotation of it holds the same five distances exactly, and there is nothing in the five to prefer one. So the picture is not a map — a map has a north, and this has a shape.

The sixth constraint chooses the rotation. It costs nothing in distances, because turning a picture cannot change a length, and in that precise sense it is free. It is not free in bearings. The turn that holds London to New York exactly leaves the worst of the other eleven bearings 165.9 degrees out, where the turn chosen to serve all twelve would have left it 132.1 — a third of a right angle worse, bought by insisting on exactness in one place.

The mirror case is the same sentence with the words exchanged. Five bearings fix the shape and leave the picture free to grow; every size of it holds the same five bearings; the sixth constraint, a distance, chooses the size. It costs nothing in bearings and it is not free in distances — the size that holds London–New York exactly makes the worst of the six distances 88.1 per cent wrong where 78.8 was available.

So what a mixed picture holds that a pure one cannot is a placement in a freedom the pure problem had no opinion about, and the exchange is exactness on one constraint against accuracy across the rest of that kind. That is not a bargain and not a swindle. It is what an exact constraint always is, and the only novelty here is that this one is being paid for in a different currency from the five beside it.

A traverse is a mixed set, and always was #

None of this is exotic in practice, which is worth saying because the enumeration above looks like a puzzle. A surveyed traverse measures a length along each leg and an angle at each station, and a network adjustment takes both kinds into one solve — so every traverse ever run is a mixed constraint set, and the budget it works against is $2n-2$ rather than $2n-3$.

A surveyor would put it the other way round and be describing the same fact: a traverse needs a starting point and a starting bearing, and the bearing is what stops the whole figure swinging. A coordinate is the output of a solve is about what such a solve produces, and where the control points are about how the arrangement of the observations decides what can be recovered at all. The count here says what the orientation is worth in the currency of exactness: one constraint, not a fraction of one and not two, however large the network.

It also explains a piece of practice that looks like belt and braces. A traverse with distances alone closes on shape and can be laid down facing any way; one with angles alone closes on shape and can be laid down at any size. Measuring both is not redundancy against error — that is a separate argument, and what a closed figure cannot see is where it lives. It is the only way to reach two of the four freedoms at once.

The picture is still as wrong as it has to be #

Six exact constraints out of eighteen is a third of them, and it buys nothing at all for the other two-thirds. The picture is wrong about Tokyo–Sydney by 6.9 per cent and about the bearing from Tokyo to Sydney by 169 degrees, which is very nearly a reversal.

That is not a defect of the mixture. Four cities that cannot be drawn to scale settled that these four places have no exact flat picture at all, and every count since has been a count of how much of an impossible thing can be had exactly. Six is more than five, and six is not seven-eighteenths of a correct picture; it is six true statements about a drawing that is otherwise as wrong as any flat picture of four world cities must be. The error belongs to a few of the places is the measurement that says where that wrongness concentrates, and nothing here moves it.

What each number was checked against #

The pure counts must come back at 2n − 3. The control is that this count, which is made by a quite different calculation from either of the two earlier ones, must reproduce both: five distances and five bearings at four places, seven and seven at five, thirteen and thirteen at eight. It does. A construction that got either wrong would be measuring itself rather than the places.

No pure set of six may be independent. All 924 sets of six bearings and the single set of six distances are dependent, exhaustively.

Some mixed set of six must be, and must be solvable. 7,596 are independent, and the first such set found solves to a worst held error of 1.9 × 10⁻¹¹ — the constraints are not merely independent to a linearised test, they are met.

Nothing of seven may be independent, in any mixture. Nought of 31,824, enumerated rather than sampled.

And the largest independent set must not depend on the order it is found in. A greedy sweep over the constraints returns a maximum-size independent set whatever order it uses, which is a property of these gradients rather than a hope; eight seeded orders are run at every set size and all eight agree, at four, five and eight places.

What a count of independent sets does not say #

Independence is decided at one picture. The rank of the constraint gradients is computed at the best flat picture of the places. A set independent there is independent at almost every picture, since dependence is a condition of measure zero — but a set could in principle be dependent at some special configuration and not at this one, and nothing here surveys the configurations.

The count is local, and one ambiguity survives it. Independence of the gradients says the picture is pinned against every small movement. It says nothing about discrete alternatives, and there is one here that no constraint in this budget removes: five distances fix a shape up to reflection, and a bearing fixes a turn rather than a handedness, so a set of five distances and one bearing has two solutions — a picture and its mirror, each turned to hold the same arrow. Both hold all six constraints exactly. Choosing between them needs a seventh fact, and the budget above says there is no room for one; in practice the choice is made by knowing which way round the world goes, which is knowledge about the places rather than a measurement of them.

Independent is not the same as drawable. The bearing-only problem has a second obstruction beyond the count: a solution may put a place on the wrong side of the one whose bearing was meant to point at it, and the pair that cannot be held is not the one the geometry names is entirely about that. A mixed set is subject to the same failure, and the 7,596 counted here are counted for independence, not for having all their arrows forwards.

The mixture has been enumerated at four places only. The maximum at five and eight is found by a greedy sweep rather than by trying every subset, which is exact for the maximum and says nothing about how many sets reach it.

And the two costs are measured on one pair. The rotation figure spends its freedom on London to New York and the scale figure on the same pair. A different sixth constraint chooses a different turn and a different size, and how much the choice matters across all eighteen is not something two curves can say.

Still open: whether a third kind of measurement buys a fourth freedom #

The count here has an obvious continuation and it is not obvious how it ends. Four freedoms, two of them reachable — a distance reaches size and a bearing reaches turning — and position is reached by neither, because nothing about how places stand to one another can say where on the paper to put them.

But a measurement that is about the paper would. A stated position for one place — this dot goes here — is a constraint of a third kind, blind to none of the four freedoms, and two of them would pin the position that neither distances nor bearings can reach. The budget would then be $2n$, exhausting the picture entirely.

Whether that is a real extension or a piece of bookkeeping depends on something this argument cannot settle: whether fixing a dot on the paper is a statement about the places at all. A map’s registration to a sheet is a real operation with real error, and it is the operation every projection performs; whether the two numbers it spends belong in this budget, and whether a picture holding $2n$ constraints of three kinds behaves like the ones counted here or is a different problem wearing the same arithmetic, is a question four cities and two kinds cannot ask.