The pair that cannot be held is not the one the geometry names

Five bearings of twelve, and only eight ways to draw them enumerated every way of asking a flat picture of four world cities to hold five of their bearings exactly. Of the 192 choices — six ways of leaving a pair out, thirty-two ways of choosing a direction on each of the five kept — eight can be drawn with every held arrow pointing forwards, and all eight leave London and New York unheld.

That was found by trying everything, and it ended on an admission. For sixteen cities there are a hundred and twenty pairs and no enumeration is possible, so the question became whether the pairs a drawable picture must avoid could be named from the geometry instead: a quantity computed from the places that says, before any picture is attempted, which pairs will not survive.

There is an obvious candidate, and it is the quantity the whole problem rests on. A flat picture has one direction between two places, and the Earth has two measured the bearing asymmetry — how far a pair’s bearing there and its bearing back are from being reverses of each other — and showed that it sets the floor on what any flat picture can do. New York to Tokyo and back miss by 128 degrees. It is the reason the problem is hard, so it ought to be the reason a particular pair is impossible.

It is not. It predicts something else, and predicts it cleanly.

What the count leaves room for #

The arithmetic is worth restating because everything below is a consequence of it. A flat picture of $n$ places has $2n-3$ numbers a bearing can be spent on — two coordinates each, less the two of a translation and one of a rotation, with scale free because a bearing does not see it. Each pair offers two bearings, one measured at each end, and they are different numbers on a sphere. So a choice is a set of $2n-3$ pairs together with a direction on each: $\binom{n(n-1)/2}{2n-3} \times 2^{2n-3}$ of them, which is 192 at four places and 15,360 at five.

That count is the same one five distances of six, and never more arrives at for distances, and the coincidence is not one. A picture of $n$ dots has $2n$ numbers in it; a distance is blind to where the picture sits and how it is turned, which removes three, and a bearing is blind to where it sits and how large it is, which removes three as well. Two quite different quantities, counted on the same page, buy the same $2n-3$.

They differ in what happens next, and the difference is the whole of this. A set of $2n-3$ distances that a picture can hold exactly has a solution reached by ruler and compasses, one place at a time, and the only thing that can go wrong is a reflection. A set of $2n-3$ bearings has a solution reached by solving linear equations, and the thing that goes wrong is subtler.

Almost none of those choices produces a picture anybody would accept. The equations that place the dots hold a line’s slope, and a line has the same slope pointing either way, so a solution may put a place on the wrong side of the one whose bearing was supposed to point at it. Such a picture satisfies every equation and has an arrow drawn backwards, which is not a picture holding that bearing at all.

At four places, five of the six pairs are in all eight drawable choices and one is in none. The one is London–New York. Its bearing asymmetry is 57.1 degrees, which is fourth of the six — behind New York–Tokyo at 127.9, London–Tokyo at 124.6 and London–Sydney at 78.5, and ahead only of New York–Sydney at 20.4 and Tokyo–Sydney at 0.2.

So the first answer is negative and it is clean: the pair that has to be let go is not the pair whose two bearings disagree most, nor the pair whose two bearings agree most. It is the shortest pair in the set, at 5,570 kilometres, which is a fact about these four cities and not obviously a rule about anything — and four cities that cannot be drawn to scale chose this set for its spread rather than for any property of London and New York.

It is worth recording what else it is not, because two other candidates suggest themselves and both fail on the same table. It is not the pair that the best distance picture gets worst — how wrong a flat picture has to be and the essays after it put that burden on the pairs involving Sydney, and London–New York involves neither. And it is not a pair the enumeration singles out weakly: five pairs are in eight choices of eight and one is in nought of eight, with nothing between, so whatever decides it decides it absolutely.

Why no count can decide it #

Before looking for a geometric predictor it is worth ruling out a combinatorial one, because a combinatorial answer would be far stronger and there is a well-known theory that supplies them.

A set of direction constraints determines a picture up to scale exactly when its graph is tight in a countable sense: $2n-3$ edges in total, and no $k$ of the places spanning more than $2k-3$ of them. That condition is what separates a picture that is pinned from one with a floppy part and a doubly-constrained part, and it is decided by the graph alone — no coordinates enter it.

It cannot be the discriminator here, and the reason is immediate. At four places, every one of the six ways of leaving a pair out produces the same graph: four vertices and five edges, which is the complete graph on four vertices with one edge removed, and all six removals give isomorphic results. A condition that reads only the graph sees six identical objects and must give them all the same answer. The enumeration gives one of them eight drawable choices and the other five none.

So whatever decides it is in the numbers rather than in the structure, and it has to be looked for among quantities computed from the places themselves. That is why bearing asymmetry was the candidate: it is the only quantity in this subject that is a property of a pair, is invisible to any flat picture, and is large.

What asymmetry does predict #

The eight drawable choices all hold the same five pairs. They differ only in direction, and they do not differ freely.

Two of the five arrows are forced. London to Tokyo points that way in all eight drawable choices, and so does New York to Tokyo; reverse either and no picture can be drawn. The other three appear four times each way, which is exactly the factor of eight: $2^3$ free directions on an otherwise fixed set of pairs.

The two forced pairs are the two of largest bearing asymmetry — 124.6 and 127.9 degrees — and the largest asymmetry among the three free ones is 78.5. The ordering is not approximate: there is no pair between.

Extended to five cities, where seven of ten pairs are held and the one-sidedness can take intermediate values, the separation survives. Every pair below 78.5 degrees of asymmetry is free in both directions in every drawable choice that holds it. Every pair with any one-sidedness at all is at 78.5 degrees or above. New York–Tokyo, the largest asymmetry in the set, is the most one-sided at 0.71; London–Tokyo and Sydney–Cape Town, next in asymmetry, are at 0.16; London–Sydney, at 78.5 degrees, is at 0.077 among five cities and free among four.

The mechanism is not mysterious once the two quantities are put side by side. A pair’s asymmetry is the amount by which its two bearings fail to be reverses of one another, and the choice between the two directions is the choice between two different equations that differ by exactly that amount. Where the two equations are nearly the same — an asymmetry of a fifth of a degree, as between Tokyo and Sydney — the picture barely notices which was used, and both work. Where they differ by more than two right angles, the two equations describe genuinely different constraints, and one of them is consistent with the rest of the picture while the other is not.

So bearing asymmetry has an exact job and it is not the one it was proposed for. It says how much a pair’s direction matters. It says nothing about whether the pair can be held.

That is a useful thing to know rather than a consolation. A picture built by hand — choosing pairs and directions in some order and checking as it goes — spends most of its effort on directions, since a set of $2n-3$ pairs comes with $2^{2n-3}$ ways of orienting them and the enumerations here find that only a small share work. Knowing that the low-asymmetry pairs will take either direction, and that the high-asymmetry ones will not, cuts the search by whatever share of the pairs falls below the threshold: six of the ten pairs of five cities, and four of the six pairs of four.

At five places there is no forbidden pair #

The question that prompted all this presupposed that a forbidden pair exists — that a picture holding twenty-nine of sixteen cities’ bearings leaves ninety-one out, and some of those ninety-one are forced. One place further up the count, that presupposition fails.

Of the ten pairs of five cities, every single one is held by some drawable picture. The shares run from 37 per cent — New York–Tokyo and New York–Sydney, each in 56 of the 152 — up to 100 per cent for London–Tokyo and Sydney–Cape Town, which are in every drawable choice there is.

That reverses the shape of the question. At five places the interesting pairs are not the forbidden ones, because there are none; they are the mandatory ones, which no drawable picture can do without. And the mandatory pairs are not the extreme ones on any of the axes available: London–Tokyo is second in asymmetry and Sydney–Cape Town third, while New York–Tokyo, first in asymmetry, is one of the two least held.

The four-place result now reads differently. Five constraints on six pairs leaves one pair out, and with so little slack the enumeration finds exactly one arrangement that works — so “the pair no picture holds” was a statement about a system with one degree of freedom in its choice, not a property of London and New York. Add one city and the choice has three degrees of freedom, and every pair finds some picture to live in. The error belongs to a few of the places found a genuine such property for distances — Sydney alone holds a picture of sixteen cities at its worst, and leaving it out halves the error — and the natural expectation was a counterpart for bearings. There is not one. The counterpart is about arrows, not about places.

Why it gets harder with more places, when it should get easier #

One more measurement, because the shape of the difficulty is now visible and it is the opposite of what the counting suggests.

A picture of four places is asked to hold five of six bearings, five-sixths of them. A picture of eight places is asked to hold thirteen of twenty-eight, under a half. The fraction demanded falls steadily, and the share of choices that can be drawn falls with it rather than against it — 9.2 per cent at four places, 2.75 at five, 0.68 at seven, 0.070 at eight.

The reason is in the second measurement. What goes wrong is arrows, and the number of them that come out backwards grows with the picture: 1.51 on average at four places, 4.92 at eight. Every place added brings two more held bearings, and a little under one more of them, on average, comes out the wrong way. A picture is drawable only when that tally is exactly nought, and a count whose mean is climbing is nought less and less often.

The best flat picture is not a map made the point for distances that the collapse between four places and sixteen is a matter of counting — $2n-3$ free numbers against $n(n-1)/2$ demands — and the same counting is at work here with an opposite-looking symptom. There the picture gets relatively worse because it has proportionately fewer numbers to spend; here it gets harder to draw at all even though it is asked to satisfy proportionately fewer constraints, because the constraints it is asked to satisfy each carry a sign condition that has nothing to do with counting.

Nothing in that is about a pair being impossible. It is about a tally having to come out empty, which is a different kind of difficulty — a conjunction of many small conditions rather than one large obstruction — and a conjunction has no single member to blame.

What each number was checked against #

Four places must have exactly one unheld pair, and it must be London–New York. The enumeration of all 192 choices gives eight drawable ones, and the set of pairs held by none of them has exactly one member.

Five places must have none, and this is the control the whole essay rests on: if some pair of five cities were held by no drawable picture, the four-place result would have been a property of the geography after all, and every conclusion here would reverse. All ten are held by at least 56 of the 152.

The forced arrows at four places must be the two largest asymmetries. They are, as sets rather than approximately, and every other held pair must be free in both directions — each of the three is, with a one-sidedness below $10^{-9}$.

The separation at five places must have nothing between the groups. The smallest asymmetry showing any one-sidedness must exceed the largest showing none, and must do so above seventy degrees; it is 78.5 in both roles, carried by one pair that is one-sided in the larger set and free in the smaller.

And the drawable share must fall from four places to five. It must fall by at least a factor of three under exhaustive enumeration, and falls from 4.17 per cent of 192 to 0.99 per cent of 15,360.

What an enumeration of five places leaves out #

The structure is ruled out and nothing replaces it. The tightness argument above shows that no condition reading the graph alone can decide which pair is unheld, which is a proof of what the answer is not. It leaves the space of candidate predictors wide, and this essay has tested one of them.

Five is not sixteen. Everything positive here is an exhaustive statement about four and five places. The share falling to 0.070 per cent at eight is a sampled estimate over a structured sampler, and at sixteen places nothing exhaustive is available at all — which is the difficulty that prompted the question and which this has not removed.

The sampler at six places and above is not uniform. Choices are grown a place at a time, two bearings each, which produces only sets that are rigid by construction; a uniform sample over all subsets would include many that are not rigid at all and would give a lower share for a reason that has nothing to do with arrows.

“Drawable” is a strict condition. A picture with one arrow backwards is rejected outright, and for a reader it might be a serviceable picture with one line labelled the wrong way round. Softening the condition would change every number here and is a different question.

The mandatory pairs are a property of drawability, not of accuracy. A pair in every drawable choice is not the pair a picture most needs to get right; the seven bearings held are exact by construction and the three left out carry errors up to 142 degrees. A place with a size can be drawn to scale is the neighbouring case where the exactness of a constraint is relaxed instead of its membership, and a relaxed bearing problem might well have no mandatory pairs at all.

And the places are five real cities. The asymmetry threshold of about seventy-eight degrees is where these particular sets happen to separate. Nothing here establishes a threshold that would survive a different set, and the mechanism argued above — that two equations differing by a large angle are genuinely different constraints — predicts a separation without predicting where.

Still open: what makes a pair mandatory #

Two pairs of ten are in every drawable picture of five cities and eight are not, and nothing here says why those two. They are not the largest asymmetries, not the longest or shortest, and not the pair joining the two places furthest from the rest. London–Tokyo joins two of the three northern cities and Sydney–Cape Town joins both of the southern ones, which suggests the two halves of the set have to be tied together somewhere and that these are where — but that is a story rather than a measurement.

What would settle it is a quantity computed from the places that ranks the ten pairs in the order the enumeration does, and then predicts the same order for a set the enumeration cannot reach. The share held is available exactly for four and five places, so a candidate can be tested properly on ten pairs and six, and the interesting test is whether it survives at six places and seven, where an exhaustive count is still just possible and the sampled share has already fallen by a factor of thirteen.