Cartographic Projection
every map is wrong, and by exactly how much
Figure

A pin moves the whole curve along by one and changes nothing else

Drawn here at the parameters it defaults to, with every essay that calls it.
A pin moves the whole curve along by one and changes nothing else. Each pinned share against the unpinned share of the set one size smaller for each pin held, at five places. If the pins and the place constraints do not interact, every point lies on the diagonal, because the factor the pins contribute cancels in a share. The worst departure is 0.0048, at a set of 8 holding one pin — which is sampling noise on 20,000 draws and not a structure. The exhaustive count at four places makes the same statement exactly.

Each pinned share against the unpinned share of the set one size smaller for each pin held, at five places. If the pins and the place constraints do not interact, every point lies on the diagonal, because the factor the pins contribute cancels in a share. The worst departure is 0.0048, at a set of 8 holding one pin — which is sampling noise on 20,000 draws and not a structure. The exhaustive count at four places makes the same statement exactly.

It is one of 73 figures drawn by the same construction, and the drawing above is this one with nothing changed.

1 essay calls it. Every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Changing this changes every one of these figures.

A picture of four places, and the one constraint that is about the sheet. The best flat picture of four world cities, with the six distances it could be asked to hold drawn between the dots and the twelve bearings implied by them. Every one of those is a statement about how two places stand to one another, so every one is blind to where the whole picture sits on the paper. The cross is a constraint of a third kind: a stated position for one place — this dot goes here — which is two numbers and is blind to nothing. Adding both of them takes the largest independent set from 6 to 8. The impossibility

Two numbers about the paper buy nothing about the places

A picture of n places has four freedoms, and a constraint holds only what its own blindness leaves: distances reach 2n − 3, bearings 2n − 3, the two together 2n − 2. Position is reached by neither, because nothing about how places stand to one another says where on the sheet to put them. A stated position does reach it, and two of them take the budget to 2n exactly. What the count conceals is that the extra two buy nothing: a largest picture holds six place constraints with the pins and six without, the same 7,596 of them, holding the same pairs at the same rates. The constraint system is a direct sum, and a pin is a different currency.

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