Three maps make a plane, and the band's best map lies outside their triangle

Two maps and the line between them do the work of every exponent used a property of conformal maps that the essays before it had never put to work. A conformal map of a region is a series of complex terms, and its log scale factor ln k at every point is a fixed term plus the real part of that series, so ln k is linear in the coefficients. Blend the coefficients of two maps and the blend is a conformal map whose ln k is the same blend of theirs. The p = 1 map, which minimises the mean departure from true scale, and the least worst map, which minimises the largest, are two solved maps, and every map on the line between them is a third for nothing.

Searched for the map that keeps the most of a region inside a surveyor’s band, that line came within a fifth of a point of the best of twelve exponents on every compact region. It fell short on the thin ones, by 0.82, 1.13 and 1.48 points on ellipses of axes 1 : 0.3, 1 : 0.2 and 1 : 0.12, and the reason was measured: the exponents between the two ends lie well off the line, up to two thirds of their own departure from the p = 1 map. The essay ended by proposing a third end. The mean-square map is the exponent family’s p = 2 member, it has a closed-form fit, and with it the line becomes a triangle.

What follows takes the proposal literally and then takes the step it implies. Three solved maps do not make a triangle of maps. They make a plane.

Three solved maps are a plane of maps #

Write $\ln k_1$ for the p = 1 map’s field, $\ln k_\infty$ for the least worst map’s and $\ln k_2$ for the mean-square map’s. Every field of the form

$$\ln k(s, u) = \ln k_1 + s\,(\ln k_\infty - \ln k_1) + u\,(\ln k_2 - \ln k_1)$$

is the field of the map whose coefficients are blended in the same proportions, and so is a conformal map of the same series. Nothing restricts s and u to lie between 0 and 1. The triangle with the three maps at its corners — the region where both are positive and their sum is at most one — is the part of the plane that interpolates; everywhere else extrapolates, and an extrapolated map is as much a conformal map as an interpolated one.

So there are three searches to compare against the twelve exponents, each over the same stored fields and each costing nothing but arithmetic once the three maps are solved. The line is one number: u held at zero. The triangle is two numbers held inside it. The plane is two numbers anywhere, here out to eight times each difference in either direction. Each is searched for the map keeping the most of the region inside a band ±τ, at sixteen bands from 0.05 to 1.0 of the region’s least worst departure w*, on the same fourteen regions the line was measured on: a cap, six ellipses from 1 : 0.8 to 1 : 0.12, a spherical triangle, square and hexagon, all 25° in their long radius, and four with a bay cut into one side.

The square at half its least worst departure shows all four answers at once. The best exponent keeps 66.4 per cent of the square inside the band, the line 66.5, the triangle 66.5: the third map adds nothing on a compact region at this band, because the line had already done what the exponents do. The best map in the plane keeps 75.8 per cent, and it sits at s = 0.72, u = −2.81 — nearly three times the mean-square map’s difference from the p = 1 map, taken in the opposite direction.

Inside the triangle the third map pays what the line owed #

The earlier essay’s question has a plain answer. Inside the triangle the third map closes most of the line’s shortfall and not all of it. On the ellipse of 1 : 0.3 the shortfall against the best exponent falls from 0.82 points to 0.14; on 1 : 0.2 from 1.13 to 0.35; on the thinnest, 1 : 0.12, from 1.48 to 0.63. The four regions with a bay go from 0.13–0.74 points to 0.00–0.21. On every compact region the triangle trails the best exponent by nothing at any band, and beats it by up to 1.4 points.

What is left on the thinnest ellipse is the part of the family the triangle still cannot reach. The members between p = 1.25 and p = 8 move out of the triangle’s plane by up to 0.13 of their own departure on that ellipse — far less than the 0.47 they lie off the line, and not nothing. A triangle-bound search therefore recovers most of the intermediate exponents and does not become them.

The family lies almost exactly in the plane #

The line could not hold the family because the family is not one-dimensional. Projected onto the line, the middle exponents leave between a third and two thirds of their departure behind. Projected onto the plane through the three maps, they leave a tenth or less on ten of the thirteen regions, and 0.13 at worst, on the thinnest ellipse. To that accuracy the twelve exponent maps, each solved by its own reweighted fit, are blends of three of them.

The dashed curve on the first figure is what that looks like. The exponents leave the p = 1 corner, curve round through the mean-square corner and come back down to the least worst corner, staying close to the triangle’s edge. They are a curve on the plane, and the triangle’s edge happens to approximate it. That is why the third map closes most of the line’s gap: it was the second direction the family had been using all along.

It is also a warning about the earlier result’s reading. The line was described as the family’s ends predicting its middle. The plane says the family’s middle is a third independent direction, and that p = 2 marks it as well as any member could, because the mean-square map is the family’s member most nearly at right angles to the line.

The band’s best map is outside the triangle #

The triangle is where interpolation ends. The best map in the plane is not in it: at bands from 0.3 of w* up, on 172 of the 182 combinations of region and band its coordinates lie outside the triangle, on three they lie on its edge, and on the remaining seven — all on regions with a bay — inside. Wherever it lies outside, it keeps more than every map the triangle holds.

How much more depends on the region and the band, and it is not small. On the square the plane keeps up to 9.4 points more than the best exponent, at half the least worst departure; on the thinnest ellipse 10.8 more, at 0.4 of it. On the ellipses of 1 : 0.6, 1 : 0.45, 1 : 0.3 and 1 : 0.2 it keeps between 16.6 and 19.0 more, on the spherical triangle 14.8, and on the ellipse of 1 : 0.6 with a bay 20.4. Only on the cap and the rounder regions is the gain a few points: 2.7 on the cap, 4.6 on the ellipse of 1 : 0.8, 3.2 on the hexagon.

The gain closes as the band widens. Near the least worst departure itself every map that keeps most of the region inside the band has to be close to the least worst map, and the plane’s best walks back toward that corner. On the square the walk is smooth: at 0.3 of w* the best map sits at (1.49, −4.84), at 0.5 at (0.72, −2.81), at 0.7 at (0.78, −1.65) and at 0.9 at (0.99, −0.58), beside the least worst corner. The spherical triangle follows nearly the same path. Which way the best map leaves the triangle depends on the region’s shape. On the compact polygons it goes away from the mean-square map, u negative; on the thin ellipses it goes past it, to u between 2 and 5 at most bands, which is the direction the exponents above 2 were already taking.

One exponent names one map, and at tight bands a better one than the band usually finds continued the exponent family below p = 1 and found it gaining about five points at tight bands by p = 0.7, before its answer stopped being unique. On the square the plane goes past the p = 1 corner in close to that family’s direction: from p = 1.25 to p = 1 the exponents move by (0.14, −0.64) in the plane, and the best map at half the least worst departure lies at (0.72, −2.81), a ratio of the two coordinates within a sixth of the family’s. The plane reaches past p = 1 without solving any power below one, and further than the family’s usable stretch went.

A fourth number is worth a point, except on the hexagon #

The earlier essay asked whether a surveyor would ever need a fourth map. The measurable form of that question is how much a search over every coefficient adds once the plane has been searched. The map that keeps the most ground inside a tolerance built that search, coordinate ascent over the twelve-term series’ twenty-five numbers along an orthonormal basis, and a criterion worth using is one whose answer is not unique found that where it ends depends on where it starts. Started from the plane’s best map, at half the least worst departure, it adds 1.0 point on the square, 0.8 on the ellipse of 1 : 0.45, 0.9 on the thinnest ellipse and 0.9 on the ellipse with a bay. On those four regions every remaining coefficient together is worth about a point.

The hexagon is the exception, and it is a large one. The plane keeps only 1.8 points more than the best exponent there, and the full search started from it adds 7.0 more. One reading, which nothing here tests, is that on a region with six corners the map that keeps the most inside a tight band treats the corners differently from the edges between them in a way none of the three solved maps does, so their plane holds no direction toward it. A fourth solved map would help only if it supplied that direction, and the measurement cannot say which map would.

The other comparison in the figure is the one that matters for practice. Started from Chebyshev’s map, the least worst map that Chebyshev’s criterion names and the start from which the tolerance map was first found, the full search ends below the plane’s best on all five regions, by between 1.9 and 27.7 points. Three closed-form solves and a two-number grid search beat a twenty-five-number search started in the wrong place, and the plane’s best is a good place to start one.

The price of the extra area is a worse worst departure #

The band criterion counts area inside the band and nothing else, so a map that keeps more of the region inside the band is free to let the rest go as far as it likes. The plane’s best maps do let it go. On the square at half the least worst departure the best map’s worst departure is 3.8 times w*, and at 0.3 it is 4.7, where the p = 1 map’s is 2.2. On the hexagon it is 2.9 against 2.3; on the ellipses of 1 : 0.45 and 1 : 0.12 it is 1.9 and 1.5 against 1.6 and 1.4 — close to the p = 1 map’s, so on the thin regions much of the gain is bought cheaply.

Whether that price is acceptable is the question every essay on this criterion has had to leave to the surveyor. A band is a statement that departures inside it are acceptable and departures outside it are not, and the criterion takes the statement at its word: a point outside the band by a hair costs as much as a point outside it by a factor of four. A second clause decides the half that was already decided added a clause to break ties among maps that keep the same area, and the plane’s best maps are the case for a clause that does more — a cap on the worst departure, which would cut the plane down to the part of it where the worst departure is below the cap, and search there.

What each number was held to #

The plane contains the triangle, and the triangle the line. At every band on every region the plane’s best must keep at least what the triangle’s does, and the triangle’s at least what the line’s does, to the resolution of the grid. They do, with the line read at the earlier essay’s 201 positions and the triangle at a hundredth in each coordinate, refined to a four-hundredth.

The grid must not decide the answer. The plane is searched on a grid of 0.08 from −8 to 8 and refined twice round the best point, to a hundredth and then to a five-hundredth. On the square and the thinnest ellipse, at the band where the plane gains most over the line, a grid four times as fine over the same window finds less than half a point more.

The mean-square map must be the third corner. Projected onto the plane by the same weighted least squares as every other member, p = 2 must sit at (0, 1) and leave nothing off it. On the square and the thinnest ellipse it does, to a millionth.

The window is stated, because on two regions it is the answer. A first search out to four times each difference put the best map on the window’s edge on three regions, so the window was doubled. At ±8, on twelve of the fourteen regions the best map lies no further than 5.5 from the p = 1 corner at any band from 0.3 up. On the other two it lies on the edge of any window. The cap’s three maps are one map at three overall levels, so its plane has a single direction and the best level is reached along a whole line of equally good points, one of which the grid happens to meet at the edge. On the ellipse of 1 : 0.8 the share creeps upward as the maps go further out: 3.0 points over the exponents within ±4, 4.6 within ±8. Its figure is a lower bound, and the only one in this essay.

Where the plane stops #

Three maps chosen in advance. The three corners are the two ends of the exponent family and its p = 2 member, chosen because they are cheap and because the family runs through them. A different third map — Chebyshev’s, say, or the band criterion’s own answer at one band — would make a different plane, and on the hexagon a different plane is plainly what is needed.

Twelve terms. Every map here is a twelve-term series, as in the earlier essays. More terms would enlarge every family compared here, the plane included, and nothing measured says how the comparison moves.

One criterion. The band criterion rewards area inside the band and ignores everything outside it. The plane’s best maps are best for that criterion and show its price plainly. Chebyshev’s map is the best at its worst, and not on average is the opposite choice, and it is one corner of the plane.

Sampled regions. Every share is an area-weighted count over forty rings and a hundred and sixty azimuths, and each region’s worst departure is the worst over those samples. The nodes were evenly spaced is the reminder that where a region is sampled can matter more than how finely.

Still open: what a cap on the worst departure does to the plane #

The plane’s best maps keep up to twenty points more area inside the band than any exponent, and they do it by letting the edges of the region go to up to four times the least worst departure. A surveyor who accepts the band but not unlimited departures outside it is asking a two-clause question: the most area inside ±τ, among maps whose worst departure is below some cap.

In the plane that question has a shape that can be drawn. The worst departure is a convex function of s and u, because it is the largest of a set of absolute values of linear functions, so the maps under any cap form a convex region of the plane, and the cap cuts the contour picture above with a convex boundary. Whether the area kept falls smoothly as the cap tightens or collapses as soon as the cap excludes the far-out maps, whether the best capped map moves back into the triangle or slides along the cap’s edge, and whether a cap of twice the least worst departure leaves most of the gain in place, are questions an uncapped search cannot ask.