One exponent names one map, and at tight bands a better one

A second clause decides the half that was already decided tried to repair a specification that has too many answers. The specification asks for the conformal map of a 25° spherical square that keeps the most area inside a band of half a per cent about true scale, and it has at least forty-two local answers, keeping from 36.7 to 65.2 per cent. Adding a tie-break — among the maps keeping the most, take the one whose worst departure is least — made each of the forty-two definite and left the spread of area kept at 28.4 points. The clause decided the part of the map the first clause had already stopped asking about.

The failure was a property of both clauses. Each reads a scale field and keeps one statistic of it: the first a count of area inside a threshold, the second a single largest value. Two maps whose fields differ everywhere except at those two statistics are the same map to the specification. The average was a choice of norm had already pointed at a way out: the mean, the mean square and the worst case are three points on one family, the mean of |ln k| raised to a power p, and a specification that names a power reads the whole field.

That essay measured how much the choice of power reorders projections. The question here is narrower and practical. Named as a specification — the conformal map of this region minimising the mean of |ln k|^p — does an exponent give one answer, does any exponent keep as much ground inside a band as the band criterion does, and what does it cost at the exponent that comes closest?

One answer for every power from one up #

The maps are the same kind the band criterion searched, and the same kind solving for the map instead of choosing it introduced: a conformal map of the square built from a series of twelve complex terms in the logarithm of its derivative, twenty-five real coefficients with the level of the scale. Over the region, ln k at every sample point is an affine function of those coefficients — exactly, to second differences of 10⁻¹⁷ — and that is what makes the band criterion’s line search exact and its local optima genuine.

The same fact makes the exponent family easy. The mean of |x|^p is a convex function of x for every p ≥ 1, and ln k is affine in the coefficients, so the mean of |ln k|^p is convex in them. A convex function has one minimum. Solved by iteratively reweighted least squares — each round a weighted fit of ln k to zero with weights |ln k|^(p−2) — continued from p = 2 in small steps up and down, it converges to the same scale field from starts at opposite ends of the family, to five parts in 10¹¹.

The two named points of the family land where they must. At p = 2 the map is the mean-square map, with a root-mean-square departure of 0.804 per cent. At p = 32 its worst departure is 1.476 per cent, within a per cent of 1.486 — the worst departure of Chebyshev’s map once its level is centred between its highest and lowest scale, which is what the least worst case over the interior is.

The pictures show what the power does. At p = 32 the map spends everything on its extremes: the corners and the centre depart by nearly the same amount in opposite directions, and the band of half a per cent is a thin ring between them. At p = 1 the map lets its corners go — they depart by over three per cent — and in exchange a broad region round the middle of each side sits inside the band. The mean-square map is in between.

The less weight on the tail, the more ground inside the band #

Area kept inside the band falls steadily as the power rises: 53.7 per cent at p = 1, 49.3 at 1.25, 47.0 at 1.5, 44.2 at 2, 38.7 at 3, and a floor near 33 from 4 upward. The reason is the weight each power gives a departure. A power of one charges a departure in proportion to its size; a large power charges the largest departures overwhelmingly and is indifferent to small ones. The band criterion is indifferent to everything outside the band, however large, so the powers that care least about the tail come closest to it. The closest convex one is p = 1.

At a band of half a per cent, p = 1 keeps 53.7 per cent. The band criterion’s best answer keeps 65.2, so the convex specification gives up 11.5 points against the best the band criterion was found to reach. Against the band criterion’s forty-two answers taken as they come it does better than twenty-three of them. The band criterion is under-determined and sometimes better; the exponent is determined and in the middle of the pack.

That is the answer to the question the tie-break left, at the band the tie-break was measured at. At other bands it is a different answer.

Why the mean lets the corners go #

The reason p = 1 behaves so differently from the powers above it is visible in what its optimum has to satisfy. A power of one charges each departure by its size, so moving the map a little changes the objective by the area on one side of true scale minus the area on the other, weighted by how much each coefficient moves each point. How far each point departs does not enter at all — only which side it is on. At the optimum, the level of the scale must therefore split the square into equal areas above and below true scale, which makes it a weighted median rather than a mean, and it does: 49.96 per cent of the square lies above true scale and 50.04 below. The mean-square map splits it 51.6 to 48.4 and the map at p = 32, 52.4 to 47.6, because both are pulled towards their extremes.

A median does not care how far the outliers go, and the corners of a conformal map of a square are its outliers. Every coefficient at p = 1 is set by the same kind of balance, a count of area on each side weighted by the coefficient’s own pattern, so the corners count once each however large their departure. That is what lets the map keep a broad region near true scale and let the corners run to three per cent: it is the closest a convex criterion can come to the band criterion’s indifference to everything outside the band. Statisticians reach for the median for the same reason, as a robust estimator that a few wild values cannot drag.

The iteration that finds it is the one that failed in a different setting. Weighted by what it disagrees with reweighted each line of a celestial fix by one over its squared residual and iterated, and the fix walked away from the truth. That reweighting is the p → 0 end of the same family, where the objective is not convex and the answer depends on the start. Here the weights are |ln k|^(p−2) with p at one or above, the objective is convex, and the same kind of iteration converges to the one answer from anywhere. The difference between a reweighting that works and one that wanders is the exponent.

The best exponent depends on the band #

At a band of a tenth of a per cent, the p = 1 map keeps 19.8 per cent of the square inside it. The best of the band criterion’s forty-two answers keeps 17.8. At two tenths, 30.6 against 29.8. At tight bands the convex specification, which was not asked about the band at all, keeps more ground inside it than every answer of the specification that was.

That is not a paradox about the objective. The band criterion’s objective is the area kept, and some map keeps at least 19.8 per cent — the p = 1 map is one. What fails is the search. The objective is a count of samples inside a narrow band, flat almost everywhere and falling off cliffs where a sample crosses the edge, and coordinate ascent from forty-two starts settles on forty-two ledges, every one of them below a map that a convex solver finds in one pass. At a tight band the ledges are small and many, and the search has less room to climb.

As the band widens the order reverses. At half a per cent the band criterion’s best answer leads p = 1 by 11.5 points, and at one per cent by 5.4. At one and a half per cent the large exponents keep everything — the least worst case is 1.48 per cent, inside the band — while p = 1, which let its corners go to 3.26, keeps 91.1. So the exponent that stands in best for a band is one for bands up to about one per cent and the worst case once the band is wide enough to hold the least worst case whole. The change happens through a narrow stretch between them. At a band of 1.25 per cent every exponent from one to thirty-two keeps between 83 and 89 per cent and the family is nearly flat; at 1.35 the best is an intermediate power, p = 8, at 92.9; by 1.5 the order has reversed completely and area kept rises with the power.

As a start, the one answer beats every other #

The p = 1 map is a better starting point than any of the band criterion’s forty-two, and that suggests using it as one.

Started from the p = 1 map, the band criterion’s search reaches 21.1 per cent at a band of a tenth of a per cent, above all forty-two of its own answers from other starts; 33.2 at two tenths, again above all forty-two; 63.1 at half a per cent, above thirty-nine; and 83.5 at one per cent, above twenty-seven. It settles in three to eight sweeps.

So the two specifications are not rivals. The exponent is the one that names a map, and the band criterion is what a surveyor actually wants. A procedure that names p = 1 first and then climbs the band criterion from there is still a search with a start, and so still not a specification in the strict sense — the tie-break essay’s objection stands. But its start is a named, unique map rather than a starting point somebody chose, and from that start the search reproduces: anyone who runs it gets the same ledge. That is most of what reproducibility asks.

The price of the band’s extra ground is paid in the tail #

A band criterion ignores the tail by construction, and the forty-two answers show what that licence buys. Placed by area kept and worst departure, the exponent family traces a curve from the least worst case to p = 1, and the band criterion’s answers scatter around it. Fourteen of the forty-two sit above and to the left of the curve — some exponent’s map keeps more area with a smaller worst departure, so those answers are beaten on both counts. The nineteen answers that keep more than the p = 1 map do it with worst departures from 3.2 to 7.3 per cent: the extra ground inside the band is bought by letting the corners go further than even p = 1 does.

That is not an argument against the band criterion. A surveyor who wants the most ground inside half a per cent and is truly indifferent to the corners should take one of those maps. It is an argument that the indifference should be stated, because the band criterion lets the tail go anywhere, and two of its answers keeping nearly the same area can differ in their worst departure by a factor of two.

Below one, the answer stops being unique #

Convexity ends at p = 1, and powers below one care still less about large departures. They might come closer still to the band criterion.

They do, a little. At p = 0.8 the minimiser keeps 58.3 per cent inside the half-per-cent band and at 0.7 about 59, against 53.7 at p = 1. Down to 0.7, the three starts end within a twentieth of a per cent of one another in the objective and within half a point in area kept, so in practice there is still one answer. At 0.6 they settle at different values of the objective — different local minima, the non-convexity arriving — and at 0.5 the start decides whether the map keeps 58 per cent or 49.

So the family has a usable stretch below one, from 1 to about 0.7, where it gains another five points while still naming one map, and then it becomes the band criterion’s problem again. At the limit p → 0 the mean of |ln k|^p counts the samples where ln k is not exactly zero, which is the band criterion with a band of width nothing: the family’s far end is the criterion it was meant to replace, and it arrives there with the criterion’s defect.

Who chose an exponent before #

The exponents have a long history in cartography under other names. Airy’s criterion of 1861 minimises a mean square of the two principal scales’ departures, a p = 2 criterion. Chebyshev stated in 1856 that the conformal map of a region with the least worst departure is the one with constant scale on its boundary, the limit p → ∞, and Grave proved it in 1896. Kavrayskiy in the twentieth century used both the mean square and the worst case and argued over which a given purpose wants. The mean absolute departure, p = 1, has been used less, perhaps because it has no closed form and its minimiser is found only by iteration.

What the band criterion adds is a specification a surveyor would write: a tolerance and an area. The map that keeps the most ground inside a tolerance found it and a criterion worth using is one whose answer is not unique found its defect. The exponent family is the older idea placed beside it, and the result is that the older idea at its smallest convex power is a strong answer to the newer question at tight tolerances and a good start for it at any.

What each number was compared against #

Convexity must show as one answer. At p = 1.5, continuation from the p = 1 map and from the p = 32 map must reach the same scale field; they agree to 5 × 10⁻¹¹.

The family’s two named points must be the named maps. At p = 2 the root-mean-square departure must equal the mean-square map’s to a per cent, and is 0.804 against 0.804. At p = 32 the worst departure must be within a per cent of Chebyshev’s map’s once centred, and is 1.476 against 1.486.

The area kept must fall as the power rises from one to four. It does, at every step: 53.7, 49.3, 47.0, 44.2, 38.7, 33.8.

Every share is read on the same samples as the band criterion’s. The design — 6,400 samples in forty rings, each weighted by its area — is the one the forty-two answers were found on, so the comparisons are between maps and not between samplings. A condition read at samples can be satisfied at the samples and nowhere else, which a condition imposed at points is not a condition found for a boundary condition; the band criterion’s own maps were checked on a finer offset sample for that reason, and the exponent maps, being smooth minimisers of a mean over the whole region, have no sample gaps to shape themselves to.

Where the comparison stops #

One region. A 25° spherical square. The tie-break essay measured the same square; other shapes — a cap, an ellipse, a sliver — would move where the best exponent jumps, since the jump is set by the least worst case, which depends on the shape.

The band criterion’s answers are from one search. Forty-two starts with coordinate ascent on exact line searches. A better search might find answers above the p = 1 map at tight bands too. The result stated is that the search the criterion was measured with finds none, and that the p = 1 map is a start from which the same search does.

The powers below one are a continuation, not a global search. Three starts agreeing is evidence of one answer down to 0.7, not proof.

Still open: whether the best exponent can be named from the band #

The exponent that stands in best for a band is one for tight bands and the worst case for wide ones, and the jump comes where the band is wide enough to hold the least worst case. That makes the right exponent a function of the band and of the region, through one number — the least worst departure the region allows — and it raises the possibility of a rule that names the exponent from the band rather than asking the surveyor for either.

What that rule would look like is not obvious. Between a band of 1.25 per cent and 1.4 the family is nearly flat and its best member is an intermediate power, so the rule is a switch with a transition in it, and a rule would need to know where the transition starts and how wide it is. Whether the crossing band can be predicted from the ratio of the band to the least worst case alone, whether that ratio carries across shapes, and whether a surveyor told “p = 1 below this ratio and the worst case above it” loses anything against the best exponent at every band, are questions one region and six bands cannot answer.