The second cut goes across the sea

One cut belongs to the island that would cost most alone held a sea with two islands fixed and asked where a single cut should go. An angle-keeping map of such a sea pays a few per cent of scale variation for leaving it whole, and each cut to the open sea buys back part of it by admitting one island’s logarithm into the map. The answer had a rule: put each island alone in the sea, ask what leaving the sea whole would cost then, and cut the island whose price alone is higher. It named the better cut in eighteen of twenty arrangements, and missed only on near-ties.

It also found something the rule does not know about. Two central islands’ cuts overlap — either alone buys most of what both would — while two islands far apart each do a separate job. It ended on the question that finding makes unavoidable: with three islands and two cuts, the lone prices would say to cut the two dearest, and the two dearest may be exactly the pair whose cuts duplicate each other.

They often are. The rule that picks one cut does not extend to picking two, and the reason is geometric enough to see on the page.

Every way of spending two cuts #

The measurement is the earlier one with a third island in it. A sea 60° across — a spherical cap of radius 30° — holds three circular islands of radius 3°, 4.5° or 6°, each placed at random between 3° and 20° from the sea’s centre and at a random azimuth, kept only if the islands neither touch each other nor come within 3° of the coast. The draws are seeded, so the forty seas are the same forty every time. Their price for being left whole, uncut spread over three-cut spread, runs from 0.33 to 4.04 per cent.

For each sea every way of spending two cuts is scored. There are three pairs of cuts to the open sea, each admitting two islands’ logarithms. There are three ways of spending one cut to the sea and one between the other two islands, which admit one island’s logarithm and the difference of the other two. Two cuts that both join islands to each other leave a loop round all three islands uncrossed, admit only differences, and are not scored. Each option’s worth is its share of what three cuts buy, in logarithms of the spread, exactly as the two-island measurement scored one cut against two.

The first figure is the sea where the obvious rule is worst. Island R is dearest alone, at 1.68 per cent, and P second, at 0.97. Cut those two and the map recovers 75 per cent of what cutting all three would. Cut Q and R instead — Q is the cheapest of the three alone, at 0.72 — and it recovers 97. Seen from the centre of the sea, R and P are 69° apart. Q and R are on nearly opposite sides, 156° apart.

Five rules for choosing two #

Five rules are worth comparing, because each asks for a different amount of work. The first two need only single-island or single-cut answers. The next two need the dearest island and a fact about positions. The last needs the two-cut fits it is choosing between.

The two dearest alone names the best pair in 23 of 40 seas. When it misses it can miss badly: 2.9 points of share lost on average, 22.6 in the worst sea.

The two best single cuts does worse still, 21 of 40, with a worst loss of 38 points. A cut’s worth alone is not its worth beside another cut, which is the whole difficulty in one sentence: the two cuts that each do most on their own are likely to be doing the same thing.

The dearest island, and whichever other island is farthest from it on the sphere names the best pair in 29 seas. The dearest island, and whichever other island lies farthest round the centre from it in azimuth — the island across the sea — names it in 32, with a mean loss of one point and a worst of 11.6. The difference between those two is instructive. Distance on the sphere counts an island near the centre and one near the coast on the same bearing as far apart, and they are; but seen from the coast, where the map’s worst scale is decided, they lie in the same direction, and their cuts duplicate each other. Azimuth measures the thing that matters and distance measures something that only usually goes with it.

The best single cut, then whichever partner adds most to it, names it in 38 of 40 and loses at most 2.8 points. That rule needs the two-cut fits for two of the three pairs, which is most of the work of simply trying all three, and it is here mainly to say that choosing the first cut alone is not the hard part.

Two dearest on one side do one job #

The failures sort by one geometric fact. Where the two dearest islands are the wrong pair, they sit a median 74° apart in azimuth seen from the sea’s centre; where they are the right pair, 121°. Of the eight seas where the rule loses more than five points, seven have the two dearest within 80° of each other, and one has them within two degrees — nearly on the same bearing.

The mechanism is the one the two-island measurement found for two central islands, generalised, and it rests on what two islands are three numbers, and the proof does not survive them established about cuts: a cut admits exactly one logarithm, and the path it takes does not matter. A cut to the open sea admits one island’s logarithm, $\ln|z - a|$, into the map’s scale. Seen from far away — from the coast, where the map’s scale extremes live — two islands on the same side of the centre have nearly the same logarithm, because $|z - a_1|$ and $|z - a_2|$ differ little for $z$ on the far shore. Admitting both is nearly admitting one twice. Admitting two islands on opposite sides gives the map two genuinely different terms, one steep towards each side of the sea, and the map can use both.

A single island at the very centre is the limiting case, the ring a ring can be drawn whole, and only one way solved exactly, where one logarithm carries the whole price. That is also why the overlap between two dearest islands’ single cuts — how much the pair buys less than the two single cuts added — falls as they separate: across the forty seas the correlation between the overlap and the azimuth between them is −0.46.

Two equal islands drawn apart #

The random seas show the pattern; a sweep shows the mechanism with nothing else changing. Two equal islands, P and Q, sit 10° from the centre of the sea, a stated azimuth apart. A smaller island R sits 14° out on the far side of their mid-line. P and Q each cost the sea 2.00 per cent alone at every separation, and R 1.08, so the lone-price rule says to cut P and Q whatever the geometry.

With P and Q 20° apart, cutting them buys 56 per cent of what three cuts buy. Cutting P and R buys all of it, to the third decimal: the two neighbours’ logarithms are so nearly one term that the third cut adds nothing a second one had not. As P and Q separate the balance turns. At 95° cutting them buys 65 per cent against 74 for P and R; between 95° and 120° they become the better pair; at 170°, nearly opposite each other, they buy 94 per cent and P with R only 65. At no separation does either island’s price alone change. Everything that decides the pair is in where the islands are relative to each other, and none of it is in what each would cost alone.

The dearest island stays #

What survives of the one-cut rule is its first half. In 37 of the 40 seas the best pair of cuts includes the island that would cost the sea most alone, to within a point of share. In the other three, keeping the dearest island anyway costs between 1.4 and 2.8 points. In one of them the dearest island lies inward of a cheaper one on nearly the same bearing, so a cut to the cheaper one admits a logarithm that does the dearer island’s work as well — the same duplication, seen from the other end. In the other two the dearest island’s cut and its best partner’s overlap more than a cheaper pair’s do, for no reason simpler than the whole fit.

So the practical rule has two steps, and they need different information. The first cut goes to the island dearest alone, which a one-island problem answers. The second goes to an island across the sea from it, which a protractor answers. In thirty-two seas of forty that pair is the best; across all forty it gives up a point of share on average, and 11.6 at worst.

When the third cut is worth having #

All of this has been about which two cuts to spend, and it presumes two are nearly enough. Usually they are. The best pair recovers a median 99 per cent of what cutting all three islands would, and in half the seas more than that to within the solver’s tolerance. A cartographer who can afford two cuts on a sea with three islands has, in most seas, all the benefit three would give.

In nine seas of forty it is not so. The best pair recovers less than 95 per cent, and in the worst sea 84. Those nine are not the seas with the largest islands or the highest price for staying whole. They are the seas whose islands are spread round the centre. Take the three islands’ azimuths seen from the centre and find the largest empty arc between them: three islands bunched on one side leave an empty arc of well over 180°, and three spread evenly leave none larger than 120°. In the nine seas where two cuts fall short, the largest empty arc is a median 156°; in the other thirty-one, 224°. Across all forty, the best pair’s share and the largest empty arc correlate at 0.66.

It is the same geometry read the other way. Two islands on one side of the sea have nearly the same logarithm, so one cut does most of both islands’ work, and a sea whose three islands crowd one side has in effect fewer than three directions to serve — two cuts serve them all. A sea whose islands point three different ways has three directions, and a third logarithm the other two cannot imitate. The number of cuts a sea is worth is closer to the number of distinct directions its islands occupy, seen from the centre, than to the number of islands.

That is a claim about three islands, and it is the one what a cut buys would need to price cuts for a real archipelago, where the question is not only where the cuts go but how many are worth their length. It also says where the one-cut rule of the two-island essay came from: with two islands on opposite sides, as that measurement placed them, the two directions were distinct by construction, and the question of duplication never arose.

A cut between islands is never worth a second cut to the sea #

The two-island essay found a cut joining two islands weak, because it admits only the difference of their logarithms. With a third island there was a reason to think it might become worth spending: one cut to the sea from one island, and one between the other two, gives the map one full logarithm and one difference, and if the other two islands sat on opposite sides a difference might be what the map wanted.

It is not, in any of the forty seas. The best mixed option never beats the best pair of cuts to the sea by more than the solver’s own tolerance; in the closest sea it comes out two tenths of a point ahead, inside that tolerance, and in the median sea it is eight and a half points short. A map that wants to wind in the same direction round every island — as a map of a sea with islands in it does — has no use for a term that winds one way round one island and the other way round another, and a third island does not change that.

What each number was compared against #

Three identical islands at 120° must value their three pairs alike. They must, by symmetry; the solver gives 0.7099, 0.7099 and 0.7099.

Every pair of cuts must buy something and no more than three cuts do, to within the solver’s own tolerance of a point. Every pair share in the forty seas lies between zero and 1.003.

The finding must be there to fail. The two dearest alone must name the best pair in fewer than three seas in four, the island across the centre must do better, and the best pair must keep the dearest island in at least nine seas of ten. They do: 23, 32 and 37 of 40.

And the sweep must turn over. With the two equal islands close together, their pair must be the worst of the three; far apart, the best. It is worst at 20° and best at 170°.

The spreads themselves come from the same harmonic fit the two-island measurements used, which reproduces the stereographic’s closed form for a sea with no island and the conformal conic’s for a band of latitude.

Where the seas are a model #

The fit is a minimax found by iteration, and ties are ties. It aims at the equal-ripple optimum Chebyshev’s criterion describes and approaches it without reaching it exactly. A pair share of 1.003 — a pair of cuts slightly better than all three — is impossible, and records how close the iteration comes to the true minimum. Every comparison here treats shares within a point of each other as equal; the rules’ counts would move by a sea or two at a tighter tolerance, and none of the conclusions would.

Circles on a cap. Islands are circular and the sea is a spherical cap, because the closed form that puts the sea in the plane needs circles. A long, thin island has a logarithm that is not a point’s, and two such islands on the same side might not duplicate each other the way two round ones do.

One sea size, three island sizes. The seas are all 60° across and the islands 6°, 9° or 12°. Larger islands closer to the coast behave as dents in the coast, as the two-island measurement found, and would change which island is dearest without obviously changing the geometry of pairs.

A cut costs nothing here but itself. What a cut buys prices a cut by its length and giving up continuity by what a reader loses at it. A cut to an island across the sea from the first is usually the longer route to the coast, and nothing here weighs that.

Still open: whether the rule counts beyond three #

With three islands the second cut goes across the sea. With four or five, the same geometry suggests a rule that places cuts to spread the admitted logarithms round the centre — the dearest first, then the islands that fill the widest gaps in azimuth — and it suggests that the number of cuts worth spending is set by how many distinct directions the islands occupy rather than by how many islands there are. Two charts are enough, and one is not found a count for the sphere as a whole; a sea with many islands might have a count of its own.

Whether a greedy placement by azimuth stays within a point of the best as islands are added, whether a sea whose islands crowd one side has a best count of cuts smaller than its number of islands, and whether that count is a property of the islands’ spread seen from the centre, are questions three islands cannot ask.