A sea with five islands is seldom worth five cuts

The second cut goes across the sea held three islands in a sea and two cuts to spend. An angle-keeping map of a sea with islands in it pays a few per cent of scale variation for leaving the sea whole, and each cut from an island to the open sea buys some of it back by admitting that island’s logarithm into the map — the result two islands are three numbers, and the proof does not survive them established. The obvious rule, cut the two islands that would each cost the sea most if they were alone in it, named the best pair in only 23 seas of 40. A better one kept the dearest island and paired it with the island across the sea from it in bearing, because two islands on one bearing have nearly the same logarithm seen from the coast, and admitting both is nearly admitting one twice.

It ended on a guess about what happens with more islands: that a greedy placement by bearing would stay close to the best, that a sea whose islands crowd one side would be worth fewer cuts than it has islands, and that the count would be set by how many directions the islands occupy rather than by how many there are.

Four and five islands confirm the second guess strongly, the third in a weaker form than it was stated, and the first only for two cuts.

Every way of cutting a sea of five #

The measurement is the three-island one with the islands added. A sea is a spherical cap 60° across. It holds four or five circular islands of radius 3° or 4.5°, each placed at random between 3° and 20° from the sea’s centre and at a random bearing, kept only if no two touch and none comes within 3° of the coast. The draws are seeded: twenty seas of four islands and twenty of five, the same every time.

For each sea, every subset of cuts to the open sea is scored — sixteen subsets for four islands, thirty-two for five — by the same harmonic fit that finds the angle-keeping map’s least scale spread, the equal-ripple optimum Chebyshev’s criterion describes. Each subset’s worth is its share of what cutting every island buys, in logarithms of the spread. Cuts between islands are not scored; with three islands a cut between two islands never beat a second cut to the sea, and nothing here gives a reason to expect it would start.

Three rules are compared with the best subset at every count of cuts. The dearest alone cuts the k islands that would cost the sea most if each were alone in it, which needs only one-island answers. Dearest first, then by bearing cuts the dearest island and then, each time, the island whose bearing from the sea’s centre lies farthest from every bearing already cut — one-island answers and a protractor. Best cut, then best partner takes the best single cut and adds, each time, whichever island adds most to the cuts already chosen; it needs the fits it chooses between, and stands for what a cartographer with the full calculation would do one cut at a time.

The two dearest share a bearing #

The sea in the figure is the one on which the dearest-alone rule does worst at two cuts. Its two dearest islands, S and T, lie on the same side of the sea, 42° apart in bearing seen from the centre. Cutting T alone buys 77 per cent of what cutting all five islands would; cutting S as well buys nothing more, to the nearest point. Seen from the coast, where the map’s worst scale is decided, S and T are one object, and a second cut on that bearing admits a logarithm the map already has.

The best pair is T with Q, the island on the opposite side of the sea, which is neither large nor dear: Q would cost the sea 1.30 per cent alone, against S’s 1.96. Together T and Q buy everything that five cuts would. The rule that cuts the dearest island and then the island farthest from it in bearing finds exactly that pair, from the one-island prices and the bearings alone.

Two cuts by bearing, three by fit #

For one cut nothing has changed. The dearest island alone is the best single cut in 18 seas of 20 with four islands and 16 with five, and where it is not, it loses at most eight points of share. The one-island price that one cut belongs to the island that would cost most alone introduced still names the first cut.

For two cuts the three-island finding holds and sharpens. The two dearest name the best pair in 9 seas of 20 with four islands and 7 with five, losing 7.3 and 7.6 points on average and up to 30. Dearest first and then by bearing names it in 15 and 13, losing 1.3 and 2.0 points on average. That is as good as the rule that tries the fits — 15 and 16 — and needs none of them.

For three cuts of five the bearing rule breaks. It names the best three in only 9 seas of 20, no better than the dearest alone, and in its worst sea it loses 21 points. Best cut then best partner names the best three in 16. At four cuts of five the fitting rule is right in all twenty seas and the bearing rule in fourteen.

So the protractor answers the question it was found answering — where the second cut goes — and not the questions after it. A third cut on a five-island sea is a choice among three remaining islands whose bearings, seen against two cuts already made, are often all at middling distances, and what decides between them is the price each would have added, which the bearings alone do not carry.

Where the dearest pair loses, bearing seldom does #

Seen sea by sea, the two rules for two cuts fail in different places. The dearest pair loses heavily in a handful of seas — 30, 23, 21, 21 and 20 points — and, as on the three-island seas, its losses go with how close the two dearest islands sit in bearing: where it loses more than five points, in 19 seas, they are a median 81° apart; where it loses at most one, in 16, a median 142°. The bearing rule is right on almost all of them. It loses more than the dearest pair in five seas of forty, and in those it loses at most twelve points; its failures are seas in which the island farthest in bearing is small and cheap, so the cut it chooses admits a logarithm too weak to matter, while a nearer but dearer island would have mattered more.

That is the second rule’s own blind spot, and it is the mirror of the first’s. The dearest-alone rule knows each island’s price and not its direction; the bearing rule knows the direction of every island after the first and not its price. A rule that weighed the two — an island’s price discounted by how close its bearing is to the cuts already made — would be the obvious next refinement, and nothing here tries it.

The dearest island stops being kept #

On three islands the best pair kept the dearest island in 37 seas of 40, and the one-cut rule’s first half survived intact: cut the dearest island first. On four islands the best pair includes the dearest island in 11 seas of 20, and on five in 13.

The reason is the duplication already seen, met from the other end. With three islands, the dearest island sharing a bearing with another left a third island across the sea to pair it with, and the dearest was always worth keeping. With four or five, the dearest island often shares its bearing with a second dear island, and the best pair may be two islands that between them cover two directions better than any pair containing the dearest — each a little cheaper than it, each on a bearing nobody else holds. The bearing rule, which keeps the dearest island by construction, still names the best pair as often as it does because keeping the dearest usually costs a point or two rather than the pair, which is what the loss figure shows.

The best three are not the best two plus one #

Every rule compared here adds cuts one at a time, and so commits to its first choices. That is harmless if the best three cuts always contain the best two. They do not. On four-island seas the best pair is part of the best three in 17 seas of 20; on five-island seas, in only 10 of 20.

Where nesting fails, no rule that adds one cut at a time can be right at both counts, whatever it knows. A cartographer who has cut the best two islands and is offered a third should, in half the five-island seas, undo one of the two. That is why even the rule that tries every fit for the next cut names the best three in only 16 seas of 20. Choosing three cuts of five well is a choice of a set rather than a sequence, and the cheap rules that served for two do not extend.

How many cuts a sea is worth #

The best share against the number of cuts rises steeply and then stops. One cut buys about three quarters of what cutting every island buys. Two buy 98 per cent on the median four-island sea and 91 on the median five-island one. Three buy more than 99 per cent on both. The fourth cut on a four-island sea, and the fourth and fifth on a five-island sea, buy what is left of a point.

Measured as the fewest cuts that buy 95 per cent, one four-island sea is worth a single cut, thirteen are worth two, five three, and one is worth four. Of the five-island seas, one is worth one cut, eight two, eight three and three four, and none is worth five. Thirty-nine seas of forty are worth fewer cuts than they have islands. The cut that costs most to make — the last — is almost never the one worth making.

The count follows the gaps in bearing #

The three-island essay’s suggestion was that the count is the number of directions the islands point. Counted by grouping islands whose bearings lie within 50° of a neighbour, the cuts a sea is worth correlate with its directions at 0.52 across the forty seas, and with its number of islands at only 0.20. In 34 seas of 40 the count is no larger than the directions, and in 17 it equals them exactly. The direction count is the better predictor by a wide margin, and it is not a rule: seas with three directions are worth anything from two cuts to four.

Part of the scatter is the grouping. A threshold of 50° is a choice, and two islands 55° apart are counted as two directions though their logarithms overlap substantially. A measure without a threshold does better.

The widest arc of bearings that holds no island is the three-island essay’s measure, and it carries over. Where two cuts buy less than 95 per cent — eighteen seas of forty — the widest empty arc is a median 124°; where they buy more, 175°. Across all forty seas the share two cuts buy correlates with the widest empty arc at 0.66, and the number of cuts worth spending at −0.70: the more of the sea’s surrounding bearings the islands leave empty, the fewer cuts it needs.

The geometry is the same as for three islands. A sea whose islands crowd one side leaves a wide arc of empty bearings, and from the far coast across that arc all of them look alike; two cuts serve them, one on each flank of the crowd. A sea whose islands are spread evenly leaves no wide gap, its coast sees islands in every direction, and each direction needs a logarithm of its own. What the extra islands change is only that a sea can now be spread evenly in more ways, and the count of directions it holds can reach three or four.

A procedure for an archipelago #

Put together, the measurements give a procedure a cartographer could follow for a sea of four or five islands without fitting a single map of the whole sea, and say where it will go wrong.

Price each island alone — the one-island problem, which a ring can be drawn whole, and only one way solved exactly for an island at the centre and the earlier essays priced for one anywhere. Cut the dearest. For the second cut, take the island farthest from it in bearing; on two seas in three that is the best pair, and on most of the rest it gives up a point or two. Then look at the widest arc of bearings the islands leave empty. If it is wide — most of a half-circle — the sea is probably served: 19 of the 24 seas whose widest empty arc exceeds 150° buy more than 95 per cent with two cuts. If it is narrow, the islands point several ways and a third cut is worth having, but which island takes it is no longer a question the protractor answers reliably, and on a sea of five it is worth trying the three candidates — and worth asking whether the second cut should move once the third is placed.

The procedure’s failures are known in advance, which is the useful property. It errs where the island farthest in bearing is too small to matter, which its one-island prices will show; and it errs at three cuts of five, which is where it stops being offered as a rule.

How the cuts were checked #

Four identical islands evenly round the centre must value alike pairs alike. The four pairs of neighbours must be worth the same and the two pairs across the centre the same, to within the solver’s tolerance of a point. They are.

Every set of cuts must buy something short of all. Every one of the 960 cut sets scored lies between zero and one, to within the same tolerance.

More cuts must not buy less. On every sea, the best share at each count of cuts must be no smaller than at the count below. It never is.

The finding has to be there to fail. On seas of both sizes, dearest first and then by bearing must name the best pair at least as often as the two dearest alone. It names it in 15 and 13 seas of 20, against 9 and 7.

Where the seas are a model #

Twenty seas of each size. The three-island essay used forty; these use twenty of each, because a five-island sea costs thirty-two fits. Every count above is a count out of twenty, and differences of two or three seas between rules are within what another twenty seas could reverse. The rules’ order at two cuts, the bearing rule’s failure at three, and the nesting failure are far outside that.

Islands are circles of two sizes. Islands of 3° and 4.5° radius on a sea of 30°. A long thin island has a logarithm that is not a point’s, and two such islands on one bearing need not duplicate each other.

A threshold for directions. Fifty degrees decides what counts as one direction, and a different angle would move the scatter in the count figure. The widest empty arc has no threshold, which is why it is the stated predictor.

A cut costs nothing but itself. What a cut buys prices a cut by its length, and giving up continuity by what a reader loses at it. The island across the sea is usually the longer cut to the coast, and the count of cuts worth spending would fall further if each were charged.

Still open: whether a weighted rule chooses the set #

The bearing rule fails where the island farthest in bearing is too cheap to matter, and the fitting rule fails where the best set is not a sequence. A rule that discounts each island’s price alone by how nearly its bearing duplicates the cuts already made, and chooses the whole set at once rather than one cut at a time, would address both — and could be computed from one-island prices and bearings, without a single multi-island fit.

Whether such a rule names the best three cuts of five as often as trying every set does, whether its discount has a form the logarithms themselves dictate, and whether the widest empty arc then predicts the count of cuts worth spending as well on a sea of ten islands as on a sea of five, are questions a choice made one cut at a time cannot ask. Two charts are enough, and one is not found a count for the sphere as a whole; whether a sea’s count can be read from its islands’ bearings before any fit is made is the version of that question an archipelago asks.