The rim adds up

There is a difference between measuring a cost and measuring a rate, and it is the difference between two data points and three.

A rectangle of tessellation cut out of the plane has a certain number of free letters. The same rectangle glued into a torus has fewer. Subtracting gives a number, and every statement this collection has made about what a boundary is worth has been that number or a multiple of it.

A number obtained by subtracting two others supports exactly one claim: that the two states differ by that much. It does not support the cost is per edge, or the cost is per cell of rim, or four letters a cell — each of which is a claim about a rate, and a rate needs a third point.

What additivity means here #

The rectangle has two pairs of opposite edges. Each pair can be glued or not, independently, so there are four sheets and three intervals between them.

Write $d_x$ for the letters that gluing the left and right pair removes, $d_y$ for the letters the top and bottom pair removes, and $d_{xy}$ for the letters that gluing both removes. The claim is

$$d_x + d_y = d_{xy}$$

on every drawing and at every size.

It is not obvious. Gluing one pair might have made the second gluing cheaper — if some creases were divided by both pairs, they would be counted twice — or dearer, if joining creases created new ones to join. Neither happens, and the reason is worth having.

Why it holds #

A crease of the tessellation appears in the rectangle as one piece or several. It is divided into two pieces by each edge it crosses, and the edges it crosses are decided by the drawing rather than by which gluing is being contemplated.

Gluing the left and right pair rejoins exactly the pieces that the left and right edges divided. Gluing the top and bottom pair rejoins exactly the pieces the top and bottom edges divided. A crease crossing both a vertical edge and a horizontal one is divided into three pieces or four, and each gluing rejoins its own cuts without touching the other’s.

So the two savings count disjoint sets of cuts, and disjoint sets add. That is the whole argument, and it is an argument about the drawing rather than about folding: nothing above mentions a letter’s meaning, a fold, or a vertex.

The panels do not add, and the corner is why #

The same question can be asked of the panel count, and there the answer is no.

A panel touching the left edge and its partner at the right are one panel of the pattern; gluing that pair merges them. The same holds for the top and bottom. But the four pieces at the corners of the rectangle are all one panel of the pattern, and gluing both pairs merges four pieces into one rather than into two pairs — one merge fewer than the sum would predict.

So the panel savings are short by exactly one, on every cell, at every size, and the discrepancy is the corner. The letters have no such correction because a crease does not run through the corner — or rather, it must not, and when one does the construction has a defect that Euler’s count finds.

Why the additive claim is the one that matters #

The letters are what a search chooses among, so the letter count is what governs cost. The panel count is a description of the object.

More to the point, the additive statement is what converts an observation into a formula. The rim of this cell costs eight letters is a fact about that cell. Each glued pair costs the creases it divides, and the pairs add is a rule that computes the answer for any cell of any drawing without running anything — count the creases the vertical edges cut and the ones the horizontal edges cut and add.

That is the difference the third sheet buys.

Where the additivity stops #

Two places, and both are worth naming because the rule is easy to over-apply.

It does not say the two halves are equal. On an anisotropic drawing they are not, and the difference can be large: on the elongated triangular tiling’s twist tessellation, gluing across the pleats saves ten letters per cell and gluing along them saves two. Half the rim names a topology, not a cost, and which pair is glued turns out to matter a great deal more than that.

It does not extend to the search. Nodes of search are not additive in anything. Gluing one pair of a two-by-two square twist cell takes the cost from thirteen nodes to eleven; gluing the other takes it from thirteen to eleven as well; gluing both takes it to nine, which happens to add and does so by accident. At three cells the same measurement gives thirty, eighty-five, twenty-four and six hundred and twenty-five, and nothing about those numbers adds.

What the letters actually are #

It is worth a paragraph on what is being counted, because free letters is a phrase that can float free of its meaning.

Every crease of a pattern is a mountain or a valley in a given folded state. On a rectangle cut out of a tessellation, the two pieces of a divided crease may take different letters — nothing connects them, since they are different creases as far as the rectangle is concerned. So the rectangle has more independent choices than the tessellation does, and the extra choices are exactly the divided creases.

Those extra choices are not free in the sense of costing nothing. They are free in the sense of being unconstrained: a lettering of the rectangle can set them however it likes without contradicting anything, because the constraint they would have violated is on the other side of a cut.

That is the whole reason the rim reads as slack rather than as an obstacle. It supplies choices that cannot be wrong.

The check, run rather than recalled #

Additivity is asserted in the code that draws these figures, not merely observed once and written down, and the distinction matters because a rule of this kind is exactly the sort that holds on the cases somebody looked at.

Every time one of the figures above is drawn, the four gluings of the cell in question are built, their free letters counted, and the sum compared. A row where the two halves did not add would stop the figure from being drawn and would say by how much and on which drawing, rather than producing a picture with a wrong number in it.

That has caught something. The Yoshimura’s cell, at the corner position the search preferred, had a crease running exactly through the corner of the rectangle — and a corner is where four edges meet, so a crease piece ending there has no partner on any one of them. The letter counts then came out short on both cylinders and shorter still on the torus, and the additivity survived while Euler’s number did not, which is how the defect was found. The additive check would have passed a broken construction; the topological one refused it.

Two checks that can both fail and fail differently are worth more than either, and this is a concrete instance of why.

The same rule, on the twists #

The three families used above — the grid, the Miura and the Yoshimura — are the ones with a plane drawing simple enough to state a period for. The twist tessellations, which are most of this collection’s tessellation work, behave the same way and the numbers are larger.

On the square twist’s two-period cell: forty free letters cut out, thirty-six either way with one pair glued, thirty-two with both. Four and four make eight.

On the triangular twist’s one-period cell: thirty-four cut out, twenty-eight one way, thirty the other, twenty-four with both. Six and four make ten.

On the elongated tiling’s one-period cell: fifty-two cut out, forty-two one way, fifty the other, forty with both. Ten and two make twelve.

The last of those is the one to keep. The two halves of half the rim are ten letters and two, on the same rectangle, differing by a factor of five — and both are correct, and their sum is the whole. A drawing that is not isotropic has two different rims, and the sum is the only quantity that is a property of the cell rather than of a direction.

What a rate is good for #

Having the cost as a formula rather than as a measurement changes what can be said without computing anything.

A cell holding $n$ periods across and $n$ up divides $n$ times as many creases with each pair of edges as a one-period cell does, so the rim’s cost grows like $n$ while the panel count grows like $n^2$. The rim’s share therefore falls like $1/n$, which is the arithmetic behind most of a patch being edge at small sizes and very little of it at large ones.

It also means the cost can be read off a drawing by inspection. Count the creases a vertical line through the cell crosses, double it for the two edges — no, once, since one pair of edges divides each such crease once — and that is $d_x$. The same horizontally gives $d_y$. Nothing has to be built.

That is a modest kind of progress and it is the kind that makes the next measurement affordable, since a quantity with a formula does not have to be measured again.

A worked count on the grid #

The smallest case is worth doing in full, because every number in it can be checked with a pencil.

Take a plain grid — creases along every horizontal and vertical line of a unit lattice — and cut a cell two units square whose corner sits at a half-integer, so that its edges fall midway between creases and miss every crossing.

Inside the cell there are two vertical creases and two horizontal ones. They cross at four points, which are the cell’s four interior vertices, and each crease is divided at the crossings into three pieces: from the cell’s edge to the first crossing, between the crossings, and on to the far edge. Four creases at three pieces each would be twelve, and twelve is what the count comes to.

Now glue the left edge to the right. Each horizontal crease ran off the right-hand edge and back on at the left, so its two outer pieces are one piece; two horizontal creases, one saving each, and the count falls to ten.

Glue the top edge to the bottom instead, and the same happens to the two vertical creases: ten again, by the grid’s own symmetry.

Glue both, and both savings happen: eight. Two and two make four, and twelve minus four is eight.

The panels do it differently. Nine pieces of paper cut out; six with one pair glued, since the three pieces down one side pair with the three down the other and one of each pair goes; six with the other pair glued; and four with both. Three and three would make six, and nine minus four is five. The missing saving is the corner, where four pieces of paper are one panel and only three of them are eliminated rather than four.

Additivity elsewhere in the subject #

It is worth noticing how unusual this behaviour is, because most quantities here do not add.

Crease length adds over regions and is one of the few things that does.

Panels nearly add and are short by the corner, as above.

Layers do not add at all: two patterns side by side have a layer count that is not the sum or the product of theirs, because the ordering is global.

Search cost does not add and does not even behave monotonically — a bigger instance can be cheaper.

Flat-foldability is the extreme case: two halves of a sheet can each fold and the whole fail, which is the whole content of the local-versus-global distinction.

Against that background, a quantity that adds cleanly across two independent operations is worth remarking on. What makes it add is that it is a count of cuts, and cuts made by different edges are different cuts. Almost nothing else in the subject is a count of something that simple.

Three points and a line #

The methodological point is small and general enough to state on its own.

Two measurements of the same quantity in two states give one difference. Any function through two points is linear, so linearity cannot be tested with two; and the claim the cost is a rate is precisely a claim of linearity.

Three states, arranged so that two of the intervals are independent, test it. If $d_x + d_y$ had come out different from $d_{xy}$, the rate story would have been wrong and the amount of rim would have been the wrong variable — the cost would have depended on the shape of the boundary rather than on how much there was.

It did not come out different, on three families of drawing at three sizes each, and the check runs whenever one of these figures is drawn rather than having been performed once.

What the claim replaces #

Before the middle sheet existed, the collection’s statement about the boundary was of a particular shape, and it is worth seeing what was wrong with it.

The statement was: a cut rectangle has some number of extra letters compared with the torus, and that number is four per cell of rim. It was arrived at by taking one cell, counting both ends of the scale, subtracting, and dividing by the amount of rim the cell had.

Every step of that is sound and the conclusion is a rate obtained by dividing one difference by one quantity — which is a rate in the way that dividing a journey by its duration gives a speed, and says nothing about whether the speed was constant.

The middle sheet supplies the second interval. It shows that the difference really does split into two pieces, that the pieces are independent, and that they belong to particular pairs of edges rather than to the boundary as a whole. And it shows, on the elongated tiling, that the pieces are not equal — which is the fact the old statement’s division by amount of rim was quietly assuming away.

So the correction is not that the earlier number was wrong. It is that the earlier number was a total presented as a rate, and it happened to be a rate on the isotropic drawings it was measured on.

The general form #

Written once, for a rectangle cut from any periodic drawing:

The free letters of the cut cell exceed those of the glued sheet by the number of crease-crossings the glued edges make. Each pair of edges contributes the crossings it alone makes; the pairs contribute independently; and the total is their sum.

The panels of the cut cell exceed those of the glued sheet by the number of panel-identifications the gluing makes, which is the sum of the two pairs’ identifications less one when both pairs are glued, the one being the corner.

And Euler’s number is one before any gluing and nought after any, which is neither of those and is the check that the gluing did what it says.

Three sentences, of which the first is additive, the second is additive with a correction, and the third is not a count at all.

Where the additive count is used #

The rate matters because several statements here are built on it.

Most of a patch is edge at small sizes is an arithmetic consequence: the rim’s letters grow with the cell’s side and the panels with its area.

The rim is four letters a cell is the rate itself, and it was a total presented as one until there was a middle rung.

What the rim was doing is the reading the rate supports — that a boundary supplies slack — and it is unaffected.

What is left open #

The rectangle’s two pairs of edges are the same length as each other in every cell here, because every cell is a tessellation’s own period and periods are what they are.

So per glued pair and per unit length of rim cannot be told apart by this measurement, and the essays that say four letters a cell of rim are using the first as though it were the second. On these cells the two coincide; on a rectangle twice as long as it is wide they would not, and no such rectangle glues, because a rectangle that is not a period does not match up across the join.

Separating them would need a pattern with two different periods in the two directions and a cell built from a multiple of one and not the other — which is available, and which nothing here has done.