Eight creases close the circle too

Where a rule can close a loop ended with a prediction it had nothing to test on. Its argument was a count of positions. A repeating rule gives the two halves of a straight line through a vertex the same letter, because both halves lie in the same row or the same column of the pattern. Round a vertex of degree $d$ those two halves sit $d/2$ positions apart. A strict alternation — mountain, valley, mountain, valley, all the way round — gives positions that far apart the same letter exactly when $d/2$ is even.

So a rule can write the alternation round a vertex of degree four and cannot round a vertex of degree six. The four corrugation families swept so far bore that out exactly: the Miura, the tapered leaf and the waterbomb, which have degree-four vertices, all had rules whose letters alternated somewhere, and the Yoshimura, whose vertices are all of degree six, had none.

The statement says more than the families could test. Half of eight is four, which is even, so a rule should be able to alternate round a vertex of degree eight — and no family swept had one.

The case is built here and the answer is yes. What makes it worth an essay is what comes with the yes: a loop twice the length of any the families closed, a hundred and twelve rules that fail nowhere else, a proof that every rule passing the vertex conditions on a three-by-three patch actually folds, and a correction to what the earlier essay said about describing the waterbomb more coarsely.

A grid with eight creases at every point #

The smallest repeating pattern with a vertex of degree eight is the square grid with both diagonals drawn through every cell. Every grid point has eight creases at forty-five degrees to one another: two halves of a row line, two halves of a column line and four half-diagonals. Every cell centre has four, where the cell’s two diagonals cross at right angles.

Both kinds of vertex have all their sectors equal. That settles two of the three local conditions before any letter is chosen. Kawasaki’s angle condition asks the alternate sectors to add to a straight angle, and four sectors of 45 degrees, or two of 90, do. The smallest-sector lemma has nothing to forbid, because no sector is strictly smaller than its neighbours. Every refusal is Maekawa’s, as it was on the waterbomb.

The rule is the waterbomb’s extended to the lines the waterbomb leaves out. The row lines take a letter by the row’s parity and the column lines by the column’s, two bits each; each of a cell’s four half-diagonals takes a letter by the cell’s parity, eight bits. Twelve bits, 4,096 rules. The half-diagonals are separate bits, so a diagonal may change letter where it passes through a point. The row and column lines cannot, and that is what makes them the straight lines the parity argument is about.

One grid point, every way #

Before the sheet, the single vertex. A grid point under such a rule has its two row halves on one letter and its two column halves on one letter, and its four half-diagonals free: six bits, sixty-four letterings.

Twenty-four of the sixty-four satisfy the count. Two alternate, and they are worth writing out: the row and the column carry the same letter, and all four half-diagonals carry the other. Going round, row, diagonal, column, diagonal, row, diagonal, column, diagonal — and the letters flip at every step, because the row and column letters sit at the even positions and the diagonals at the odd ones.

The two cannot overlap. An alternation round any even number of creases has equal numbers of each letter, four and four here, and Maekawa wants them to differ by two. So no alternating lettering ever satisfies the count, at any degree, and every rule that writes one has already failed at that vertex. That is why the loops counted below are a property of failing rules, and it is why the earlier essay could say that an admissible lettering never closes the shortest circle at any vertex.

The parity argument has a cleaner statement than counting positions, and it belongs to a different subject. Draw the eight creases as dots in a ring. Join consecutive dots with a line meaning these two differ, and join each crease to the opposite half of its straight line with a line meaning these two agree. The alternation exists exactly when every dot can be given a letter that keeps every line’s promise. In 1953 Frank Harary characterised such signed graphs, which he was studying as models of attitudes in a social group: they can be satisfied exactly when every cycle crosses an even number of differ-lines. Go from a crease halfway round the ring to its opposite, $d/2$ differ-lines, and back along the agree-line: the cycle crosses $d/2$ of them. So the alternation is available exactly when $d/2$ is even, which is the parity rule, now as an instance of Harary’s balance theorem.

How many rules survive #

On a two-by-two patch 384 rules pass. On three-by-three, 144, and on four-by-four the same 144, rule for rule. The fall is the waterbomb’s lesson repeated: a unit that folds is not a tessellation found 56 of the waterbomb’s rules passing on a small patch and 32 on every larger one, because the small patch did not contain one of each kind of vertex. Here a grid point’s kind is the parity of its row and of its column, four kinds in all; a two-by-two patch contains one grid point and so tests one kind, and the rules that fail at the other three pass there.

From three-by-three upward every kind is present and nothing new can fail. The count stops moving because the patch has stopped hiding anything, not because the rules got better.

The rings of eight #

The prediction is now a count, and it can be made two independent ways.

The first reads letters. For each rule and each interior vertex, list the creases in order round the point and ask whether every consecutive pair differs. The second folds: build the sheet’s folded state under the rule, find the panels with a corner at that point, collect the arrow each crease forces between its two panels — from the one that must lie below to the one that must lie above — and ask whether those panels’ arrows go round in a ring. The loop a vertex cannot close explained why the two should agree: the paper turns over at every crease, so for the arrows to point the same way round the ring the letters have to flip at every step to cancel the turning.

438 rules alternate round some grid point, and at every one of those points the eight panels are forced into a ring. 960 alternate round some cell centre, and the four panels there form a ring every time. Not one rule closes a loop with no alternation anywhere, and not one alternates without closing one. The prediction holds at degree eight with nothing left over.

The ring is twice as long as anything the corrugation families produced. Every loop in the rule that breaks the count was four panels long, round a waterbomb cell centre. Here the loop round a grid point has eight panels, because a vertex’s ring has as many panels as it has creases. It is still the shortest loop that vertex can close. The earlier essay’s claim was always about the ring round one vertex, not about the number four.

The ring has the parity every forced loop in this collection has had. A contradiction is even collected 1,149 loops from irregular letterings across every family and found every one with an even number of panels, and none with four. A ring of eight is even, as it must be — the panels alternate between face up and face down round a vertex, so a ring has as many of one as of the other. What separates these rings from that population is where they come from. Those were circuits through several vertices, every one of them admissible. These are rings round a single vertex that has already failed the count.

And a hundred and twelve rules close their ring of eight with nothing else wrong. Of the 880 rules that fail at grid points and nowhere else — every cell centre passing — 112 alternate round a grid point. On the waterbomb a rule could only reach its loop through a cell centre, where it was failing anyway; on this grid the degree-eight loop arrives on a sheet whose degree-four vertices are all admissible. Which bits decide it can be read off without folding: the loop is in the rule found three bits deciding a grid corrugation’s loops, and here the alternation at a grid point is the row and column agreeing while the four half-diagonals carry the other letter — six bits at one point, read from the rule.

A rule that folds, and a proof that it does #

The rules that pass are the other half of the count, and passing every vertex is a weaker claim than it sounds.

Kawasaki, Maekawa and the smallest-sector lemma are conditions at single vertices, and a sheet can satisfy all three everywhere and still have no way to stack its layers. A unit that folds is not a tessellation left exactly this open for the waterbomb, and the folded objects of thirty-two rules, one object were folded positions, not stackings. So for these 144 the question was asked directly: does each have a folded state — a choice of which panel lies above which, for every pair that overlap, keeping every crease’s letter, never letting a panel pass through a crease or two creases interleave, and never closing a cycle?

All 144 have one. The search places the thirty-six panels one at a time, walking outward across creases, and at each step chooses which of the panels already placed the new one lies above. It checks every rule the instant the panels it mentions are all down, and stops at the first complete stacking.

How it found them is worth a paragraph, because the first attempt did not. Begun from the first panel, three of the 144 ran past two million steps without an answer. Begun from another panel the same three fell in between ninety thousand and eight hundred thousand steps, counting every abandoned attempt, and every one of the 144 was found within a budget that grew by four each round across a few starting panels. That is a heavy-tailed search, and a long run shows itself early is about exactly that tail in the search for letterings. The answer does not depend on where the search starts, only the price does.

So on this pattern, at this size, the vertex conditions are not merely necessary. Every rule that satisfies them at every vertex folds.

What a description can forbid #

Where a rule can close a loop noticed that the count of circle-closing rules is a joint fact about a pattern and the way its letters are described. It then made a specific claim that turns out to be the wrong way round: “A waterbomb rule with one bit per diagonal instead of four would close no circles at its cell centres.”

The waterbomb with each diagonal tied to one letter has thirty-two rules. None folds, and twenty-four alternate round a cell centre and close a circle of four there. The grid described with one letter per whole line has 256 rules: none folds, 102 alternate round a grid point and 192 round a cell centre.

The parity rule says why, and it says it in the direction the earlier sentence missed. At a cell centre two straight lines cross, so its four creases are two halves of one diagonal and two halves of the other, and the halves of each sit two apart. Tying each line to one letter gives the four creases $a, b, a, b$. If $a$ and $b$ differ that is an alternation and a circle of four. If they agree it is four alike. Neither is three and one, so a tied crossing can never satisfy the count. Tying a line’s halves forbids folding there and forbids no alternation, because at degree four, as at degree eight, an alternation wants opposite creases to agree, which is what a tie enforces.

The general form is short. A description can forbid an alternation only by forcing two creases an odd number of positions apart to agree, or two an even number apart to differ. A straight line through a vertex of degree $d$ ties two creases $d/2$ apart, which forbids the alternation when $d/2$ is odd — the Yoshimura’s case — and permits it when $d/2$ is even. The waterbomb’s four separate half-diagonal bits were never what made its circles possible; at degree four they were possible anyway.

The essay that made the claim now carries a correction.

What is drawn and what is assumed #

Repeating rules. Every count here is over rules — letterings that repeat with the parity of a row, a column or a cell. An irregular lettering of the grid can alternate round any vertex it likes, and the counts say nothing about letterings in general. The degree-eight alternation is available to rules, which is what the parity statement predicted; an arbitrary lettering never needed permission.

Small patches. The census is taken on patches up to four by four and the folded states on three by three. The census stops changing at three by three for the stated reason — every kind of vertex is already present — but the folded states were found at one size, and whether every passing rule folds on every larger patch is a separate claim, not made here.

Zero thickness and exact angles. The folded states are stackings of panels with no thickness, at exactly forty-five degrees. A grid point of degree eight is where real paper is thickest — eight panels of a folded state meet there — and the pile, not the panel is the account of what layers cost a material with depth.

The pattern. The square grid with both diagonals is nobody’s design. It is the grid a tessellation folder pre-creases before collapsing anything, and it is here because it is the smallest repeating pattern with a vertex of degree eight, not because anybody folds it lettered by a rule. Its collapses under the 144 rules are folded states that exist; whether any of them is a model somebody would want is a question about origami rather than about parity.

How the rings and the stackings were checked #

Every one of the 4,096 rules is built and put past all three conditions at every interior vertex, on patches of two, three and four cells a side, and the passing sets are compared mask by mask.

The alternations are read off the letters at each vertex with nothing folded, and the rings are found in the folded sheet: the panels with a corner at the vertex, the arrow each crease between them forces, and a walk along those arrows that must visit every panel once and return. A loop found by the separate walk over the whole sheet is required to occur exactly when some vertex alternates. The two readings share only the pattern.

The stackings are searched for, not inferred, by placing panels one at a time under every rule a stack must keep, and the search is checked against the complete list of stackings on two-by-two sheets, where that list can be made: every stacking it returns must be one of the listed ones. On a wider sample of small sheets of both this grid and the waterbomb it agreed with the list about whether any stacking exists forty-eight times in forty-eight.

Still open: where degree eight stops agreeing with degree four #

Degree eight behaved exactly as degree four did. Rings close where letters alternate, the alternation is a property of failing rules, and the rules that pass all fold. That is the prediction confirmed and nothing more. The next degree at which the parity rule makes a claim of its own is ten, where $d/2$ is five and rules should be unable to alternate as on the Yoshimura. Twelve should allow it again. A pattern with degree-ten vertices and a natural repeating rule does not come to hand on a square grid, and finding one is the first step.

The folded states raise a question the waterbomb answered and this grid has not. Thirty-two rules, one object found that all of the waterbomb’s passing rules fold to the same panels in the same places, so that counting rules counted labels. Whether the 144 rules here fold to one object, to a few or to 144 is a comparison of folded positions, cheap to make, and it would say whether this grid’s rule space is as large as it looks. Thirty-two rules, thirty-two pieces found that the waterbomb’s surviving rules, though they fold to one object, sit in thirty-two separate pieces of the space of letterings, none reachable from another without unfolding. That was a statement about a lettering’s neighbours, and the grid’s 144 rules have never had theirs counted.

The search cost is a finding of its own. Three of 144 rules defeated a search started from the first panel at two million steps and gave way at a few hundred thousand from elsewhere. Which panel to start from was found by trying, and a rule for choosing it — from the folded sheet’s structure, before searching — would be worth more than another restart schedule.

The habit worth carrying is about statements made in passing. When an argument predicts a case it cannot test, also check the sentences that described what it does not predict. The prediction about degree eight was the visible claim and it held. The aside about tied diagonals was the invisible one, and it was backwards.