The loop is in the rule

A rule that does not fold a corrugation has failed a condition somewhere, and somewhere is doing a lot of work in that sentence. There are two quite different ways for a lettering to be wrong, and separating them has taken this collection a good deal of effort.

The weaker failure is a vertex that cannot close: the letters round some point break the count, so the paper there will not lie flat. The stronger one is a circle in the arcs: the letters demand that a panel lie above a panel that lies above a panel that lies above the first, which is a proof that no arrangement of the layers exists anywhere in the pattern, however the rest of it is drawn.

Of the forty-eight repeating rules that fail on a grid corrugation, thirty-eight do the second thing and ten only do the first.

Which thirty-eight, and why, is the question. And the answer is three bits.

Where a four-panel circle comes from #

The shortest circle a crease pattern’s panels can make goes round a single interior vertex: the four panels between the four creases of a degree-four vertex, in order, back to the start.

For the arcs round that circuit to point the same way all the way round, the letters have to alternate strictly — mountain, valley, mountain, valley — because the paper turns over at every crease, so a term that flips at every step has to be cancelled by another that flips at every step. And a strict alternation has equal counts, which the counting theorem forbids. That is why no admissible lettering closes it and why every circle in a folding pattern encloses more than one vertex.

But a rule that is not admissible is under no such protection. If it breaks the count at a vertex, the alternation is available there, and the shortest circle in the whole pattern opens up.

So the question does this rule close a four-panel circle becomes: does this rule write a strict alternation round some interior vertex?

Three bits #

Under a repeating rule, an interior vertex of a grid corrugation sees three of the six bits and no others.

The two row-halves through the point are collinear and belong to the same row, so both carry the row’s letter h. The column crease above the point belongs to the row above and carries v; the one below belongs to the row below and carries w. Going round the vertex, the four letters in cyclic order are h, v, h, w.

A strict alternation of that sequence means the two hs agree with each other, which they do automatically, and the two column letters agree with each other and differ from h. So:

A four-panel circle is available at a vertex exactly when v = w ≠ h.

That is a test on three bits. It builds no pattern, places no panel, and computes no arc. A rule closes a circle somewhere if any interior vertex satisfies it, and the vertices differ only in which parity classes they draw their three bits from — so the test runs over four combinations and stops.

Why two tests rather than one #

The closed form is a derivation and derivations are where mistakes live. What makes it evidence rather than a restatement is that the other test shares no line with it.

The arrow walk builds the pattern under the rule, folds it — placing every panel by reflecting it across the creases on a path back to a fixed panel — collects one arrow per fold from the panel that must lie below to the one that must lie above, and runs a depth-first search looking for a back edge. It knows nothing about parity classes or bits. It would find a circle of eight panels or twenty just as readily as one of four.

The closed form reads three bits and returns a boolean.

They agree on a hundred and twenty-eight rules. Neither is checking the other’s arithmetic; they are two accounts of one object, and the agreement is what makes either believable.

It is worth saying what a disagreement would have meant, since the check is only worth running if it could have failed. A rule the closed form called circular and the walk called clean would mean the alternation does not in fact orient the four panels — an error in the sign rule relating a letter to an arrow. A rule the walk called circular and the closed form called clean would mean a circle somewhere other than round a single vertex, which would be a far more interesting result and would contradict the whole account. Neither happened, on either family.

The ten that fail without a circle #

The ten failing rules that close no circle are the ones where the alternation is never written, and the closed form says exactly why: wherever the columns fail to alternate, the row letter agrees with them.

The plainest members are the two constant rules. All-valley writes V, V, V, V round every vertex — the count wants three and one and gets four and nought, so it fails at all fifteen vertices — but four identical letters are not an alternation, so the arrows do not go round. All-mountain is the same the other way.

That is the counter-intuitive part of the table and it is worth dwelling on. The most uniformly wrong rules are the ones whose layer statements are consistent. A pattern lettered all-valley cannot be folded, and the reason has nothing to do with its layers; the paper simply will not close at any vertex. The rules that produce a genuine layer contradiction are the ones that got the count wrong in a structured way — half right, alternating where they should not.

Failure, in this family, is not a scalar. A rule that fails at fifteen vertices may make no contradictory layer statement at all, and a rule that fails at six may make one at each of them.

Thirty-eight and ten, counted rather than reported #

The split is called uneven above and left at that. It is worth deriving, because the closed form gives it exactly and the derivation says which ten.

Write the six bits as two row letters $h_0, h_1$ and, for each column class, the letter it carries in the band above a row and in the band below: $a_c, b_c$. A class sticks when $a_c = b_c$, and a rule fails precisely when some class sticks.

A circle appears at a vertex of a stuck class $c$ when the row letter there differs from that class’s common value $x_c$. The sweep has three rows and they are not all of one parity, so both row letters occur beside every column — and therefore a stuck class closes a circle somewhere unless $h_0 = h_1 = x_c$.

Now count the forty-eight.

One class sticks, sixteen rules each way. Two choices of $x_c$, two ways for the other class to alternate, four row patterns. Suppressing the circle pins both row letters to $x_c$, which is one row pattern of the four: $2 \times 2 \times 1 = 4$ clean, and twelve with a circle.

Both stick, sixteen rules. Four choices of $(x_e, x_o)$ and four row patterns. Suppression now needs $h_0 = h_1 = x_e = x_o$, which is two rules — and those two are exactly all-mountain and all-valley. So fourteen carry a circle.

$$12 + 12 + 14 = 38, \qquad 4 + 4 + 2 = 10.$$

The measured numbers, from three bits and no pattern built.

Which says what the ten have in common #

The decomposition also identifies them, which the count alone does not.

Eight of the ten have one stuck class, and in each the stuck class’s letter is written on both rows as well — so the sheet carries one letter everywhere except in the class that still alternates. The other two are the constant rules.

So every clean failure is a rule that is uniform over everything it gets wrong. A circle needs a disagreement between a stuck column and a row, and a rule that suppresses the disagreement everywhere has nothing left with which to contradict itself about layers — which is the general form of the observation that the most uniformly wrong rules are the consistent ones.

What the groups are #

Counting how many vertices each failing rule breaks makes the whole family legible in one line.

The pattern swept has fifteen interior vertices. The forty-eight failing rules break the count at six of them, at nine, or at all fifteen — sixteen rules in each group, and nothing in between.

The reason is the two column classes. Each class either alternates down the rows or does not. Both alternate: the rule folds. Only the class in the odd columns fails: three columns by three rows, nine vertices. Only the even class: two by three, six. Both: fifteen.

Then within each group the circle test splits on the row letters, and the split is uneven because a row letter has to agree with a stuck column class to suppress the circle, and there are two row letters and two classes to satisfy. Ten of the forty-eight manage it. Thirty-eight do not.

Doing one vertex by hand #

The whole test fits on a corner of a page, and it is worth doing once rather than taking on trust, because the cyclic order is the only part that can be got wrong.

Take an interior vertex of a sheared grid. Four creases leave it: one row-half to the left, one column-half upward, one row-half to the right, one column-half downward. That is the cyclic order — row, column, row, column — and it matters that the two row-halves are opposite rather than adjacent, which is forced by their being two pieces of one straight line.

Now write the letters in that order: h, v, h, w.

A strict alternation is a sequence where every letter differs from the one after it, going all the way round. Position one and position three are both h and are not adjacent, so they impose nothing on each other. Position two is v and position four is w; they are not adjacent either. What the alternation demands is that v differs from h, that w differs from h, and — closing the ring — that h differs from w and from v again. Two independent demands: v ≠ h and w ≠ h.

Two letters both differing from h in an alphabet of two means v = w. So the alternation is precisely v = w ≠ h, which is the closed form, arrived at without any machinery.

The step that is easy to get wrong is the cyclic order. If the two row-halves were adjacent rather than opposite — which they would be at a vertex where the row line turned — the alternation would demand h ≠ h, which is impossible, and no rule would ever close a circle. That is not a hypothetical: it is exactly what happens on the Yoshimura, whose course-halves are opposite but whose two courses at a vertex share a row and so share a letter.

The same shape, in a family the collection already had #

This is the second time the count has been caught doing exactly this, and the first time was not designed to show it.

The waterbomb tessellation’s five hundred and twelve rules were swept earlier. Thirty-two fold. A hundred and twenty of the four hundred and eighty failures close a circle, every one of them four panels long, every one round a single interior vertex of degree four, and not one of the hundred and twenty is among the thirty-two.

The waterbomb’s rule space is a nine-bit family rather than a six-bit one, its vertices come in two degrees rather than one, and its geometry is a square grid with both diagonals rather than a sheared grid. None of that changes the shape of the answer. The count refuses every failure; a subset of the failures write an alternation at a degree-four vertex; those are exactly the ones that close a circle.

Two families of different sizes over different geometries, and the same three sentences describe both. That is what makes it worth stating as a rule about repeating rules rather than as a fact about the Miura.

What the disjointness says #

One number in the waterbomb table deserves more attention than it usually gets: not one of the hundred and twenty circle-closing rules is among the thirty-two that fold. The same holds here — none of the thirty-eight is among the sixteen.

That has to be true, and seeing why is a small check on the whole account. A rule that folds satisfies the count at every vertex, so it never writes an alternation, so it closes no four-panel circle; and a longer circle would need the letters to go round a circuit enclosing several vertices, which no admissible lettering of a corrugation manages at these sizes. So folds and closes a circle are disjoint by construction rather than by measurement.

What the measurement adds is that the disjointness is not vacuous in either direction. There are thirty-eight circle-closers and sixteen folders and ten rules that are neither, so all three cells of the table are occupied, and a claim about their being disjoint is a claim about something.

It also means the cheap test is strictly weaker than the vertex conditions on this family — every rule it refuses was already refused. That is worth stating plainly because the cheap test’s usual selling point is that it catches things the vertex conditions miss, and on a tessellation patch it does. On repeating rules it never does. Which test is stronger depends entirely on what is being tested, and there is no general ordering between them.

What the closed form is good for #

It is not faster in any way that matters — the arrow walk on a pattern this size takes a millisecond — so the value is not speed.

It is that the test says which bit to change. An arrow walk reports that a circle exists and hands back a list of four panels; a folder holding a rule that does not work learns nothing actionable from that. The closed form reports v = w ≠ h at some vertex, which names the fault as a column class that failed to alternate, and the repair is to flip one bit.

That is the difference between a refusal and a diagnosis, and this collection has been on the wrong side of it before. The circle a walk finds on a tessellation patch is six to twelve panels of a hundred and fifty-seven, and it looks like a local fault and is not one; the tangle it lies in covers two thirds of the sheet and there is nothing to change. Here there is exactly one thing to change, and the reason is that a repeating rule has six bits rather than two hundred and eighty-two letters.

Where it does not reach #

The closed form is a statement about four-panel circles under a repeating rule on a grid corrugation, and each of those three qualifications is doing work.

Take away four-panel and it says nothing. A circle of six or eight panels encloses more than one vertex and no test on three bits could see it. The collection’s other populations produce circles of six, eight, ten and up to twenty-two, and the closed form is silent on all of them.

Take away repeating rule and there are no bits to read. An irregular lettering has one letter per crease and the question reverts to walking arrows — which is what every measurement of a tessellation patch’s letterings in this collection does, and why none of them comes with a diagnosis.

Take away grid corrugation and the cyclic order round a vertex changes. The Yoshimura’s vertices have six creases, two of which are course-halves that a repeating rule gives the same letter and which sit opposite one another — so the alternation is unavailable to any rule at all, which is why none of its thirty-eight failures closes anything.

Within those limits it is exact, and the exactness is checkable: a hundred and twenty-eight rules, two tests, no disagreements.

Where the ladder goes next #

The next question is the one the Yoshimura raises by being different, and it is a question about vertex degree rather than about letters. A repeating rule can only write an alternation where the pattern’s own structure lets it, and on a degree-six vertex whose opposite creases are forced equal it cannot. Where a rule can close a circle is that measurement across all four families.

And there is a question this rung leaves open which nothing here answers. The closed form was found by looking at the cyclic order round a vertex and reading off what an alternation would need. Whether the same reading gives a closed form for six-panel circles — the ones that go round two adjacent vertices — is not obvious, and would be the first test of whether this is a technique or a special case.