A grid that will not close

The square grid is the least exotic pattern in this subject. It is what a sheet of paper looks like after being folded in half a few times in each direction, it is the substrate a whole design method is built on, it is the object the oldest open problem in the field is about, and there is nothing about it that is hard.

Join a piece of one to itself and it stops folding, half the time, for a reason that is nothing to do with the grid.

The construction #

Take the plane ruled into unit squares, with a crease along every line. Cut a rectangle $n$ squares across and $n$ up, placing its corners at half-integers so that the rectangle’s edges fall midway between creases and miss every crossing. Then declare its left edge to be its right edge and its top to be its bottom.

The result is a torus of paper with a grid on it: $n^2$ vertices, each of degree four, and $n^2$ panels.

At $n = 2$ it folds. At $n = 3$ it does not. At $n = 4$ it folds, at $n = 5$ it does not, and so on.

Why #

A path running once round the torus in one direction crosses $n$ creases. It cannot be shrunk to a point, so the parity of that count is not forced by anything happening at a vertex, and it becomes a condition on the whole sheet: crossing a crease exchanges which face of the paper is up, so the path returns the paper flipped exactly when $n$ is odd.

A sheet on which some closed path returns the paper flipped has no consistent two-colouring of its panels, and a sheet with no two-colouring has no flat folded state. That is the same argument a loop of paper makes, on a sheet with two loops instead of one.

The two computations #

The parity can be read two ways and the collection computes both, deliberately without sharing any code between them.

The first draws a path straight across the cell and counts the creases it crosses. It is arithmetic on the flat drawing and it never mentions folding. The path is swept over every height that misses a vertex, because a single line can slip through a crossing and read two creases as one, and what has to be constant across the sweep is the parity rather than the count — a path across a Yoshimura cell crosses more diagonals at one height than at another and always crosses an even number.

The second folds the cell flat and compares the motions of two panels that the gluing says are the same panel. If their motions differ in the sign of the determinant, the fold has turned the paper over between one cell and the next, and the identification cannot be satisfied.

They agree everywhere, which is the check that matters. A claim about a parity is exactly the shape of claim that a single dropped sign produces, and the defence against that is two computations that cannot share a sign to drop.

What is not the reason #

Three explanations suggest themselves and all three are wrong, which is most of what makes the result worth an essay.

It is not that the grid is hard. A grid is the easiest pattern this collection measures. Cut out of the plane, a cell of it settles in exactly one node of search per panel, at every size, which is the reading the whole family of repeating patterns shares.

It is not that the cell is small. One period refuses, three refuse, five refuse; two, four and six fold. Size is not the variable and the sequence does not settle down.

It is not a defect in the identification. Euler’s number comes out at nought on every one of these sheets, the vertex count is unchanged by the gluing, and the letters and panels behave exactly as they do on the cells that fold. The construction is producing a perfectly good torus; the torus has no flat folded state.

A grid with an odd number of columns #

It is worth translating the result out of the torus and into something a person could hold, because there is a version of it in the hand.

Roll a sheet of paper into a tube and crease it lengthwise. The creases run along the tube; a path round the tube crosses each of them once; and the tube flattens exactly when the number of creases is even. Two lengthwise creases flatten a tube — that is how a paper straw is pressed — and three do not.

The torus is that argument twice over, in two directions at once. Rolling the tube’s ends together as well adds a second loop with its own count, and a sheet needs both counts to be even.

The tube version is the one a folder meets. The torus version is the one this collection can build, and it is the same parity.

The paper version, done properly #

The tube is worth actually making, because it turns an abstraction into something with a definite feel.

Cut a rectangle of paper about twenty centimetres by ten. Crease it once down the long direction, at the halfway line. Now roll it into a tube by bringing the two long edges together and taping them.

That tube has two creases running along it: the one that was marked and the one the tape makes. It presses flat immediately, into the shape a flattened drinking straw has, with the two creases becoming the two edges of the flattened strip.

Now start again with a fresh rectangle and crease it twice, at a third and two thirds of the way across. Roll and tape. That tube has three creases along it, and it will not press flat: pushing two of them together forces a fourth to appear somewhere, and the fourth crease is the paper solving the problem the only way it can, by making the count even.

The tape counts. That is the part that surprises people and it is exactly right — the seam is a line the flattened tube has to bend along whether or not anyone creased it, so it belongs in the tally with the rest.

Where the tape does not count #

There is a genuine subtlety hiding in that last paragraph and it is worth separating out, because the seam of a cylinder and the seam of a Möbius band behave differently and this essay’s seam is neither.

On a tube pressed flat, the seam becomes one of the flattened edges, and it is a crease of the folded state whether or not it was creased in advance. So it counts among the creases the loop crosses.

On the torus above, the cell’s edges are not creases. They are ordinary points in the middle of the paper: the identification joins a crease piece running off one edge to the piece arriving at the other, and the join is in the interior of a crease rather than at a fold. A path crossing the cell’s edge crosses no crease at all there.

That is why the count for the torus is the number of creases inside the cell rather than that number plus one. It is a small distinction and getting it wrong shifts every parity by one, which turns a correct result into its exact negation.

The count at other cell shapes #

The cells above are square — $n$ periods across and $n$ up. Nothing requires that, and taking them apart shows which loop is doing what.

A cell three across and two up has a loop crossing three creases in one direction and two in the other. The first is odd, so the sheet is refused; and it is refused by the across gluing specifically. Gluing only the top and bottom edges, leaving the left and right as rim, gives a cylinder whose loop crosses two creases, which is even, and that cylinder folds.

So the two directions are independent conditions and a sheet can satisfy one and fail the other. On the square grid the two conditions are the same condition because the two directions are alike, and on anything anisotropic they are not: the Miura crosses one crease per period in one direction and four in the other, so one of its directions has a parity condition and the other cannot have one at any size.

That asymmetry is the reason the two cylinders of a cell are kept apart throughout rather than being treated as one case with half the rim.

The condition, stated for any repeating pattern #

The grid is the clean case and the statement it makes is general.

Take any pattern that repeats, cut a rectangular cell of it whose edges miss every vertex, and identify opposite edges. A path running once round the cell in one direction crosses some number of creases; the parity of that number is the same for every such path, and a sheet whose path returns the paper flipped has no flat folded state.

So for each of the two directions there is one bit, and both have to come out even. The bits are properties of the cell rather than of the pattern, since doubling the cell doubles the count and therefore fixes any parity that was odd. Every pattern that repeats therefore has a torus that folds — take an even number of periods in each direction — and half of its cells do not.

What varies between patterns is only the crossing count per period. The grid crosses one per period in each direction, so its odd cells refuse. The Miura crosses one across and four along, so one direction has a parity and the other never fails. A twist tessellation crosses an even number per period in both directions, so it never fails either way, which is why the collection’s earlier measurements never met the case.

That last one is the useful sentence. A pattern whose pleats come in pairs has no parity problem, and a pattern whose creases are singletons does.

Where this leaves the search #

A refusal on orientation grounds arrives before any lettering is tried, and it is worth being clear that this is not an optimisation.

A search over letterings asks which assignment of mountains and valleys is consistent. That question presupposes an assignment exists to be consistent — and on a sheet that returns the paper flipped there is no consistent assignment of anything, because the arcs recording which panel lies over which cannot be given a direction. Running the search anyway does not fail cleanly: it explores a graph in which one relation has been recorded twice with opposite senses, and it terminates with a verdict about nothing.

That is what was happening before the orientation was checked separately, and it produced a genuine wrong answer: a three-period grid cell reported as having a lettering in one direction and not in the other, on a drawing symmetric between the two.

The repair is to ask the orientation question first and refuse. It costs one comparison of two motions per identified pair, and it makes the difference between a search that answers the wrong question and one that is never asked.

Why nobody meets this #

A parity condition that refuses half of all sizes ought to be famous, and it is not, and the reason is that nobody folds a torus.

The sheets people fold are discs. A disc has no loop that cannot be shrunk, so every closed path bounds, so the parity is forced by what happens at the vertices inside the path, and the condition has no independent content. It is satisfied automatically wherever the vertex conditions hold, which is why the two-colouring is normally mentioned once as a consequence and then left alone.

The sheets engineers fold are tubes, and there the condition does have content — but the count is small and fixed early, and a designer who has made one tube that flattens will make the next one the same way. The condition is met, satisfied and never named.

The torus is a mathematical object here and not a physical one: a rectangle of paper with both pairs of edges identified does not fit in space without stretching. So the second loop, and with it the second parity, has never had a reason to be noticed by anyone folding anything.

What was actually measured #

For the record, since the claim is a negative one and negatives are easy to get wrong.

Cells of the grid at one, two, three, four and six periods in each direction were built, glued across, along, and both ways, and each was asked two questions: how many creases a loop crosses, and whether the folded motions of two identified panels differ in orientation.

The two answers agree at every size and every gluing. Odd cells: the loop crosses an odd number, the motions differ in orientation, and the sheet is refused before any lettering is considered. Even cells: the loop crosses an even number, the motions agree in orientation and differ by a translation which is the folded sheet’s own lattice vector, and the search for a lettering proceeds and finds one.

The refusals are not searches that failed. They are refusals issued before a search starts, because the sheet has no flat folded state to find a lettering for, and running a lettering search on such a sheet would be asking which arrangement of letters folds an object that does not fold.

Two ways a gluing fails, and this is the first #

The grid refuses for a reason with a name, and the name is worth keeping separate from a second reason that looks similar in a table and is not.

A gluing can bring the paper back the other way up, which is what happens here. There is no two-colouring, there is no flat folded state, and the count that decides it is a parity.

A gluing can also bring the paper back turned through an angle — the fold carrying one cell onto the next by a rotation rather than a slide. That is a different failure, it has no parity in it, and the Yoshimura is the family it happens to.

Until the two were separated they were one number — a count of identified panel pairs whose folded motions disagreed — and that number was reporting two different facts about the sheet under one heading.

The condition among the subject’s others #

It is worth placing the parity beside the conditions a reader will already know.

Kawasaki’s and Maekawa’s conditions are at a point and hold here. The big-little-big lemma is at a point and holds. The two-colouring is the condition that fails, and on a disc it is implied by the others rather than adding anything.

So the grid on a torus is the case where the subject’s one global condition is doing work, and every local condition it has is satisfied.

What it says about the collection’s other results #

Every measurement this collection has made on a glued cell was made on a twist tessellation, and every twist tessellation’s cell turns out to be orientable at every size tried. So nothing already published is affected, and the reason is worth knowing rather than being a relief.

A twist tessellation’s cell has an even number of creases across it by construction: the pleats come in pairs, one entering and one leaving each polygon. The parity is therefore satisfied automatically and never became visible.

The grid has no such structure. Its creases are one per unit in each direction, so the count round a loop is just the number of units, and half of those are odd.

What a reader should take #

That a sheet’s own shape can refuse a pattern the pattern has no trouble with.

Everything about the square grid is easy. It is developable, it satisfies Kawasaki at every vertex by symmetry, it has a well-known flat folded state, and searching it costs one node per panel. None of that survives being glued into an odd torus, and none of it is what fails.

What fails is a count that the pattern does not control and the sheet does. The collection has met that shape of argument before — a loop of paper with three creases passes every vertex condition and cannot be folded — and the grid is the same shape with the sheet doing the refusing rather than a hole.