Flat-folding

A loop that goes somewhere

Every crease says which of its two panels lies above the other, and a loop in those statements is a proof that the pattern has no flat folded state. On a sheet with no edge that sentence is false. The loops of a periodic pattern carry a lattice step each, and a loop that ends one cell to the right is not a contradiction — it is a stack of paper with no bottom layer.

Assumes A proof in one pass and What the rim was doing.

The cheapest true thing anybody can say about a crease pattern with letters on it is this. Each crease separates two panels, and the letter says which of them ends up above the other when the sheet is folded. That is one relation per crease, obtained by reading the pattern once, and if those relations contain a cycle — this panel above that one, that one above a third, the third above the first — then the letters describe no stacking of paper at all.

It is a genuine proof and it costs one pass over the crease list, which is why it is the only negative answer available on a pattern large enough to be interesting. Enumerating stackings gives up at a dozen panels; a cycle can be found on a pattern of two hundred.

The sentence has a hypothesis in it that has never been written down, and this collection has been applying it for its whole existence to the patterns least entitled to it.

Where the hypothesis is

The relations are a partial order in the making. Panel a below panel b, b below c, and so on; a cycle says the order is impossible, and no cycle says the order can be completed to a stacking of all the panels from bottom to top.

“All the panels” is the hypothesis. A sheet of paper has finitely many panels because it has an edge, and a finite acyclic set of relations can always be extended to a list. Remove the edge and the panels do not stop.

A periodic pattern — a twist tessellation, a corrugation continued for ever, anything that repeats — has one panel for every panel of one period and every step of the lattice. Its relations are infinite too. What can be written down is the quotient: the panels of one period, with each relation carrying the lattice step it takes. A relation might join two panels inside the same period, or join a panel to one in the period next door.

One period of the square twist tessellation, with its edges joinedThe crease pattern of a single repeating cell of a twist tessellation on the square grid, drawn on the rectangle it repeats in. The rings mark where a crease meets a side of the cell: each one on the left is the same crease as one on the right, and each on the bottom the same as one on the top. Joined that way the 40 pieces are 32 creases, the 25 drawn panels are 16, and all 16 vertices are interior.one period of the square grid's twist tessellationa ring is where a crease leaves and returns on the far side40 crease pieces → 32 creases25 drawn panels → 16 panels16 vertices, every one interiorV − E + F = 0mountainvalleyraw edge
Fig. 1 One period of a square twist tessellation. Sixteen panels of the sheet, sixteen vertices, thirty-two creases — and eight of those creases run out of one side of the rectangle and back in at the other, so the relations they force reach into the next period.

There is a modest bookkeeping problem in writing that down, and it is worth naming because it is the only technical thing in the essay. The panels of one period are not quite the panels of a rectangle drawn on the pattern: a rectangle cuts some panels in two, and the two halves — one at the left edge, one at the right — are the same panel of the sheet. So the quotient’s panels are the rectangle’s panels with those pairs identified, and each identification records that one half sits a period away from the other. A two-period square rectangle shows twenty-five panels and the sheet has sixteen.

The identifications have to agree with each other. Two different routes from one panel to another must record the same displacement, and if they do not, the drawing was not a period of anything. On every cell here they agree to within a part in a hundred million million, which is the arithmetic saying the rectangle really does repeat.

What a cycle in the quotient is

Follow a chain of relations round the quotient until it comes back to the panel it started from. Down in the pattern itself, that chain is a chain of relations too — but each step also moves by whatever lattice step its relation carried, so the chain ends not at the panel it started from but at the copy of that panel some number of periods away.

A closed walk in the quotient is a cycle of the pattern only if its lattice steps add to nothing.

A walk that comes back to the same panel one period to the right has not closed. It says: this panel is below the panel one cell to its right, which is below the panel two cells to its right, and so on for ever. That is not a contradiction. It is a stack with no bottom, and an infinite sheet of paper is entitled to one.

The certificate for the square cell's loopsEach row is one step of the argument that no closed walk in this lettering's layer arcs has its lattice steps adding to zero. A direction on which no loop descends removes every arc with slack to spare; what remains splits into smaller strongly connected pieces and the next direction is asked of those. 3 directions empty it.ruling out the square cell's loops, one direction at a timedirection (0, -1)62 arcs go, 6 remainwhat is left splits42 arcs go, 4 remaindirection (-1, 0)11 arcs go, 1 remainwhat is left splits01 arcs go, 0 remaindirection (1, 0)11 arcs go, 1 remainwhat is left splits01 arcs go, 0 remainthe bar is how many arcs are still in play after the step
Fig. 2 Ruling out the loops of the smallest square cell, one direction at a time. Each round removes the relations that no zero-summing walk could use; what is left splits, and the next direction is asked of the pieces.

The square tessellation says it plainly

Take the square twist tessellation, one period across and one up, and glue its edges. Four panels, eight creases, four vertices.

Apply the rule that a cycle is a contradiction and the search exhausts in three steps — not runs out of budget, exhausts: a proof that no lettering of this pattern passes. Apply the rule that only a zero-summing walk is a contradiction and a lettering turns up in three steps as well.

At two periods the numbers separate. The first rule exhausts in thirty-five steps and finds nothing. The second finds a lettering in nine. At three periods it is three thousand four hundred and fifty-five against six hundred and twenty-five, and at four the first rule does not finish inside two hundred thousand while the second returns a lettering.

Every one of those exhaustions is a proof, and every one of them is of something false.

Two tests on a sheet with no edgeFor each tiling, one 2×2 glued cell searched twice. The middle column applies the collection's own rule that a cycle in the layer arcs is a contradiction, and it exhausts with nothing found. The right column asks instead whether a cycle's lattice steps add to zero, and finds a lettering.the same 2×2 glued cell, searched under two rulesa cycle is a contradictiona cycle whose steps add to zero isand what the loops dothe square gridnothing, in 359 nodesevery loop travels (2 directions)the triangular gridnothing, in 12,143455 nodesevery loop travels (2 directions)the honeycombnothing, in 9,6191,043 nodesevery loop travels (3 directions)the elongated triangular tilingnothing, in 9,123162 nodesevery loop travels (5 directions)the rhombille tilingunfinished at 200,000unfinished at 200,000“nothing, in n” is an exhausted search: a proof that the pattern has no consistent lettering, which is false
Fig. 3 The same glued cell searched under the two rules, on five tilings. The middle column proves there is nothing to find; the right column finds it.

The two proofs are different animals

It is worth dwelling on what an exhausted search is, because the phrase does most of the work in the paragraph above.

A search that runs out of budget has learned nothing: it says the answer was not found in the time allowed. A search that exhausts has visited every possibility the conditions permit and rejected all of them, which is a proof in the ordinary mathematical sense. Thirty-five steps is not a small number of tries; it is the whole of a tree, closed.

So the first rule does not merely fail to find the lettering. It produces a certificate that the lettering does not exist, and the certificate is valid reasoning from a false premise — the premise being that a loop among these sixteen panels means what a loop among the panels of a piece of paper means.

That is a more interesting kind of error than a bug. Nothing in the code is wrong; a sentence in the mathematics is missing.

Which way the loops go

The lettering the second rule returns has loops in its quotient relations — that is why the first rule rejects it. What those loops do is travel.

On the two-period square cell the sixteen panels fall into two groups of eight. Inside one group every closed walk ends one period to the right of where it began; inside the other, one period to the left. Neither group contains a walk that ends where it started, and no walk crosses between them and comes back.

Read as paper: going one cell to the right takes the reader one layer up, for ever, in one half of the pattern, and one layer down in the other. There is no bottom sheet and no top sheet. Every panel has paper under it and paper over it, and at any particular point of the plane only finitely many panels lie over that point, so the reader looking at the folded sheet sees an ordinary finite stack everywhere and an order with no least element overall.

Why a single direction is not the certificate

The obvious way to prove that no walk closes is to find a direction in which every walk climbs: a compass bearing such that each loop’s lattice steps, added up and measured along it, come to something strictly positive. Then no sum of loops can be zero, and there is nothing to argue about.

That is sound and it is not enough, and the smallest cell of the square tessellation is the counterexample worth keeping.

Its four panels are all reachable from each other. Among their loops there is one that steps a period to the left, one that steps a period to the right, and one that steps a period down. No direction is positive on both the first two. And there is still no walk that adds to zero — because the only way to get from the left-stepping loop to the right-stepping one and back is a walk that steps down, and nothing anywhere steps up to pay it back. The arrangement is a little like a ring of relations that a construction’s own suggested lettering forces seen from outside: what looks locally like a closed circuit is, in the sheet it belongs to, a spiral.

So the certificate is a sequence rather than a single bearing. Choose a direction on which no loop descends; every relation with room to spare under it can be discarded, because no zero-summing walk could have used it. What is left falls apart into smaller pieces, and the next direction is asked of those. Three directions empty the smallest square cell; two empty the two-period one.

The certificate for the triangular cell's loopsEach row is one step of the argument that no closed walk in this lettering's layer arcs has its lattice steps adding to zero. A direction on which no loop descends removes every arc with slack to spare; what remains splits into smaller strongly connected pieces and the next direction is asked of those. 3 directions empty it.ruling out the triangular cell's loops, one direction at a timedirection (-1, 0)186 arcs go, 18 remainwhat is left splits126 arcs go, 12 remaindirection (0, 1)31 arcs go, 3 remainwhat is left splits03 arcs go, 0 remaindirection (0, -1)71 arcs go, 7 remainwhat is left splits07 arcs go, 0 remainthe bar is how many arcs are still in play after the step
Fig. 4 The same argument on the triangular tessellation’s smallest cell, which also needs three directions. A single bearing settles neither.

The check that has nothing to do with tori

An argument this abstract deserves a witness that owes it nothing, and there is one available: write the periodic lettering onto an ordinary square patch and hand it to the instruments that read sheets of paper with edges.

Those instruments are the four vertex conditions and a folded sheet rebuilt from the coordinates and walked for a circle. Neither has any notion of a period, a quotient or a lattice step. On patches of one, four and nine periods, on four tilings, up to fifteen hundred creases and seven hundred and twenty vertices, every vertex condition holds and no patch has a loop.

The lettering that was proved impossible, checked on paper with an edgeEach bar is one clipped patch carrying the periodic lettering, its length the number of creases. Every patch passes all four vertex conditions and has no forced loop in its layer order, on 4 tilings and at 3 sizes.the impossible lettering, on ordinary patchessquare ×140 creases16 vertices · every condition holds · no forced loopsquare ×2144 creases64 vertices · every condition holds · no forced loopsquare ×3312 creases144 vertices · every condition holds · no forced looptriangular ×1116 creases48 vertices · every condition holds · no forced looptriangular ×2424 creases192 vertices · every condition holds · no forced loophexagonal ×1116 creases48 vertices · every condition holds · no forced loophexagonal ×2424 creases192 vertices · every condition holds · no forced loophexagonal ×3924 creases432 vertices · every condition holds · no forced loopelongated ×1184 creases80 vertices · every condition holds · no forced loopelongated ×2688 creases320 vertices · every condition holds · no forced loopthe bar is the crease count; the note is what the ordinary checks said
Fig. 5 The lettering the first rule proved impossible, on ordinary clipped patches. The instruments checking it are the ones this collection uses for a sheet of paper.

That is exactly what the theory predicts. A finite patch of an infinite acyclic structure is a finite acyclic structure, so if the loops of the periodic pattern all travel, no square cut out of it can contain a closed one. The check could have come back the other way and did not.

The counts, and the direction they run in

Two things about the glued cell are worth setting beside each other, because they make the failure of the old rule feel inevitable rather than surprising.

The glued sheet has fewer creases than the rectangle it came from — thirty-two against forty at two periods, a hundred and twenty-eight against a hundred and forty-four at four — because the cut’s severed pairs are reunited. And it has more structure, because every panel now has neighbours on all sides. A rectangle’s rim panels have paper on one side and nothing on the other, so the relations peter out there; on the glued sheet they wrap round and keep going.

Relations that wrap round are exactly the relations that make loops. So the glued sheet has loops for the same reason it has no rim, and a rule that treats every loop as fatal was always going to reject it. What is surprising is not that the rule rejects the glued cell but that nobody had noticed the rule had a domain, and the reason nobody had noticed is that until the sheet with no rim could be built there was nothing outside the domain to try it on.

What cutting a sheet out of a tessellation addsEach bar counts the creases that a rectangular cut divides, which become two independently lettered creases on the cut sheet and are one crease on the glued one. The note gives the two crease counts and the number of vertices, which is the same either way: the cut runs between the vertices and changes no condition asked of any of them.what a cut adds, in letterssquare ×148 creases become 12 · 4 vertices either waysquare ×2832 creases become 40 · 16 vertices either waysquare ×31272 creases become 84 · 36 vertices either waytriangular ×11024 creases become 34 · 12 vertices either waytriangular ×22096 creases become 116 · 48 vertices either waytriangular ×330216 creases become 246 · 108 vertices either wayhexagonal ×11024 creases become 34 · 12 vertices either wayhexagonal ×22096 creases become 116 · 48 vertices either wayhexagonal ×330216 creases become 246 · 108 vertices either waythe bar is how many creases the cut divides; nothing else about the two sheets differs
Fig. 6 The severed creases, counted. Each one is a pair of letters on the cut sheet and a single letter on the glued one, and each is a relation that wraps round rather than petering out.

What the collection had, and what it was

The test being corrected is not a small piece of machinery. It is the argument behind a proof that reads the crease list once, behind the loop that a construction’s own suggested lettering forces, and behind the whole distinction between letters that agree at every vertex and letters that describe a sheet — which is the difference this collection returns to most often.

None of that is wrong. Every pattern those essays are about is a finite sheet with an edge, and on a finite sheet with an edge acyclicity is exactly right, because “no cycle” and “extends to a stacking” are the same statement about a finite partial order.

What was missing was the sentence saying so. The rule was imported as a cycle means no folded state rather than as a cycle in the relations among finitely many panels means no folded state, and the difference only becomes visible on an object this collection could not previously build.

What replaces it, and what it costs

The replacement is not free, and the honest accounting is worth giving.

Deciding whether some closed walk sums to zero is much dearer than deciding whether any closed walk exists. Acyclicity is one sweep; the other question needs the walk structure taken apart direction by direction. So the search asks the cheap question first — a lettering with no loop at all in its quotient has no loop in the pattern either, so acyclicity passing is sufficient and settles nearly every step — and only what the cheap question rejects costs the expensive one.

On the two-period square cell that is five of nine steps. On the four-period cell it is fifty thousand five hundred and forty-six of fifty-six thousand seven hundred and seventy-two. The proportion rises with size, because on a larger cell almost every partial lettering has a loop in it somewhere.

What it costs to prove the wrong thingThe bar is how many nodes the collection's own consistency rule takes to exhaust its search of a glued cell — that is, to prove that no lettering of it is consistent. The note gives what the rule that reads each arc's lattice step cost instead, on the same cell, to find one.proving the glued square cell has no lettering1×1, 4 panels3proved there is none · the other test found one in 32×2, 16 panels35proved there is none · the other test found one in 93×3, 36 panels3,455proved there is none · the other test found one in 6254×4, 64 panels200,000still running at the budgeta bar at the budget is a search still running, not a proof
Fig. 7 What the discarded rule costs when it is applied anyway: the nodes it takes to prove a glued cell has no consistent lettering, against what the right rule takes to find one.

The same reading on the other tilings

The square tessellation is the easiest to describe and it is not a special case.

The triangular tessellation’s period is a rectangle one tiling unit across and 3\sqrt3 up, holding twelve panels, twenty-four creases and twelve vertices at its smallest. The first rule exhausts it in seven steps; the second finds a lettering in eight, and its loops need three directions to rule out. The honeycomb’s cell has the same counts and behaves identically. The elongated triangular tiling’s cell — twenty panels, forty creases, twenty vertices — is exhausted in three steps and lettered in eleven.

At two periods the gap opens on all of them. The triangular cell is exhausted in twelve thousand one hundred and forty-three steps and lettered in four hundred and fifty-five; the honeycomb’s in nine thousand six hundred and nineteen against one thousand and forty-three.

Five tilings, four of them settled, and on every one the same shape: the first rule proves there is nothing, and there is something.

One period of the hexagonal twist tessellation, with its edges joinedThe crease pattern of a single repeating cell of a twist tessellation on the honeycomb, drawn on the rectangle it repeats in. The rings mark where a crease meets a side of the cell: each one on the left is the same crease as one on the right, and each on the bottom the same as one on the top. Joined that way the 34 pieces are 24 creases, the 23 drawn panels are 12, and all 12 vertices are interior.one period of the honeycomb's twist tessellationa ring is where a crease leaves and returns on the far side34 crease pieces → 24 creases23 drawn panels → 12 panels12 vertices, every one interiorV − E + F = 0mountainvalleyraw edge
Fig. 8 The honeycomb’s twist tessellation, one period. Twelve panels, twelve vertices, twenty-four creases — and nine of those creases leave the rectangle and come back on the far side.

What is still unsettled

The two answers this gives — a lettering whose loops all travel, or a walk that genuinely closes — do not exhaust the possibilities in principle. A pattern could have loops that neither peel away nor close within any window examined, and the honest report on such a case is that it is open rather than that it is fine. Nothing here has produced one, which is a fact about the patterns tried rather than a theorem.

And the rhombille tessellation, which is the tiling whose vertices are not all alike, does not oblige at two periods: the search does not finish under either rule, so what its glued cell is remains unknown. That is the same tiling that has been the exception to every measurement in this thread, and it is not obviously a coincidence.

What a travelling loop looks like on paper

The abstract statement has a concrete consequence a reader can check on a patch, and it is worth having because it is the only part of this essay that touches ink.

If every loop travels, no panel of the pattern has nothing below it — follow the chain downward and it never terminates. So on a patch of that pattern, the panels with nothing below them can only be the ones whose neighbours below were removed by the cut.

They are. On twelve patches over four tilings, holding between twenty-five and seven hundred and ninety-three panels, every panel with nothing under it touches the paper’s edge and none is in the interior. The count grows with the rim rather than with the sheet: one, two and three on the square patch as it goes from one period to nine.

The bottom of the stack sits at the paper's edgeFor each patch carrying a periodic lettering, the bar counts the panels with nothing below them in the order the letters force — the bottom of the stack. The note gives the panel count, how many panels touch the paper's edge, and where the minimal ones are. On all 10 patches every one of them is at the edge.panels with nothing below them, and where they aresquare ×1125 panels, 16 of them touching the edge · all 1 at the edgesquare ×2281 panels, 32 of them touching the edge · all 2 at the edgesquare ×33169 panels, 48 of them touching the edge · all 3 at the edgetriangular ×1369 panels, 39 of them touching the edge · all 3 at the edgetriangular ×25233 panels, 79 of them touching the edge · all 5 at the edgehexagonal ×1469 panels, 39 of them touching the edge · all 4 at the edgehexagonal ×27233 panels, 79 of them touching the edge · all 7 at the edgehexagonal ×310493 panels, 119 of them touching the edge · all 10 at the edgeelongated ×12105 panels, 48 of them touching the edge · all 2 at the edgeelongated ×23369 panels, 96 of them touching the edge · all 3 at the edgethe sheet these letters belong to has no such panel at all
Fig. 9 The consequence, counted: panels with nothing below them, on patches of a lettering whose loops all travel. Every one of them is at the edge.

What the picture cannot show

A drawing of the quotient’s relations is a drawing of a graph, and the lattice step each relation carries is a label rather than a direction on the page. So a figure can show the walks and it cannot show the plane the walks are moving through, which is where the whole argument lives. The reader has to hold the picture of an infinite sheet in mind while looking at a finite one.

Nor is a stack with no bottom something a photograph of paper could ever show. Any actual sheet is finite, so any actual folded model has a bottom layer, and it sits at the edge — which is the subject of the essay that asks where the bottom layer of a patch is. The unbounded stack is a property of the idealised infinite pattern, and the idealisation is the thing every claim about a tessellation has always been about.

There is a third thing no figure here shows, and it is the reason the essay is about a test rather than about paper. Whether a set of relations with no closed walk can actually be completed to a stacking of infinitely many panels is a question about orders, not about folding, and the answer is yes for reasons that have nothing to do with creases. What a picture of a crease pattern can show is which relations there are; what it cannot show is why their acyclicity is enough, which is the same gap between a necessary condition and a sufficient one that runs through the whole subject.

What this makes readable

Essays that name this one as a prerequisite.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

What links here

The 8 essays that link to this one and share the most of its objects, of 14 that link here.

The objects this essay names

Each one links to every other essay that touches it.

AssignmentBoundaryCrease assignmentExhaustive searchInterior vertexLayer orderLayer orderingPanelPeriodicityTessellation