Theme

Local rules, global behaviour

Two conditions at a single vertex decide whether it folds flat. Whether a whole sheet does is a different question, and a much harder one.
VMMM60°90°120°90°Kawasaki60° + 120° = 180°90° + 90° = 180°both 180° — satisfiedMaekawa3 mountains, 1 valleysdifference 2exactly 2 — satisfiedangles sum to 360°which is what a flat sheet requiresmountainvalley Flat-folding

Two conditions at a point

Whether a single vertex folds flat is decided completely by two tests — one on the angles, one on the assignment. They are independent, they are easy to check, and together they settle the case entirely.

MVMwalk the folded edge and count the turns:each mountain turns +180°, each valley −180°the walk closes, so the total is ±360° — which forces |M − V| = 2the sheet must come back to where it started Flat-folding

Why the difference is two

Maekawa's theorem says mountains and valleys differ by exactly two at every flat-foldable vertex. The constant is not empirical — it is a full turn, and the theorem is about winding rather than about paper.

MVMM40°foldsopposite across the small sectorMMVM40°does not foldthe same on both sidesboth satisfy Kawasaki and Maekawa — the angles and the counts are identical Flat-folding

The smallest sector decides

Two assignments can satisfy both flat-folding theorems and only one of them folds. What separates them is a condition about the smallest angle, and it is the first rule in the subject that is not about counting.

6 interior vertices, every one satisfying both theoremswhat the local tests seeangles at each vertexassignment at each vertexwhat they cannot seewhether layer 3 passes through layer 7whether a flap has room to existwhether the order is consistent everywhereBern and Hayes, 1996: NP-hardso this pattern is checked, not proved Flat-folding

Local is not global

Every vertex can satisfy every condition and the sheet still not fold. Deciding whether a whole crease pattern folds flat is NP-hard, which means no figure will settle it and no algorithm will scale.

taco-tacoallowedforbiddentwo folds at the same place may nest or stand clearthey may not interleavetaco-tortillaallowedforbiddena flat layer may pass outside a foldit may not pass through onea crease pattern can satisfy every vertex condition and still break one of these Flat-folding

Which layer goes on top

The mountain-valley assignment says which way each crease turns. It says nothing at all about which sheet ends up above which, and that second question is a different object with its own rules — and all of the difficulty.

MVMV123455 segments, 4 creases12345the stack, solvedassignmentsMVMVvalid stacks1decided byexhaustive searchover the orderingsthe folded positions come from the crease spacing; the assignment only decides which way each turn wraps Flat-folding

A strip is decidable

Take the same problem down one dimension and it stops being hard. The reason is not that strips are small — it is that overlaps on a line form a chain, and chains cannot contain the cycles that make the two-dimensional question intractable.

at every vertexthree of one, one of the other15 interior vertices, all identicalwhat the sheet gainsone degree of freedom, not manyit opens and closes in both directions at oncea negative Poisson's ratio22 mountain and 16 valley creases · 6.2 sheet-widths of foldingmountainvalleyraw edge Tessellations

One vertex, repeated

Take a single flat-foldable vertex and tile the plane with it. The sheet stops being a sheet and becomes a material — with a stiffness, a packing behaviour and a Poisson's ratio that the paper never had.

nearly flatwidth ×0.91 length ×0.98ν = -0.22half closedwidth ×0.66 length ×0.88ν = -0.52nearly packedwidth ×0.45 length ×0.55ν = -2.93both dimensions shrink together — pulling it open in one direction opens it in the other Tessellations

A sheet with one freedom

A Miura-folded sheet can move in exactly one way. Pull it open in one direction and it opens in the other — a negative Poisson's ratio, arriving entirely from the crease pattern and not at all from the paper.

4 corners, all alikesectors 90°, 90°, 90°, 90°two equal pairs, so no sectoris strictly the smallestthe assignment256 of 4096 fold6 mountain, 6 valleythe ring takes two lettersthe panels can be orderedwhat is checked4 interior verticesand not the tilinga twist of radius 0.17 sheet-widths12 creases, 4.70 sheet-widths of foldingmountainvalleyraw edge Tessellations

A square that turns

A twist is a small polygon that rotates as the sheet closes around it. The geometry is forced rather than designed — Kawasaki fixes one sector, the big-little-big lemma forbids a strictly smallest one, and what is left is the pattern Ron Resch was drawing in the 1960s.

00.20.40.60.80123how far the sheet is closed−ν, so every curve shown is a negative ratioMiura, slant 0.25Miura, slant 0.42Miura, slant 0.6accordionexactly zerothe same paper,three behaviours,chosen by the creasepattern alone Tessellations

A material made of creases

A folded sheet has a Poisson's ratio, a stiffness and a packing behaviour that the flat sheet did not. None of them belong to the paper — they belong to the pattern, and changing the pattern changes them without changing the material at all.

found rather than designedcrushed drink cans, tree bark,deployable boomsat every interior vertexsectors 60°, 60°, 60°, 60°, 60°, 60°two courses and four diagonalstwo of one letter, four of the otherwhy the height is not freea steeper diagonal makes the topsector strictly smallest, flankedby two of the same letter20 interior vertices · 23 mountain and 56 valley creasesmountainvalleyraw edge Tessellations

Patterns nobody designed

Crush a thin cylinder and it folds into a diamond lattice. Nobody chose the pattern — it is the buckling mode with the lowest energy, and it satisfies the flat-folding theorems because it just folded.

050100150-100100how far the vertex is drivenfold anglesectors60° 90° 120° 90°crease 1: Mcrease 2: Vcrease 3: Mcrease 4: Mcrease 2 is the odd onethree agree, one does notKawasaki holdsand it reaches flatfound by the linkage,not by the theorem Rigid folding

What the vertex does on the way

A four-crease vertex is a linkage on a sphere. Solving its closure gives the fold angles at every moment, and two theorems that are usually proved about the flat state turn up in the answer without being put there.

degree-4 vertex4 of 1625.0% · 4 creasespreliminary base112 of 25643.8% · 8 creasesmiura 2×28 of 1650.0% · 4 creasesmiura 3×232 of 12825.0% · 7 creasesmiura 3×3256 of 4,0966.3% · 12 creasesevery count enumerated, none estimatedthe share falls as the pattern grows, and the count still rises Flat-folding

How many assignments fold

The local conditions throw away most of the ways a pattern could be creased. They throw away a smaller and smaller fraction as the pattern grows, and what survives grows faster than what is discarded — which is why a strong filter is not a decision procedure.

the shrink5 intermediate outlines drawneach edge moved inward by thesame distance — computed, not drawnthe tracesstraight, because every edge movesat one rate along its own normalhow far it can go0.5391 sheet-widthsfound by bisection on the outline'sown area, not by inspectionthe construction that always works — which is what universal means here Designing a base

The last free parameter

Once the packing is fixed, one number is left in the whole design: how far a leftover polygon can be shrunk before it stops being a polygon. Everything else about the crease pattern has already been decided.

-0.02-0.02-0.03-0.04-0.06-0.10-0.19-0.49-0.05-0.06-0.07-0.10-0.15-0.23-0.43-1.01-0.10-0.13-0.16-0.21-0.30-0.45-0.81-1.79-0.20-0.24-0.30-0.39-0.53-0.79-1.36-2.93-0.34-0.41-0.50-0.64-0.85-1.24-2.10-4.41-0.54-0.65-0.79-0.99-1.31-1.88-3.13-6.48-0.80-0.94-1.14-1.42-1.87-2.64-4.36-8.950.120.200.300.420.550.700.850.080.200.320.440.560.680.800.90panel slanthow far the sheet is closedwhat the map saysevery cell negativefrom -0.02 to -8.95a factor of 535.6a material has one value;this has a working point,and it is chosendarker means more negative Tessellations

A property you can dial

Steel has one Poisson's ratio. A Miura-folded sheet has a surface of them, and where on that surface it sits is set by the panel shape and by how far it happens to be folded — which is why it is a mechanism rather than a material.

60° / 90°all 4 reached30° / 120°all 4 reached45° / 45°2 of 8 reached50° / 70°all 4 reached80° / 55°all 4 reachedsectorseach square is one assignment the theorems allowfilled — a rigid motion arrives there · open — a flat state with no path to itthe gap opens where two sectors are equal, and nowhere else on this listbig-little-big has nothing to forbid there — the linkage still does Rigid folding

A state no motion reaches

Flat-foldability asks whether a folded state exists. Rigid-foldability asks whether there is a path to it. The two sets are different, and the difference can be counted on a single vertex.

4 creases, assignment MVMVthe strip0.200.200.200.200.20MVMV3 availableafter crimp 10.200.200.20MV1 availableafter crimp 20.20nothing left2 crimps, each removing two creases4 creases is an even number, and that is not a coincidencethe merged segment measures outer minus middle plus outer What it costs to know

A machine that can only crimp

Change the atom and the whole picture changes. A machine whose single move folds two adjacent creases at once reaches strips no simple-fold machine reaches, is defeated by strips they handle easily, and cannot fold an odd number of creases at all — for reasons that are pure arithmetic.

6 creases, 7 segments, assignment MVMVMVDoes it fold flat?at most 5,040 orderings, and it may stop earlyyesas far as the first legal oneHow many ways?every one of them, because the last is as likely as the first15,040 orderingsWhat are they?the same search, paying a second time for what it keeps1 stackings, written out5,040 orderings, and the answer as wellCan a machine make it?a different search, over sequences of folds rather than over stackingsno1,275 statesthe four are not four difficulties of one problem — they are four problemsthe cost is work rather than time — a clock reading would differ on every build What it costs to know

Four questions about one sheet

Deciding, counting, listing and optimising are not four difficulties of one problem. They are four problems, and folding is the subject that proves it: a ruled map is trivial to decide and unsolved to count, while a general crease pattern is the other way round.

state 0state 1V M M V — the same pattern in both2 valid stackings, found by enumerationwhat a junction would addthree wires meeting, with the layer orders forced to disagree —which is a clause, and which is where the reduction gets its powernot drawn and not verified: nothing here decides layer order in two dimensions What it costs to know

Hardness is about the worst one

Flat-foldability is NP-hard, and every crease pattern on this site is decided in under a second. Both are true, and holding them together is the difference between using the result and repeating it: hardness is a statement about the worst instance a family contains, and nobody folds the worst one.

2468101222.22.42.62.833.23.4stampsratio to the term beforeodd terms, from aboveeven terms, from belowfilled: computed here, to 9 stamps · hollow: 10 and 11 and 12, computed once and quoted4,536 foldings at 9 stamps What it costs to know

Where the exponent comes from

The number of ways a strip of stamps folds grows exponentially, and the base of the exponential is a number nobody has proved exists. The ratio of one term to the last climbs past three and is still climbing where the computation stops — which is the only structural handle anybody has on the sequence.

developableKawasakiMaekawabig-little-bigsectors that do not alternatefour creases turning the same waya small sector flanked by one lettera 4×3 Miura, every vertexthe last row passes all four tests at all 6 of its vertices, and passing is not a proofthe tests are conditions at a single vertex; whether the layers can be stacked is a condition on the whole sheetno arrangement of vertex tests decides that, which is what NP-hardness means when it is spelled out What it costs to know

What a checker cannot check

Every crease pattern on this site is run past four theorems before it is allowed onto a page, and passing all four proves nothing. The gap is not a bug to be closed: it is the NP-hardness result, arriving as a property of a hundred lines of code.

polygonsectors at the twist vertexKawasakitiles the plane3-gon60.0 · 60.0 · 120.0 · 120.0180.0° = 180.0°yes — 6 round a point4-gon90.0 · 90.0 · 90.0 · 90.0180.0° = 180.0°yes — 4 round a point5-gon108.0 · 108.0 · 72.0 · 72.0180.0° = 180.0°no6-gon120.0 · 120.0 · 60.0 · 60.0180.0° = 180.0°yes — 3 round a point7-gon128.6 · 128.6 · 51.4 · 51.4180.0° = 180.0°noevery one of these twists satisfies the local theorems and folds flat on its ownthe interior angle has to divide 360° for the twists to meet, which only 3, 4 and 6 dobeyond 7 sides the assignment search runs out — 21 free creases, and the enumerator refuses above 22 Tessellations

Which polygons twist

Twist tessellations come in three kinds — triangle, square, hexagon — and it is natural to read that as a fact about twists. It is not. A twist can be built around any regular polygon and every one of them folds; what stops at three is the tiling, and the tiling is a fact about the plane.

the alternating-angle conditionHusimi, 1979Kawasaki, 198910 yrmountains minus valleys is twoHusimi and Maekawa, 1979Justin, 19867 yrthe big-little-big lemmaJustin, 1986the lemma, 19948 yrone fold solves a cubicBeloch, 1936Huzita, 199155 yrthe diamond buckling patternYoshimura, 1951Yoshimura, 196918 yrthe bi-directional foldMiura, 1970Miura-ori, 199525 yr1940196019802000mean lag 21 years · longest 55proof Who found it, and when

The name is not the date

Kawasaki's theorem is in Husimi's book ten years before Kawasaki's paper. Maekawa's is Justin's too. The mean gap between a result in this field and the name it is known by is twenty-two years, and it runs in one direction.

what the shell produced14 interior vertices, all alike17 mountain, 40 valley57 creases carrying a letterand it folds flatchecked, not asserted11.0 sheet-widths of crease, chosen by a buckling loadmountainvalleyraw edge Who found it, and when

Found before it was designed

Crush a thin cylinder and it falls into a diamond lattice. That pattern was published in aeronautics in 1951, twenty years before anybody designed with it — and what the buckling load chose was not only the creases but the mountain-and-valley assignment, which is the part a designer gets wrong.

degree-4 vertex, 60/90/120/90°4 of 164 creases · 25.0% surviveone degree-6 vertex8 of 646 creases · 12.5% survivethe preliminary base112 of 2568 creases · 43.8% surviveand these are only the local tests — a pattern can pass every vertexand still collide once the layers stack, which is the hard part Who found it, and when

The same vertex, found four times

A degree-four vertex with a three-to-one assignment turns up in a buckled cylinder, in a Miura fold, in a Resch tessellation and in a crumpled sheet. It is not a coincidence and it is not influence: the flat-folding conditions are restrictive enough that a small set of vertices is nearly all there is.

a route the search found123456121110987131415161718242322212019helices 24colours 12 : 12a route is not forbiddenscaffold used 21%48 staples of 3224 helices · 1536 bases · 48 staples · colours 12 : 12 Folding nobody designed

A sheet that routes itself

DNA origami folds one long strand into a shape by holding it against itself with a few hundred short ones. There is no sheet and no crease — what has to be designed is a route — and the first thing that can go wrong is a counting argument crease patterns already know under another name.

246810012345units (creases, or joints)log₁₀ statesa chain, 3 states per jointa sheet, two letters per creaseat one vertex, 4 of 16 assignments survive four local conditionsthe sheet's count has a local test that removes 75% of it · the chain's has none Folding nobody designed

Two things called folding

A protein folds and a sheet folds, and the word is the same word by accident. Both have exponentially many states and that is not the difference. The difference is that one of them can be filtered by four conditions checked at a single point, and the other cannot be filtered by anything local at all.

20 panels, two coloursno crease has the same colour on both sidesall 12 interior vertices carryan even number of creasesthe colour is which side of the paperthat panel shows when the sheet is foldedmountainvalleyraw edge Flat-folding

The sheet has two sides

Read a crease pattern as a set of panels rather than a set of lines and a condition appears that no vertex theorem states: the panels take two colours, no crease has the same colour on both sides, and the colour is which face of the paper each panel ends up showing.

-3-2.5-2-1.5-1-8-6-4-20tolerance (log₁₀ radians)fraction inside it (log₁₀)1 vertex · slope 1.002 vertices · slope 2.013 vertices · slope 3.0140,000 random vertices, none of them constructed to fold and none of them folding Flat-folding

Almost every pattern fails

Kawasaki's condition is one equation for each interior vertex, and a drawing satisfies an equation with probability zero. Every pattern on this site folds because it was constructed to, and the fraction that would fold by accident can be measured.

1 row0 interior verticespasses every condition2 rows4 interior verticesfails Kawasaki4 rows12 interior verticesfails Kawasakia pattern that folds is not a pattern whose enlargement folds — the conditions arrive with the interior Flat-folding

Where the paper stops

Every flat-folding theorem is a statement about a full turn of paper, so a vertex at the edge of the sheet is subject to none of them. Cutting a patch out of a pattern removes conditions rather than preserving them, and a small enough patch has almost none left.

562 × 24 cells323 × 39 cells324 × 416 cells325 × 525 cellsrepeating rules that pass every condition, out of 51224 rules pass on the smallest patch and on none of the others Tessellations

A unit that folds is not a tessellation

Of the 512 repeating rules for the waterbomb tessellation, 56 pass every condition on a two-by-two patch and 32 pass on every larger one. The twenty-four that die were never foldable — the small patch simply contained one of the four kinds of vertex the pattern makes, and the failures were at the other three.

sectors 80°, 55°, 100°, 125° in every one of them, and 4 assignments fold in every onelongest ÷ shortest 1.00footprint 0.806longest ÷ shortest 3.09footprint 0.911longest ÷ shortest 3.33footprint 0.623longest ÷ shortest 4.00footprint 1.782every one of them folds; their folded footprints differ by a factor of 2.86 Flat-folding

The lengths are free

Kawasaki reads angles, Maekawa counts letters, and the big-little-big lemma compares one sector with its neighbours. Not one condition in the subject mentions how long a crease is — so a single vertex is not a pattern but a whole family of them, every member folding, no two folding into the same shape.

creases at 0.25, 0.50, 0.75, marked MMM3 legal stackings of 4 segments, read from the bottom of the pile up12341: 2 · 1 · 4 · 3assignment12342: 4 · 2 · 1 · 3taco-taco12343: 2 · 4 · 3 · 1taco-tacothe paper lands in the same place every time — only the order through the pile differs Flat-folding

More than one way to lie flat

A crease pattern with its mountains and valleys marked is spoken of as though it named a folded object. It does not. The legal stackings can be counted exactly in one dimension, the count is routinely more than one, and its size is a property of the pattern that nobody quotes.

interior vertices, by number of creases meeting therenone3odd594evennone5odd326evennone7odd18even8 patterns, 92 interior vertices, and not one of them with an odd number of creases Flat-folding

Nothing meets at three

Every interior vertex of a flat-foldable pattern carries an even number of creases and at least four. So a crease cannot stop in the middle of the sheet, three creases cannot meet anywhere, and every crease pattern in the subject ends up looking the same way — all crossings and no stars.

20° a crease0.50 turns of papernothing touching anything34° a crease0.85 turns of papernothing touching anything36° a crease0.90 turns of paper1 pair through one another50° a crease1.25 turns of paper5 pairs through one anotherone strip of 10 panels, seen end-onit laps itself at 36.0° a crease, which is where its cross-section closesevery panel is the same length in every frame; the only thing changed is how far each crease is turned Rigid folding

Paper through paper

Every test the subject has for rigid folding is a statement about a neighbourhood, and a neighbourhood cannot see the far side of the sheet. So a pattern can satisfy all of them while driving one panel straight through another, and the sharpest witness has no interior vertex in it at all.

what the construction produced9 twists, 36 interior verticesturned 24.1° from the tiling's edgespleats 0.118 to 0.118 wide2.20× smaller once the pleats are taken upevery vertex passes all four conditionsmountainvalleyraw edge Tessellations

Any tiling makes a twist

A twist tessellation is usually drawn, admired and copied. It can be derived instead: hand the construction any tiling of the plane and it returns a crease pattern that folds flat, with the twist polygons' shapes forced by the tiling's own angles and nothing left to choose but how large and how turned.

what the construction produced23 twists, 122 interior verticesturned 24.1° from the tiling's edgespleats 0.068 to 0.068 wide1.58× smaller once the pleats are taken upevery vertex passes all four conditionsmountainvalleyraw edge Tessellations

Where two twists share a pleat

Every twist tessellation the tradition draws has one size of twist, because every tiling it is drawn on has one kind of vertex. Hand the construction a tiling with two, and the pleat between a large twist and a small one turns out to fix their sizes exactly — three to one, and nothing else folds.

30°60°90°0.250.400.550.700.85how much of the room between two vertices the twists takeno paper leftno assignment existstwist angleboth curves are measured rather than plotted from a formula Tessellations

Fenced at both ends

The twist angle of a tessellation looks like a free dial, and it is fenced twice. Turn too far and the pleats have no paper left. Turn too little and something stranger happens: every angle condition in the subject goes on holding and the pattern loses its mountain-valley assignment entirely.

1 × 4161 × 5501 × 61442 × 282 × 3602 × 43203 × 31,3684 × 4300,608filled — counted here, by exhaustive search over stacking ordersopen — Lunnon's published count, quoted rather than computed What it costs to know

Two directions that will not separate

A map has rows and columns, and a strip of stamps is a map with one row. The obvious hope is that the two-dimensional count is built from the one-dimensional one — fold the rows, then fold the columns. It is not: a two-by-three map folds 60 ways against a product of 12, and the discrepancy grows from a factor of two to a factor of thirty-eight over the counts anybody has.

the opposite crease, ×1the odd crease, ×0.26794910how far the vertex has foldedsectors 60° · 90° · 120° · 90°constant to 8.2e-13 over 199 points of the motionand equal to cos((α+β)/2) ÷ cos((α−β)/2), which the solver never forms Rigid folding

The vertex is geared

A rigid four-crease vertex has one degree of freedom, which says that one number decides everything and not how. The how is a fixed ratio: the tangents of the half fold angles at two creases stay in constant proportion for the whole of the motion, and the proportion is a function of the sector angles and nothing else.

a disc, with a vertexa ring, with noneone interior vertex, 3 creases at itodd degree, so they do notno interior vertices at alland the panels still do notboth refuse: two routes round the sheet leave a panel 1.87 sheet-widths apart Flat-folding

Even is not enough

Every vertex theorem in the subject is a statement about one point, and the two-colouring of the panels looks like the exception. It is not — on a square of paper it is a parity at each vertex and nothing more. Cut a hole and the two come apart: a loop of paper with three creases has no interior vertices at all, satisfies every theorem there is, and cannot be folded flat.

102°60°78°120°one smallest sectorthe lemma constrains one pair4 foldable assignmentsof the 16 markings120°60°60°120°two smallest sectors equalthe lemma constrains nothing8 foldable assignmentsof the 16 markings Flat-folding

Where the lemma says nothing

The big-little-big lemma asks for a sector strictly smaller than both its neighbours, and the word doing the work is strictly. At a vertex whose two smallest sectors are equal the lemma has no opinion at all — and those are the vertices origami actually uses. The count of markings the conditions admit doubles, discontinuously, at exactly the angles everybody folds.

4 patterns, one folded profile1/122/124/127/1225311/123/126/128/1225312/124/125/127/1225312/125/127/128/122531foldedthe layer counts under each band are the same in every row, and so are the widths Flat-folding

The shadow does not name the pattern

A photograph of a folded model carries an outline and a thickness at every point of it, and that is the whole of what it carries. It is not enough. Crease patterns in genuinely different places fold to identical outlines with identical layer counts, and nearly a third of the folded objects a short strip can reach are reached by more than one pattern.

what the vertex conditions settle once one crease is chosenpatternsettled, against what is therepreliminary base1 of 81 vertices still choosingmiura 6×41 of 3815 vertices still choosingwaterbomb 4×41 of 7625 vertices still choosingyoshimura 6×51 of 8422 vertices still choosingsquare twist grid3 of 14464 vertices still choosingtriangular twist grid2 of 236104 vertices still choosingKawasaki was settled by the angles before a letter was written; the letters are what is left, and they are nearly all left Flat-folding

How little the conditions decide

Local is not global is a statement about sufficiency: every vertex can pass and the sheet still fail. There is a sharper complaint available, and it is about strength. Fix one crease of a tessellation and propagate every condition the subject has to a fixed point: three creases out of a hundred and fifty-eight follow, and sixty-six vertices are still holding more than one answer.

a disc, with a vertexa ring, with noneone interior vertex, 3 creases at itodd degree, so they do notno interior vertices at alland the panels still do notboth refuse: two routes round the sheet leave a panel 1.87 sheet-widths apart Curves and material

A cut that removes no paper

Cuts in this subject are graded. Take a wedge out and the angle at a point falls by exactly the wedge; take twice as much and it falls twice as far. A hole is not like that. Its effect on what the sheet can do is the same whether it is a tenth of the paper or a ten-thousandth, and it is the same because it is not a quantity at all.

parallel columnsKawasaki to 3e-14°columns fanning by 5.2°Kawasaki to 5e-14°columns fanning by 9.2°Kawasaki to 5e-14°the mountain-and-valley letters are read off the motion rather than drawn, and then put past Maekawa Rigid folding

The family the Miura belongs to

Move one vertex of a Miura and the sheet has no rigid folded position at all — which leaves the obvious question unanswered. What else moves? A row of paper reflected in each of a fan of lines is flat-foldable for nothing at all, and whether it also folds rigidly turns out to be a condition on a table of cosines: it has to be a column of numbers times a row of numbers.

the ratio of each column's cosine to the first column'scol 1col 2col 3col 4col 5the sheet that folds1.0000-1.08221.0000-1.08221.00001.0000-1.08221.0000-1.08221.00001.0000-1.08221.0000-1.08221.0000row 1row 2row 3one row moved by 14 per cent1.0000-1.08221.0000-1.08221.00001.0000-1.13971.0000-1.13971.00001.0000-1.08221.0000-1.08221.0000row 1row 2row 3worst disagreement between rows: 4e-16 before, 0.0575 after Rigid folding

Where an error goes

A misplaced crease in a folded sheet has to be paid for somewhere, and this subject has two answers already — the error is folded too, and the hinge is where it ends up. There is a third. In a quadrilateral mesh a mistake in one row has no consequence in that row at all: it is felt by the columns, which is to say by every other row on the sheet.

0.287267creases in the moleculewhere the corner sits7 creases6 creasesat the marked shape three of the polygon's edges vanish at the same instant, and either side of it they vanish one at a time Designing a base

The skeleton changes its mind

The universal molecule fills any convex polygon, always, which is what makes it the part of the tree method with no special cases. It does not fill it continuously. Slide one corner along its edge and the number of creases in the molecule sits at six, jumps, and sits at seven — so two designs a hairsbreadth apart have crease patterns that are not small variations on one another.

parallelstraight creases spread 0.0°flat, they spread 0.0°fan 5.2°straight creases spread 25.5°flat, they spread 30.9°fan 9.2°straight creases spread 46.6°flat, they spread 55.0°fan 13.8°straight creases spread 72.4°flat, they spread 82.5° Tessellations

The corrugation that curves

A Miura is a flat sheet that becomes a flat slab. Open its straight creases into a fan and the same construction gives a corrugation that wraps a cone — exactly a cone, with every straight crease passing through one point to fifteen decimal places, at every moment of the fold, with the apex travelling as the sheet closes.

how many ways each map foldsa strip of five50of 120a plus120= 5! — every stackinga tee120= 5! — every stackinga two-by-three60of 720a two-by-three, one gone40of 120one corner gone848of 40320the middle gone8016of 40320the full square1368of 362880 What it costs to know

The map that is not a rectangle

Take one square out of a three-by-three map and the number of ways it folds does not go down by an eighth. It goes up — to 848 if the square came from a corner, and to 8,016 if it came from the middle. Two maps of eight squares in the same box, differing by nearly a factor of ten, and no function of the box tells them apart.

the waterbomb tessellation's odd vertexdegree six, and this site prints nine of them on one sheetevery assignment64passes all four conditions30has a flat folded state1812 labellings satisfy every condition the subject has and have no flat folded state Flat-folding

Crimp it away and ask again

Four conditions decide whether a vertex folds flat, and they decide it exactly at a vertex whose sectors are all different sizes. Everywhere else they over-count: two markings of every tied four-crease vertex, twelve of the degree-six vertex this site prints nine of on one sheet. What decides the case is not a fifth condition but a procedure — fold the smallest sector away and ask the smaller vertex.

of the markings that fold, how many folded objects each one makesmarkings that foldexactly one objectthe most any one makesthe preliminary base4 creases · four equal sectors, the first vertex anybody folds881a halved four-crease vertex4 creases · degree four with its two smallest sectors equal — the case the lemma is silent at661the waterbomb tessellation's odd vertex6 creases · degree six, and this site prints nine of them on one sheet18126a Yoshimura vertex6 creases · degree six with every sector equal, and twenty-two of them on the printed pattern30122the preliminary base's centre8 creases · degree eight, and the vertex at the middle of the first base anybody folds112164a vertex at no particular angles6 creases · degree six, drawn from the census and rounded to a tenth of a degree881at four creases the marking names the object; above it, it need not Flat-folding

One marking, many objects

A crease pattern with every mountain and valley written on it is spoken of as though it named a folded model. At four creases it does. At six it need not, and at the eight-crease vertex in the middle of the first base anybody folds, a single marking can be folded into four genuinely different objects — same creases, same letters, four answers.

what the outline and the thickness leave open, and what the order closesprofilesambiguousthe order settlesand does not3 creases on 12ths69166103 creases on 16ths1844712354 creases on 12ths23371692the last column is patterns that fold to the same object, so no better photograph reaches them Flat-folding

The order does not name it either

A photograph of a folded model carries an outline and a layer count, and that is not enough to recover the pattern. Hand the observer the layer order as well — everything the object physically is — and most of the ambiguity goes. Most. What is left are pairs of genuinely different crease patterns that fold to the same object, which no better photograph reaches.

flat-foldable at every vertex, all threeworst Kawasaki residual 0e+0 radiansone vertex moved -30 per centloop residual 2.4e-1the Miuraloop residual 5.8e-14one vertex moved 40 per centloop residual 3.6e-1every one of these is developable and flat-foldable at every interior vertex, exactly Rigid folding

The condition that is not flat-foldability

Take away the assumption that one crease family runs straight through every vertex and ask what makes a quadrilateral mesh move. It is not flat-foldability. There is a one-parameter family of meshes, every one of them developable and flat-foldable at every vertex to machine precision, and exactly one member of it folds — the Miura. Slide a single vertex along the ray that keeps every condition exact and the sheet stops moving, first order in the displacement.

one crease decided, and how much of the sheet followsthe flat-folding conditions, propagated2 of 12the rigid-folding conditions, propagated12 of 12and the rigid propagation leaves 1 consistent set of fold angles Rigid folding

One crease decides the sheet

Fix one crease of a flat-folding problem, propagate every condition the subject has, and three creases out of a hundred and fifty-eight follow. Fix one fold angle of a rigid one and every crease on the sheet follows, with a single consistent answer. The same experiment, two questions, opposite answers — and it is why a self-folding sheet needs one biased vertex rather than one per vertex.

the equation on an edge, at both of its endsas drawnevery vertex moved 12 per centratio − 1round a loopratio − 1round a loopthe square grid000.200.90the triangular grid000.170.75the honeycomb000.351.73the rhombille tiling2.0002.992.14the elongated triangular tiling000.171.27the rhombille's tiles are not regular, its ratios are three and a third, and they still multiply to one Tessellations

The propagation that never had to work

The twist construction carries one equation per edge of its tiling and propagates the twist sizes outward from a seed. On every tiling anybody has drawn a twist on, every one of those equations is satisfied trivially — both ends of an edge read the same two numbers, because a regular polygon has one interior angle. The construction has been running and doing nothing, and the one tiling where it did something is the one whose tiles are not regular.

what the construction produced7 twists, 60 interior verticesturned 17.2° from the tiling's edgespleats 0.123 to 0.123 wide1.76× smaller once the pleats are taken upevery vertex passes all four conditionsmountainvalleyraw edge Tessellations

The dial that decides nothing

Turn a twist tessellation's angle from one fence to the other and every measurable thing about it changes: the smallest sector goes from 88 degrees to under one, the pleats swallow a quarter of the sheet and then almost none of it, the folded footprint changes by a third. The number of ways it can be creased does not change at all — sixteen, at every angle tested — because the lemma reads which sector is smallest and never how small.

every flat-foldable vertex whose sectors are multiples of 45°6 of them, to degree 8passfoldone objectthe most45·45·135·135866145·90·135·90444190·90·90·90888145·45·45·45·90·90302012245·45·90·45·45·90301812645·45·45·45·45·45·45·451121121643 of the 6 carry markings the conditions accept and the paper refuses Designing a base

The whole alphabet of a grid

Box pleating is defended as a trade — give up packing efficiency, buy creases that land where they should. There is a third thing it buys and it is much stronger than either: on a forty-five degree grid there are exactly six kinds of interior vertex a flat-foldable design can contain, ever. On a thirty degree grid there are thirty.

every condition holds here6 creases4 creasesevery condition holds at the vertex on the paper — and one crimp later the smallest sector has the same letter on both sidesthe four conditions all hold · a stacking does not exist What it costs to know

A short reason to say no

When a folding question comes back yes it brings an object anybody can check. When it comes back no it usually brings nothing but the assurance that a search looked everywhere. At one vertex that is false: a refusal comes with a witness one or two steps long, out of a search space of a hundred and twelve, and the witness is a vertex the crease pattern does not contain.

degreevertices visited per letteringcrimps needed48 of 16 fold32630 of 64 fold1038112 of 256 fold41410420 of 1024 fold2065121584 of 4096 fold12376The work grows by a factor of about 6.0 for every two creases added; the necessity grows by one. Flat-folding

A tie is not a decision

The crimp reduction decides a vertex by folding its smallest sector away, and where two sectors tie for smallest it has no forced move and must try each of them. That search is not rare — on the vertex at the centre of the first base anybody folds it happens for fourteen of the sixteen letterings — and it has never once changed the answer.

40°95°25°110°60°30°2 strictly smallest sectors, at 25° and 30°every vertex with the same shading admits exactly the same letterings Flat-folding

The order decides the count

Ask how many mountain-and-valley letterings a vertex admits and the answer looks as though it should depend on the angles. It does not. Three of the four conditions never see an angle at all, and the fourth asks only which sector is smallest — so the count is a function of a combinatorial arrangement, and a walk round the cycle that never looks at a vertex reproduces it exactly.

what the observer is givenambiguities separated, of 71the outline alonewhat a silhouette carries0the outline and the colourwhat a photograph of duo paper carries, from both sides0the complete layer orderwhat taking the model apart carries69the middle row is what anybody can actually see, and it is the top row Flat-folding

Which side is showing

Two earlier rungs asked what a folded object records about the pattern that made it, first from its outline and then from its complete layer order. Neither observation is one anybody can make. A photograph of duo paper carries the outline, the thickness and the colour showing at every point — and over the whole census the colour separates nothing at all.

mismatch 0.066 radiansmismatch 8.5e-14 radiansevery vertex of both is developable and Kawasaki-exact to the last bit a double holds Rigid folding

Solving every face at once

A quadrilateral mesh that folds rigidly has to close round every one of its faces, and the rung that built the general mesh could close one. Four of them at once resisted a descent that drove each free length to its own root, because closing a loop is a condition on several lengths together — and solving them jointly finds a sheet with no two vertices alike that folds, and a surface of them sixteen dimensions wide.

creaseworst amplification of an error in itc:0:11.76c:0:21.77c:0:31.74c:1:11.76c:1:21.77c:1:31.74c:2:11.76c:2:21.77c:2:31.74c:3:11.76c:3:21.77c:3:31.74r:1:01.00r:1:11.00r:1:21.00r:1:31.00r:2:01.58r:2:11.58r:2:21.58r:2:31.58r:3:01.65r:3:11.65r:3:21.65r:3:31.65the best crease is 1.77 times better than the worst, and it is on the sheet's edge Rigid folding

Which crease to push

Deciding one fold angle settles every other one on a quadrilateral mesh, which is what makes a self-folding sheet buildable with a single actuator. It leaves a question that sounds like an afterthought: which crease. Driving each of a mesh's twenty-four in turn gives twenty-four different answers to how far an error in it travels — and on the sheet that repeats one vertex, it gives several answers to what shape the sheet takes.

23456789101105101520flapshow much the requirement costs, per cent14.61.70.0-0.05.60.510.810.44.72.8 Designing a base

The shapes the optimum has

Requiring a circle packing to be its own mirror image halves the number of coordinates a search has to find, so the same effort covers a much smaller space. Whether that helps depends on something the search cannot know in advance — whether the best packing was symmetric — and measured flap count by flap count the answer alternates without a pattern anybody could use.

populationpassfoldhave a gapbranchcut twice at random51 vertices, 13 kinds9.68.020%0%whole multiples of 45°60 vertices, 1 kinds30.019.3100%69%whole multiples of 30°60 vertices, 13 kinds19.713.177%26%a named vertex, jittered60 vertices, 4 kinds8.08.00%0% What it costs to know

Which vertices are the random ones

Every measurement on this site that begins 'over 373 random degree-four vertices' is a statement about a population nobody declared. There is no canonical way to pick a crease pattern at random, four defensible ways of doing it disagree about the same three questions by factors rather than by margins, and the disagreement reaches a sentence this site has published as though it were general.

the edge of the paperMVM40°60°20°60°the same sectors, in a lineMVM40°60°20°60°this lettering folds4 of 8 letterings foldVMV MMV VVM MVMno vertex theorem applies here at all— the sectors do not close, and there is no cycle to alternate round Flat-folding

The vertices nobody checks

Every figure on this site is gated on four conditions evaluated at every interior vertex, and the word interior has been carrying the whole sentence. On the printed patterns there are 105 vertices on the edge of the paper against 92 inside it, not one of them has ever been examined, and the condition that decides them has been available since the second phase of the collection.

8 letterings fold · 1 piece under any two creasesflip two creases anywhere round the vertex, which is the smallest change Maekawa allowsMMVVVVMMMMVVMVVVMVMVMMMVVMVVVMVMMMVMVVVVMMVVMMMMsectors 43° · 110° · 121° · 57° · 16° · 13°one piece: every folding is reachableevery crease at once: stays inside its own piece Flat-folding

Walking between two foldings

The letterings a vertex folds in are always counted and never navigated. Counting says a generic degree-six vertex has eight of them; navigating says that changing any two creases turns any one into any other, and that changing two neighbouring creases does not — and that the vertices which come apart are the ones with no coincidences in them, which is the opposite of what every other measurement here would suggest.

a single vertex is always one piece; a pattern with more is notand the number of pieces is decided by the creases that never reach the edge of the paperThe preliminary base1 vertices inside the paper112 letterings admitted1 piece of 1120 creases buried2^0 = 1The square twist4 vertices inside the paper256 letterings admitted16 pieces of 164 creases buried2^4 = 16The hexagon twist6 vertices inside the paper4096 letterings admitted64 pieces of 646 creases buried2^6 = 64Fold and cut — the triangle1 vertices inside the paper30 letterings admitted1 piece of 300 creases buried2^0 = 1 Flat-folding

The creases that cannot move

One vertex's foldings are always joined up. A pattern's are not, and the number of pieces they fall into is exactly two to the power of the number of creases with an interior vertex at each end — four on a square twist, six on a hexagon twist, none at all on a preliminary base. The creases a local change cannot reach are the creases that never reach the edge of the paper.

5 of 6 solved meshes are solid at every angle sampledthe bar is the deepest interpenetration found anywhere in the motion, in panel widthsmesh 11closes to 9e-14solid at every anglemesh 17closes to 1e-121.18 — panels 3:0 and 3:2, 2 steps apartmesh 19closes to 6e-14solid at every anglemesh 23closes to 5e-14solid at every anglemesh 27closes to 2e-12solid at every anglemesh 71closes to 4e-12solid at every angle Rigid folding

Closing is not building

A quadrilateral mesh solved so that every loop closes to within a millionth of a radian is a mesh whose fold angles are consistent. It is not necessarily an object. One of the six solved here drives a panel through another at every angle of its motion — there is no part of the fold at which it could be made of solid panels — and the pair that crosses is two steps apart in the sheet, where nothing evaluated at a vertex could see it.

uncut: 4 vertices inside the paper, 6.3% of letterings admittedcut a crease with an interior vertex at each end25.0% admitted2 vertexes released · 4× the share · 4 such creases, all alikecut a crease that already reaches the edge12.5% admitted1 vertex released · 2× the share · 8 such creases, all alikea released vertex is one the four conditions no longer reach, and each is worth a factor of two Curves and material

A cut is a licence

What a cut buys is usually described in words — freedom, release, a shape a fold cannot reach. It can be counted, and the unit is vertices. Cutting one crease of a square twist turns two interior vertices into vertices no theorem applies to, and the share of letterings the pattern admits goes up by a factor of two for each vertex released: exactly, on every cut tried.

the four cheap tests are polynomial in the drawing; the fifth is notreading across a row is one pattern put to all fivecrease pairsverticespanelscreasessearch nodesthe square twist6649127,565the Miura fold703152438refusedthe waterbomb sheet2,850255276refusedthe Yoshimura3,655226586refuseda square patch3,486364984refuseda rhombille patch39,621126157282refuseda refused search is a pattern about which the expensive test says nothing at all, at full price What it costs to know

The cost is in the coincidences

How big an instance is, is what a hardness statement is about, and it is the weaker predictor of what deciding one costs. Hold the degree fixed and vary only how many of a vertex's sectors are equal: the work of deciding it rises by a factor of nearly three, against a factor of two for doubling the number of creases. The expensive instances are the ones a designer draws on a grid.

the bar is the creases with an interior vertex at each enda folder holding one of these patterns is in one piece of the count on the right, and cannot leave it107 of 862 moves survive across the shelf · 0 touch a buried creaseThe preliminary base0 buried · 1 piecesThe Miura fold22 buried · 4,194,304 piecesThe square twist4 buried · 16 piecesThe hexagon twist6 buried · 64 piecesThe Yoshimura pattern48 buried · 2.81 × 10^14 piecesFold and cut — the triangle0 buried · 1 piecesThe tapered corrugation27 buried · 1.34 × 10^8 piecesThe waterbomb tessellation42 buried · 4.39 × 10^12 pieces Flat-folding

The pieces without the list

The letterings a pattern folds in fall into pieces no folder can cross, and the count was found by writing every lettering down — which stops at eighteen creases. The Miura has thirty-eight, the Yoshimura eighty-six, and the number of pieces can be read off the drawing without listing anything: four million and two hundred and eighty-one million million.

the bar is every legal stacking; the dark part is the ones with no move out of thema swap is legal when the result is still a stacking — the two layers need not be joined by a crease3 segments2 of 6 isolated · 4 legal moves4 segments16 of 16 isolated · 0 legal moves5 segments34 of 50 isolated · 16 legal moves6 segments144 of 144 isolated · 0 legal moves7 segments366 of 462 isolated · 96 legal moves Flat-folding

Nothing slides past anything

A marked strip has several legal stackings and this site has counted them at length. Nobody asked whether a folder holding one can reach another by lifting a flap over its neighbour: five hundred and sixty of six hundred and seventy-two stackings have no such move at all, and whether any exists depends on the parity of the segment count.

the pale bar is the share that folds in a ring, the dark one in a line40 vertices from each population at each degreeangles at random, degree 425% in a ring · 70% in a lineangles at random, degree 613% in a ring · 46% in a linemultiples of 45°, degree 432% in a ring · 83% in a linemultiples of 45°, degree 630% in a ring · 87% in a linemultiples of 30°, degree 429% in a ring · 79% in a linemultiples of 30°, degree 623% in a ring · 70% in a line Flat-folding

A ring and a line

A vertex has a certain amount of paper at it, and the paper either closes round or it does not. Holding the sectors fixed and changing only that: the ring has twice as many letterings to choose from and folds in a quarter of them, the line has half as many and folds in seven-tenths, and cutting a ring open has never once cost a lettering.

the pale bar is the published count, the dark one the objectsneither operation ever fixes a folding; doing both sometimes does, and that is why it is not a quarter2 stamps2 labelled · 1 objects · 2 fixed by doing both3 stamps6 labelled · 2 objects · 2 fixed by doing both4 stamps16 labelled · 5 objects · 4 fixed by doing both5 stamps50 labelled · 14 objects · 6 fixed by doing both6 stamps144 labelled · 38 objects · 8 fixed by doing both7 stamps462 labelled · 120 objects · 18 fixed by doing both8 stamps1392 labelled · 353 objects · 20 fixed by doing both What it costs to know

The count counts labels

One, two, six, sixteen, fifty, a hundred and forty-four: the oldest sequence in the subject counts foldings of a strip of numbered stamps. A folded strip of blank paper has no first stamp and no top side, and neither of those operations ever leaves a folding alone — so the count of objects is 1, 2, 5, 14, 38, 120, and it is not the count over four.

the pale bar is every folded state; the dark one is the states the machine reachescounted over every marking of the strip that folds at all3 equal stamps12 of 12 reached4 equal stamps32 of 32 reached5 equal stamps100 of 100 reached6 equal stamps288 of 288 reachedcreases at .13 .31 .62 .780 of 24 reached — 24 missedcreases at .08 .24 .28 .35 .720 of 48 reached — 48 missed What it costs to know

Where the machine catches up

The weakest machine in the subject folds every layer at once and is stopped by a strip with two creases in it. On a strip of equal stamps it is stopped by almost nothing: every one of the 288 folded states a six-stamp strip has is reachable by a sequence of all-layers folds, and on every unevenly creased strip tried it reaches none of them. At seven stamps the completeness ends, and finding out where it ended is what checking it past six was for.

a 3 × 3 patch: 42 creases, 18 of them buriedthe bar is on a log scale, because the two numbers differ by four orders of magnitudepieces the 3×3 patch has262,144pieces containing a repeating rule32, one eachevery one of the 32 rules is in a piece no other rule is in Tessellations

Thirty-two rules, thirty-two pieces

The waterbomb tessellation's surviving repeating rules fold to one object — same panels, same places, same areas. Put them in the space of letterings the patch admits and they occupy thirty-two different pieces of a quarter of a million, so no two of them can be reached from one another without unfolding the sheet.

a strand of 7249 bases at 64 to a helixthe bar is the largest member of the family that is still buildablea square blockstopped by the strand's length100 helices · 88% of the stranda single rowstopped by the strand's length113 helices · 100% of the stranda comb of teethstopped by the routing5 helices · 4% of the stranda plusstopped by the routingno size works at allan L with equal armsstopped by the strand's length113 helices · 100% of the strand Folding nobody designed

Two ceilings

A DNA origami is limited by the length of one viral strand and by whether its helices can be visited once each in a single pass. Grown one step at a time, a square block runs into the first at a hundred helices and a plus runs into the second at five — so which limit a shape meets is decided by the shape and not by the chemistry.

the bar is the pairs of panels that lie over one anotherThe preliminary base288 panels · 12 rules · an ordering existsThe Miura fold22824 panels · 228 rules · not decidedThe square twist369 panels · 48 rules · an ordering existsThe hexagon twist6613 panels · 96 rules · an ordering existsThe Yoshimura pattern205565 panels · 1187 rules · not decidedFold and cut — the triangle217 panels · 15 rules · an ordering existsThe tapered corrugation28228 panels · 351 rules · not decidedThe waterbomb tessellation92652 panels · 654 rules · not decideda pattern with no bar has no two panels over one another, and its order is not a question Flat-folding

No height to swap

A folded strip is a permutation of segments, and the smallest change a hand can make to it is a swap of two heights: 672 stackings, 560 of them isolated. A folded sheet has no height. Its layers are ordered by statements about which panels share ground, and on every printed pattern the search can finish, the answer is one stacking and no way out of it.

the square twist, sieved three timesevery lettering4,0962 to the 12passes every vertex2566.3% of themletters are consistent2524 force a loop of panelshas a folded state80.20% of themthe bars are on one scale, so the last one is the size of the answer against the size of the question Flat-folding

The lettering that folds nowhere

The conditions at a vertex admit 256 letterings of the square twist. Eight of them have a folded state. The other 248 satisfy developability, Kawasaki, Maekawa and the big-little-big lemma at every vertex of the pattern and cannot be folded by anyone — and this site printed one of them for years, at true scale, with instructions to fold it first.

the bar is the draws whose letters do not contradict themselvesa loop of panels is a proof that no flat folded state exists, and it costs one passone square twist39 of 409 panels · 12 creasesone hexagon twist40 of 4013 panels · 18 creasesa small square tiling24 of 4049 panels · 72 creasesthe square tiling7 of 4049 panels · 84 creasesthe patch a propagation returns first is not a draw and has no reason to be among these Tessellations

The tiling the unit could not promise

Every twist on this site carries the same caveat: the unit is verified and the plane is not, because deciding a whole pattern is intractable. There is one thing about a whole pattern that costs a single pass over its crease list, and it says no. The square twist tiling was drawn with a lettering that contains a loop of twenty-eight panels, so the patch on this site had no flat folded state at all — and only seven of forty independent redraws avoid one.

the bar is the number of foldings, on a logarithmic scaleboth routes give the number printed; a disagreement anywhere would be a defect in one of them2 × 122 letterings · 1 creases3 × 164 letterings · 2 creases4 × 1168 letterings · 3 creases5 × 15016 letterings · 4 creases6 × 114432 letterings · 5 creases2 × 288 letterings · 4 creases3 × 26032 letterings · 7 creases4 × 2320128 letterings · 10 creases3 × 31,368256 letterings · 12 creasesa strip of stamps is the one-row case, and the classical sequence 2, 6, 16, 50, 144 is the top of the table What it costs to know

The map counted from the layers

The classical map-folding counts are computed from a rule that never places a panel: work out which edge of the folded square each fold wraps around, and refuse the orderings that interleave two folds at one edge. Place the panels instead and order them by the general non-crossing rules, and the same numbers come out — 2, 6, 16, 50, 144, 8, 60, 320, 1368 — on nine sizes, by machinery that shares no line of code with the first.

the bar is the share of the population with a folded stateevery pattern in all four passes every condition at every interior vertexthe printed patterns4 of 80 cannot be placed · 0 cannot be ordered · 4 undecidedtwist tessellations2 of 125 cannot be placed · 2 cannot be ordered · 3 undecidedquadrilateral meshes2 of 60 cannot be placed · 4 cannot be ordered · 0 undecidedfold-and-cut patterns5 of 70 cannot be placed · 0 cannot be ordered · 2 undecidedundecided is a real answer here and is not rounded toward either side What it costs to know

The patterns a checker is tested on

This site keeps four populations of crease patterns and runs its checkers over them, which is what makes a claim about typical instances measurable rather than rhetorical. Asked whether the members actually fold, the populations answer: thirteen of thirty-three do, six place and cannot be ordered, five cannot be placed at all, and nine are past what the search will finish.

the bar is the share of the population with a folded stateevery pattern in all four passes every condition at every interior vertexthe printed patterns4 of 80 cannot be placed · 0 cannot be ordered · 4 undecidedtwist tessellations2 of 125 cannot be placed · 2 cannot be ordered · 3 undecidedquadrilateral meshes2 of 60 cannot be placed · 4 cannot be ordered · 0 undecidedfold-and-cut patterns5 of 70 cannot be placed · 0 cannot be ordered · 2 undecidedundecided is a real answer here and is not rounded toward either side Rigid folding

A collision is an order

Paper passing through paper is treated here as a thing that happens during a motion and is caught by watching for it. At the flat state it is not an event at all: it is the absence of an ordering, and it can be proved rather than observed. Four of the six quadrilateral meshes this site solves for rigid folding place perfectly and admit no ordering of their nine panels — so every one of them must pass through itself, and none of them was ever driven to find out.

the bar is the number of foldings, on a logarithmic scaleboth routes give the number printed; a disagreement anywhere would be a defect in one of them2 × 122 letterings · 1 creases3 × 164 letterings · 2 creases4 × 1168 letterings · 3 creases5 × 15016 letterings · 4 creases6 × 114432 letterings · 5 creases2 × 288 letterings · 4 creases3 × 26032 letterings · 7 creases4 × 2320128 letterings · 10 creases3 × 31,368256 letterings · 12 creasesa strip of stamps is the one-row case, and the classical sequence 2, 6, 16, 50, 144 is the top of the table Axioms and construction

The grid a division makes

Dividing a square into thirds in both directions is a construction: four creases, each exact, each landing on a rational the ladder can name. The object it leaves behind is a three-by-three map of stamps, and how many ways that folds is the oldest open problem in the subject — 1,368 at three, 300,608 at four, and unknown at five.

the bar is the letterings with a folded statethe row is how many times the letter changes going round the central polygonthe ring reads as one letter032 pass every vertex · 28 have no order2 changes round the ring8192 pass every vertex · 184 have no order4 changes round the ring032 pass every vertex · 32 have no ordera twist looks like a twist when the ring reads as one letter, which is why this was never checked Who found it, and when

Taught with a wrong reason

Four mountains and four valleys is what the preliminary base's symmetry suggests and Maekawa forbids it; a twist looks like a twist when its central ring reads as one letter, and no such lettering folds; a tessellation is verified because its unit is, and a forty-nine-panel patch of one had no folded state at all. In each case the conclusion taught is right and the reason offered for it is not, and the site that repeats them is this one.

the bar is the vertices the drawing has and the list does notThe preliminary base09 listed · panels closeThe Miura fold035 listed · panels closeThe square twist016 listed · panels closeThe hexagon twist022 listed · panels closeThe Yoshimura pattern045 listed · panels closeFold and cut — the triangle011 listed · panels closeThe tapered corrugation040 listed · panels closeThe waterbomb tessellation041 listed · panels closethe square grid, assembled064 listed · panels closethe triangular grid, assembled1282 listed · panels 1.73 apartthe honeycomb, assembled1884 listed · panels 2.00 apartthe rhombille tiling, assembled12138 listed · panels 1.86 apartthe elongated triangular tiling, assembled576 listed · panels 1.73 apartevery pattern with a bar has panels that cannot be placed, and every pattern without one places exactly Flat-folding

The vertex the list does not have

Every condition this collection checks is asked at a vertex of a crease pattern, and a crease pattern is handed to the checker as a list of points and segments. A reader is handed ink. Read the same patterns the second way and eight printed sheets gain nothing at all — while four tessellation patches gain 12, 18, 12 and 5 vertices that nobody wrote down, every one of them a place where two creases were drawn across each other.

the same tessellation on the same square, cut out of the plane two waysassembled from whole unitsclipped from the plane12 crossings · panels 1.73 apart0 crossings · panels closemountainvalleyraw edge Tessellations

Cutting a patch out of a plane

A tessellation is infinite and a sheet is not, so every picture of one is a decision about where the paper stops. Assembling whole twist units on a square and running the outstanding pleats to the rim puts 12, 18, 12 and 5 creases across other creases on four of five tilings; generating the pattern over a larger region and clipping it puts none. The panels then place exactly — and what is waiting behind the repair is a different refusal that could not be asked about before.

the bar is the average number of layers over the folded footprint0.5 of the sheet3.891 whole of 9 · 89% cut by the rim0.42 of the sheet3.641 whole of 9 · 89% cut by the rim0.34 of the sheet4.049 whole of 9 · 0% cut by the rim0.28 of the sheet4.309 whole of 21 · 57% cut by the rim0.22 of the sheet4.349 whole of 25 · 64% cut by the rim0.18 of the sheet4.7925 whole of 45 · 44% cut by the rima bulk property arrives as the boundary leaves, and neither a page nor a sheet of paper reaches the end of it Tessellations

The property a patch does not have

A folded corrugation is described as a material — a packing ratio, a stiffness, a Poisson's ratio — and every one of those is a statement about an unbounded medium. Fold the same tiling at six sizes on the same square and the compaction climbs from 3.89 layers to 4.79 as the share of units the rim cuts falls from nine tenths to four, and it has not settled at the fine end. The number a patch gives is the material's number minus its own boundary.

the sectors are 40°, 140°, 40°, 140° — each row is one way the vertex can start to foldcrease 1crease 2crease 3crease 4mode 10.7100.7102 of the four creases movemode 200.7100.712 of the four creases movethe numbers are the four fold angles' ratios to one another as the vertex leaves the flat state Rigid folding

Two mechanisms at one point

Two creases drawn across each other cannot fold flat — Maekawa's count refuses them at every angle. They move perfectly well as rigid panels, and they move in two ways: bend along one line while the other stays flat, or the reverse. Every other developable vertex of degree four has two ways too, and in both of them all four creases move together at a fixed ratio. The crossing is the case where the two motions have nothing to do with each other.

the outline, shrunk — and the one corner that moves outwardwhat the shrink finds5 edges, 1 of them meeting at a corner that turns back3 skeleton nodes7 arcs traced by the cornersthe last of them forms at 0.181 of a sheet Designing a base

The corner that splits the shrink

The universal molecule fills a convex polygon by shrinking it, and at a corner that turns back the shrink does something no convex polygon does: the region breaks in two. That event can now be computed — the skeleton of a non-convex outline is available here for the first time, and it is what lets one straight cut reach a star. It does not give the molecule back, because a molecule needs the shrinking region to stay one piece and a split is exactly the moment it stops.

the bar is how many of the 33 patterns each refusal is the first to catchtwo creases cross5one sweep over pairs of creasesa vertex condition fails0one pass over the verticesthe panels do not place0one walk over the panelsthe letters force a loop0one pass over the crease listno ordering exists6every ordering of the panels22 of the 33 are refused by none of these and are folded, undecided, or waiting on a search too large to run What it costs to know

The order the refusals come in

This collection can say no to a crease pattern in five ways, and they cost wildly different amounts: a sweep over pairs of creases, a pass over the vertices, a walk over the panels, a pass over the crease list, and an enumeration of every ordering of the panels. Run all five over the thirty-three patterns in the four test populations and the cheapest refuses five, the most expensive refuses six, and the three in between refuse nothing at all.

the bar is how far apart two routes to one panel end up, before the cut3 creases1.751.75 apart · cut open, 0e+05 creases1.591.59 apart · cut open, 0e+07 creases1.431.43 apart · cut open, 0e+09 creases1.321.32 apart · cut open, 0e+011 creases1.241.24 apart · cut open, 0e+0after one cut from the hole to the rim, every one of them places to rounding — with no crease changed Curves and material

A cut that reaches the edge

A ring of paper with three creases running from its hole to its rim satisfies every condition the subject has — vacuously, because it has no interior vertex at all — and cannot be folded: its panels take no two colours and the two routes to one of them end up 1.75 sheet widths apart. One cut from the hole to the edge, crossing no crease and changing no letter, and it folds exactly. The cut removes an adjacency, which is the one thing neither a fold nor an edge can do.

34560%20%40%60%80%100%creases in the stripreachable by simple folds72%30%17%13%the basic symbolsa dashed line — valleya dotted line — mountainan arrow — fold it nowand what they missreverse, squash, sink,petal — every one of thema move no dashed linecan ask forevery assignment of 68 seeded spacings Who found it, and when

A file has no paper

The field's interchange format is three arrays — where the vertices are, which pairs of them an edge joins, and a letter for each edge — and that is exactly the object every computation on a crease pattern starts from. A list of edges cannot say that two of them must not cross, because crossing is a property of the drawing and the list has no drawing in it. So a pattern that no paper could carry is a perfectly well-formed file, and four of this collection's own were.

the bar is the share of draws whose letters agree among themselvesa draw that disagrees is a proof that the pattern has no flat folded state with those lettersthe preliminary base200 of 2008 panels · 8 creases · 0 contradict themselvesthe square twist198 of 2009 panels · 12 creases · 2 contradict themselvesthe Yoshimura190 of 20065 panels · 86 creases · 10 contradict themselvesthe Miura fold181 of 20024 panels · 38 creases · 19 contradict themselvesa square twist patch26 of 20049 panels · 84 creases · 174 contradict themselvesa hexagonal patch2 of 20077 panels · 142 creases · 198 contradict themselvesa rhombille patch0 of 200157 panels · 282 creases · 200 contradict themselvesthe sampler returns solutions rather than a uniform draw over them, so these are shares of what it found Flat-folding

A proof in one pass

Deciding whether a crease pattern has a flat folded state is hard, and the search that decides it gives up at twenty-four panels. One line of the same machinery does not search at all: each crease says which of the two panels it joins lies above the other, and a circle in what those statements demand is a proof that no folded state exists. It costs one pass over the crease list, and on a tessellation patch of a hundred and fifty-seven panels it answers in milliseconds.

the bar is the letterings that pass every condition at the vertexnone of them forces a loop, because the one lettering that would is the one Maekawa forbidsdegree 48 pass · 0 loop16 letterings · 8 admissible · the alternation fails Maekawa alonedegree 630 pass · 0 loop64 letterings · 30 admissible · the alternation fails Maekawa alonedegree 8112 pass · 0 loop256 letterings · 112 admissible · the alternation fails Maekawa alonechecked at equal sectors and at a skew of 0.18 radians, so the count is not a fact about a symmetry Flat-folding

The loop a vertex cannot close

A crease pattern's letters can contradict themselves, and the contradiction is never local. Enumerate every mountain-valley labelling of a single interior vertex at degree four, six and eight — a hundred and fifty pass every condition the subject has — and not one of them sends its panels round in a circle. The one labelling that would is refused by Maekawa, alone: Kawasaki holds on it and so does the big-little-big lemma.

shaded is every panel that lies on some loop49 panels · 1 tangle · biggest 3535 panels on some loop — 71.4% of the patch52 of 84 arcs run inside it, so one cut removes one of them Flat-folding

The loop is not the tangle

A search that finds a contradiction in a pattern's letters reports the first circle it meets, and on a tessellation patch that is eight to twelve panels of forty-nine. It reads as a local fault. Decompose the same arrows a second way and the set of panels that lie on some circle is thirty-five of forty-nine on the square patch and ninety-nine of a hundred and fifty-seven on the rhombille — which is why the smallest available repair does not reach it, and cannot be tried on most of the creases at all.

the square twist, sieved three timesevery lettering4,0962 to the 12passes every vertex2566.3% of themletters are consistent2524 force a loop of panelshas a folded state80.20% of themthe bars are on one scale, so the last one is the size of the answer against the size of the question Flat-folding

Consistent is not foldable

The square twist has 4,096 mountain-valley labellings. Two hundred and fifty-six satisfy every condition at every vertex; two hundred and fifty-two of those have letters that do not contradict themselves; and eight have a folded state. So the cheap proof that reads the letters in one pass accounts for four of the two hundred and forty-eight failures, and the other two hundred and forty-four are refused by a search over orderings that nothing shorter replaces.

each arrow points from the lower panel to the higher one9 panels · 12 creases · 12 arcsa loop of 8 panels — no order existsthe arrows are the whole of the test — nothing here asks which panels lie over which Tessellations

The ring is the loop

The square twist's central polygon is four creases enclosing one panel, and a lettering that gives all four the same letter has no folded state. That was established by enumerating the orderings of nine panels. It can now be read off the crease list in one pass, because the eight panels the letters send round in a circle are exactly the ring — the twist's own defining feature, contradicting itself.

two kinds of vertex, both forced16 of degree 490°, 90°, 90°, 90°9 of degree 690°, 45°, 45°, 90°, 45°, 45°40 mountain and 36 valley creases14.3 sheet-widths of foldingmountainvalleyraw edge Tessellations

The rule that breaks the count

The waterbomb tessellation has five hundred and twelve repeating rules for its letters and thirty-two of them fold. A hundred and twenty of the other four hundred and eighty send four panels round in a circle — the shortest circle a crease pattern can have — and every single one of those hundred and twenty has broken Maekawa's count at the very vertex the circle goes round. The theorem that closes the shortest circle, caught doing it, a hundred and twenty times.

the bar is how many of the 38 patterns each refusal is the first to catchtwo creases cross5one sweep over pairs of creasesa vertex condition fails0one pass over the verticesthe panels do not place0one walk over the panelsthe letters force a loop1one pass over the crease listno ordering exists6every ordering of the panels26 of the 38 are refused by none of these and are folded, undecided, or waiting on a search too large to run What it costs to know

The refusal that reads the list once

There are five ways of saying no to a crease pattern here, and their costs are two hundred and eighty-two, a hundred and twenty-six, a hundred and fifty-seven, thirty-nine thousand six hundred and twenty-one — and a search that is refused outright. On the largest patch the four cheap tests together do less work than one of them looks like it should, and the fifth cannot be started. A refusal that reads the crease list once is the only kind that scales.

the bar is the mean share of redraws that agree with themselvesas the populations stand, every member is consistent and the refusal fires on none of themthe printed patterns96.7%8 of 8 could be asked · worst member 90%twist tessellations55.0%7 of 12 could be asked · worst member 7%quadrilateral meshes96.9%6 of 6 could be asked · worst member 82%fold-and-cut patterns100.0%7 of 7 could be asked · worst member 100%a member with no folded state has no letters to redraw and is counted as not asked rather than as passing What it costs to know

A population that cannot fail

Thirty-three crease patterns are kept here to run the checkers over, and every one of them has letters that agree with themselves. That is not a property of the patterns. It is a property of how they were made: each came from a construction that returns a lettering, so a test looking for letters that contradict themselves has nothing to fire on. Reletter the same thirty-three and the failure is available at once — on one member, four of sixty redraws.

the bar is the nodes the ordering search visitedthe letters are consistent on every one of these, so the one-pass test says nothing about any of themmesh 37,4739 panels · 7,473 nodes · no order existsmesh 58,0079 panels · 8,007 nodes · no order existsmesh 89,3469 panels · 9,346 nodes · no order existsmesh 111,0159 panels · 1,015 nodes · an order existsmesh 141449 panels · 144 nodes · an order existsmesh 199,0629 panels · 9,062 nodes · no order existsa red bar is a pattern with no folded state, found only by visiting every ordering it might have had Rigid folding

Two refusals that refuse differently

Four of the six developable quadrilateral meshes this collection solves have no ordering of their nine panels — they must pass through themselves, and a search over every ordering proves it. On all four, the letters agree with themselves perfectly. The linear proof and the exponential search are not a fast test and a slow one: they answer different questions, and neither contains the other.

each arrow points from the lower panel to the higher one9 panels · 12 creases · 12 arcsa loop of 8 panels — no order existsthe arrows are the whole of the test — nothing here asks which panels lie over which Who found it, and when

The first thing about layers

A folder is taught four conditions at a vertex, or is taught nothing at all, and neither one says anything about the layers — which is where most of what goes wrong actually goes wrong. There has never been a rule about layer order simple enough to teach, because the question is global and every answer to it was a search. A chain of panels whose arrows all point the same way is the first one that fits on a finger.

18 interior vertices26 mountains · 19 valleyscolumns taper 2.44 : 1packs to 11.2% of flatmountainvalleyraw edgethe taper is in the columns, because Kawasaki does not mention their widthtapering the rows instead puts the alternating sums at 186.4° and 173.6° Folding nobody designed

The taper decides nothing

A leaf's corrugation narrows toward its margin, and the taper is what the pattern is for. It has no effect whatever on how often the pattern's letters agree with themselves: four width profiles from perfectly even to strongly tapered give a hundred and seventy-four consistent letterings of two hundred, identically. What moves the number is the count of rows, and on that measure a leaf tracks a Miura rather than the corrugation it most resembles.

the bar is how many times the search took a letter backand every one of those was the arcs closing a loop, never a vertex running out of labellingsthe square patch126 nodes · 1 refused by the arcs · 0 by the vertex conditionsthe elongated patch335 nodes · 3 refused by the arcs · 0 by the vertex conditionsthe hexagonal patch241 nodes · 2 refused by the arcs · 0 by the vertex conditionsthe triangular patch747 nodes · 7 refused by the arcs · 0 by the vertex conditionsthe rhombille patch246561 nodes · 246 refused by the arcs · 0 by the vertex conditionsthe vertex conditions are propagated rather than tested, so they narrow the choice instead of refusing it Flat-folding

Which condition does the refusing

A search for a lettering carries five conditions: developability, Kawasaki, Maekawa, the big-little-big lemma, and the demand that the arcs the letters force have no circle in them. Run it on five tessellation patches and count what makes it take a letter back. The four everybody checks refuse nothing at all. Every single backtrack is the fifth.

the bar is how many moves survive the conditions at a vertexa move flips two creases meeting at one point, which is what pushing a vertex through doesthe square patch0216 pairs tried at each of two letterings · 0 legal · 0 leave the verdict alonethe elongated patch6270 pairs tried at each of two letterings · 6 legal · 6 leave the verdict alonethe hexagonal patch8360 pairs tried at each of two letterings · 8 legal · 8 leave the verdict alonethe triangular patch8360 pairs tried at each of two letterings · 8 legal · 8 leave the verdict alonethe rhombille patch16756 pairs tried at each of two letterings · 16 legal · 16 leave the verdict aloneevery one of them leaves the lettering on the side of the question it was already on Flat-folding

Every move leaves the verdict

The only change a folder can make to a lettering without breaking it is to push one vertex through, flipping two creases at once. Try every such move on five tessellation patches, from two different letterings each: nineteen of two thousand nine hundred and sixty-four survive the conditions, and not one of the nineteen turns a lettering that agrees with itself into one that does not, or the other way about.

the 16 repeating rules that fold, written outrows first, then the two column classes — and every one of them alternates down the columnrows · columns above|below20VV · MV|MV21MV · MV|MV22VM · MV|MV23MM · MV|MV24VV · VM|MV25MV · VM|MV26VM · VM|MV27MM · VM|MV36VV · MV|VM37MV · MV|VM38VM · MV|VM39MM · MV|VM40VV · VM|VM41MV · VM|VM42VM · VM|VM43MM · VM|VMfour ways of writing the rows times four ways of alternating the columns is sixteen, and there is nothing else Tessellations

Sixty-four rules, sixteen fold

The Miura fold's letters are usually given as a recipe: rows one way, columns changing at every row. Write down every rule of that shape — the letter on a crease depending only on which row and which column it is in — and there are sixty-four. Sixteen fold flat. They are exactly the ones whose columns change at every row, the row letters do not matter at all, and every one of the forty-eight refusals is the counting theorem's alone.

the bar is how many rules the two tests agree aboutone reads three bits of the rule; the other folds the sheet and walks the arcsthe Miura fold64 of 6438 rules predicted to close a loop · 0 disagreementsthe tapered leaf64 of 6438 rules predicted to close a loop · 0 disagreementsthe closed form says a loop is available exactly where the columns fail to change letter and the row disagrees with them Tessellations

The loop is in the rule

Of the forty-eight repeating rules that do not fold a grid corrugation, thirty-eight send four panels round in a circle and ten merely fail the count. Which is which can be read off three of the rule's six bits, without building the pattern, folding it or walking a single arrow — and the closed form agrees with the arrows on all sixty-four rules of both grid families.

the bar is the middle run of a hundred and twentysame pattern, same code — only the order the letters are tried in differsthe square patch2725 at best · 27 at the middle · 36 at worstthe elongated patch3432 at best · 34 at the middle · 39 at worstthe hexagonal patch4339 at best · 43 at the middle · 51 at worstthe triangular patch4539 at best · 45 at the middle · 53 at worstthe rhombille patch16684 at best · 166 at the middle · 48 of 120 unfinished at 20000an unfinished run is left out of the middle rather than counted as its budget Tessellations

Four easy patches and one that is not

Run the same search a hundred and twenty times on each of five tessellation patches, changing nothing but the order the letters are tried in. Four of them answer in between twenty-five and fifty-three steps every single time. The fifth answers in eighty-four steps at best, a hundred and sixty-six in the middle, and does not answer at all in forty-eight runs of the hundred and twenty.

the bar is how many rules send four panels round in a circleout of the rules that already fail the count at some vertexthe Miura fold3848 refused · every loop four panels · vertices of degree 4the tapered leaf3848 refused · every loop four panels · vertices of degree 4the Yoshimura pattern038 refused · not one closes a loop · vertices of degree 6the waterbomb tessellation120480 refused · every loop four panels · vertices of degree 4 and 6a loop of four needs the letters to alternate round one point, and only a degree-four vertex lets a repeating rule do that Tessellations

Where a rule can close a loop

Three corrugation families have repeating rules whose letters send four panels round in a circle, and one has none at all. The one that has none is the one whose vertices are all of degree six — and the reason is that a straight line through a point carries a single letter under any repeating rule, while a strict alternation round six creases needs the two halves of that line to differ.

the bar is the second-smallest sector at a typical vertexthe triangular patch at seven turns, with the same panels and the same creases at all of themturn 0.261.92°smallest sector 60.00° · next 61.92° · no lettering exists, proved by exhaustionturn 0.2160.71°smallest sector 60.00° · next 60.71° · no lettering exists, proved by exhaustionturn 0.215560.06°smallest sector 60.00° · next 60.06° · no lettering exists, proved by exhaustionturn 0.21660.00°smallest sector 60.00° · next 60.00° · a lettering existsturn 0.2260.00°smallest sector 59.52° · next 60.00° · a lettering existsturn 0.2560.00°smallest sector 56.10° · next 60.00° · a lettering existsturn 0.3560.00°smallest sector 46.15° · next 60.00° · a lettering existsthe verdict changes exactly where that sector passes sixty degrees and stops being the second smallest Flat-folding

Where a sector crosses sixty

Turn the twist polygons of a tessellation patch a hundredth of a radian further and the pattern goes from having no mountain-valley labelling at all to having one immediately. Nothing about its graph changes across the transition — the same eighty-three panels, the same hundred and forty-two creases, the same four labellings at every one of its sixty vertices. What changes is which sector at a vertex is the smallest one.

the bar is how many letterings pass every condition at every vertexand the note is how many of those close a loop in the arcs2 by 122 panels · 2 letterings pass every vertex · 0 close a loop3 by 143 panels · 4 letterings pass every vertex · 0 close a loop4 by 184 panels · 8 letterings pass every vertex · 0 close a loop5 by 1165 panels · 16 letterings pass every vertex · 0 close a loop2 by 284 panels · 8 letterings pass every vertex · 0 close a loop3 by 2326 panels · 32 letterings pass every vertex · 0 close a loop4 by 21288 panels · 128 letterings pass every vertex · 0 close a loop3 by 32569 panels · 256 letterings pass every vertex · 4 close a loopa map's difficulty is not here — it is in the rules about which panels may lie between which What it costs to know

The test that never fires on a map

The cheapest refusal this collection has reads a crease list once and reports that no arrangement of the layers exists. Enumerate every labelling of every map from two panels to nine and it fires on four of the four hundred and fifty-four — all four on the largest map, none at all below it. On the oldest open problem in the subject, the cheap test has essentially nothing to say.

the 16 repeating rules that fold, written outrows first, then the two column classes — and every one of them alternates down the columnrows · columns above|below20VV · MV|MV21MV · MV|MV22VM · MV|MV23MM · MV|MV24VV · VM|MV25MV · VM|MV26VM · VM|MV27MM · VM|MV36VV · MV|VM37MV · MV|VM38VM · MV|VM39MM · MV|VM40VV · VM|VM41MV · VM|VM42VM · VM|VM43MM · VM|VMfour ways of writing the rows times four ways of alternating the columns is sixteen, and there is nothing else Who found it, and when

Half the recipe is decoration

Every account of the Miura fold gives its letters as two instructions: the rows go one way, and the columns change letter every time they cross a row. Enumerate all sixty-four repeating rules and the second instruction is the whole of the condition — all four ways of writing the rows appear among the sixteen that fold, in every combination. The first instruction has never constrained anything.

the bar is how many repeating rules fold, and it is the same bar six timesthe same corrugation redrawn at six geometries, every one of them swept in fullas printed1648 refused, all by the count · 38 of them close a circle of foureven columns1648 refused, all by the count · 38 of them close a circle of foura one-sided ramp1648 refused, all by the count · 38 of them close a circle of foura violent taper1648 refused, all by the count · 38 of them close a circle of foursix taller rows1648 refused, all by the count · 38 of them close a circle of foura steeper zigzag1648 refused, all by the count · 38 of them close a circle of fourno vertex condition reads a column width, a row height or a row count, so none of them can move the table Folding nobody designed

The leaf's rules are the Miura's

A plicate leaf packs into a corrugation that is broad in the middle and narrow at both ends. Enumerate every repeating mountain-valley rule it admits and the table is the Miura fold's table, rule for rule, number for number — and it stays that table when the leaf is redrawn with even columns, a violent taper, taller rows or a steeper zigzag. The plant's geometry cannot reach its own letters.

each point is one patch, searched twice002020404060608080square · 26elongated · 32hexagonal · 39triangular · 39rhombille · 80nodes, mountain firstnodes, valley firstthe dashed line is y = x, and nothing has been fitted to anything Flat-folding

The order that is its own mirror

Trying a mountain first and trying a valley first are two different searches, and on a hundred and forty-two crease patterns they cost the same number of steps — not on average, not nearly, but identically, pattern for pattern. The reason is a symmetry of every condition the subject has, and it is four lines long.

each cell is one patch, searched to a verdictgreen: a lettering exists · magenta: none exists, by exhaustion0.150.250.350.50.70.91.11.3turn angle, in radianssquare2626262626262626elongated1515323231313232hexagonal1515394545464545triangular1515393939373737the number in a cell is the nodes the search visited; 6 of 32 patches have no lettering at all Flat-folding

A region with no lettering

One turn angle at which a tessellation patch has no consistent lettering was found by sweeping a dial. Sweeping two dials finds nine patches with none, across three tilings, filling a corner of the parameter space — and never touching the square tiling, whose sectors have no sixty degrees to cross.

each cell is one patch, searched to a verdictgreen: a lettering exists · magenta: none exists, by exhaustion0.150.250.350.50.70.91.11.3turn angle, in radianssquare2626262626262626elongated1515323231313232hexagonal1515394545464545triangular1515393939373737the number in a cell is the nodes the search visited; 6 of 32 patches have no lettering at all Tessellations

The dial and the tiling that is not alike

Four of the five tilings a twist tessellation can be built on behave identically under every dial the construction has. The fifth has two kinds of vertex, and everything about it is different: it is the only one whose search has a tail, the only one whose shallow patches take minutes to draw, and the only one where a distance has to be solved rather than assumed.

each point is one pattern: panels across, nodes up00100100200200one node per panelnodes visitedpanels2 by 2 to 16 by 16, and not one backtrack anywhere in the family Tessellations

A corrugation never backtracks

As a box-pleating grid goes from two divisions to sixteen, the share of random letterings that agree with themselves falls from a hundred in a hundred to one. The cost of finding one that does stays at exactly one step per panel — four, nine, sixteen, twenty-five, and two hundred and fifty-six — with not a single wrong guess anywhere in the family.

the bar is how many letterings of the mesh can have their panels stackedout of every labelling of its twelve creases, enumeratedmesh 3016 pass every vertex · 16 agree with themselves · arrived refusedmesh 5032 pass every vertex · 32 agree with themselves · arrived refusedmesh 8832 pass every vertex · 32 agree with themselves · arrived refusedmesh 11832 pass every vertex · 32 agree with themselves · arrived foldablemesh 141416 pass every vertex · 14 agree with themselves · arrived foldablemesh 19416 pass every vertex · 16 agree with themselves · arrived refusedtwo of the meshes have none at all, and two more were refused only at the lettering they came with Rigid folding

A search with nothing to reorder

One search on a crease pattern costs eighty steps or fifteen thousand depending on the order it takes its decisions in. The other search on the same crease pattern costs 1,188,571 steps whatever order it is given — twelve permutations of the panels, twelve identical counts. The difference between them is one line of code that neither has and one has.

16 × 16 gridevery crease on a grid line, or at 45°which is why a 64-grid design can be folded at allmountainvalley Designing a base

Ninety-nine in a hundred pass

A designer checks a box-pleated pattern the way every text teaches: vertex by vertex, counting mountains and valleys, watching the smallest sector. At sixteen divisions that check passes a hundred letterings in a hundred, and one of them folds. The check that separates them costs a single sweep over the crease list and is in no recipe anywhere.

the bar is the share of the twists on the paper that the paper's edge cuts0.5 of the sheet86%1 whole · 6 cut by the edge0.42 of the sheet55%5 whole · 6 cut by the edge0.34 of the sheet59%7 whole · 10 cut by the edge0.28 of the sheet70%7 whole · 16 cut by the edge0.22 of the sheet37%17 whole · 10 cut by the edge0.18 of the sheet49%23 whole · 22 cut by the edgea patch is a picture of a tessellation, and the smaller the unit the less of the picture is edge Designing a base

The edge is what makes it hard

Grids, crumples, leaves, corrugations and fold-and-cut patterns all give up a consistent lettering at one step per panel with no wrong guess anywhere. The one family that does not is a tessellation clipped to a square, and what separates it from the others is not disorder, not size and not irregularity. It is having a rim.

18 interior vertices26 mountains · 19 valleyscolumns taper 2.44 : 1packs to 11.2% of flatmountainvalleyraw edgethe taper is in the columns, because Kawasaki does not mention their widthtapering the rows instead puts the alternating sums at 186.4° and 173.6° Folding nobody designed

The plant's pattern is not a hard case

A hornbeam leaf packs into its bud by corrugating, and the pattern it uses gives up a consistent lettering at nine, twelve, fifteen, eighteen, twenty and twenty-four steps on nine, twelve, fifteen, eighteen, twenty and twenty-four panels. Nothing about the plant's problem is combinatorially difficult, and saying so is worth as much as finding a case that is.

the same drawing, cut out of the plane and glued upnodes, log scale, against periods across the sheet10100100010⁴10⁵1×12×23×34×4glued upcut outan open mark is a search that ran out of budget rather than a cost What it costs to know

What the rim was doing

One rectangle of a twist tessellation, cut out of the plane in the ordinary way, gives up a consistent lettering in forty-eight steps. Join its opposite edges so that no crease is divided and the same drawing, at the same vertices, under the same conditions, takes fifty-six thousand seven hundred and seventy-two. The edge of the paper was never the difficulty. It was the slack.

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 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.

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 Flat-folding

The lettering that was proved impossible

A search closed its whole tree on a glued square tessellation and reported that no mountain-and-valley assignment of it is consistent. Written onto ordinary patches of one, four and nine periods and handed to the four vertex theorems and a folded sheet rebuilt from scratch, the assignment it says cannot exist passes every check, on four tilings, up to fifteen hundred creases.

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 wayelongated ×11240 creases become 52 · 20 vertices either wayelongated ×224160 creases become 184 · 80 vertices either wayelongated ×336360 creases become 396 · 180 vertices either wayrhombille ×11248 creases become 60 · 24 vertices either wayrhombille ×224192 creases become 216 · 96 vertices either wayrhombille ×336432 creases become 468 · 216 vertices either waythe bar is how many creases the cut divides; nothing else about the two sheets differs Flat-folding

The rim is four letters a cell

Cut a rectangle out of a tessellation and it asks exactly the vertices the tessellation asks, exactly the same questions. What it adds is four free letters for every period of edge — the creases the cut divides, which become two independently answerable creases instead of one. Eight letters on a two-period square, sixteen on a four-period one, and nothing else about the two objects differs at all.

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 Flat-folding

The bottom layer is at the rim

A hundred and sixty-nine panels of folded tessellation, and three of them have nothing underneath. All three touch the paper's edge, and the same is true on every tiling at every size measured. Which panel is at the bottom of a stack turns out to be a fact about where the sheet was cut rather than about the pattern, and the pattern itself has no bottom at all.

labellings a vertex keeps, against nodes a panel costs0.000.250.500.751.00481530the box-pleating gridthe tapered leafa crumple, deepeningthe waterbombthe Yoshimura, as drawnthe Yoshimura, tiltedthe twist patcheslabellings the conditions leave at a vertexthe dashed line is one node a panel, which four of these families sit on exactly Flat-folding

One step per panel is a table size

Four families of crease pattern search at exactly one step per panel — a grid at nine sizes, a leaf, a Miura, six crumples — and it was read as a law about patterns that fill their own sheet. It is a number: the conditions at each of their vertices admit eight labellings. Where the conditions admit four, the cost is half. Where they admit thirty, it moves again, and the same pattern at two proportions demonstrates it with everything else held still.

the Yoshimura, as drawn: nodes against panels050100one a panel0 panels119every vertex of this family keeps 30 labellings Flat-folding

Six creases and the same straight line

The one family here whose vertices are degree six was said to break the arithmetic that every other family obeys, on the strength of a single pattern. Built as a family — six sizes from twenty-one panels to a hundred and nineteen — the Yoshimura is exactly as linear as a grid, with no decision ever withdrawn. What degree changes is the constant, and it changes it in both directions depending on one angle.

the Yoshimura at 6 by 5, at nine proportionsrow height 1.257 nodes30 labellings a vertex · 0.88 nodes a panelrow height 1.557 nodes30 labellings a vertex · 0.88 nodes a panelrow height 1.757 nodes30 labellings a vertex · 0.88 nodes a panelrow height 1.732050857 nodes30 labellings a vertex · 0.88 nodes a panelrow height 1.732050919 nodes8 labellings a vertex · 0.29 nodes a panelrow height 1.7419 nodes8 labellings a vertex · 0.29 nodes a panelrow height 1.819 nodes8 labellings a vertex · 0.29 nodes a panelrow height 219 nodes8 labellings a vertex · 0.29 nodes a panelrow height 2.519 nodes8 labellings a vertex · 0.29 nodes a panelthe equilateral Yoshimura is drawn at √3 = 1.732050808, on the dear side Flat-folding

A knife edge nine decimals wide

Draw the Yoshimura with its rows 1.7320508 half-columns tall and each vertex admits thirty labellings and the pattern costs fifty-seven steps. Draw it at 1.7320509 and each admits eight and it costs nineteen. The number between them is √3, which is the proportion everybody draws — and below it the sectors are unequal and the lemma is still silent, because the small ones sit next to each other.

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 What it costs to know

Pruning on proofs alone

A search that discards a branch it cannot prove wrong is not a search. Deciding whether a periodic pattern's layer relations really contradict themselves is far dearer than the disc's one-pass test, so the cheap test is asked first — it is sufficient, so it settles almost everything — and the expensive one runs only on what the cheap one rejects. Five of nine steps on a small cell, fifty thousand of fifty-seven on a large 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 What it costs to know

The cost of proving something false

A search closing its whole tree is the strongest result this collection can produce, and on a glued tessellation it produces one that is wrong. What it costs to reach is three steps at one period, thirty-five at four, three thousand four hundred and fifty-five at nine, and more than two hundred thousand at sixteen — growing far faster than the cost of finding the lettering it says does not exist.

the twist patches: nodes against panels050100150one a panel0 panels157every vertex of this family keeps 4 labellings Tessellations

The most decided vertex here

Sixteen ways to letter four creases; Maekawa allows eight; the big-little-big lemma allows four. A twist polygon's corner is one of the few vertices in this collection where the second cut applies, so it keeps four labellings where a grid, a leaf, a Miura and a crumple all keep eight — and the family the collection long called difficult turns out to be the one whose conditions decide the most.

clipped tessellation patches, nodes per panel0.000.250.500.751.00one node a panelthe square gridthe triangular gridthe honeycombthe elongated triangular tiling0 panels413 panelsthe family the collection called hard is the one below the line Designing a base

The edge was not what made it hard

Five families of pattern searched at one step per panel and a tessellation patch did not, and the property left standing after four alternatives were killed was having a rim. Measured under a fixed letter order the patches cost between a half and two-thirds of a step per panel, at every tiling and every size — below the line rather than above it, and the rim is why.

the box-pleating grid: nodes against panels0100200one a panel0 panels256every vertex of this family keeps 8 labellings Designing a base

The designer's grid is the dearest thing here

Two hundred and fifty-six panels of box-pleating grid take two hundred and fifty-six search steps to letter — exactly one per panel, at every size from two divisions to sixteen, with not one decision withdrawn. That is the most any pattern in this collection costs per panel of paper. A twist tessellation costs half of it, and a tilted corrugation a quarter.

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 Rigid folding

An order with no least element

Enumerating every way a folded pattern can be stacked works by building upward from a panel with nothing below it. The smallest square twist patch has exactly one such stacking and takes eleven thousand steps to find it. The pattern that patch was cut from has no panel with nothing below it at all, so the enumeration has nothing to start from — and the sheet is perfectly well stacked anyway.

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 Who found it, and when

A test imported without its hypothesis

The rule that a loop in a folded sheet's layer relations proves the pattern cannot fold arrives from the layer-ordering literature, where the sheet is a disc and the panels are finitely many. This collection took the rule and not the sentence that says which sheets it is about, then applied it for years to patterns whose whole interest is that they repeat.

the tapered leaf: nodes against panels01020one a panel0 panels24every vertex of this family keeps 8 labellings Folding nobody designed

Nothing grown was cut out of anything

A leaf's corrugation costs twelve steps on twelve panels, sixteen on sixteen, twenty on twenty, twenty-four on twenty-four — exactly one per panel at every geometry, which is the most any pattern here costs. A tessellation patch costs half that, and the reason is that somebody cut it out of something. A leaf's creases stop at the margin because the plant stopped there.

3 creases on a Möbius bandthe panels take two coloursseamthe same seam123the right edge onto the left, turned over3 creases, 3 panelsinterior vertices: 0two-coloursoff by 4.000 of a widthand turn the paper the right waymountainvalleyraw edge Flat-folding

The seam carries a sign

A loop of paper folds flat when it has an even number of creases round it. A Möbius band folds flat when it has an odd number. The drawing is the same in both cases, the creases are the same creases, and what changed is a factor of minus one contributed by the sheet rather than by anything drawn on it.

5 creases on a Möbius bandthe panels take two coloursseamthe same seam12345the right edge onto the left, turned over5 creases, 5 panelsinterior vertices: 0two-coloursoff by 4.000 of a widthand turn the paper the right waymountainvalleyraw edge Flat-folding

The band that needs an odd number

A Möbius band is the first sheet in this collection with one side, and the consequence is sharper than a reversed parity. Mountain and valley are defined relative to a side, so on a sheet with no consistent side a crease has no letter — and Maekawa's condition survives the loss while the assignment it is about does not.

the composition, and what it has to equal3 reflections, in order[ 1.000 0 ][ 0 -1.000 ]+ ( -1.732, 1.000 )=?the gluing map of a Möbius band[ 1.000 0 ][ 0 -1.000 ]+ ( -1.732, 1.000 )they agree to rounding, so the band foldsand both turn the paper the same way, so the parity is righton a disc the right-hand side is the identity, which is why nobody writes it down Flat-folding

Closure is not the identity

Walk a folded state from panel to panel, composing a reflection at every crease, and come back to where the walk started: the composition has to be the identity. That is the rule everybody states, and it is a special case. On a sheet whose edges are glued the walk does not come back to where it started, and what the composition has to equal is the gluing map.

the composition, and what it has to equal5 reflections, in order[ -1.000 0 ][ 0 1.000 ]+ ( 2.000, 0 )=?the gluing map of a Möbius band[ 1.000 0 ][ 0 -1.000 ]+ ( -2.000, 1.000 )they differ by 4.000 of a width, so it does notand both turn the paper the same way, so the parity is righton a disc the right-hand side is the identity, which is why nobody writes it down Flat-folding

Parity is not enough

A Möbius band needs an odd number of creases round it. Give it three, square across the strip, and it does not fold — nor does five, nor seven, nor any odd number at all. The counting argument is necessary and it is not close to sufficient, and the thing it cannot see is which way the creases point.

the angles that admit a thirdφ₁ − φ₂ + φ₃ a multiple of a straight angle30°30°60°60°90°90°120°120°150°150°60°, 120°the first crease's angle, against the secondevery other pair of angles folds nothing,at any length and any positions Flat-folding

An alternating sum of angles

Kawasaki's condition says the sectors round a vertex alternate to a straight angle. A glued band has no vertices and obeys a condition of exactly the same shape: the crease angles have to alternate to a multiple of a straight angle. Two different quantities, two different sheets, one arithmetic — and in both cases what is being said is that a product of reflections came back the right way.

the Miura, 2×2 cellsone drawing, four sheetscutoutgluedacrossgluedalonggluedboth waysvertices8888free letters22182016panels1510128V − E + F1000the vertex row is the control: identifying edges can neither make nor destroy a vertexand Euler's number is the cheapest check that the gluing did what it says Flat-folding

Half a rim

A rectangle of tessellation cut out of the plane has four edges; glued into a torus it has none. Gluing one pair and leaving the other gives the middle of the scale — the same drawing, the same vertices, the same conditions asked of them, and exactly half the rim. What the rim costs turns out to be measurable per edge rather than only at the ends.

letters saved by gluing, and the two halves of itthe grid ×121 across + 1 along = 2 · 4 letters cut, 2 gluedthe grid ×242 across + 2 along = 4 · 12 letters cut, 8 gluedthe Miura ×132 across + 1 along = 3 · 7 letters cut, 4 gluedthe Miura ×264 across + 2 along = 6 · 22 letters cut, 16 gluedthe Yoshimura ×164 across + 2 along = 6 · 12 letters cut, 6 gluedthe Yoshimura ×2128 across + 4 along = 12 · 36 letters cut, 24 gluedone comparison says the rim costs something; four say the price is per edge Flat-folding

The rim adds up

What one glued pair of a cell's edges saves in free letters is what the other pair saves, and gluing both saves the sum. That is a rate rather than an observation, it is the form of the claim two objects could never support, and it is what makes 'the rim costs four letters a cell' a statement about tessellations rather than about one drawing.

the Yoshimura, 2×2 cellsone drawing, four sheetscutoutgluedacrossgluedalonggluedboth waysvertices8888free letters36283224panels29202416V − E + F1000the vertex row is the control: identifying edges can neither make nor destroy a vertexand Euler's number is the cheapest check that the gluing did what it says Flat-folding

Euler counts the gluing

Vertices minus creases plus panels comes to one on a rectangle of paper and nought on any gluing of it. That is the cheapest check that an identification did what it says, it costs three counts already being made, and it is what found a crease running exactly through the corner of a cell — a case the corner search could not see and no other check would have noticed.

the period cell of the gridone period, with its neighbours round it1 interior vertices in the cell4 crease pieces drawnperiod 1.000 × 1.000one square, because a grid repeats at every linethe cell is a rectangle of ordinary paper until somebody says its edges are one edge Flat-folding

A grid that will not close

Take the simplest crease pattern there is — a square grid — and join a cell of it into a torus. With an even number of squares across it folds. With an odd number it has no flat folded state at all, and the obstruction is a parity that has nothing to do with the pattern being difficult, because a grid is not difficult.

nodes of search per panel, as the rim goesthe grid ×1 cut1.004 nodes · 4 panels · 4 lettersthe grid ×2 cut1.009 nodes · 9 panels · 12 lettersthe grid ×2 cyl x1.177 nodes · 6 panels · 10 lettersthe grid ×2 cyl y1.177 nodes · 6 panels · 10 lettersthe grid ×2 torus1.506 nodes · 4 panels · 8 lettersthe grid ×3 cut1.0016 nodes · 16 panels · 24 lettersthe Miura ×1 cut1.006 nodes · 6 panels · 7 lettersthe Miura ×1 cyl y1.255 nodes · 4 panels · 6 lettersthe Miura ×2 cut1.0015 nodes · 15 panels · 22 lettersthe Miura ×2 cyl x1.1011 nodes · 10 panels · 18 lettersthe Miura ×2 cyl y1.0813 nodes · 12 panels · 20 lettersthe Miura ×2 torus1.2510 nodes · 8 panels · 16 lettersthe Miura ×3 cut1.0028 nodes · 28 panels · 45 lettersthe Miura ×3 cyl y1.0425 nodes · 24 panels · 42 lettersthe Yoshimura ×1 cut0.9110 nodes · 11 panels · 12 lettersthe Yoshimura ×1 cyl y1.139 nodes · 8 panels · 10 lettersthe Yoshimura ×2 cut0.9728 nodes · 29 panels · 36 lettersthe Yoshimura ×2 cyl y1.0024 nodes · 24 panels · 32 lettersthe Yoshimura ×3 cut0.9653 nodes · 55 panels · 72 lettersthe Yoshimura ×3 cyl x1.0042 nodes · 42 panels · 60 lettersthe Yoshimura ×3 cyl y0.9445 nodes · 48 panels · 66 lettersthe Yoshimura ×3 torus1.0036 nodes · 36 panels · 54 lettersfewer panels to divide by, and the same argument to settle Tessellations

One node per panel, with the rim gone

A rectangle of repeating pattern cut out of the plane costs exactly one node of search per panel, on every family and at every size. Take the rim away and the total falls and the cost per panel rises, because the letters that were removed were the ones that could not be wrong.

the folded period of the Yoshimuradrawn periods across the top123456the turnsame way upslides240°yesno120°yesnoyesyes240°yesno120°yesnoyesyesa turn of 240° comes back to nothing after three of them, and that is the folded periodthe drawing repeats every one, which is what makes it a tessellation Tessellations

The turn a column costs

The Yoshimura's drawing repeats every column. Folded flat, it does not: the fold carries one column onto the next by a turn of two hundred and forty degrees, so the folded state repeats every third column and not before. A pattern has two periods and only one of them has ever been written down.

the folded period of the Yoshimuradrawn periods across the top1234567the turnsame way upslides240°yesno120°yesnoyesyes240°yesno120°yesnoyesyes240°yesnoa turn of 240° comes back to nothing after three of them, and that is the folded periodthe drawing repeats every one, which is what makes it a tessellation Tessellations

The period nobody measured

Every repeating pattern in this collection has its drawn period recorded, because a drawing cannot be generated without one. Its folded period is recorded nowhere, and on one of the families measured the two differ by a factor of three — which means the number that has always been quoted is the wrong one for anything about the folded object.

what each sheet costs, per panel — a square twistcut out ×10.5565 nodes on 9 panels · 12 lettersglued across ×10.6674 nodes on 6 panels · 10 lettersglued along ×10.6674 nodes on 6 panels · 10 lettersglued both ways ×10.7503 nodes on 4 panels · 8 letterscut out ×20.52013 nodes on 25 panels · 40 lettersglued across ×20.55011 nodes on 20 panels · 36 lettersglued along ×20.55011 nodes on 20 panels · 36 lettersglued both ways ×20.5639 nodes on 16 panels · 32 letterscut out ×30.61230 nodes on 49 panels · 84 lettersglued across ×32.02485 nodes on 42 panels · 78 lettersglued along ×30.57124 nodes on 42 panels · 78 lettersglued both ways ×317.361625 nodes on 36 panels · 72 lettersthe letters go down as the rim goes and the cost per panel goes up What it costs to know

Half the slack

Gluing one pair of a cell's edges removes half the free letters and costs almost nothing. Gluing the second pair removes the other half and costs three orders of magnitude. The letters go linearly and the search does not, and the reason is that the last free letter is worth more than all the others.

the Yoshimura, 2×2 cellsone drawing, four sheetscutoutgluedacrossgluedalonggluedboth waysvertices8888free letters36283224panels29202416V − E + F1000the vertex row is the control: identifying edges can neither make nor destroy a vertexand Euler's number is the cheapest check that the gluing did what it says What it costs to know

Which pair is glued

A cell's two cylinders have the same Euler number, the same amount of rim and the same name. On a symmetric drawing they have identical counts of letters, panels and vertices — and searching them costs twenty-four nodes one way and eighty-five the other. Half the rim is a description of the topology and not of the object.

what each sheet's shape costs in conditionsa square0 loopsχ = 1 · no loop that cannot be shrunka slit from the rim0 loopsχ = 1 · the same paper, topologicallyone hole1 loopχ = 0 · one parity conditiona cylinder1 loopχ = 0 · the same sheet as one holetwo holes2 loopsχ = -1 · two independent conditionsa torus2 loopsχ = 0 · two conditions, no rim at alla slit inward from the rim changes nothing, and a closed cut changes everything Curves and material

A cut is surgery

Two cuts that look identical on the paper do completely different things to the sheet. A slit run inward from the rim changes nothing at all; a closed cut in the middle removes a disc and leaves a sheet carrying a condition it did not have before. What separates them is not the length of the cut or how much paper it removes.

two holes, two conditionsone loop is odd — the sheet refuses2 out to the left, 1 to the right, 1 betweenround the left hole: 3 creases, oddround the right hole: 2, evenround both: 3, oddno two-colouring exists0 interior verticesa loop round one hole says nothing about a loop round the other Curves and material

Two holes are two conditions

One hole in a sheet of paper gives one loop that cannot be shrunk and one parity to satisfy. Two holes give two, and they are independent: an arrangement of creases can satisfy the condition round one hole and fail the condition round the other, and the sheet refuses on the strength of the one it failed.

which bands foldcreases across the strip123456nofoldsnofoldsnofoldsfoldsnofoldsnofoldsnocylinderMöbius bandthe gluing map of a cylinder is a slide and of a Möbius band a slide with a flipand a composition of k reflections turns the paper over exactly when k is odd What it costs to know

A proof in no nodes at all

A parity refuses a sheet before any search begins. It costs one addition, it is certain, and it says nothing about why — while a search that exhausts on the same sheet costs thousands of nodes and produces a proof of the same fact. Two proofs of one thing, and the cheap one is available only where somebody has noticed the invariant.

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 What it costs to know

The cost of asking the wrong sheet

A test written for a sheet with an edge, run on a sheet without one, does not fail. It exhausts — proving, at three, thirty-five and three thousand four hundred and fifty-five nodes, that no lettering exists — and the letterings it proved impossible fold, on the collection's own machinery, at every size they were tried at.

the period cell of the gridone period, with its neighbours round it1 interior vertices in the cell4 crease pieces drawnperiod 1.000 × 1.000one square, because a grid repeats at every linethe cell is a rectangle of ordinary paper until somebody says its edges are one edge What it costs to know

A map with no edges

Counting the ways a rectangular map folds is the oldest open problem in the subject, and every version of it assumes the map has an edge. Join the map's opposite edges and the question changes shape: half the sizes have no folded state at all, and the ones that do have no bottom layer to count from.

the case the corner search cannot seeedges clear of every vertex, and a crease through a corner anywaythe corner is where four edges meeta crease piece ending there has no partneron any one of themand Euler's count comes out −1the cure is a nudge along a gap the vertex search had already cleared Tessellations

The seam that is not a symmetry

Gluing a cell's edges looks like a symmetry of the drawing and is not. It is an instruction about which points of the paper are the same point, the drawing has to agree with it along the whole of a glued edge, and a rectangle that is not a period of the pattern does not glue at all — which turns out to be the only real restriction on which cylinders exist.

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 Flat-folding

A bottom layer on half a rim

The bottom of a folded stack lives at the paper's edge, which is why a sheet with no edge has an order with no least element. A cylinder has half a rim, so it has a bottom — and the count of panels that could be it falls with the rim, which makes the claim a measurement rather than a boundary case.

ruling out the square cell's loops, one direction at a timewhat is left splits248 arcs go, 24 remaindirection (1, 0)102 arcs go, 10 remainwhat is left splits010 arcs go, 0 remaindirection (-1, 0)102 arcs go, 10 remainwhat is left splits010 arcs go, 0 remainthe bar is how many arcs are still in play after the step Flat-folding

The arc that arrived twice

Which of two panels a crease calls its near one is decided by the order a face walk happened to number them, and the mirrored record is the same relation. Except on one sheet, where it is not — and that sheet turned out to be the one whose folded state comes back the other way up, which is how a duplicate in a graph became a diagnosis.

one rectangle, glued four waysa disc, two cylinders and a torus — from one drawing4 edges lefta disc2 edges lefta cylinder, across2 edges lefta cylinder, alongno edges lefta torusthe same rectangle and the same creases in all four, and nothing in the drawing says which is whichmatching arrowheads mean the two edges are one edge of the paper Flat-folding

The drawing does not say what is glued

One crease pattern, four sheets, four different answers to whether it folds — and nothing in the drawing distinguishes them. The identification is data the picture cannot carry, and the picture is the object this collection has been treating as complete.

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 loopthe bar is the crease count; the note is what the ordinary checks said Designing a base

The symmetry a gluing adds

A patch of a tessellation has whatever symmetry its outline allows — a few reflections, a rotation or two. Glue its edges and it acquires translations, and a lettering of the glued sheet has to be invariant under them. That is a much stronger requirement than a lettering of the patch, and it is why one answer covers every patch at once.

four conditions with nowhere to holddevelopabilityholds, at 0 verticesKawasakiholds, at 0 verticesMaekawaholds, at 0 verticesbig-little-bigholds, at 0 vertices6 of 12 of these bands have no flat folded stateand only the panel colouring can see it Curves and material

A crease with no vertex to belong to

Crease density is measured as length of line per area of paper, and everything else about a crease is measured at the vertex it runs into. A band of paper has creases that run from one edge to the other and meet nothing, so it has density and no vertices at all — and it still refuses to fold.

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 Rigid folding

Two panels that are one panel

Paper cannot pass through paper, and every test for it compares pairs of panels. On a glued sheet two pieces of the drawing can be the same piece of paper — so a test that does not know the identification either reports a collision between a panel and itself, or misses one where the sheet meets itself round the loop.

the period cell of the Miuraone period, with its neighbours round it2 interior vertices in the cell7 crease pieces drawnperiod 1.000 × 2.000one column wide and two rows high, because the zigzag returns after twothe cell is a rectangle of ordinary paper until somebody says its edges are one edge Rigid folding

A mechanism that closes on itself

A rigid-foldable pattern is a mechanism: panels as rigid plates, creases as hinges, and a motion counted by degrees of freedom at each vertex. Close the sheet into a tube and the mechanism has to come back to itself after a circuit — a constraint that is not at any vertex and that the degree-of-freedom count does not see.

a disc, with a vertexa ring, with noneone interior vertex, 3 creases at itodd degree, so they do notno interior vertices at alland the panels still do notboth refuse: two routes round the sheet leave a panel 1.87 sheet-widths apart Who found it, and when

A theorem with an unstated hypothesis

Maekawa's and Kawasaki's conditions are quoted everywhere without saying which sheet they are about, and they do not need to be — they are conditions at a point and every point is the same. The two-colouring is quoted the same way and it is not a condition at a point, and the omission there is not harmless.

0204060801001200510152025foldssurface heldpeak at 64 foldsbox of side 1 · sheet thickness 0.01 · most surface at 64 folds · 128 folds fills the box with sheet alone Folding nobody designed

Nothing grown has a seam

A gut is a tube and a leaf is a disc, and neither was made by joining anything. The sheets this collection builds by identifying a rectangle's edges are the same objects a body grows, reached by an operation no organism performs — and the difference shows up in where the boundary is and in what has to close.

spacing rulelayers across the folded footprintratio reacheduniform12 panels, longest 1.0012.00the only one at the topgeometric-1.112 panels, longest 2.857.5038% shortgeometric-1.312 panels, longest 17.924.1565% shortalternating-212 panels, longest 2.009.0025% shortone-long12 panels, longest 3.004.6761% short12 panels · ratio = Σℓ ⁄ max ℓ · uniform is the unique maximiser, and every other rule pays for its longest panel Folding nobody designed

The census returns one

The rung below this one asked for a census: every pattern reaching a stated packing ratio while opening from a single input, with its crease density. The census is makeable for the corrugations and it comes back with one member. A corrugation piles its panels over a footprint as wide as its longest panel, so its ratio is the total length divided by that longest one — and that equals the panel count only when every panel is the same.

0204060801000510152025foldssurface heldno supply — 50 foldsδ = 0.01 — 25 foldsδ = 0.03 — 12 foldsδ = 0.09 — 5 foldsbox of side 1 · sheet thickness 0.01 · optimum at S ⁄ 2(t + δ), so supply and sheet are charged the same way Folding nobody designed

The surface has to be supplied

The curve that turns over does so because the sheet's own thickness fills the box it is folding into. A surface in a body has to be reached as well as fitted, and the channel that reaches it takes depth out of the same box on exactly the same terms — so the best fold count and the surface it delivers both fall by the ratio of the sheet's thickness to the sheet and its supply together.

0204060801000510152025foldssurface held1 surface — 50 folds2 surfaces — 25 folds3 surfaces — 17 folds4 surfaces — 12 foldsbox of side 1, thickness 0.01 · a ceiling goes as depth², so m sharers of one depth reach one m-th of it between them Folding nobody designed

Two surfaces in one box

A body folds several surfaces into one volume and each of them does a different job. Dividing the depth between them looks like a fair split costing nothing overall, and it is not: the area a single surface can reach goes as the square of the depth it has, so m surfaces sharing a depth reach a total of exactly one m-th of what one of them would have reached alone.

wavesamplitudefits inside2∫κ² = 170.03980.043∫κ² = 390.02660.045∫κ² = 1080.01590.04, 0.028∫κ² = 2760.01000.04, 0.02, 0.0112∫κ² = 6210.00660.04, 0.02, 0.0120∫κ² = 17240.00400.04, 0.02, 0.01depth 0.04 needs 2 waves or more · depth 0.02 needs 4 waves or more · depth 0.01 needs 8 waves or moreexcess 6.0% · amplitude × waves = 0.0797 throughout · floor = that constant ⁄ depth Folding nobody designed

The container picks the member

Two large waves and eight small ones are the same metric, and geometry has nothing to say about which. Bending has nothing to say either — it rises at every step of the family, so a least-bending rule always answers the fewest waves and never anything else. What decides is the container: amplitude times wave count is constant across the family, so a ceiling on the height is a floor on the number, exactly inversely.

growth profilelargest |K| it carries∫K dAgrown by the same factor everywhereflat — it can be laid in a plane0.0000.0e+0 — nothinggrown more at the rimnot flat at any radius1.400-3.258grown more at the centrenot flat at any radius7.8426.767the growth of a spherical capnot flat at any radius0.3911.118the growth of a hyperbolic discnot flat at any radius0.391-1.360grown so that the curvature cancelsnot flat at any radius1.4004.8e-5 — nothing∫K dA = −2πR (ln Ω)′(R) — the total is decided at the rim, so the interior cancels out of it Folding nobody designed

The test measures the rim

Flattening the specimen is the right test and the measurement anybody actually makes on a flattened specimen is a boundary one — how far the margin overruns its chord. Total curvature is a boundary quantity too: it equals minus two pi R times the growth profile's slope at the rim, and nothing else about the interior survives into it. So a sheet can be curved everywhere and integrate to nothing, and the test reports it flat.

how much choice there is in the cuttingevery subset of the joins, tested for connectivitygridcranesjoin pointssubsetsconnectedfewest joins2 × 250% of them work41211 of 1one way only3 × 36.3% of them work941614 of 4one way only4 × 44.1% of them work169512215 of 9one way only5 × 51.2% of them work251665,53678510 of 1650 wayseach interior lattice point holds four cranes at once · every subset of them tried, and the piece has to come out in one piece Who found it, and when

Which cranes can stay joined

The 1797 book slits a square into a grid and leaves the cranes attached at the interior lattice points, each of which holds four of them at once. Of the sixteen ways to choose which of a three-by-three's four points to leave joined, exactly one leaves the piece in a single object — and it is the one that uses all four. The cutting is very nearly forced rather than chosen.

how many witnesses each claim has5 of 15 have one, and a loss removes them from the record rather than weakening themclaimsurviving sourcesafter one lossPaper is made in China21 leftPaper reaches Japan1nothing attests itPaper is made in Europe32 leftFolded paper is used ceremonially in Japan43 leftPaper is folded for amusement in Japan1nothing attests itThe thousand cranes1nothing attests itThe pajarita is folded in Spain21 leftPaper folding is taught as geometry43 leftOne fold solves a cubic1nothing attests itThe diamond pattern in a crushed cylinder21 leftThe conditions at a flat-foldable vertex32 leftThe dashed-and-dotted diagram notation21 leftThe Miura fold32 leftA five-pointed star from one straight cut1nothing attests itAny straight-line drawing, from one straight cut21 leftby kind of source: artefact 2 · manuscript 2 (1 single) · printed 9 (3 single) · secondary 2 (1 single)the average overrun falls from 357 years to 193 without them, and the median from 201.5 to 184.5 — the effect is two rows, not a tendencywitnesses counted from the record itself · Paper is folded for amusement in Japan and The thousand cranes are the two largest overruns and have one document each Who found it, and when

One lost source and the story changes

Five of the record's fifteen claims rest on exactly one surviving document. Take those away and the field's most-quoted statistic — how far ahead of its evidence a popular date runs — falls from three hundred and fifty-seven years to a hundred and ninety-three, while the median hardly moves at all. The effect is not a tendency spread through the record; it is two documents, and both of them are single-witness.

-2.5-2-1.5-1-0.5050100150200250depth given to the inner level, log₁₀ of the boxsurface over the flat sheetone level, the whole boxreaches 250.0two levels, any splitnever above 62.5box depth 1 · sheet 0.001 · one level reaches 250.0, and a nest of two reaches 62.5 however the depth is shared Folding nobody designed

A nest pays four a level

A corrugation folded inside the panels of another looks like the arrangement that multiplies surface rather than dividing it. It multiplies the factors and divides the depths, and the depths cancel: a packed level can hold at most its depth over four times what it folds, the thing it folds is as thick as the depth the level below was given, and so a nest of L levels reaches at most the box over 4ᴸ sheet thicknesses. One level with the whole box beats any nest of two by exactly four.

the bar is how many of 200 random spring settings give two or more resting statessprings set at random fold angles, the same settings for every rowvertex 60·90·120·901982 with 1 · 198 with 2vertex 45·100·135·801964 with 1 · 196 with 2vertex 80·95·100·851964 with 1 · 126 with 2 · 62 with 3 · 8 with 4a corrugation0200 with 1a vertex's configurations are two branches through the flat state; a corrugation's are one line Folding nobody designed

A corrugation has one resting state

A folded wing held short of shut stores energy in its hinges, and a wing that could stay both open and folded with nothing holding it would need that energy to have two bottoms. A corrugation cannot provide them: every crease in it folds by one angle, so the energy of any set of crease springs is a parabola in that angle and has exactly one resting state, however much the springs disagree. A single degree-four vertex has two branches through the flat state, and the same springs give it two resting states on almost every setting tried.

which blocks a single strand can routecolumns are the block's width and rows its height, both counted in helicessquare lattice1234567891012345678honeycomb lattice12345678910a route through every helixthe count allows it, and no route existsa column the rows never reach Folding nobody designed

A row the route cannot leave

Every rectangular block of helices up to ten by eight routes on the square lattice. On the honeycomb, the lattice a double helix's pitch prefers, twenty of the eighty do not — and every block odd in both directions fails for a reason visible along one row: every second helix on the top row has no neighbour off it, the two corners have one neighbour each, and a route forced through them runs the length of the row and ends. The colour count passes all of them, and a degree count along a single row refuses them.

00.20.40.60.81-11234radius of the three marks, on the flat sheetmiss against flat (%)grown by the same factor everywhere · −0.00%grown more at the rim · +4.27%the growth of a spherical cap · −1.71%grown so that the curvature cancels · +3.52%a centre mark and three a third of a turn apart · on a flat sheet the three sit √3 times their radius apart Folding nobody designed

Three marks see nothing

The measurement a rim cannot make is an interior one, and the obvious interior measurement — two marks a known distance apart, measured again after growth — cannot detect curvature at all, because a uniformly enlarged sheet changes that distance and stays flat. Three marks cannot either: any three distances obeying the triangle inequality are the sides of a flat triangle. Four marks give six distances, and six distances are not free on a flat sheet. The growth profile a rim measurement reads as flat misses by three and a half per cent with four marks at the rim.

00.20.40.60.81-3-2-11radius of the cut, on the flat sheet∫K dA inside the cutR ⁄ √2grown by the same factor everywheregrown more at the rimthe growth of a spherical capgrown so that the curvature cancelsa cut at radius r reads the total curvature of the disc inside it, −2πr (ln Ω)′(r) — the rim measurement moved inward Folding nobody designed

A cut reads a slope

A rim measurement returns one number for a whole grown disc, and a family of growth patterns share it. Cut the disc in a circle and the piece inside has a rim of its own, and its reading is minus two pi r times the slope of the log of the growth at the cut — so a cut reads a slope, a set of cuts reads the slope at each radius, and the slopes add up to the growth profile itself. The one thing no cut can recover is how much the whole sheet was enlarged, which is the one kind of growth that curves nothing.

two ways to come aparta crane left hanging, or every crane held and the piece still in islandsgridsubsetsnothing hangingholds togetherfailures localheld, of those passing3 × 34 joins1611100.0%100.0%4 × 49 joins512322197.8%65.6%5 × 516 joins65,5361,21578599.3%64.6%6 × 625 joins33,554,432260,625141,62199.6%54.3%a crane hangs from nothing when none of the joins at its four corners is kept — one crane, four points, no search Who found it, and when

Nearly every cutting fails at one crane

Six by six connected cranes have twenty-five joins and thirty-three million ways to keep some of them, and an exhaustion over all of them takes a fifth of a second. Of the 33,412,811 that fail, 99.64 per cent fail at a single crane — one left holding none of the joins at its corners — which a maker can check by looking at each crane in turn. The arrangements that pass that check hold together less often as the grid grows: all of them at three by three, 54 per cent at six by six.

the rules that fold, against the clauses they all obeya clause says an odd or even number of some chosen letters are mountainsfamilyrulesfoldparity clausesthey allowa same-or-differ recipethe Miura fold6416216yesthe tapered leaf6416216yesthe Yoshimura pattern6426132nothe waterbomb tessellation51232364noa recipe of same and differ clauses allows exactly the rules its clauses allow; that is the test Who found it, and when

A recipe needs degree four

The Miura's letters are taught as a recipe of same-and-differ clauses, and the recipe is exact: the sixteen repeating rules that fold are precisely the rules two clauses allow. The Yoshimura's twenty-six cannot have such a recipe, because twenty-six is not a power of two. The waterbomb's thirty-two is a power of two and still has none — its clauses allow sixty-four rules and half of them fail. The difference is one vertex: at degree four the counting theorem leaves a parity, and at degree six it leaves 'not all alike', which no clause of that kind can say.

the bar is the share of sound cuttings that hold the cranes in one piecesound means every crane keeps at least one join at its corners3 × 3100.0%2⁴ subsets of 4 joins · 1 states carried4 × 465.6%2⁹ subsets of 9 joins · 7 states carried5 × 564.6%2¹⁶ subsets of 16 joins · 23 states carried6 × 654.3%2²⁵ subsets of 25 joins · 52 states carried7 × 751.1%2³⁶ subsets of 36 joins · 123 states carried8 × 846.7%2⁴⁹ subsets of 49 joins · 294 states carried9 × 943.6%2⁶⁴ subsets of 64 joins · 714 states carried10 × 1040.5%2⁸¹ subsets of 81 joins · 1,758 states carried11 × 1137.7%2¹⁰⁰ subsets of 100 joins · 4,380 states carried12 × 1235.1%2¹²¹ subsets of 121 joins · 11,024 states carriedcounted one join at a time, carrying only which cranes on the frontier are already joined Who found it, and when

The border is where the cranes come apart

Counted one join at a time rather than one subset at a time, the slit grid of connected cranes runs to twelve by twelve, where there are 2¹²¹ ways to keep some of the joins. Among the cuttings that hold every crane by something, the share that also hold together keeps falling — 54.3 per cent at six by six, 35.1 at twelve — and from eight by eight on it falls by the same factor at every size. A constant factor is the signature of the border: at six by six, 99.5 per cent of the sound cuttings that come apart do so through a stray piece touching the outermost ring of joins.

-40-20204060801000510152025degrees of the driven crease from the flat sheetenergy in the springsthe road from one resting state to the other · sectors 60° · 90° · 120° · 90°, springs set to mixedbranch one's stateenergy 12.40branch two's stateenergy 17.65the flat sheetenergy 26.16to leave the deeperclimb 13.76to leave the shallowerclimb 8.51left of the middle is branch one, right of it branch two; they meet only at the flat sheet Folding nobody designed

The wall is the flat sheet

A sprung degree-four vertex usually has two resting states, one on each branch of its motion, and the branches meet in one place a sheet can pass through: the flat state. So the only road from one resting state to the other crosses the flat sheet, and on every setting of the springs tried on two vertices the flat sheet is the highest point of that road. Its energy is each spring's stiffness times its rest angle squared, summed, which does not contain the vertex's sector angles at all — the same springs put on four different vertices give a wall of exactly the same height. The geometry decides only how far below the wall each state sits, and the shallower one sits a median of five per cent below it.

the clauses a recipe needs, family by familyeach recipe is checked to allow exactly the rules that fold, and nothing elsefamilyrulesfoldparitiesprohibitionsdegree-six kindsclauses in allthe Miura fold64162002the Yoshimura pattern64261223the waterbomb tessellation512323245a parity says an even or odd number of some letters are mountains; a prohibition says some letters are not all alike Who found it, and when

Two sentences for the Yoshimura

No recipe made only of same-and-differ clauses picks out the Yoshimura pattern's twenty-six folding rules, because its vertices have six creases. A recipe allowed one other kind of sentence does, and it is short: an even number of the four zigzag classes are mountains, and no course carries the letter all four zigzags share. That is three clauses, one parity and a prohibition for each of the pattern's two kinds of vertex, and it allows exactly the twenty-six. The waterbomb tessellation, with four kinds of six-crease vertex, needs three parities and only two prohibitions, because its parities do half the prohibiting.

the bar is the size of the smallest shape the tests pass that no route reacheseach row adds one more cheap test to the ones above itsquares: the colour count, the ends and the cuts9 helices64 shapes of that size pass and have no routesquares: and the steps the ends force11 helices68 shapes of that size pass and have no routesquares: and the colour of every forced end11 helices32 shapes of that size pass and have no routehoneycomb: the colour count, the ends and the cuts12 helices18 shapes of that size pass and have no routehoneycomb: and the steps the ends force15 helices12 shapes of that size pass and have no routehoneycomb: and the colour of every forced end16 helices6 shapes of that size pass and have no routeevery shape of every smaller size is either refused by the tests or routed by the search Folding nobody designed

Every cheap test misses a shape

A strand routed through a bundle of helices has to visit each once, and whether a shape allows that is hard to decide — so the cheap tests that refuse shapes are necessary and never sufficient, and for every set of them there is a smallest shape they pass and no route reaches. Listing every connected shape and searching the ones the tests let through finds it: nine helices on the square lattice for the colour count, the ends and the cuts, eleven once the steps a route's ends force are added, and still eleven once the ends' colours are checked. On the honeycomb the same three stages give twelve, fifteen and sixteen. Each test pushes the smallest unroutable shape out or leaves it where it is; none removes it.

the bar is the median number of resting states, over the same kind of random springsa chain shares one crease between each vertex and the next1 vertex2 states2 branch combinations · 198 of 200 settings rest on every one · 197 of 198 cross at the flat sheet2 vertices4 states4 branch combinations · 193 of 200 settings rest on every one · 200 of 200 cross at the flat sheet3 vertices8 states8 branch combinations · 181 of 200 settings rest on every one · 199 of 200 cross at the flat sheet4 vertices16 states16 branch combinations · 134 of 200 settings rest on every one · 200 of 200 cross at the flat sheetevery combination of branches holds a resting state, and every switch between them goes over the whole flat sheet Folding nobody designed

A chain of vertices switches all at once

One sprung degree-four vertex rests in two states, one on each branch of its motion, and switches between them only by passing through the flat sheet. Chain vertices together by sharing a crease between each and the next, and a branch can be chosen at every vertex: two, four, eight and sixteen combinations for chains of one to four. The median spring setting rests once on every combination. And every combination's curve of configurations passes through the same single point — the whole chain flat at once — and meets no other anywhere else, so every switch, even of one vertex's branch, takes the whole chain back to flat. The wall that switch climbs is every spring's flat energy added up, growing by a crease's worth for every crease, and a chain's second state sits several times further below it than a single vertex's does.

-3-2-1123050100150turn in each hinge (radians, signed by branch)energy storedrests hereand hereflat: 179the barrier isthe flat state8 folds, hinge 0.05creased to 1.6the two branches are the same sheet folded opposite ways, and the flat sheet is the only state they share Folding nobody designed

Springs that disagree do not offer a choice

A corrugation of hinges that remember different angles was expected to have more than one position in which nothing pushes. It has exactly one, at the stiffness-weighted mean of what they remember, because a sum of parabolas in one variable is a parabola. The second resting place comes from somewhere else entirely — the mirror pattern — and the flat sheet is the barrier between them, which is also why a creased sheet cannot be pulled flat at all.

012340200400600800clearance above the basesurface, as a multiple of the basewalls: 2c ⁄ τplies: c ⁄ 4τeight times lesssheet thickness 0.01 · both lines are straight and their ratio is eight everywhere, so no clearance makes the stack competitive Folding nobody designed

Standing up beats lying down by eight

A level that fills its clearance with plies lying flat makes every ply share the clearance, and the sharing costs a factor of four. A level that fills the same clearance with walls standing on the base gives every member the whole height and charges them only for footing. The ratio is exactly eight, at every clearance and every thickness — and it is the difference between a cost charged against depth and a cost charged against the space beside it.

a body of three segments — two internal edges, 0.6 and 0.9the segments are the pairs at their limit — 4 of 8 of them measured through the body rather than around it Designing a base

What the condition does not decide

A tree of seven leaves imposes twenty-one separations and eight of them bind. The rest are slack, the eight pin six of the seven leaves against the sheet's own edges, and the seventh can be moved half a per cent of the sheet for nothing. The requirement that looks quadratic is doing linear work, and what it leaves undecided is the part a designer is actually choosing.

00.20.40.60.811.21.41.600.10.20.30.4arc from the pole (radians)excess, as a share of the circle7 rings5.0% stretchfirst at 0.35last at 0.98of the way to the rima ring goes in wherever the residual excess would otherwise pass what the material takes Curves and material

Where a ring of divisions belongs

A pattern that divides the circle everywhere as finely as its rim requires is over-divided for most of its radius, because the excess grows from nothing. Putting a ring of new divisions in wherever the residual would otherwise pass what the material takes gives seven rings on a hemisphere at five per cent of stretch, at 0.35, 0.50, 0.62, 0.72, 0.81, 0.90 and 0.98 of the way out — and the first of those sits where a completely different criterion put its first tuck start.

how many folded states each crease of a four-by-four Miura leavesdriven to 0.6 radians, with every consistent assignment enumerated rather than the first eight1 state4an actuator belongs on one of these2 states82 states, so the sheet has a choice4 states44 states, so the sheet has a choice8 states88 states, so the sheet has a choicethe four that leave one are c:3:2, c:3:3, r:3:2, r:3:3 — all of them at the same corner of the sheet Rigid folding

Only four creases decide a Miura

Driving one crease of a rigid quadrilateral mesh settles every other one — except that on the pattern everybody builds it often does not. Enumerated properly, four of a four-by-four Miura's twenty-four creases leave exactly one folded state and the other twenty leave two, four or eight. A mesh whose vertices all differ leaves one from every crease. The ambiguity is not a property of quadrilateral meshes; it belongs to the symmetry.

the two things a domain has to be long enough forthe bar starts where the demand is first met and runs to the rightunique in the whole design11 bases and upstill paired at 45 degrees17 bases and upwhere a honeycomb crossover may sitfirst one inside both bands: 21 baseswhere a square-lattice crossover may sitfirst one inside both bands: 24 bases4812162024283236a domain runs between two crossovers, so its length is a whole multiple of the period Folding nobody designed

A domain too short to be unique

A staple holds the scaffold by pairing with a stretch of it, and two arguments decide how long that stretch has to be. One is combinatorics — a stretch of seven bases has about four hundred other places in a 7,249-base strand it would also match. The other is thermodynamics, and it is the one that binds: a duplex that is unique at eleven base pairs still comes apart at the temperature the design is held at, and staying paired takes seventeen. Rounded up to the crossover period, that is three periods on both lattices, and the lattice the helix prefers is the one whose three-period domain leaves some of its staples unattached.

the bar is the size of the smallest shape the tests pass that no route reacheseach row adds one more cheap test to the ones above itsquares: the colour count, the ends and the cuts9 helices64 shapes of that size pass and have no routesquares: and the steps the ends force11 helices68 shapes of that size pass and have no routesquares: and the colour of every forced end11 helices32 shapes of that size pass and have no routesquares: and the colour count on a stretch a cut has fenced off12 helices12 shapes of that size pass and have no routehoneycomb: the colour count, the ends and the cuts12 helices18 shapes of that size pass and have no routehoneycomb: and the steps the ends force15 helices12 shapes of that size pass and have no routehoneycomb: and the colour of every forced end16 helices6 shapes of that size pass and have no routehoneycomb: and the colour count on a stretch a cut has fenced off16 helices6 shapes of that size pass and have no routeevery shape of every smaller size is either refused by the tests or routed by the search Folding nobody designed

A test that only knows one lattice

The cheapest argument that refuses the smallest shape no cheap test could refuse was read off that shape: cut at one helix, find the piece with no end in it, and count the colours of the stretch the route is then forced to cross. Added to the census it refuses every one of the square lattice's thirty-two eleven-helix survivors and pushes the smallest survivor to twelve, where twelve placements of two shapes survive out of half a million. On the honeycomb it refuses none of the six at sixteen. A test inherits the lattice of the witness it was read off, and the staircase is two staircases.

the pale bar is every folded state the strip has; the dark one is what the machine reachesthree machines on the same strips, and none of them is the flat-folding theorem3 equal stamps · takes every layerall 123 equal stamps · takes one layer4 of 123 equal stamps · takes any block reaching an edgeall 124 equal stamps · takes every layerall 324 equal stamps · takes one layer4 of 324 equal stamps · takes any block reaching an edgeall 325 equal stamps · takes every layerall 1005 equal stamps · takes one layer4 of 1005 equal stamps · takes any block reaching an edgeall 1006 equal stamps · takes every layerall 2886 equal stamps · takes one layer4 of 2886 equal stamps · takes any block reaching an edgeall 288counted over every marking of the strip that folds flat at all What it costs to know

Deciding is not making

Four earlier essays here ask which machines can flatten a strip at all, and the answer sorts them into a lattice with one column full and three with holes in it. Asked instead what each machine can produce, the three sort completely differently: the machine that may choose its block reaches every folded state of every strip tried, the machine that takes one layer reaches exactly four whatever the strip is and however long, and the machine that takes the whole pile is the only one whose answer depends on the spacing at all.

spending the bird's length where the arrangement prices it lowesta price is the scale lost per unit of length added to every edge of a group, re-solved at every stepstepscaledearestpricecheapestpricetight pairs00.2651head0.092tail-0.000510.2753head0.102wings0.034520.2748legs0.081wings0.012530.2781body0.072wings0.037640.2817head0.081wings0.054350.2778tail0.074head0.000560.2780head0.076tail0.0026a price holds for as long as the dearest and the cheapest group stay the same groups Designing a base

A price holds until the arrangement moves

Every edge of a subject's tree has a price — the scale lost per unit of extra length — and the obvious use of a price list is to spend a fixed total of limb where it is cheapest. Done a tenth of a unit at a time, re-pricing at every step, it works and then stops: the bird's scale rises 6.3 per cent in four steps and no further. But the prices do not hold while it happens. The bird's free tail stops being free after the first tenth, and its legs nearly treble in price without being touched. The lizard's prices hold for four steps, because its arrangement keeps the same three pairs at their limit for four steps. A price is a statement about which pairs are at or near their limit, and it lasts as long as they stay there.

the bird rounded to a grid two waysunits in the order body, wings, head, legs, tail; the drawn tree unrounded has size 0.2651gridnearesterrorsizecheap wayerrorsize4 units1 4 2 2 316%0.26751 4 2 2 423%0.28076 units2 6 3 3 511%0.27241 6 3 3 545%0.27848 units3 8 5 4 715%0.26312 8 4 4 720%0.2782size is the scale times the sheet length one unit of the subject's own length receives; error is the worst limb's Designing a base

Rounding in the cheap direction

A tree spelled on a grid has every limb rounded to a whole number of units, and the rounding is chosen to keep the subject's proportions. Each rounding is also a small move of length between edges, and the edges have prices. Rounding the bird's dearer edges down and its cheaper ones up gives the largest model of every rounding tried, on grids of four, six and eight units — 2 to 6 per cent larger than rounding to the nearest unit, and larger than the unrounded bird itself on all three. The proportions pay for it, by five points of error on eight units and by thirty-four on six, which is the trade the grid had been making silently in whichever direction the arithmetic happened to fall.

the layer-order field for the printed patterns, filled in where it can bepairs is how many signs the field holds; varying is how many of them differ between folded statespatternpanelspairsstatesvaryingto recordThe preliminary base82810noneThe Miura fold24228not listed24 panelsThe square twist93610noneThe hexagon twist136610noneThe Yoshimura pattern652055not listed65 panelsFold and cut — the triangle721261.00 bitsThe tapered corrugation28282not listed28 panelsThe waterbomb tessellation52926not listed52 panelsa pattern past eighteen panels is not listed here, and those are the patterns anybody folds Who found it, and when

The field is empty where it would say nothing

The interchange format for crease patterns has a field for the layer order and nothing ever fills it in. Filling it in where the folded states can be listed — four of the eight printed patterns, and Miura patches to twelve panels — finds that the preliminary base and both twists have exactly one folded state, so every one of the field's signs follows from the crease pattern and the field would record nothing a reader could not compute. The fold-and-cut triangle has two states. The Miura is different: every patch with three or more columns has several — three, six and eleven on the three-by-two, three-by-three and four-by-three — so on the pattern that gets built the field carries information from six panels up, and the field's size had been measured as log₂ of the panels' orderings, which on the preliminary base is fifteen bits for an object that has zero.

a 4-by-4 Miura, every crease driven in turn, at 10 angles along the motionevery consistent assignment enumerated at each crease, with both configurations found at every vertexanglecreases × states they leavethe creases that decide it0.24×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:30.44×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:30.64×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:30.84×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:31.04×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:31.24×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:31.64×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:32.04×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:32.44×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:32.84×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:3every row is the row above it: the census is a property of the pattern, not of how far it has folded Rigid folding

The deciding set does not move

A driven Miura leaves several folded states from most of its creases and exactly one from a few, and those few are where an actuator belongs. It was reported that the few change along the motion — four of twenty-four at 0.6 radians, fourteen at 0.8 — and that a five-by-five sheet had a crease leaving fifteen states where every other count was a power of two. Mapped at twenty angles from 0.1 to 3.0 radians on three sizes of sheet, neither survives. Every crease leaves the same number of states at every angle, every number is a power of two, and the same creases decide the sheet throughout. The changes were the vertex solver losing one of a vertex's two configurations on 138 of 8,640 solves, and the configurations it lost can be carried exactly from an angle where it finds both.

a corrugation remembering 0.8 rad, held at 2.4 and later opened to 0.4torques as shares of the one total both ends of the story share; the barrier as a multiple of its creased valueheld for, τremembered turncontainer suppliesopening needsbarrier0.00.80080%20%1.00×0.51.43049%51%3.19×1.01.81129%71%5.13×2.02.18311%89%7.45×3.02.3204%96%8.41×5.02.3891%99%8.92×the container's share and the opening's share always sum to the whole, whatever the holding time Folding nobody designed

Holding a fold moves the force

A creased hinge held at an angle slowly comes to remember that angle, so a leaf or a wing packed in a bud for a season is gradually holding itself and the bud has less to do. The force is not used up in the process. The torque the container must supply falls as e^(−T⁄τ), the torque later needed to open the structure rises by exactly the same amount, and the two sum to the same total at every moment of the holding. Held for three relaxation times, a corrugation creased to 0.8 radians and packed to 2.4 needs 4 per cent of the total from its container and 96 per cent from whatever opens it — and the barrier to its mirror image has grown more than eightfold. A packing that lasts buys independence from its container with a harder unfolding.

four vertices round one panel, as a chain and as a loopa combination survives the loop only if going round it brings every fold angle back to where it startedthe faceopen chain of fourclosed loopthe same at every anglea Miura face164yesa face with no two vertices alike, seed 11161yesdriven at 0.3, 0.6, 1, 1.4, 1.8 radians · a combination counts when every crease is folded and the loop closes Folding nobody designed

A loop takes choices away

A chain of four sprung degree-four vertices has sixteen combinations of branches, each a resting state, and switches between them only through the flat sheet. Close the chain into a loop round one panel and the combinations must agree when the fold angles come back round. On a face whose four vertices all differ, one combination survives; on a Miura face, four. The count is the same at every angle the face is driven to, and two surviving assignments at the same driven angle are never closer than one and a half times that angle — so they separate as the face folds and meet only when it is flat. A loop does not create the junction a region would need to switch on its own. It removes choices and leaves the switch as global as before.

do the pleat equations close round every loop, as drawn and under three linear mapstrivially means every edge's equation is one at both ends; otherwise the largest disagreement round a loopas drawnshearedstretchedgeneralthe square gridcloses, triviallycloses, triviallycloses, triviallycloses, triviallythe triangular gridcloses, triviallycloses, triviallycloses, triviallycloses, triviallythe honeycombcloses, triviallycloses, triviallycloses, triviallycloses, triviallythe rhombille tilingclosesoff by 1.76off by 2.18off by 2.96the elongated triangular tilingcloses, triviallyoff by 3.14closes, triviallyoff by 4.77the maps: a shear of 0.3, a stretch of 1.5 along one axis, and the matrix [1.3, 0.4; −0.2, 0.9] Tessellations

Closing the loops is not folding

The twist construction propagates one equation along every edge of a tiling, and it can only work where the equations agree round every loop. Asked which irregular tilings pass, a linear map gives a clean answer: the square grid, the triangular grid and the honeycomb pass under every shear and stretch tried, because each edge has a half-turn symmetry that makes its equation exactly one at both ends, and a half-turn survives any linear map. The rhombille passes only as drawn. But passing is not folding. On every one of those images — including the ones whose loops close exactly — the construction produces a pattern that fails the angle condition at every turn tried. The loops were a necessary condition all along, and the construction needs something the tilings' images do not give it.

does the construction fold, with one side distance a vertex and with one a sideat a turn of 0.42 radians; the second construction's pleats run at -0.5 radians from their own edgesas drawnshearedstretchedgeneralone a vertex / one a sidethe square gridfolds/foldsno/foldsno/foldsno/foldsthe triangular gridfolds/foldsno/foldsno/foldsno/foldsthe honeycombfolds/foldsno/foldsno/foldsno/foldsthe rhombille tilingfolds/foldsno/foldsno/foldsno/foldsthe elongated triangular tilingfolds/foldsno/foldsno/foldsno/foldsthe maps: a shear of 0.3, a stretch of 1.5 along one axis, and the matrix [1.3, 0.4; −0.2, 0.9] Tessellations

One number where the corners wanted four

The twist construction gives a vertex a single side distance, and every account of these patterns does the same — it is what rotate-and-shrink means. The conditions never asked for it. Written out, the corner condition is one linear equation per pleat crease in the distances taken one per edge, so a degree-four vertex carries four unknowns against two independent equations. Given them back, the twelve sheared and stretched tilings that refused to fold all fold.

each side of the twist, divided by the edge it facesone vertex of the the square grid under [1.3, 0.4, -0.2, 0.9]1.0520.3871.0520.387the weightsthe same edges, scaled by them, end to endthe 4 weighted edges close to 3e-16 of their own total length, and the two ends of every edge agree to 3e-15 Tessellations

Every twist writes an equilibrium

Divide each side of a twist polygon by the length of the edge it faces. The polygon closing says those numbers, weighted onto the edges, balance at the vertex; the pleat matching says the two ends of an edge agree on the number. Together they are a positive equilibrium stress — the thing a tiling has when it is the plan of a spider web — and the construction has been writing one at every vertex without being asked for it.

how far each of a curved fold's two surfaces reaches before its rulings crossas a share of the crease's own tightest radius of curvatureone surfacethe othershare boundeda circular crease at 0.40.389100% / 0%a circular crease at 0.70.644100% / 0%a circular crease at 10.841100% / 0%a circular crease at 1.30.964100% / 0%an elliptical crease at 0.40.389100% / 0%an elliptical crease at 0.70.644100% / 0%an elliptical crease at 10.841100% / 0%an elliptical crease at 1.30.964100% / 0%a parabolic crease at 0.40.389100% / 0%a parabolic crease at 0.70.644100% / 0%a parabolic crease at 10.841100% / 0%a parabolic crease at 1.30.964100% / 0%a wave at 0.40.3890.38950% / 50%a wave at 0.70.6440.64450% / 50%a wave at 10.8410.84150% / 50%a wave at 1.30.9640.96450% / 50%a dash is a surface whose rulings never converge, which is a surface with no boundary of this kind at all Curves and material

Only one side can run out

A curved fold has two surfaces and every reach ever computed here has been one of them. The closed form's denominator is the crease's curvature plus the rate the ruling angle turns at, and crossing to the other surface negates both — so at any point of any crease at most one of the two surfaces can have its rulings converge. A crease that never changes the way it bends therefore has a surface with no such boundary at all, anywhere along it.

the cost of one lettering, by size and by how the cell is gluednodes of search, under one fixed branch orderperiodsfree lettersa discone cylinderthe othera torustorus over discthe square grid1×11254430.62×240131212131.03×384262428532.04×4144456244116926.05×52207066209292741.8the honeycomb1×1341291080.72×2116403439952.43×324676285386418655.1the triangular grid1×13412101080.72×2116373134166845.13×324691570524!12000131.9the rhombille tiling1×160226218160.72×2216!12000!120001009!120001.0a plus sign is a search that ran out of budget rather than out of possibilities; the free letters are the cut sheet's What it costs to know

Each drawing has its own threshold

Gluing a cell's edges was measured once, at one size, and found to cost three orders of magnitude — which cannot tell a threshold from a slope, nor say whether a cut sheet has one further out. Swept from one period to five on four tilings, every sheet starts at about a third of a node per free letter and every drawing leaves that behaviour at a size of its own: four periods on the square grid, three on the honeycomb, two on the triangular grid and two on the rhombille, where even the cut sheet crosses.

what ten branch orders cost on the same four sheetsnodes of search; the sheets are the same drawings as the sweep abovethe square grid, 4×4, glued69 to 24636, 1 gave upthe square grid, 4×4, cut42 to 55the square grid, 3×3, glued20 to 731the square grid, 3×3, cut25 to 32each bar runs from the cheapest of 8 branch orders to the dearest, on a logarithmic scale; a dot is the middle one What it costs to know

The route, not the sheet

Every cost measured for a glued sheet has been one number from one branch order, and a backtracking search's cost belongs to the pair. Asked under eight orders instead of one, a cut cell's cost barely moves — 42 to 55 nodes — while the torus over the same drawing runs from 69 to 24,636, with one order giving up entirely. The glued sheet's best order costs less than twice the cut sheet's, so most of what a single order charged to the gluing belongs to the route through it.

what happens when the ruling angle is not constantthe share of the crease each surface is bounded over, and how far it reaches thereturning rateone surfaceits reachthe otherits reach0.00100%0.29560%0.25100%0.23500%0.50100%0.19090%0.75100%0.14840%0.90100%0.11990%1.00100%0.09930%1.1086%0.077314%2.95101.2580%0.041320%1.17461.4075%0.000025%0.7250the crease is a circular crease and the angle runs 1.4 plus the rate times a sine, so the rate is how fast it turns against how fast the crease bends Curves and material

An angle that turns faster than the crease

Which of a curved fold's two surfaces runs out is decided by a sum of two rates — how fast the crease bends and how fast the ruling angle turns — and every measurement so far has set the second to zero. Let it turn and it carries the sign on its own: past a rate of exactly one, a crease of unchanging curvature bounds both of its surfaces, which no constant angle on that crease can do. Below that rate the turning costs reach without changing anything else.

234567024681012columns (rows, for the two-column patches)folded statestwo rows high: 1, 3, 5, 7, 9, 11three rows high: 6, 11two columns wide: 1, 1, 1, 1, 1a strip two rows high has 2c − 3 states; every patch's states are one choice with that many answers Who found it, and when

One choice with eleven answers

A folded state was proposed as a short list of free choices — which way a flap lies, where a rim panel sits — with the layer-order field's signs following from them. Listed exhaustively on every Miura patch small enough, the choices are never independent: every sign that varies is tied to every other through a shared panel, so the states are one choice with many answers. And there are more answers than the record said. The overlap test had a blind spot a third of a panel wide, and with it corrected the three-by-three Miura has six folded states, not one, and the four-by-three eleven, not five.

nodes per free letter, cheapest route against the middle onecheapest of eightmiddle of eight× where no route of that kind finished · periods along the bottom0.3110100×2345the square grid, glued×123the triangular grid, glued××1234the honeycomb, glued0.3110100×123the elongated triangular tiling, glued××12the rhombille tiling, glued×123the rhombille tiling, cut What it costs to know

The cheapest route crosses later

A search for a consistent lettering has a threshold: below it the letters propagate and the cost is a third of a node per crease, above it the search backtracks and the cost explodes. The threshold was measured with one branch order. Measured with eight, the cheapest route never starts searching before the typical one, and on most sheets it starts a period or two later — so part of every threshold on the record belongs to the route. And the one cut sheet past its threshold, the rhombille's, spreads across nearly three orders of magnitude of cost, which moves the spread off the gluing and onto the threshold.

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