Field

Flat-folding

When a crease pattern collapses flat — two local theorems, one global problem, and the gap between them.
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

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

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

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

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

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

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.

the cut line3 straight edgesthe pattern3 skeleton arcs3 perpendiculars1 interior vertexassignments that fold30 of 646 creases in allarcs one way, perpendicularsthe other: fails Maekawaequidistance off by 1.9e-16mountainvalleyevery node sits the same distance from each edge that formed it,which is why one fold can carry several edges onto the line at once

One straight cut

Any drawing made of straight lines can be folded so that the whole drawing lands on a single line, and one cut releases it. The construction is a shrinking process, and it explains itself the moment the shrinking is drawn.

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

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.

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

The gadgets that make it hard

Flat-foldability is NP-hard, and the proof is a construction rather than an obstruction: a machine for turning any satisfiability problem into a sheet of paper that folds exactly when the problem has an answer.

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

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.

the patternthe panels, foldedsheet 12.000footprint 1.966 · 6.11 layers on average · 12 at the deepest1.966 × 6.11 = 12.007, which is the sheet

The paper is all still there

A folded sheet is smaller than it was and none of it has gone anywhere. How much smaller it is and how many layers deep it is are not two properties of a pattern — they are one number, and their product is the sheet.

-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

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

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.

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

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

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

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.

0.5°1%2%3%8%10°16%20°33%how far each sector would have to moveshare that would fold40,000 random four-crease verticesthe median vertex is 31.31° per sector from folding, the mean 33.91°none of them folds, and almost none of them nearly does either

A near miss is nearly as rare

Flat-foldability is a coincidence of measure zero, which is usually where the argument stops. Measure how far a random vertex is from folding rather than whether it does, and the answer is thirty-one degrees a sector — so the tolerance real paper has does not buy back anything at all, and a pattern that nearly folds had to start near one that did.

patternraw edge against crease, by lengthpreliminary base8 panels29.3% rawMiura, 5 by 420 panels20.8% rawsquare twist9 panels26.7% rawYoshimura, 6 by 565 panels6.7% rawthe shaded part is the sheet's own edge; the rest of the outline is creasemeasured with a step of 0.001 of the sheet, and checked across a tenfold sweep of itevery length here is summed over the layers, so a buried edge counts for nothing

The outline is mostly crease

The edge of a folded model is what a reader looks at, and almost none of it is the edge of the paper. Measured across five patterns, the sheet's own boundary accounts for between nothing and a third of the exposed edge; the rest is fold, and on a waterbomb tessellation the raw edge does not reach the outside at all.

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

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

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

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

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.

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

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

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

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.

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.

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

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

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.

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

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

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

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.

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

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

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

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 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

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 upper bar is the rim, the lower is the middlethe value is how many other panels an average panel of that kind lies overThe Miura fold18.0 · 21.016 at the rim, 8 away from itThe square twist8.0 · 8.08 at the rim, 1 away from itThe hexagon twist10.0 · 12.012 at the rim, 1 away from itThe Yoshimura pattern61.6 · 64.021 at the rim, 44 away from itThe tapered corrugation19.0 · 22.218 at the rim, 10 away from itThe waterbomb tessellation31.0 · 37.716 at the rim, 36 away from itthe difference is small and it has the same sign every time

The rim lies over less

A folded sheet's boundary is usually discussed as the place the theorems stop applying. It is also visible in the pile: a panel carrying a raw edge of the paper lies over fewer of the other panels than one that does not, on every printed pattern that has both kinds — 18.0 against 21.0 on a Miura, 31.0 against 37.7 on a waterbomb tessellation, and never once the other way round.

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

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 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

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 vertex nobody listedthe two lines meet at 22.9°sectors 157.1° 22.9° 157.1° 22.9°alternating sums 314.2° and 45.8°Kawasaki fails — it holds only at a right angle2 mountain and 2 valleyMaekawa fails — a crossing can only be 4–0, 2–2 or 0–4mountainvalleyraw edge

Two creases that cross

A crossing is four creases at a point, so the four conditions of the subject apply to it — and three of them can be satisfied. It is developable at every angle, it satisfies the big-little-big lemma whenever its two lines carry different letters, and it satisfies Kawasaki's condition when the lines meet squarely. Maekawa's refuses it always, at every angle and under every lettering, because a crossing's four spokes belong to two creases and can only be four and none, two and two, or none and four.

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

Most of a patch is edge

Between 34% and 91% of the vertices in the crease patterns drawn here sit on the edge of the paper rather than inside it, and on the tessellation patches — the figures that are meant to show what a repeating pattern looks like — it never falls below a third. A boundary is one unit deep whatever the unit is, so the share falls like one over the number of units across and reaches nothing at any size a page can carry.

the cut line10 straight edgesthe pattern10 skeleton arcs0 perpendiculars1 interior vertexassignments that fold420 of 102410 creases in allarcs one way, perpendicularsthe other: fails Maekawaequidistance off by 1.9e-16mountainvalleyevery node sits the same distance from each edge that formed it,which is why one fold can carry several edges onto the line at once

One cut for a star

The fold-and-cut construction here could reach a triangle, a pentagon and a house, and refused everything that turned back on itself, because shrinking an outline with a reflex corner needs an event the shrink did not implement. With split events it reaches a five-pointed star — ten creases through one point, four hundred and twenty letterings that fold, and every edge of the outline landing on one line to a part in 10^16.

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

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

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

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 bar is how many circles of that many panels were found726 circles, from 6 panels to 32, over every pattern family measured here4 panels0round one vertex — Maekawa forbids it5 panels0odd — the two-colouring forbids it6 panels21129.1% of the circles measured7 panels0odd — the two-colouring forbids it8 panels21129.1% of the circles measured9 panels0odd — the two-colouring forbids it10 panels8211.3% of the circles measured11 panels0odd — the two-colouring forbids it12 panels9513.1% of the circles measured13 panels0odd — the two-colouring forbids it14 panels304.1% of the circles measured15 panels0odd — the two-colouring forbids it16 panels314.3% of the circles measured17 panels0odd — the two-colouring forbids it18 panels172.3% of the circles measured19 panels0odd — the two-colouring forbids it20 panels141.9% of the circles measured22 panels81.1% of the circles measured24 panels152.1% of the circles measured26 panels71.0% of the circles measured28 panels20.3% of the circles measured30 panels20.3% of the circles measured32 panels10.1% of the circles measuredthe empty rows are not rare cases — they are lengths that cannot occur, and each has its own reason

A contradiction is even

A crease pattern's letters can demand a circle of panels each of which lies below the next, which is a proof that the sheet has no folded state. Every such circle found here — one thousand one hundred and forty-nine of them, across every family of patterns this collection draws — has an even number of panels in it, and none has four. Both facts are theorems rather than observations, and they come from opposite ends of the subject.

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

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.

a lettering of the patch that agrees with itselffound by testing the arcs while the letters were chosen, not after561 nodes · 246 backtracks · verified against a rebuilt folded sheet157 panels · 282 creasesits own lettering sends its panels round in a circle0 of 200 random letterings agree with themselvesthis one was found in 561 nodes and 246 backtracksit differs from the drawn lettering on 155 of 282 creasesthe drawing is the pattern; nothing here is a picture of the folded object

The lettering nobody could draw

Two hundred letterings drawn at random from the rhombille tessellation patch, and not one of them agrees with itself. Two thousand, and still not one. The patch was left as an open question — and it has an answer, found in five hundred and sixty-one steps by a search that tests the arcs while it is choosing the letters instead of after it has chosen them all.

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

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 creases the found lettering writes differentlymeasured against the lettering the pattern's own construction producedthe square patch4545 of 84 creases · 31 of them buriedthe elongated patch6666 of 106 creases · 42 of them buriedthe hexagonal patch6767 of 142 creases · 45 of them buriedthe triangular patch8787 of 142 creases · 65 of them buriedthe rhombille patch155155 of 282 creases · 117 of them burieda buried crease has an interior vertex at each end, and no legal move ever changes one

One solution of a search nobody ran

A crease pattern arrives with its letters already on it, and they look like part of the drawing. They are not. Every construction here ends in a propagation, a propagation ends wherever its first guess took it, and the lettering that comes out differs from the one a search finds on between a half and three-fifths of the creases — on patterns whose own letters are perfectly good.

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

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 bar is the shortest crease in the pattern, on a scale of powers of tenthe hexagonal patch at four turns of its polygons, everything else heldturn 0.21.4e-2130 creases · every one carries an arc · 2.25 mm on a 160 mm sheetturn 0.357.9e-6142 creases · 12 of them carry no arc · 1.3 µm on a 160 mm sheetturn 0.57.5e-3154 creases · every one carries an arc · 1.20 mm on a 160 mm sheetturn 0.72.8e-2154 creases · every one carries an arc · 4.53 mm on a 160 mm sheetone turn of one patch drops four orders of magnitude below the others, and it is the turn this collection prints

Twelve creases a micrometre long

A patch this collection has drawn for a long time carries a hundred and forty-two creases and a hundred and thirty arcs, and nobody had asked what the other twelve were. They are fragments left where the clip caught a pleat almost exactly at a corner — between one and nine micrometres long on a printed sheet, at one turn angle out of four, and it is the turn the collection prints.

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

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 dot is one run's cost, ranked; the rule is the constant order1001e+31e+4nodes visited40 seeds, ranked by cost80 nodes, every seed15 unfinished at 20,000same pattern, same conditions, same test at every node — the only difference is which letter is tried first

The difficulty was in the coin

One tessellation patch, one search, one test at every node — and a cost that runs from eighty-six steps to fifteen thousand depending on nothing but the starting seed. The heavy tail is real, it was measured carefully, and it was made by a single line of the search that nobody had thought of as a choice at all.

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

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.

the bar is how many DIFFERENT letterings 20 runs returneda coin at every choice1414 of 20 runs found onea constant, with the coin only on the creases no vertex constrains120 of 20 runs found onea constant at every choice120 of 20 runs found oneon the rhombille patch, 157 panels and 282 creases

One witness or forty

Taking the randomness out of a search made it three orders of magnitude cheaper in the worst case and cost it thirty-nine of its forty answers. The compromise everybody reaches for — randomise only the choices that cannot matter — recovers four of the forty on two patches and none on the other three, because the diversity was never where it looked.

each circle holds a crease shorter than a thousandth of the sheet12 fragments7.9·10⁻⁶ of a sheet · M5.9·10⁻⁵ of a sheet · M7.9·10⁻⁶ of a sheet · M5.9·10⁻⁵ of a sheet · V7.9·10⁻⁶ of a sheet · V5.9·10⁻⁵ of a sheet · Mand 6 more1018× to draw the longest142 creases and 60 interior vertices, 12 of the first and six of the second invisible

The crease the drawing cannot show

Twelve creases on a printed crease pattern are eight millionths of a sheet long. They are in every count the collection takes of that patch, they pass every theorem, and no printer resolves them and no hand folds them. They are also the only thing holding the folded sheet together.

one row per pitch, with the printed setting markeda fragment is a crease shorter than a thousandth of the sheetpitchcreasesinterior verticesfragments0.320154660.330154660.335142600.34014260120.345130540.350130540.36013054130 creases and 54 vertices is what deleting the fragments was expected to give, and one step of pitch gives it with a sheet that folds

A patch on a knife edge

The tessellation patch this collection prints has twelve creases nobody can see. Move the pitch of its tiling by five thousandths and they are gone — and so is a whole ring of twists. The patch sits exactly on the moment a ring of the pattern passes through the edge of the sheet, and the blemish is what that moment looks like.

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

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.

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

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

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.

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

A sheet with no edge

A twist tessellation repeats, so a rectangle of it is a description of the whole plane rather than a piece of paper. Joining the rectangle's opposite sides makes that explicit and produces an object every gate in this collection can read: twenty-five drawn panels become sixteen, forty crease pieces become thirty-two, sixteen vertices are all interior, and the three counts add to nothing.

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

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

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

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

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

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.

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

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

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

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

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.

3 creases on a Möbius bandthe panels take two coloursseamthe same seam123the right edge onto the left, turned over3 creases, 3 panelsinterior vertices: 0two-coloursthe reflections closeand turn the paper the right waymountainvalleyraw edge

The triangle a strip becomes

A Möbius band of paper folds flat into an equilateral triangle, and the shortest strip that will do it is √3 times its own width. The number is not put in: the crease angles come out of a condition on their alternating sum, the positions come out of two linear equations, and the length is where the drawing stops fitting.

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

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 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

How rare a band that folds is

Almost every crease pattern fails to fold flat, and the usual way of saying so is a count over discrete choices. A glued band fails for a reason that no count can reach: its crease angles have to satisfy an equation, and a set defined by an equation has no volume in the space it sits in.

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

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

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

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

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.

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

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

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

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.

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