The collection

Every essay

One idea per essay, ordered so that the earlier ones set up the later ones — but nothing here depends on being read in sequence.

Axioms and construction · Flat-folding · Designing a base · Tessellations · Rigid folding · Curves and material · What it costs to know · Who found it, and when · Folding nobody designed

Axioms and construction

What a single fold can do, and why folding reaches numbers that a straightedge and compass cannot.

axiom 1through two pointslinearaxiom 2point onto pointlinearaxiom 3line onto linelinearaxiom 4through a point, square to a linelinearaxiom 5point onto a line, through a pointquadraticaxiom 6two points onto two linescubicaxiom 7point onto a line, square to a linelinearthe degree each axiom can solve — one of them is why paper beats the compass

One fold at a time, and there are exactly seven of them

A fold is specified by bringing points and lines into coincidence. There are seven ways to do that, the list is provably complete, and one of the seven does something no compass can.

6 figures
hh/263°42.0°21.0°The foldone fold, made so that the cornerreaches the lower crease at the sameinstant as the point above it reachesthe ray — two conditions, one creaseWhat it produces63° divided into21.00° and 42.00°a third of 63° is 21.00°— measured off the fold, not drawnA cubic, so no compass reaches it.mountainvalley

Folding beats the compass, by exactly one degree

Straightedge and compass solve quadratics. A single fold solves cubics. That one-step difference settles two problems Greek geometry could not, and leaves a third exactly as impossible as it was.

6 figures
axiom 1through two pointslinearaxiom 2point onto pointlinearaxiom 3line onto linelinearaxiom 4through a point, square to a linelinearaxiom 5point onto a line, through a pointquadraticaxiom 6two points onto two linescubicaxiom 7point onto a line, square to a linelinearthe degree each axiom can solve — one of them is why paper beats the compass

Why the list stops at seven

The seven axioms are not seven useful folds somebody collected. They are every fold there is, and the proof is an exercise in counting degrees of freedom that takes about a minute.

7 figures
3 equal partsestimated by eyethe left-hand divisions are exact — a consequence of the fold, not of carethe right-hand ones are a guess, and the error compoundsvalleymountain

Dividing without measuring

A square can be divided into any whole number of equal parts by folding alone — exactly, with no ruler, and with no error to accumulate. The construction is one fold and a theorem nobody expected.

6 figures
compassfolding3242546276the first gap8496104111012413121461581681716186191820821122210232224825202612n across the top, φ(n) underneathcompass: φ(n) a power of two — Gauss, and Wantzel's proof that nothing else worksfolding: φ(n) with no factor above three, because one fold solves a cubic

The heptagon a compass cannot reach

Which regular polygons a tool can build is a condition on a single number. The compass needs it to be a power of two; a fold needs only that it has no factor above three — and seven is the first place the two answers differ.

7 figures
guessoff by 0.16667fold 1off by 0.08333fold 2off by 0.04167fold 3off by 0.02083fold 4off by 0.01042fold 5off by 0.00521fold 6off by 0.00260solid mark — the fold is aiming at 1/3; dashed — at where 1/3 has gonehalving word R L — period 2, from 2^2 − 1 = 3 × 1every fold halves the error exactly, so 6 folds divide it by 64

Folding a strip into thirds

A third cannot be constructed by the axioms, so it is not constructed. It is guessed, and then halved into place — an algorithm rather than a construction, with an error that falls by exactly half at every fold.

6 figures
startend8x³ + 4x² − 4x − 1legs 1.000, 0.500, -0.500, -0.125each turn a right angle3 real rootsx = -0.900969x = -0.222521x = 0.623490each ray lands on the end to 1.1e-16the launch angle's negative tangentis the root — which is the folda right-angled bounce off two linesat once is exactly what one fold does

Where the cubic comes from

Folding solves cubics, and the usual explanation stops at the sixth axiom. The reason is older and better: a right-angled bounce along a path of coefficients is a root-finder, and one fold is exactly such a bounce.

6 figures
pp − 1, factoredcompassfoldingthe regular p-gon32both tools52 · 2both tools72 · 3folding only112 · 5neither132 · 2 · 3folding only172 · 2 · 2 · 2both tools192 · 3 · 3folding only232 · 11neither292 · 2 · 7neither312 · 3 · 5neither372 · 2 · 3 · 3folding only11 is the first prime out of a fold's reach — 11 − 1 = 2 · 5the factor of five is the obstruction, and no arrangement of folds produces onea compass needs a power of two; a fold needs nothing above three

The eleven-sided one nobody can fold

Folding reaches the heptagon, which a compass cannot. It does not reach the hendecagon, and the obstruction is a single prime factor: ten has a five in it, a fold solves cubics, and no arrangement of cubics produces a five.

7 figures
nφ(n)its prime factorscompassone foldtwo at once322422542 · 2622762 · 3842 · 2962 · 31042 · 211102 · 51242 · 213122 · 2 · 31462 · 31582 · 2 · 21682 · 2 · 217162 · 2 · 2 · 21862 · 319182 · 3 · 32082 · 2 · 221122 · 2 · 322102 · 523222 · 112482 · 2 · 2the 11-gon is the first a single fold misses, and two simultaneous folds reach itthe 23-gon is the first that needs more than two, because 22 has an 11 in it

Two creases at once

The seven axioms describe what one fold can do, and the restriction to one fold is a rule somebody imposed rather than a property of paper. Allow two creases to be made simultaneously and the reachable degree rises — and the hendecagon nobody could fold becomes foldable.

9 figures
tilted 15.00°side 1.0352846.41% of the sheetand the same answer twicea corner construction givesside 1.03528from a quadratic, sharing no codethe width maximisation and the closed form agree to nine figures

The largest triangle in a square

The biggest equilateral triangle a square sheet holds is tilted by exactly fifteen degrees and uses 46.4% of the paper. Both numbers come out of a quadratic — which means a compass reaches this optimum too, and folding's advantage is not needed here at all.

9 figures
sidesshare of the sheet the largest one usestilt346.4%15.00°4100.0%45.00°567.4%9.00°669.6%15.00°772.9%6.43°882.8%22.50°975.1%5.00°1075.3%9.00°1176.3%4.09°1280.4%15.00°1376.9%17.31°1476.9%6.43°1577.3%3.00°1679.6%11.25°the 8-sided polygon is the peak, and every one of the 8 polygons after it does worse

The biggest one that can also be folded

Which regular polygon uses a square sheet best, and which of them a fold can actually construct, are two questions with completely different pedigrees. Answered side by side over sixteen polygons, they turn out to agree — and the reason is that both are questions about the arithmetic of the same number.

9 figures
the bare sheet4 references · 4 linesnothing has been foldedafter 1 fold9 references · 12 lineshalves, and nothing elseafter 2 folds565 references · 92 lineshalves, thirds, fifths — and worsea fold is an alignment, and an alignment needs something already on the paper to align565 references after 2 folds, and the count is finite however many folds are allowed

A fold needs something to align

Every axiom names things that must already be on the paper — a point to fold onto a point, a line to bring to a line. So what a folder can build is bounded by what they can refer to, and that set is finite at every depth: nine references after one fold, several hundred after two, and every one of them computable in advance.

8 figures
proportion 1.3two shapes, in turn1.3001.5381.3001.5381.300proportion √2 = 1.4142one shape, throughout1.4141.4141.4141.4141.414halving turns a proportion of r into one of 2/r, and those are the same number only at √2the 1.3 sheet is a different shape after every fold; the √2 sheet is the same shape after all of thema square metre at √2 is 840.9 × 1189.2 mm, which is the 841 × 1189 printed on a sheet of A0

The rectangle that keeps its shape

Halving a rectangle across its long side turns a proportion of r into one of 2/r, so almost every sheet comes out of the fold a different shape from the one that went in. Exactly one does not, and it is not a shape anybody chose.

8 figures
1/25 folds to reach 1/5, and every mark on the way is exactthe solid line is the one fold used at every step; the dashed lines are the stepsnothing here converges — each crossing lands on its fraction and stops

One crossing, and then another

Folding a strip into thirds by Fujimoto's method halves the error at every fold and never reaches a third. There is a construction that arrives instead: cross the square's diagonal with a line through the mark you already have, and the crossing lands on the next fraction exactly — one fold per step, all the way down.

8 figures
1 : √3 = 1.7321cut into 3, each part is 1.7321 — the same rectanglethe family1 : √2 = 1.4142 → 2 parts, 1 folds to build1 : √3 = 1.7321 → 3 parts, 2 folds to build1 : √4 = 2.0000 → 4 parts, 3 folds to build1 : √5 = 2.2361 → 5 parts, 4 folds to build1 : √6 = 2.4495 → 6 parts, 5 folds to buildA0 is printed at 1.413793and halves into 1.414634, which is a different rectangleevery ratio here is checked against a square root the construction never takes

One member of a family

A4 halves into A5 and keeps its shape, which is the one thing everybody knows about paper sizes. The property is not about halving and not about two: a rectangle in the ratio √n divides into n copies of itself, for every n, and every one of those rectangles can be folded out of a square one diagonal at a time.

8 figures
the bare sheet4 references · 4 linesnothing has been foldedafter 1 fold9 references · 12 lineshalves, and nothing elseafter 2 folds565 references · 92 lineshalves, thirds, fifths — and worsea fold is an alignment, and an alignment needs something already on the paper to align565 references after 2 folds, and the count is finite however many folds are allowed

Cheap where it reaches

Two folds from a bare square put marks at a half, a third, a quarter, a fifth, a sixth, an eighth and a twelfth — and at no seventh, ninth or eleventh at all. A rule that reaches every fraction takes n folds to reach one nth. The systematic route and the short one disagree everywhere, and neither of them knows about the other.

8 figures
distinct fold lines this axiom specifies and no other doesaxiomafter 4 pointsafter 9 pointsafter 565 pointsA1 — through two points08121054A2 — one point onto another08142649A3 — one line onto another4564994A4 — perpendicular through a point001661distinct lines in all1292274300the four operations name 38 folds at the first round and draw 12 lines with them

What each axiom is worth

The list of seven folds is complete, and the proof of that says nothing at all about whether its members are independent or equal. Measured on a bare square, one of the four elementary axioms supplies every fold the others cannot and the other three supply nothing. Two rounds later the ranking has inverted, and the one that carried the first round is the least productive of the four.

8 figures
the degree of the equation, and what it is made ofnumberdegreemade ofwhere it comes from½11a fold in half√222^1the diagonal of the squareφ22^1the silver rectangle's cousin∛233^1doubling the cube2 cos(2π/7)33^1the regular heptagon∜242^2a square root of a square root∛2 · √262^1 · 3^1a product of two of them2^(1/5)5not twos and threesa fifth root2 cos(2π/11)5not twos and threesthe regular hendecagonchecked by exhaustion: no number here satisfies a rational equation of lower degree with coefficients up to 6

The numbers a fold reaches

Folding solves cubics, which is one fact about one fold. The reason the subject has a theory rather than a bag of tricks is a second fact about all of them: the lengths a folder can mark are closed under addition, subtraction, multiplication, division, square roots and cube roots. Constructions can therefore be built out of constructions — and no tower of them ever arrives at a fifth root.

8 figures
what each round of folding reaches, and how close together it ison a sheet 150 mm squaremarksclosest pairmedian gapwithin 0.2 mmone fold, every axiom975.000 mm75.000 mm0.0%two folds, every axiom5650.520 mm2.700 mm0.0%three folds, point onto point only5538230.000 mm0.058 mm94.4%a crease in ordinary paper is about that wide, so the last column is the share of marks a folder cannot separate

Closer than a crease is wide

One fold from a bare square leaves nine marks, seventy-five millimetres apart. Two folds leave five hundred and sixty-five, the closest pair half a millimetre apart. Three folds — using one axiom of the seven — leave half a million, and ninety-four per cent of them have another mark within a fifth of a millimetre. What bounds a folder is not what the axioms reach; it is what the paper can tell apart.

8 figures
the same count, with more than one fold made at a timefreedomsoperationsalignments at onceone fold272two at once4224three at once6506four at once8958five at once1016110the second column is what the algebra gains; the third is what a pair of hands has to hold

Seven, and then twenty-two

The seven axioms are not seven useful folds somebody collected; they are the number of ways to spend a fold line's two degrees of freedom, and the count can be derived. Run the same derivation for two folds made at once and it gives twenty-two, for three fifty, for five a hundred and sixty-one — while the number of coincidences a pair of hands has to achieve in the same instant goes two, four, six, ten.

8 figures
4681012141600.10.20.30.40.50.60.7parts the strip is divided intohow far the crease lands out, mmthe exact ladderFujimotothey cross at 4

Exact is not accurate

This site has two ways of dividing a strip into equal parts: a ladder that lands on the fraction as a rational number, and Fujimoto's method, which never arrives. Read as mathematics that settles it. Read as instructions for somebody with a sheet of paper it settles nothing, and past four parts the method that never arrives is the one whose crease lands nearer the mark.

9 figures
2 folds from a bare sheet, every proportion at the same areamarks separated by at least 0.3 mm on a 150 mm sheetthe squarewhat origami paper is sold as565 marks · 92 distinct foldsthe A serieshalves into itself45,705 marks · 752 distinct foldstwo squaresa square cut the long way26,155 marks · 540 distinct foldsthe 1 : √3 rectanglethirds into itself42,746 marks · 732 distinct foldsthe golden rectanglenot in the halving family43,233 marks · 732 distinct folds

The sheet decides which points exist

Every measurement of what folding can locate has been made on a square, because origami paper is sold square. Hold the area fixed and change the proportion: one fold reaches nine marks on a square and twenty-nine on the A-series rectangle, and two folds reach 565 against 45,705. The square is the worst of five proportions at both depths, and the reason is its own symmetry.

8 figures
one round of folds through two points and folds placing one point on anothera crossing that lands inside the hole is not a reference and is not drawnwith a hole: 212 referencessolid: 9from 8 corners and 8 edgesfrom 4 corners and 4 edges

A hole is an edge

A folder's first fold has to be specified by aligning things that are already there, and what is already there is the sheet's outline. Cut a square hole in the middle and the outline doubles: one round of alignments reaches nine references on a plain square and two hundred and twelve on a holed one — more than the plain square reaches in two rounds.

8 figures
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

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.

8 figures
the bar is how many folds the alignment names, averaged over the trialsa1 — the fold through two points1.000.0% none · 0.0% twoa2 — one point onto another1.000.0% none · 0.0% twoa3 — one line onto another2.000.0% none · 100.0% twoa4 — a perpendicular through a point1.000.0% none · 0.0% twoa5 — a point onto a line, through a point1.5223.8% none · 76.2% twoan alignment with no fold is not a failed construction; it is an alignment the paper cannot make

An axiom may name no fold

The seven operations are stated about points and lines in a plane, and a plane has no edges. On a square, three of them always name exactly one fold and always land it on the paper; placing a line on a line names two, and 13.3% of the folds it specifies are creases the sheet never reaches; and placing a point on a line through a second point names two folds, one, or — 23.8% of the time — none at all.

6 figures
the bar is the share of each round's new references that lie on the sheet's own edgeafter one fold80%4 on the edge · 1 inside · 12 fold linesafter 2 folds9%48 on the edge · 508 inside · 92 fold linesan edge is a line a fold can cross twice, so it yields marks with the folds and not with their pairs

The edge was there first

A folder's first fold has nothing to align to but the sheet's own outline, and it shows in where the marks land: four of the five references the first fold adds are on the paper's edge. By the second fold the edge holds forty-eight of five hundred and fifty-six new ones. The rim is where references are cheap and it fills up, because an edge is a line a fold can cross twice while two folds inside the paper cross once each.

6 figures
the axiom is satisfied by both, and says nothing about whichthe two are square to one another, alwaysthe thin lines are the two the axiom is given; the two heavy ones are its answers

The axiom that names two folds

Bring one line onto another and the operation is satisfied by either of two folds, always exactly perpendicular to one another, creasing the paper in completely different places. Over four thousand random line pairs on a square, both folds land on the sheet two thousand nine hundred and sixty-five times, and the statement of the axiom does not say which one is meant. The fifth axiom is worse: its two answers are at any angle at all, from half a degree apart to square.

8 figures
565 references · worst gap 0.089 of a sheet · under 0.092the disc is where a folder has the least to align againstthe faint lines are the folds themselves; a reference is where two of them cross on the paper

How far from the nearest reference

One fold puts nine reference points on a square sheet and two folds put five hundred and sixty-five. That is sixty-three times as many points, and it brings the worst-covered spot on the paper from a third of a sheet away to a twelfth — four times closer. A count of references is not a measure of what a fold buys, because a set of points can be arbitrarily crowded and still leave most of the sheet out of reach.

9 figures
the bar is the largest gap between anywhere on the sheet and a referencea third fold specifies more folds than can be listed, so a sample of them is taken insteadnone of them0.089565 references, from two folds50 of them0.0763,498 references · 0.02% of the round100 of them0.0508,056 references · 0.04% of the round200 of them0.04723,480 references · 0.07% of the round400 of them0.02274,694 references · 0.15% of the round800 of them0.014270,882 references · 0.29% of the roundevery row is a lower bound on what the whole round would buy, because leaving folds out can only make the gap larger

The third fold cannot be listed

Two folds from a bare square reach five hundred and sixty-five reference points. The third round specifies three hundred and seventy-eight thousand folds, of which two hundred and seventy-four thousand are distinct — and the crossings of those with each other run to the tens of billions. The closure stops being computable at exactly the depth a folder starts working at, and what can be said instead is a bound rather than a list.

6 figures
axiom 1through two pointslinearaxiom 2point onto pointlinearaxiom 3line onto linelinearaxiom 4through a point, square to a linelinearaxiom 5point onto a line, through a pointquadraticaxiom 6two points onto two linescubicaxiom 7point onto a line, square to a linelinearthe degree each axiom can solve — one of them is why paper beats the compass

Which of the seven survive

The seven axioms are the complete list of ways one fold can be specified by aligning marked things. Every one of them names points and lines on a sheet, three of them quietly assume that a line has two sides, and on a closed sheet a line need not — so the list is complete for a disc and shorter for anything else.

8 figures
the bar is the share of each round's new references that lie on the sheet's own edgeafter one fold80%4 on the edge · 1 inside · 12 fold linesafter 2 folds9%48 on the edge · 508 inside · 92 fold linesan edge is a line a fold can cross twice, so it yields marks with the folds and not with their pairs

A reference on a sheet with no corner

Every construction in this subject begins from the sheet's own boundary: two edges meet at a corner, a corner is a point, and a point is what an axiom takes as input. A cylinder has two circles of edge and no corners at all, so a construction on one has nothing to start from and the seam is not a mark.

6 figures
guessoff by 0.16667fold 1off by 0.08333fold 2off by 0.04167fold 3off by 0.02083fold 4off by 0.01042fold 5off by 0.00521fold 6off by 0.00260solid mark — the fold is aiming at 1/3; dashed — at where 1/3 has gonehalving word R L — period 2, from 2^2 − 1 = 3 × 1every fold halves the error exactly, so 6 folds divide it by 64

Dividing a loop into n

Fujimoto's method divides a strip into any number of equal parts by folding badly and then folding the error in half, over and over. It converges because each step halves what is left over. On a closed loop there is no edge for the leftover to sit against, and what replaces the edge is the loop's own closure.

7 figures
1 : √3 = 1.7321cut into 3, each part is 1.7321 — the same rectanglethe family1 : √2 = 1.4142 → 2 parts, 1 folds to build1 : √3 = 1.7321 → 3 parts, 2 folds to build1 : √4 = 2.0000 → 4 parts, 3 folds to build1 : √5 = 2.2361 → 5 parts, 4 folds to buildA0 is printed at 1.413793and halves into 1.414634, which is a different rectangleevery ratio here is checked against a square root the construction never takes

The proportion a band asks for

√2 is a shape: a rectangle either has it or does not, and what it buys is that halving returns the same shape. √3 is what a Möbius band needs, and it is a different kind of number — a minimum rather than a shape, with every longer strip working and no shorter one.

6 figures
degree is how far out the number is; height is what the construction costsnumberdegreesteps in the towerwhat the steps are½10 stepsa fold in half√221 stepthe diagonal of the squareφ21 stepthe silver rectangle's cousin∛231 stepdoubling the cube2 cos(2π/7)31 stepthe regular heptagon∜242 stepsa square root of a square root∛2 · √262 stepsa product of two of them2^(1/5)5no tower reaches ita fifth root2 cos(2π/11)5no tower reaches itthe regular hendecagon2^(1/8)83 stepsthree square roots2^(1/9)92 stepstwo cube roots2^(1/12)123 stepstwo squares and a cube2^(1/16)164 stepsfour square roots1 pairs invert: 2^(1/9) is of degree 9 and 2 steps, 2^(1/8) of degree 8 and 3a degree 2^a · 3^b is a steps of square root and b of cube root, so the height is a + b and nothing else

Reachable is not cheap

The closure is what makes folding a theory rather than a bag of tricks: constructions can be built out of constructions. What that also means is that constructions have lengths and the lengths compose, so every reachable number has a height as well as a degree — the number of extension steps the shortest tower to it must take. The two orderings disagree, and a ninth root is a shorter tower than an eighth.

6 figures
what the plane specifies, and what the paper carries2 rounds from a bare sheet, every proportion at the same rulesheeton the paperlost to the edgesquare1 × 1.000565 references72%1,440 off the paperA-series1 × 1.414114,927 references47%102,162 off the paper3 : 21 × 1.500135,281 references48%123,872 off the paperdouble square1 × 2.00044,673 references59%65,294 off the paper2 rounds of a1, a2, a3, a4 · a crossing is a reference only where the paper is · square loses the largest share, at 72%

The field has no edge

Origami numbers are a field on the unbounded plane and a folder has a piece of paper. A fold line runs forever; a crossing of two of them is a number in the field wherever it lands, and it is a reference somebody can put a finger on only where there is paper under it. Counted rather than assumed, two rounds of the four linear axioms on a square put seventy-two per cent of their crossings off the sheet.

6 figures
050100150200250300350400020406080100120140160sides, up to npolygons reachablea fold — 155a compass — 40both sets computed by division to 400 · 4 Fermat primes and 13 Pierpont primes below it

How many polygons a fold reaches

The heptagon is what the extra axiom buys and it is one polygon. What it actually buys is a density: to a thousand sides a compass reaches fifty-two regular polygons and a fold reaches two hundred and seventy-five, and the ratio between them is still widening. The compass has five usable primes in the whole of arithmetic and may use each once; a fold keeps acquiring new ones and may repeat the factor of three as often as it likes.

6 figures
how squarely the folds that fix a reference crossand what a crossing at that angle does to an error in the foldingthe four linear axioms565 references · worst 36.9°with the conic axiom16,890 references · worst 6.3°60° to 90°error × 1.264.6%54.4%45° to 60°error × 1.418.2%21.8%30° to 45°error × 2.017.2%17.1%15° to 30°error × 3.90.0%5.2%8° to 15°error × 7.20.0%1.2%under 8°error × 14.30.0%0.3%the linear axioms bottom out at 36.9° — a multiplier of 1.67 — and the conic axiom reaches 6.3°, a multiplier of 9.12 rounds on a square · a reference priced at 1 ⁄ sin of the widest angle its own folds make · shares, so the two sets are comparable

What buys the reach costs the accuracy

A reference is a crossing of two creases, and a crossing transmits a folding error multiplied by one over the sine of the angle the creases make. Measured across the whole closure on a square, the four linear axioms never produce a crossing shallower than thirty-seven degrees — and the conic axiom, the one that sends a point onto a line and reaches the heptagon, produces crossings under seven.

6 figures
the count that was made, and the count that was notboth are floors: neither enumeration tracks which fold an alignment attaches tofreedomspaper onlywith simultaneous creasesneeding oneone fold277two at once4228664three at once650296246four at once895791696five at once1016117921631at two folds the omission is 64 operations of 86 — 74% of them, and none can be described without naming the other creasea pair of hands cannot make a condition between two creases it is making; a jig holding two lines can

Twenty-two is a floor

The enumeration that gives seven single-fold axioms spends each fold line's two degrees of freedom on alignments to points and lines already on the paper, and its own account says what it leaves out — an alignment may refer to a crease being made in the same instant. Adding those back leaves the single-fold count at seven and takes the two-fold count from twenty-two to eighty-six, of which sixty-four cannot be stated in terms of the paper at all.

5 figures
the same census, on four sheetsshare of the sheet used, at each polygon's own best rotationsidessquareA series3 : 26 : 5346.4%40.8%38.5%48.1%4100.0%70.7%66.7%83.3%567.4%51.4%48.4%60.5%669.6%61.2%57.7%72.2%772.9%53.5%50.5%63.1%882.8%58.6%55.2%69.0%975.1%54.4%51.3%64.1%1075.3%57.4%54.2%67.7%1176.3%54.8%51.6%64.5%1280.4%56.8%53.6%67.0%best after the square: square 8 · A series 6 · 3 : 2 6 · 6 : 5 6each polygon at its own best rotation on each sheet · the square uses all of a square and 70.7% of the next sheet along

The square is in the answer

The largest regular polygon a square sheet holds is not increasing in the number of sides, and the octagon's win is the striking part: it uses 82.8% of the paper against the twelve-gon's 80.4% and the hexagon's 69.6%. Run the same census over rectangles and the octagon's advantage is gone — on every proportion tried the hexagon leads, and the order among the even-sided polygons reverses outright.

5 figures
the same fold, on five sheetsa corner brought to the midpoint of the far edge, and what comes outsheetleft edgeright edgethe crossingwhat happenedsquare1.000 × 1.0003/87/82/3a third, exactlyA series, tall1.000 × 1.4147/1611/162/7a number, and not a thirdA series, wide1.414 × 1.0001.2500the crease leaves the paper3 : 2, tall1.000 × 1.5004/92/31/4a number, and not a third3 : 2, wide1.500 × 1.0001.3438the crease leaves the paper2 sheets answer and are wrong; 2 refuse — and the difference between the two is a right anglethe alignment does not know what shape the paper is, and neither does the folder following it

A construction assumes its sheet

Haga's fold gives exactly two thirds on a square. Run the same alignment on an A-series sheet held tall and it gives exactly two sevenths, with the crease meeting the vertical edges at seven sixteenths and eleven sixteenths — every one of them a clean fraction, none of them what the recipe promised. Turn the same rectangle through a right angle and the crease leaves the paper instead, which is the loud failure rather than the quiet one.

5 figures
seven constructions on five sheetswhere each construction's point lands, as a fraction of the sheet, against the squaresquareA series, tallA series, wide3 : 2, talldouble, widehalve it, edge onto edgea fold along the stretch1/2, 0the samethe samethe samethe samea third, from two crossing linescrossings only1/3, 1/3the samethe samethe samethe samea fifth, by repeated crossingscrossings only1, 1/5the samethe samethe samethe samea third, by Fujimoto's halvingsfolds along the stretch0.333, 0the samethe samethe samethe sameHaga's fold, corner to midpointa fold across a slant1, 2/31, 2/7off the paper1, 1/4off the papera corner halved, edge onto edgea fold across a slant1, 11, 0.7070.707, 11, 2/31/2, 1a corner onto the opposite cornera fold across a slant1, 0off the paper3/4, 0off the paper5/8, 0a stretch along the edges keeps crossings, midpoints and folds along the edges; it does not keep a fold across a slant

A stretch keeps crossings

A rectangle is a square stretched along its edges, and a stretch along the edges keeps straight lines straight, crossings as crossings, midpoints as midpoints and the fraction a point divides a segment into. So a construction made only of those — halve an edge, cross two lines — lands at the same fraction of every rectangle, and four standard constructions do. A fold across a slanted line is a reflection the stretch does not keep, and every construction that uses one — Haga's, a corner halved, a corner brought to its opposite — returns a different point on some rectangle, or none.

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which tool the best polygon on each sheet needseach polygon at its own best rotation; the rank counts every polygon from three sides to twenty-foursheetbest polygonbest a compass buildsbest only a fold buildsbest no fold buildssquare8-gon · 82.8%8-gon21-gon · 7th · 77.9%23-gon · 6th11 : 108-gon · 75.3%8-gon14-gon · 4th · 72.6%22-gon · 9th6 : 56-gon · 72.2%6-gon14-gon · 5th · 66.6%22-gon · 9thA series6-gon · 61.2%6-gon14-gon · 5th · 56.5%22-gon · 9th3 : 26-gon · 57.7%6-gon14-gon · 5th · 53.3%22-gon · 9th2 : 16-gon · 43.3%6-gon14-gon · 5th · 39.9%22-gon · 9th3 : 16-gon · 28.9%6-gon14-gon · 5th · 26.6%22-gon · 9ththe square itself is left out; a fold builds an n-gon when the totient of n has no prime factor above three

Every even polygon beats every odd one

Crossed with what each tool can build, the census of the largest regular polygon a sheet holds gives the same verdict on every proportion from a square to three to one: the best polygon is one a compass already builds, and the best polygon only a fold can build places fourth at best. The ranking itself stops moving at a proportion of 1.1284, where the hexagon overtakes the octagon. On every longer sheet only the short side holds a polygon, each polygon's share is a fixed constant divided by the length, and the constant — its area over the square of its least width — comes down to the circle's π⁄4 for even polygons and climbs up to it for odd ones. So every even polygon beats every odd one.

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three enumerations of the same thinga fold line has two freedoms, so m of them have 2m — the question is whether the budget is one pool or m pursesfolds at oncepaper onlypooled, with crossingseach alignment attachedthe ratio1777× 1.022286105× 1.23502963042× 10.3495791145,211× 183.6516117929,782,771× 5459.1the last column is what tracking which fold an alignment names is worth, and it grows because the naming itself grows

Each fold needs its own two

The enumeration that gives seven axioms spends a fold line's two degrees of freedom on alignments; run for m folds it spends 2m from one pool, and a pool can be spent three on one line and one on the other, which determines neither. Attaching every alignment to the fold it constrains repairs that, and two other things — and the two-fold count goes from twenty-two to a hundred and five, of which only twenty-eight have to be made at one instant.

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first pointsecond pointthe sixth axiom, at its full countthe first foldthe second foldthe third foldthree folds, each carrying one point onto one line and the other point onto the other — the most any single fold can offer

Counting operations is not counting power

The catalogue of simultaneous-fold operations runs from seven to nearly ten million between one fold and five. What a construction can reach does not: each fold admits at most three lines, because two parabolas have three proper common tangents and not four, so m folds admit at most three to the m — and the largest polynomial degree they actually settle is smaller again, at twice m plus one. Three counts of the same subject, growing at three speeds.

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throughthroughlands on the other creaseand so does this onea cyclic operation, solvedeach crease is described in terms of the other, so neither can be made first and no order exists

A crease that does not exist yet

Simultaneous folding is usually described as a problem of dexterity — several coincidences to be achieved in the same instant. The reference graph says otherwise: of the hundred and five two-fold operations, twenty-eight need no simultaneity and forty-nine can be done in an order, leaving twenty-eight whose folds each name the other. Those are not hard to hold. They are hard to know, and a loop that guesses and re-solves finds them at eight per cent of the error a pass.

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every degree a fold reaches up to 200, as a square root count against a cube root counta point at (a, b) is the degree 2 to the a times 3 to the b, and a tower to it takes a + b steps0123456701234square rootscube roots13927812618541624123610882472164814432966419212825 of the first 200 degrees, and they are the lattice points under a line of slope minus log 2 over log 3the pale points are the degrees a compass reaches as well

Twos and threes run out

A fold reaches a number exactly when the degree of its equation is a product of twos and threes, which sounds like a large set because it is infinite and because it is so much larger than the compass's. Counted, the reachable degrees are the lattice points under a straight line, so there are about half a log-squared of them: twenty of the first hundred, a hundred and forty-two of the first million. The share falls from a fifth to one part in seven thousand, and the factor by which folding beats the compass rises at every decade without ever settling.

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how many extension steps the shortest tower to each polygon takesa polygon of n sides needs the degree of two cosine of a turn over n, which is Euler's totient halved3 sides0degree 1 · square roots only, so a compass reaches it4 sides0degree 1 · square roots only, so a compass reaches it5 sides1degree 2 · square roots only, so a compass reaches it6 sides0degree 1 · square roots only, so a compass reaches it7 sides1degree 3 · 1 cube root8 sides1degree 2 · square roots only, so a compass reaches it9 sides1degree 3 · 1 cube root10 sides1degree 2 · square roots only, so a compass reaches it12 sides1degree 2 · square roots only, so a compass reaches it13 sides2degree 6 · 1 square root and 1 cube root14 sides1degree 3 · 1 cube root15 sides2degree 4 · square roots only, so a compass reaches it16 sides2degree 4 · square roots only, so a compass reaches it17 sides3degree 8 · square roots only, so a compass reaches it18 sides1degree 3 · 1 cube root19 sides2degree 9 · 2 cube roots20 sides2degree 4 · square roots only, so a compass reaches it21 sides2degree 6 · 1 square root and 1 cube root24 sides2degree 4 · square roots only, so a compass reaches it26 sides2degree 6 · 1 square root and 1 cube root27 sides2degree 9 · 2 cube roots28 sides2degree 6 · 1 square root and 1 cube root30 sides2degree 4 · square roots only, so a compass reaches it32 sides3degree 8 · square roots only, so a compass reaches it34 sides3degree 8 · square roots only, so a compass reaches it35 sides3degree 12 · 2 square roots and 1 cube root36 sides2degree 6 · 1 square root and 1 cube root37 sides3degree 18 · 1 square root and 2 cube roots38 sides2degree 9 · 2 cube roots39 sides3degree 12 · 2 square roots and 1 cube root40 sides3degree 8 · square roots only, so a compass reaches itthe pale bars are the polygons a compass reaches, and they are not the cheap ones — 4 of the one-step polygons need a cube root

Gauss's polygon is the expensive one

Which regular polygons a fold reaches is a condition on the factorisation of Euler's totient, and every polygon that passes it also has a height — the number of extension steps the shortest tower to it takes. Read that column instead of the verdict and the field inverts: the heptagon, which no compass reaches, costs one step; the seventeen-sided polygon that made Gauss famous costs three, the most on the list; and the polygons a compass finds easy are the ones a folder pays most for.

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what each axiom set specifies, and what of it the sheet carrieson a unit square, from its four corners and four edgesaxiomshow manyroundsfold lineson the paperoff itlosta line through two points116500.0%a line through two points126500.0%and a point onto a point218900.0%and a point onto a point22321336432.5%and the other two linear ones411291257.1%and the other two linear ones42925651,44071.8%and a point onto a line through a point51283310475.9%and a point onto a line through a point52past the capthe conic axiom loses three quarters of its crossings at one round, and cannot be run a second

The axiom that reaches furthest wastes most

The field of origami numbers is defined on an unbounded plane and a folder has a square. Counted axiom set by axiom set on the same sheet, the share of crossings that land off the paper rises with every axiom added: nothing at all from the first two, fifty-seven per cent from the four linear ones at a single round, and seventy-six per cent from the conic axiom at a single round — more, in one round, than the linear four lose in two. The instrument that reaches furthest into the field delivers the smallest share of what it specifies.

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the proportion at which the hexagon passes each polygonmeasured on the share curves, and computed as √3⁄2 + ½√(8K⁄3√3 − 1) with K the polygon's constantsidessearchedclosed formdegreeodd primesthe tool it needs71.0696431.069643243one fold81.1284411.1284418nonea compass91.0802961.080296123one fold101.1163331.11633316nonea compass111.0852961.085296405two folds at once121.1097491.1097494nonea compass131.0880641.088064483one fold141.1057721.105772243one fold151.0897621.08976216nonea compass161.1031871.10318716nonea compassthe sheet is one wide; each proportion is where the hexagon's share equals the polygon's, found two ways

The crossing is as hard as the polygon

Lengthen a square sheet and the largest hexagon it holds turns, pressed against all four edges, until it overtakes the polygons held by the short side alone. Every one of those overtakings happens at a proportion with a closed form, √3⁄2 + ½√(8K⁄3√3 − 1), and the number that comes out is exactly as hard to mark as the polygon being overtaken is to build. The octagon's 1.1284 is a compass number of degree eight. The heptagon's 1.0696 has degree twenty-four and needs a fold. The hendecagon's 1.0853 has degree forty and needs two folds at once.

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the best rank a tool's polygons reach, on a few sheets and on all of thema proportion between the named sheets is where each tool's best polygon does bestthe polygonon seven sheetson every sheetits own sheetonly one fold builds it14 sides · 4th14 sides · 3rd1.0257only two folds at once build it22 sides · 9th22 sides · 5th1.0103no two folds build it47 sides · 12th46 sides · 11th1.0023ranks among every polygon from three sides to 48; the seven sheets run from a square to three to one

The sheet a polygon fits exactly

A regular polygon with 4k + 2 sides has flat edges along one axis and corners along the other, so there is one sheet, 1⁄cos(π⁄n) long, that it touches on all four edges at once. On that sheet it is beaten only by the multiples of four with fewer sides, and so it ranks exactly (n − 2)⁄4. That puts the fourteen-gon, which only a fold builds, third rather than fourth; the twenty-two-gon, which needs two folds at once, fifth rather than ninth; and the forty-six-gon, beyond two folds, eleventh. Seven sheets from a square to three to one had missed all three.

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creases to mark each fraction of an edgeportable: every rectanglesquare onlyA-series sheet only0123456creases11112112313124131234515/2/3/4/5/6/7/8/9/10/11/12the portable column is one number for every rectangle; each sheet's column is true of that sheet alone

What the square saves

A construction made only of crossings, midpoints and folds along the edges lands at the same fraction of every rectangle, so its cost is one number for all of them. Counted crease by crease against every fold the first four axioms allow, that portability costs about a crease and three quarters a fraction up to twelfths — and the square is not the cheapest sheet to give it up for. An A-series sheet, where Haga's fold goes silently wrong, marks two sevenths in two creases; the square needs three, and any sheet at all needs five.

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11.051.11.151.20.850.90.9511.05the sheet's length, with its width oneshare, over the share on a square6 sides · +7.7%10 sides · +2.6%14 sides · +1.3%22 sides · +0.5%each curve is one polygon's share over its share on the square; the dot is its own sheet

Turning is uphill all the way

A regular polygon of 4k + 2 sides on a sheet a little longer than a square cannot lie flat: it turns, pressed against all four edges, until the sheet is exactly its own. Its share on the way has a closed form, and the closed form's slope is proportional to h² − 1 for every such polygon — flat on the square, rising all the way to the own sheet, and falling after it. So the own sheet is exactly the peak, the gain from the square to it is the average of one and the sheet's length, and the rank the census measured for polygons of this kind, (n − 2)⁄4, is now a theorem.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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-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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

9 figures
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.

8 figures
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.

8 figures
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.

8 figures
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.

9 figures
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.

8 figures
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.

8 figures
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.

8 figures
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.

8 figures
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.

8 figures
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.

7 figures
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.

6 figures
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.

8 figures
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.

8 figures
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.

8 figures
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.

8 figures
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.

8 figures
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.

7 figures
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.

6 figures
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.

6 figures
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.

6 figures
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.

6 figures
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.

6 figures
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.

8 figures
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.

9 figures
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.

8 figures
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.

8 figures
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.

8 figures
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.

8 figures
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.

9 figures
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.

9 figures
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.

8 figures
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.

9 figures
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.

9 figures
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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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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Designing a base

Getting from a shape somebody wants to a crease pattern that produces it, by packing circles.

Levery point within L is spentthe flapLL = 0.28 of the sheet's side, so the disc costs πL² = 24.6% of itthe circle is not a metaphor — it is the paper the flap consumesso designing a base is packing circles

A flap costs a circle

A flap of a given length uses up every point of the sheet within that distance of it. Two flaps whose circles overlap are asking for the same paper twice — and that one observation turned origami design from an art into an algorithm.

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leglegarmarmheadthe checkclosest approach 0.0000no overlap — the packing is validcircles use 71% of the sheetthe rest becomes the bodyefficiency is how much of thesquare the circles can claim,and it is an open problemthe dashed skeleton is the subject; the circles are what it costs

Packing is the hard part

Once a subject is a set of circles, designing the model is fitting them into a square. That step has no general algorithm, no known optimum, and it is where every remaining difficulty in origami design now sits.

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riverwidth 0.3legarmheadlegtailthe check10 pairs testedtightest by 0.0800(leg and arm)the ruledistance on the sheetat leastdistance through the treetwo nodes, and an edge between themthe extra width is thebody the flaps hang fromthe discs are what each flap costs; the strip is what joins the two halves of the subjectand both are the same condition, read off different pairs of leaves

What joins the flaps

Circles are the rule for flaps that all hang from the same point. As soon as two groups of flaps hang from different places, the paper between them has to be paid for too — and the payment is a strip whose width is the distance between them.

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2 discs53.9%r = 0.29293 discs61.0%r = 0.25434 discs78.5%r = 0.25005 discs67.3%r = 0.20716 discs66.3%r = 0.18767 discs66.9%r = 0.17448 discs72.8%r = 0.17029 discs78.5%r = 0.1667hexagonal density 90.69%every bar is the best a seeded search found, not a proved optimum —which is the honest state of the problem for all but the first few values of n

How much paper is wasted

The efficiency of a design is the fraction of the sheet its flaps can claim, and for almost every number of flaps nobody knows the best possible value. The bars in these figures are the best a search could find, which is not the same thing.

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32 × 32 gridevery crease on a grid line, or at 45°which is why a 64-grid design can be folded at allmountainvalley

Designing on a grid

Box pleating gives up the efficiency of a free packing and buys creases that land where they are supposed to. For a design with hundreds of folds that is not a compromise — it is the only thing that makes it foldable.

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the treeleg0.62leg0.62body0.34arm0.50head0.78the basethe axis0.620.620.340.500.78the flaps are the tree's edges, at the tree's lengths, all square to one lineso the base's shadow along the axis is the tree, and nothing else can be designed this waywhich is the restriction the circle argument quietly depends on

Every flap on one axis

The tree method does not design a shape. It designs a base whose flaps all lie along a single line — and that restriction, which is almost never stated aloud, is what makes the circle argument true.

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what the packing gives5 discs, 4 contactsno two overlapping, checkedhinge creasesone per contact, perpendicularto the line of centresridge creasesalong the axial lines, dividingthe paper between the flapsthe packing is the hard part;this part is a constructioncorner discs of radius 0.28, the middle one 0.427 — every contact measured

From a packing to a crease pattern

The circles say where the flaps are. They do not say where to fold, and the step in between is a construction rather than a search — two families of crease, both determined by the packing, neither of them visible in the picture of the discs.

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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

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.

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quadrilateralconvexmolecule builtpentagonconvexmolecule builtL, one reflex cornerone reflex cornerconstruction refusedthe polygon admits a shrinkingdart, one reflex cornerone reflex cornerconstruction refusedthe polygon admits a shrinkinga convex polygon shrinks inward and stays a polygon; a reflex corner is a wall the shrink runs intoso a non-convex region is split into convex pieces first, and choosing the split is a searchwhich is where a construction that always works hands the difficulty to whatever comes before it

The molecule that does not exist

The universal molecule fills any convex polygon, always, which is what makes it the part of the tree method with no special cases. Hand it a reflex corner and it does not produce a worse pattern — it produces nothing, and the difficulty moves backwards to whoever chose the polygons.

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00.10.20.30.40.500.20.40.60.81flap width, as a fraction of the sheetarea showingthe two are equal at a thirdfront showingreverse showingtotal facemeasured on the folded state at 4 flap widths, and the marks are those measurements

Bringing the other side to the front

Paper has two sides and most models show one. A colour change shows the other, and it is not a crease problem — which panels can show the reverse is settled by the pattern's two-colouring, and what it costs is twice what it shows.

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2345678910-0.15-0.1-0.0500.050.10.15discsshortfall of the symmetric search14.6%1.7%0.0%-0.0%5.6%0.5%10.8%10.4%4.7%symmetry: mirror · both searches at 90 restartsneither number is a proved optimum — this compares two searches

When symmetry costs

Design software and designers both reach for symmetry, and for a good reason: it makes the search enormously easier. It is a heuristic and not a theorem, and how much it gives away can be measured — including the case where the optimum is symmetric about an axis nobody imposed.

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123456760%65%70%75%80%sheet, longer side to shorterfraction of the sheet claimedbest at 7.0 to 17 flaps · every sheet the same areathe sheet's shape is a design variable that origami paper hides by being sold square

The square is a choice

Every packing on this site has been into a square, because origami paper is sold square. Hold the area fixed and vary the shape instead and the efficiency turns out to be spiky rather than smooth — with the same peak value at every proportion that is a ratio of two factors of the flap count, and nowhere else.

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the design as it stood20.00 of paper · 5 creases meet the cutthe same design, one strip wider21.80 of paper · the strip is 0.45 acrossthe band is 0.45 × 4 = 1.8000, and that is the entire difference between the two patternsall 44 creases away from the cut keep their length; the 5 that cross it are longer by 0.45 and by nothing elsea design grows by accretion because the arithmetic of growing it is this short

Paying in paper

A feature added to a finished design costs exactly the paper inserted for it. Cut the crease pattern along a line, slide in a strip, and every existing crease continues across it unchanged — so the bill is the strip's width times the length of the cut, and there is no second term.

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1½¼what one flap costsin the middle · a whole disc0.2463 of the sheeton an edge · half of one0.1232 of the sheetin a corner · a quarter0.0616 of the sheeteach one integrated over the sheetrather than taken from the fractionat 0.28 sheet-widths a flap costs 0.2463 of paper in the middle, 0.1232 on an edge and 0.0616 in a cornerso an efficient design fills the boundary first, and the edge of the sheet is the cheapest paper on it

The corner is worth four times the middle

A flap consumes every point of paper within its own length of its tip — but only the paper that is actually there. On an edge of the sheet that is half a disc, and in a corner a quarter, so the same flap costs four different prices depending on where it stands and the boundary is the cheapest paper on the sheet.

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flapsfree search4×4 grid8×8 grid30.25430.2500 −1.7%0.2500 −1.7%40.25000.2500 −0.0%0.2500 −0.0%50.20710.1768 −14.6%0.1768 −14.6%60.18760.1250 −33.4%0.1398 −25.5%the 6-flap case on the finest lattice here is one of 3.25e+8 arrangements, and the bound settles all of themthe free optimum is unknown for most of these counts and the lattice optimum is known for all of thema finer lattice costs less and asks for more creases, which is the trade a designer actually makes

What the grid settles

Box pleating is usually defended as a trade: give up efficiency, buy creases that land where they should. There is a second thing it buys and nobody quotes it — on a lattice the best possible packing is a finite question with an answer, while off the lattice nobody knows the best packing of six circles in a square and probably never will.

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7 flaps, packed as tight as they will goradius0.174457630loose flaps1room to move0.1127the inner ring is where that flap's centre may sit; the algorithm reports one point of it and stops

The flap nobody holds

An optimal packing is presented as an answer: here are the circles, here is where they go. For some numbers of flaps that is not what it is. The best arrangement of seven discs in a square leaves one of them free to wander over an eleventh of the sheet without changing the answer at all — and the algorithm reports one point of that region and stops.

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42:3:4:363:4:6:583:4:6:5125:7:11:8165:7:11:8248:11:17:13each limb roundedthe best whole numbershow wrong the worst limb isgrid units across the longest limbthe numbers under the axis are the best whole-number limbs at that resolution

Spelling a tree on a grid

Box pleating asks every limb of a design to be a whole number of grid squares, which sounds like rounding and is not. Rounding each limb to its own nearest whole number is one way to choose the numbers, and at most resolutions it is not the best way — the best whole-number version of a subject is often a coarser one, with fewer squares and a shape twice as close.

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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

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.

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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

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.

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one graft, then a second across itstrips 0.3 and 0.3 wideas designedone stripand one across itthe first strip adds 0.300the second adds 0.390two separate bills would be 0.600the sheet gained 0.690the excess is 0.090and the strips cross over 0.090the crossing rectangle is 13.0 per cent of everything the two features cost

The second term

A feature grafted into a finished design costs exactly the paper slid in for it, and the bill has one term. Add a second feature across the first and it has two: the rectangle where the strips cross is paper both features are charged for and neither uses. At a strip a fifth of the sheet wide it is nine per cent of the bill; at four fifths it is nearly a third.

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patternfootprint with both colours over itlayers thereThe preliminary base8 panels, 4 one way up and 4 the other100%8.0The Miura fold24 panels, 12 one way up and 12 the other100%9.2The square twist9 panels, 5 one way up and 4 the other65%4.1The hexagon twist13 panels, 7 one way up and 6 the other72%4.1The Yoshimura pattern65 panels, 32 one way up and 33 the other100%60.0Fold and cut — the triangle7 panels, 4 one way up and 3 the other4%6.4The tapered corrugation28 panels, 14 one way up and 14 the other100%8.4The waterbomb tessellation52 panels, 26 one way up and 26 the other100%31.1

Decided before the design

A colour change brings the reverse side of the paper to the front, and the usual account is that the two-colouring of the panels decides which panels are available. Measured on the site's own printed patterns, availability is not the constraint: both sides lie over more than ninety-nine per cent of most folded footprints. The other side is not scarce. It is under eight layers of paper.

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

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.

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sectorsletterings that branchdecided by the choice22.5° 22.5° 157.5° 157.5°4/16022.5° 45° 157.5° 135°0/16022.5° 67.5° 157.5° 112.5°0/16022.5° 90° 157.5° 90°0/16045° 45° 135° 135°4/16045° 67.5° 135° 112.5°0/16045° 90° 135° 90°0/16067.5° 67.5° 112.5° 112.5°4/16067.5° 90° 112.5° 90°0/16090° 90° 90° 90°14/160

The other grid

Box pleating is drawn at forty-five degrees, and the twenty-two-and-a-half-degree grid is usually described as the same thing done finer. It is not a refinement, it is a different alphabet: five kinds of vertex become fifty-six, and the share of letterings whose decision needs a search falls from 60 per cent to 22. A finer grid is a larger vocabulary and a less ambiguous one.

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6 flaps at radius 0.1875883 contact graphs among the runs that agree about itone run5 contacts · 5 against the paper's edgeanother run4 contacts · 5 against the paper's edge

Two packings, one radius

A packing search reports a number, and the number is not the design. What a crease pattern is built from is the graph of which discs touch which — and at five and six flaps, runs of the same search that agree about the best radius to four decimal places come back with contact graphs that are provably not the same graph. The answer an optimiser gives has not determined the pattern it is supposed to have found.

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The preliminary base: 8 symmetries, 112 letteringsthe bar is the share of letterings the symmetry carries to themselvesa quarter turnnone of 112 — this symmetry cannot be foldeda half turnnone of 112 — this symmetry cannot be foldedthree quarters of a turnnone of 112 — this symmetry cannot be foldeda mirror across the sheet12 of 112a mirror up the sheet12 of 112a mirror in one diagonal12 of 112a mirror in the other diagonal12 of 112

The symmetry the letters cannot keep

Every pattern in this subject is drawn symmetric and the symmetry is always quoted of the drawing. A folded object is a drawing and a lettering together, so a symmetry survives only if the letters keep it — and the preliminary base loses every rotation while the square twist, drawn with the same eight, loses the other half.

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the pale bar is the solid sheet, the dark one the sheet with a holelower is better: it is the average share of a flap's disc that has to be paid forflap 0.060.9293 against 0.9496flap 0.10.8826 against 0.9137flap 0.150.8296 against 0.8714flap 0.220.7586 against 0.8159flap 0.30.6760 against 0.7496

A hole is cheap paper

A flap claims every point of paper within its own length of its tip, but only the paper that is actually there — so an edge is half price and a corner a quarter. A hole in the middle of the sheet is more edge, and holding the area fixed, a square with a hole in it is cheaper paper than a solid square at every flap length measured.

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the bar is the share of the footprint showing the side that started face upthe rest of it shows the other side, and neither is chosen by anybodyThe preliminary base0.3%2 of 8 panels in viewThe square twist47.4%7 of 9 panels in viewThe hexagon twist90.8%10 of 13 panels in viewFold and cut — the triangle100.0%1 of 7 panels in view · 2 statesa panel out of view from above is not hidden — it is under the pile, and turning the sheet over shows a different set

Which side arrives

A colour change is described as a choice: bring the reverse of the sheet to the front where the design wants it. On a pattern whose panels can be ordered, nobody chooses. The preliminary base shows the side that started face up over one part in a thousand of its own footprint, and two of its eight panels are the only ones in view at all.

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two tips 0.6 of a sheet width apartthe bar is how long a flap each can carrysolid paper0.300the discs meet halfwaya slot 0.1 wide0.35719% further than solid papera slot 0.2 wide0.40435% further than solid papera slot 0.3 wide0.46655% further than solid papera slot 0.4 wide0.50067% further than solid papera slot 0.5 wide0.55184% further than solid paperthe tips do not move; only the paper between them does

Spending the cheap paper

A hole makes a sheet cheaper by the square inch, because a flap against its rim claims only half a disc. Whether a design can spend that was left open, because no packing search here could express a region that is not paper. It can now, and the mechanism turns out not to be the discount at all: across a hole, two flaps may overlap, and two tips three fifths of a sheet apart carry flaps of 0.500 rather than 0.300.

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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

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.

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the bar is how much cheaper the paper is than a plain square of the same areaflaps of 0.061.40%notch 0.60% · hole 1.40%flaps of 0.12.73%notch 1.31% · hole 2.73%flaps of 0.154.59%notch 2.28% · hole 4.59%flaps of 0.226.76%notch 3.20% · hole 6.76%flaps of 0.39.85%notch 4.54% · hole 9.85%same paper removed, twice the saving — a hole has four sides of rim and a notch has three

A notch is not a hole

Remove the same rectangle of paper from the middle of a square and from its edge, and the two sheets are not worth the same. The hole is cheaper paper at every flap length measured — 4.71% cheaper than a plain square of equal area against the notch's 2.58% — because what a cut is worth is rim with paper on both sides of it, and a notch spends one of its four sides on an edge the sheet already had.

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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

A tree cannot argue

A molecule fills a polygon with creases taken from its straight skeleton, and a straight skeleton is a tree. So a molecule's panels have almost no closed chains for its letters to contradict themselves round — one to three, against thirty-six on the smallest tessellation patch. Two hundred and eighty independent letterings across seven outlines, including an L and a five-pointed star, and not one of them disagrees with itself.

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the bar is how many of a hundred random letterings agree with themselveson the orthogonal grid a box-pleated design is drawn on, at five sizes4 by 4949 interior vertices · 16 panels · found in 16 nodes6 by 67325 interior vertices · 36 panels · found in 37 nodes8 by 85849 interior vertices · 64 panels · found in 65 nodes10 by 103681 interior vertices · 100 panels · found in 100 nodes12 by 1215121 interior vertices · 144 panels · found in 145 nodes16 by 161225 interior vertices · 256 panels · found in 261 nodesevery interior vertex is a four-panel circuit, so the number of places a contradiction could sit is the number of vertices

What a grid costs in circuits

Box-pleating puts every crease on a square grid, and a square grid is the shape with the most short circuits per panel that this collection draws. On the sixteen-by-sixteen grid a designer actually works on, one mountain-valley labelling in a hundred agrees with itself. A search still finds one in two hundred and sixty-one steps.

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

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.

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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

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.

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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

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.

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

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.

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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

A sheet with two edges

Design in this subject starts from a square, and the square's four edges are where every flap ends and every construction begins. A cylinder has two edges instead of four, and what a designer loses turns out to be measurable in the same letters a search counts — and what they gain is that the sheet is what gets manufactured.

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a flap of 0.12 of the sideless is cheaper: a flap claims only the paper that is actually therethe open middle of the sheet1.0000 of the discagainst an edge0.5000 of the discin a corner0.2500 of the discagainst the hole0.5000 of the discin the hole's outside corner0.7500 of the disc

The corner premium, with no corners

A square's corners are its most valuable paper: a flap placed there is claimed by a quarter-disc rather than a whole one, so the corner goes four times as far as the middle. A closed sheet has no corners at all, and the accounting that ranks sheet shapes by their corners has nothing left to rank.

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a flap of 0.12 of the sheet's side, on two sheets of the same areadarker is cheaper: less of the flap's disc is paper that has to be paid forwith a holesolid, same areamean claim 0.8646mean claim 0.8959

A base needs an edge to point at

Every flap of a uniaxial base ends in a point, and every point is made of paper that came from the sheet's boundary. A sheet with no boundary has nowhere for a point to come from, and a sheet with half a boundary can only have points at the half it has left.

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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 glued

Box pleating is the designer's grid: every crease on a line, every angle a right angle or forty-five degrees, and a whole design method built on the convenience of it. Roll the grid into a tube and half the column counts stop folding, on a pattern whose whole selling point is that it always works.

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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

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.

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the lines a graft could useevery gap between vertex columns, and every gap between vertex rows, triedpatternvertical lineshorizontal linesgraftsThe preliminary base0 of 20 of 2nowhereThe Miura fold6 of 130 of 4one wayThe square twist0 of 60 of 6nowhereThe hexagon twist0 of 42 of 11one wayThe Yoshimura pattern0 of 120 of 5nowhereFold and cut — the triangle2 of 82 of 8both, in margin onlyThe tapered corrugation7 of 150 of 4one wayThe waterbomb tessellation0 of 80 of 8nowherea 4 by 4 grid4 of 44 of 4both waysthe grid, one square creased diagonally3 of 43 of 4both waysa line is admissible when every crease it crosses is square to it; the entry is admissible of the lines between vertex columns or rows

A graft needs a square line

A strip can be slid into a finished crease pattern only along a line every crossed crease meets square, because only such a crease continues across the strip as itself. Tried on every line between the vertex columns and rows of eight printed patterns, four take no strip in either direction and three take one in a single direction. The eighth, the fold-and-cut triangle, appears to take strips both ways, and every line it admits runs through blank margin. No printed pattern takes a strip across its creases in both directions — and one diagonal crease in a plain grid removes exactly the row and the column it sits in.

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a body with four legs and a tail — one internal edge of 0.8circles of radius m·ℓ, and every pair's requirement drawn — the ones through the body ask for more than the two circles do

Every pair, not every circle

A uniaxial base is designed by packing a circle for each flap, and the circles are not the condition. The condition is that every pair of the subject's extremities be separated on the sheet by the distance between them through the tree — which for two flaps meeting at one point is the sum of their lengths, and for two flaps across a body is more. Circles are the case with no body in it, so a design read off circles alone is promised a base sixteen to thirty per cent larger than the sheet can give.

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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

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.

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what lengthening each edge of a body with wings, legs, a head and a tail coststhe scale falls from 0.2651 by this much per unit of extra length, measured by re-solving the arrangementchest–head0.0786a flap, 5.9% of the scale per 0.2rump–ll0.0471a flap, 3.6% of the scale per 0.2chest–rump0.0429the body, 3.2% of the scale per 0.2chest–wl0.0361a flap, 2.7% of the scale per 0.2chest–wr0.0264a flap, 2.0% of the scale per 0.2rump–tail0.0123a flap, 0.9% of the scale per 0.2rump–lr0.0120a flap, 0.9% of the scale per 0.2an edge no tight pair passes through is an edge the design can spend freely, and the condition says which

The price of a limb is not its length

Lengthening an edge of a subject's tree costs the design some of its scale, and the amount can be measured by re-solving the arrangement. It is not proportional to the edge, and it is not the body that is dearest. On a bird whose wings are twice its legs, the head — nine tenths of a unit against the wings' one and six — costs three times what a wing costs, and two edges of a lizard cost nothing at all.

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several strips each way, and what the sheet gainedwidths 0.14, 0.14, 0.14 across and 0.18, 0.18 down3 strips across, totalling 0.422 strips down, totalling 0.36charged separately: 0.780the sheet gained: 0.931the excess: 0.1512the two totals multiplied: 0.15126 rectangles where they crossthe crossing term is 16.2 per cent of everything the features cost, and it is one term

Six rectangles and one term

Two grafted strips crossing leave one rectangle both features are charged for and neither uses. Three strips crossing two leave six, and the obvious budget adds them up. It does not have to: the six rectangles sum to the product of the two families' total widths, exactly, so a design with any number of features is priced by two numbers rather than by a double sum — and a family of strips in one direction alone carries no crossing term at all, however many of them there are.

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a 6 by 6 grid, and the lines it still admits a strip onthe same diagonals placed three ways — the count is of lines, not of creases05100123456diagonal creases in the patternlines a strip could be slid into, both directionsno two in a row or a columnall in one rowdropped at randoma diagonal costs the row and the column it sits in, so a design that keeps its diagonals in a few rows keeps its lines clear

A design that keeps its lines clear

A strip can be slid in only along a line every crossed crease meets square, so a diagonal crease spends the lines it crosses — and the census of eight printed patterns found four taking no strip in either direction. What decides how fast a design spends them is not how many diagonals it has but which rows and columns they sit in: six diagonals on a six-by-six grid leave twelve clear lines when they share a row and none at all when no two do, from the same six creases and the same amount of paper.

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a 8 by 8 grid, spacing 0.125 — what each strip width does to itthe bar is how many times finer the grid becomes — one means unchanged0.1251x1 spacings — the grid is unchanged0.251x2 spacings — the grid is unchanged0.3751x3 spacings — the grid is unchanged0.06252xthe grid becomes 0.0625, 2 times finer0.15xthe grid becomes 0.0250, 5 times finer0.1325xthe grid becomes 0.00500, 25 times finer0.25xthe grid becomes 0.0250, 5 times finera width four per cent away from a spacing divides the grid by twenty-five; a width half a spacing away divides it by two

The width is charged in grid

A grafted strip may be any width at all and the bill in paper is exactly width times length, with no second term — which is what the first of these essays established, and is a statement about area. The grid is a different property of the same pattern and it is not conserved: a strip whose width is not a whole number of spacings puts every vertex past the cut onto a finer grid, and a width four per cent away from a spacing divides the grid by twenty-five where a width half a spacing away divides it by two. A width nearly right costs far more than one plainly wrong.

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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

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.

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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

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.

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the bar is the footprint, split by how deep the other side liesone layer downtwo layers downnot under this point at allThe preliminary base8 panels · 1 stateThe square twist9 panels · 1 stateThe hexagon twist13 panels · 1 stateFold and cut — the triangle7 panels · 2 statesa 2 × 2 Miura patch4 panels · 1 statea 3 × 2 Miura patch6 panels · 3 statesa 3 × 3 Miura patch9 panels · 6 statesa 4 × 3 Miura patch12 panels · 11 statesthe letter fold3 panels · 2 statescounted over every folded state and from both faces, so nothing here is one lucky pile

One sheet down

A colour change has been priced by how deep the pile is — eight layers over every point of the preliminary base, six on a small Miura. Ordered, the piles say something else: wherever the other side of the paper lies under a point, it is the next sheet down on every pattern with one folded state, and never more than two down on any. And relettering the same creases cannot reach it. Of 112 letterings of the preliminary base that fold, every one shows either the printed face or the whole face turned the other colour; the square twist's eight only turn its face round.

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