Field

What it costs to know

Deciding, counting, listing and optimising are four different questions about the same sheet, and folding answers them at four wildly different prices.
1 × 4161 × 5501 × 61442 × 282 × 3602 × 43203 × 31,3684 × 4300,608filled — counted here, by exhaustive search over stacking ordersopen — Lunnon's published count, quoted rather than computed

The oldest open problem

In how many ways can a map be folded? The question needs no notation to state, the answer is a small integer for small maps, and after sixty years there is still no formula — only a list of numbers, each one found by searching every possibility.

creases at 0.40, 0.50 — assignment MVflat1 layerno sequence of all-layers folds finishes this pattern — the search exhausted 5 states

The fold a machine can make

A theorem that says a folded state exists says nothing about getting there. A machine that folds every layer at once is stopped by a strip with two creases in it — one that folds flat perfectly well, and that a pair of hands folds in about four seconds.

evenly spaced — 4 creasesany flat folding16 of 16some-layers16 of 16all-layers16 of 16one-layer2 of 16crimping only6 of 16uneven — 4 creasesany flat folding8 of 16some-layers8 of 16all-layers0 of 16one-layer2 of 16crimping only0 of 16a machine that takes fewer layers is weaker, not more patientthe paper is joined, so what it declines to hold it also cannot move

The patient machine is the weak one

A machine that folds one layer at a time sounds like a machine with more freedom, not less. It has less, and the reason is the most ordinary fact about paper there is: it is joined, so whatever a machine declines to hold it also cannot move.

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

A machine that can only crimp

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

evenly spaced — 4 creasesany flat folding16 of 16some-layers16 of 16all-layers16 of 16one-layer2 of 16crimping only6 of 16one short segment — 4 creasesany flat folding4 of 16some-layers4 of 16all-layers0 of 16one-layer2 of 16crimping only4 of 16uneven — 4 creasesany flat folding8 of 16some-layers8 of 16all-layers0 of 16one-layer2 of 16crimping only0 of 16a machine that takes fewer layers is weaker, not more patientthe paper is joined, so what it declines to hold it also cannot move

The machine that may choose

Three restricted machines lose patterns that fold perfectly well. Give one of them a choice — any block of layers, top or bottom — and the loss vanishes: over a hundred and seventeen spacings, every flat folding of every strip became reachable. Being forced was the whole problem.

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

Four questions about one sheet

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

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

Hardness is about the worst one

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

mapflat foldingsand what it took2 × 284 cells, computed here2 × 3606 cells, computed here2 × 43208 cells, computed here3 × 31,3689 cells, computed here2 × 51,9800.6 s3 × 415,55254 s4 × 4300,608not reached herethe 1 × n case is the strip, and it is the only row of this table with a fast methodnobody has a formula for any entry, and nobody has proved there is none

The answer is bigger than the question

A twelve-square strip of stamps is twelve numbers of input and 146,376 objects of output. No algorithm writes that faster than it can be written, so 'efficient' has to be measured against the answer rather than against the question — and in folding that is the normal case.

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

Where the exponent comes from

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

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

What a checker cannot check

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

4681012010203040sides of the polygoncreasescreases in the patternperpendicularsskeleton arcsa 12-sided outline needs 36 creases and one skeleton nodea convex outline is the cheap casea reflex corner splits the shrinking front, and this solver refuses those rather than guessing

What universality costs

The fold-and-cut theorem says any straight-line drawing can be flattened onto a single line. It says nothing about how much crease pattern that takes, and the amount is a measurable quantity — computed here by running the construction rather than by estimating it.

discsfoundproved bestshort by20.292880.292890.00%30.254310.254330.01%40.250000.25000matched50.207050.207110.03%60.187580.187680.05%70.174360.174460.06%80.170220.170540.19%90.166670.16667matchedworst shortfall 0.19% of the radius, at 8 discsthe search never consults the published values, so the comparison measures the searchbeyond nine discs there is nothing to compare against, because nothing has been proved

Getting close instead of getting it right

When the best answer is out of reach the question stops being what it is and becomes how much is lost. For packing discs into a square the loss is measurable: a seeded search in this repository comes within a fifth of a percent of the best radius anybody has proved, and proves nothing.

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

A no costs more than a yes

When a folding question comes back yes, it comes back with an object: a labelling, a stacking, a folded state that anybody can check in one pass. When it comes back no, it comes back with nothing but the assurance that a search looked everywhere — and that assurance is the first thing to break.

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

Two directions that will not separate

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

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

The map that is not a rectangle

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

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

A short reason to say no

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

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

Which vertices are the random ones

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

each row is an exhaustive count over the patterns that construction producedon the edgedeepest piletimes smallercrease densitythe printed patterns8 patterns62%19.415.5×7.9twist tessellations12 patterns52%10.02.7×12.9quadrilateral meshes6 patterns67%8.74.8×5.2fold-and-cut patterns7 patterns86%10.41.2×2.2

Four ways to draw a pattern

Every sentence here of the form over some crease patterns is a statement about a construction nobody declared, and it is worse than the same problem at a vertex because a pattern has a shape as well as angles. Four ways of producing a pattern that satisfies every condition disagree about how far it shrinks by a factor of twelve, about how much creasing it costs by a factor of six, and about how much of it is edge by a factor of two.

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

The cost is in the coincidences

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

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

The count counts labels

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

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

Where the machine catches up

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

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 map counted from the layers

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

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

The patterns a checker is tested on

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

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

The order the refusals come in

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

the bar is the share of random drawings with at least one crossing in them2 segments23.1%0.23 crossings on average3 segments51.2%0.69 crossings on average4 segments73.5%1.36 crossings on average6 segments95.2%3.48 crossings on average8 segments99.4%6.53 crossings on average12 segments100.0%15.30 crossings on average20 segments100.0%43.76 crossings on averageevery crease pattern in this collection has none, and none of them was drawn at random

Drawn by the same hand

Two straight segments dropped on a square cross about 23% of the time; four of them cross 74% of the time; twelve cross with certainty, about fifteen times over. Every crease pattern in this collection's four test populations has none — not because the checkers were catching them, but because the same rules that drew the patterns were incapable of producing one, and nothing looked until a construction finally did.

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

The refusal that reads the list once

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

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

A population that cannot fail

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

the bar is what the whole job costs if every attempt is stopped thereon the rhombille patch, read off 120 measured runsstop at 10051219% of runs finish by thenstop at 20053033% of runs finish by thenstop at 500105435% of runs finish by thenstop at 1000162442% of runs finish by thenstop at 2000263847% of runs finish by thenstop at 5000569749% of runs finish by thenstop at 100001060450% of runs finish by thenstop at 200001629160% of runs finish by thena run that never finished counts as above every cutoff, so the tail is read conservatively

Stopping is cheaper than finishing

A search whose cost varies by a factor of two hundred with nothing but the order of its guesses should not be waited out. Give up after a hundred steps, reseed and start again, and the whole job costs five hundred and twelve steps in expectation; run each attempt to twenty thousand and it costs sixteen thousand two hundred and ninety-one. Patience is thirty-two times more expensive than impatience.

the bar is how many patterns the population holdseach one sampled forty times and then searched, to see whether the two methods ever disagreethe printed patterns80 never lettered by 40 draws · all 8 settled by search · worst 60 nodestwist tessellations70 never lettered by 40 draws · all 7 settled by search · worst 19 nodesquadrilateral meshes60 never lettered by 40 draws · all 6 settled by search · worst 6 nodesfold-and-cut patterns70 never lettered by 40 draws · all 7 settled by search · worst 14 nodesthey never do here — the patterns that separate them are not in any of these four

Four populations with nothing to separate

This collection keeps four standing populations of crease patterns to test its machinery against. Twenty-eight patterns, sampled forty times each for a lettering that agrees with itself and then searched for one — and on every single member the two methods return the same verdict in the same breath. The patterns that separate them are in none of the four, and the reason they are not is what the populations are for.

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

The test that never fires on a map

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

heavier means the crease lies on more independent circuits157 panels, 282 arcs, circuit rank 126; circuits run from 4 to 26 arcs

Which choice the cost lives in

A backtracking search takes two decisions at every step — which thing to decide, and what to decide about it. The literature is almost entirely about the first. On these crease patterns the whole of the cost was in the second, and the structural improvement everybody reaches for first makes matters worse on fifty-two patterns out of eighty-seven.

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

The order that proves nothing exists

Twelve crease patterns with no consistent lettering at all. Proving it takes fifteen steps under one rule and half a million under another — and on three of the twelve the two rules swap places, so neither is the good one. The cost of a negative is two to the power of how many free choices sit above the contradiction.

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 population nobody chose

Five crease patterns were measured over and over because somebody had drawn five. Ninety-six drawn from a stated grid of tiling, turn and pleat width say something the five could not: nine of them have no consistent lettering at all, and the phenomenon the collection had spent so long measuring belongs to the one tiling the grid leaves out.

the curve is stop-and-restart; the rule is a constant letter order1001e+31e+41001e+31e+4563 at a cutoff of 10080 nodes, deterministic, nothing to restartexpected nodes in totalcutoff, in nodes

Restarting what cannot be restarted

Stopping a search early and starting it again with a fresh seed costs five hundred and twelve steps in expectation against sixteen thousand for patience. Every number in that is right. The distribution it is right about was made by the search's own coin, and taking the coin out costs eighty — with nothing left to reseed.

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

What the rim was doing

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

proving the glued square cell has no lettering1×1, 4 panels3proved there is none · the other test found one in 32×2, 16 panels35proved there is none · the other test found one in 93×3, 36 panels3,455proved there is none · the other test found one in 6254×4, 64 panels200,000still running at the budgeta bar at the budget is a search still running, not a proof

Pruning on proofs alone

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

proving the glued square cell has no lettering1×1, 4 panels3proved there is none · the other test found one in 32×2, 16 panels35proved there is none · the other test found one in 93×3, 36 panels3,455proved there is none · the other test found one in 6254×4, 64 panels200,000still running at the budgeta bar at the budget is a search still running, not a proof

The cost of proving something false

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

sliding the cut across one period of the square tessellation36 vertices at every position, and a different set of creases divided at each0102030cut at the start of a periodone period alongnodes; the axis starts at zero, and the whole spread is inside a factor of 1.32

Where you cut hardly matters

Slide the same rectangle across one whole period of the same tessellation and every position gives a different patch: different creases divided, different half-panels round the edge, panel counts from forty-nine to sixty-one. The cost of lettering them runs from twenty-five steps to thirty-three. Whether a cut is made changes the answer by three orders of magnitude; where it falls changes it by a third.

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

Half the slack

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

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

Which pair is glued

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

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

A proof in no nodes at all

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

the same 2×2 glued cell, searched under two rulesa cycle is a contradictiona cycle whose steps add to zero isand what the loops dothe square gridnothing, in 359 nodesevery loop travels (2 directions)the triangular gridnothing, in 12,143455 nodesevery loop travels (2 directions)the honeycombnothing, in 9,6191,043 nodesevery loop travels (3 directions)the elongated triangular tilingnothing, in 9,123162 nodesevery loop travels (5 directions)the rhombille tilingunfinished at 200,000unfinished at 200,000“nothing, in n” is an exhausted search: a proof that the pattern has no consistent lettering, which is false

The cost of asking the wrong sheet

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

the grid, 2×2 cellsone drawing, four sheetscutoutgluedacrossgluedalonggluedboth waysvertices4444free letters1210108panels9664V − 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

One population, four sheets

A population of patterns is a way of asking what is typical, and it has always been a population of drawings. Put the same drawings on four different sheets and the verdicts move — not because the drawings changed but because the sheet did, which means a population has two halves and only one of them was ever chosen.

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 map with no edges

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

the pieces that are one panelleft and right edges identified — 9 pieces, 6 panels9 pieces on the drawing6 panels on the sheet10 creases, 4 verticeskeeps the sidetwo pieces of one shade are one piece of paper, a cell apart

The tube a map makes

Join one pair of a map's edges and the result is a tube — a real object, foldable in the hand, and neither the strip's problem nor the torus's. It has one loop that cannot be shrunk instead of two, it keeps its bottom layer because it keeps half its rim, and half its sizes are refused by a parity the flat map does not have.

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

Deciding is not making

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

how deep into the pile the machine has to be allowed to reach before it reaches every stateone layer is the patient machine and the full depth is the machine that may choose3 equal stamps22 of a possible 3 · 12 states4 equal stamps33 of a possible 4 · 32 states5 equal stamps44 of a possible 5 · 100 states6 equal stamps55 of a possible 6 · 288 statescreases at .20 .55 .7022 of a possible 4 · 8 statescreases at .15 .40 .50 .8522 of a possible 5 · 16 statescreases at .13 .31 .62 .7844 of a possible 5 · 24 statescreases at .40 .50 .62 .7233 of a possible 5 · 12 statescreases at .08 .24 .28 .35 .7255 of a possible 6 · 48 statesthe even strips are the ones that need the most, and they are the ones the machine that takes everything does best on

The easiest strip needs the deepest reach

The patient machine and the machine that may choose are the two ends of one number: how many layers of the pile a machine is allowed to hold. At one it reaches four states whatever the strip; at the pile's full depth it reaches everything. In between it is a machine nobody has defined, and measuring where completeness arrives inverts these essays' own ordering — the evenly creased strip, which the machine that takes everything folds perfectly, needs the deepest reach of all, and one uneven strip is complete at two.

the shortest sequence of folds to each state, for the machine that may choose its blockfewest, mean and most over every state of every marking; the last column compares the machine that takes everythingstripstatescreasesfewestmeanmostall layers4 equal stamps32322.753the same5 equal stamps100433.404the same6 equal stamps288534.045the samecreases at .20 .55 .708333.003reaches nonecreases at .15 .40 .50 .8516444.004reaches nonecreases at .13 .31 .62 .7824444.004reaches nonecreases at .40 .50 .62 .7212444.004reaches nonecreases at .08 .24 .28 .35 .7248555.005reaches nonea fold uses at least one crease, so no sequence is longer than the crease count

A shallow machine pays in states, not folds

A machine allowed to take only a few layers of the pile at a time reaches fewer folded states, and the natural fear is that it also reaches the ones it does by much longer sequences. Walked breadth first, so that every state's shortest sequence is found, it does not. On unevenly creased strips every state takes exactly one fold per crease at every depth, because no two creases ever lie on one line. On strips of equal stamps a shallower machine needs one fold more for a minority of states and two more for eight of the 924 states at seven stamps — and never more than the crease count, which no machine can exceed.

the pile 0 6 1 2 3 4 5 and its turns, which the all-layers machine cannot foldMVMVMMVVMVMMMMVMVMVVVMVMVMMMVMVMVVVMVMVMMM

Fourteen states are one pile

A machine that folds every layer at once reaches every folded state of a strip of six equal stamps and misses fourteen piles at seven. The fourteen are not fourteen things. Taking a pile's bottom stamp and putting it on top maps foldings to foldings, so the 462 piles of seven stamps fall into 33 classes of exactly fourteen, and the missed piles are one whole class: the pile 0 6 1 2 3 4 5 — an accordion of five stamps with the last stamp wrapped round it and slid into the fold that holds the first — seen from each of its seven stamps. At eight stamps the machine misses 64 piles, and they are exactly the piles that leave that one when an end stamp is removed.

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

Each drawing has its own threshold

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

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

The route, not the sheet

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

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

The cheapest route crosses later

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

All essays