Enumeration — where it appears
Named by 29 essays across 6 fields — each of them below, with the objects they name alongside it.
How many assignments fold
The local conditions throw away most of the ways a pattern could be creased. They throw away a smaller and smaller fraction as the pattern grows, and what survives grows faster than what is discarded — which is why a strong filter is not a decision procedure.
The 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.
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.
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.
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.
The shadow does not name the pattern
A photograph of a folded model carries an outline and a thickness at every point of it, and that is the whole of what it carries. It is not enough. Crease patterns in genuinely different places fold to identical outlines with identical layer counts, and nearly a third of the folded objects a short strip can reach are reached by more than one pattern.
What 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.
The map that is not a rectangle
Take one square out of a three-by-three map and the number of ways it folds does not go down by an eighth. It goes up — to 848 if the square came from a corner, and to 8,016 if it came from the middle. Two maps of eight squares in the same box, differing by nearly a factor of ten, and no function of the box tells them apart.
The condition that is not flat-foldability
Take away the assumption that one crease family runs straight through every vertex and ask what makes a quadrilateral mesh move. It is not flat-foldability. There is a one-parameter family of meshes, every one of them developable and flat-foldable at every vertex to machine precision, and exactly one member of it folds — the Miura. Slide a single vertex along the ray that keeps every condition exact and the sheet stops moving, first order in the displacement.
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.
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.
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.
The ring is the loop
The square twist's central polygon is four creases enclosing one panel, and a lettering that gives all four the same letter has no folded state. That was established by enumerating the orderings of nine panels. It can now be read off the crease list in one pass, because the eight panels the letters send round in a circle are exactly the ring — the twist's own defining feature, contradicting itself.
The rule that breaks the count
The waterbomb tessellation has five hundred and twelve repeating rules for its letters and thirty-two of them fold. A hundred and twenty of the other four hundred and eighty send four panels round in a circle — the shortest circle a crease pattern can have — and every single one of those hundred and twenty has broken Maekawa's count at the very vertex the circle goes round. The theorem that closes the shortest circle, caught doing it, a hundred and twenty times.
The 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.
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.
Two refusals that refuse differently
Four of the six developable quadrilateral meshes this collection solves have no ordering of their nine panels — they must pass through themselves, and a search over every ordering proves it. On all four, the letters agree with themselves perfectly. The linear proof and the exponential search are not a fast test and a slow one: they answer different questions, and neither contains the other.
Sixty-four rules, sixteen fold
The Miura fold's letters are usually given as a recipe: rows one way, columns changing at every row. Write down every rule of that shape — the letter on a crease depending only on which row and which column it is in — and there are sixty-four. Sixteen fold flat. They are exactly the ones whose columns change at every row, the row letters do not matter at all, and every one of the forty-eight refusals is the counting theorem's alone.
The test that never fires on a map
The cheapest refusal this collection has reads a crease list once and reports that no arrangement of the layers exists. Enumerate every labelling of every map from two panels to nine and it fires on four of the four hundred and fifty-four — all four on the largest map, none at all below it. On the oldest open problem in the subject, the cheap test has essentially nothing to say.
The 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.
Refused at one lettering
Four of six quadrilateral meshes here have no arrangement of their nine panels — established by searching every ordering, at the labelling each mesh arrived with. Enumerate every labelling instead and two of the four fold perfectly well at a different one. What was reported as a fact about four meshes is a fact about two meshes and two labellings.
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.
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.
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 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.
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.
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.
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.
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.
Named alongside it
The objects these essays reach for when they reach for this one.
AssignmentLayer orderingNecessary conditionMap foldingStamp foldingCombinatorial explosionThe counting problemLayer orderMultifoldOperation setAxiomsDecision procedure