Nothing slides past anything

A crease pattern with its mountains and valleys written on it does not name a folded object. It names several, and at the eight-crease vertex in the middle of the first base anybody folds, a single marking can be folded four genuinely different ways.

That has always been reported as a count. A count says how many there are and nothing about how they sit relative to one another, and there is an obvious question underneath it that nobody here has asked: given one of them, can a folder get to another?

The question is not idle, and it is not the same question as the one asked of letterings. A folder who has finished a model and does not like how a flap sits does not unfold it. They lift the flap over the one above it and put it back down. That is a move — the smallest one the object admits — and this essay is what happens when the legal stackings of a marked strip are put through it.

What the move is #

Take a folded strip. Number the layers from the bottom. Choose a height, lift the piece of paper at that height and the piece above it, and exchange them, leaving every other layer exactly where it is.

The result is either a legal stacking or it is not. It is not when a crease is made to wrap the wrong way — a mountain and a valley are the same U-turn seen from opposite sides, and the direction of travel decides which of the pair is on top — or when a sheet is made to pass through a sheet.

Two things about that definition are worth stating because both were choices and both could have been made the other way.

The two layers do not have to be joined by a crease. A move that only exchanged the two ends of a fold would be a much smaller thing, and it would make the answer below trivially yes. What is being asked is whether the object can be rearranged at all, so any adjacent pair is a candidate.

A swap of two segments that never lie over one another is not a move. Two pieces of paper at opposite ends of the folded image are not stacked in any observable sense, and exchanging them changes nothing a photograph could record. Counting such a swap as a rearrangement would report bookkeeping as paper, and the answers below would be dominated by it.

The answer, on the strips the counts are famous for #

An evenly creased strip is a strip of stamps, and its foldings are the sequence — 2, 6, 16, 50, 144, 462 — that nobody has a formula for. Those are the strips to ask first, because their counts are known independently and because every segment lies over every other, so no swap is dismissed as invisible.

Three things in that table, and the third was not what anybody was looking for.

From four segments up, every marking whose stacking is not forced has stackings that cannot be reached from one another. Six of six, fourteen of fourteen, thirty of thirty, sixty-two of sixty-two. There is no marking anywhere in the range where the legal stackings form one connected set.

Most of them are isolated outright — 560 of the 672 stackings measured have no legal swap of any kind out of them. Reaching another means unfolding.

And the fourth column has a parity in it. At four segments and at six, not one swap is legal anywhere, at any marking, over 768 tried. At five and at seven, some are.

Why the parity is there #

The strip’s two ends are free edges. Walking from one end, the direction of travel reverses at every crease, so segment i is travelled forwards when i is even and backwards when it is odd — which means the two end segments are travelled the same way when the segment count is odd and opposite ways when it is even.

The rule deciding which of two consecutive segments sits higher is the product of the crease’s letter and the direction of travel, so a pair of segments at the top or the bottom of the pile can only be exchanged when the constraint binding them is slack, and on a strip whose ends run opposite ways there is no such pair anywhere.

That is the mechanism, and it is worth being careful about what it licenses. It explains the two rows that are zero. It does not explain why the odd rows are as small as they are — sixteen legal swaps out of two hundred, ninety-six out of 2,772 — and nothing here derives that.

What a stacking is, and why the count is of states #

The measurement is over states rather than over total orders, and the distinction is not a nicety — it decides three of the numbers in the table above.

A total order of the segments says more than a folded object does. If two segments never lie over one another, exchanging them leaves every layer of paper exactly where it was, and the two orders are one object. On an evenly creased strip the distinction is empty, because every segment lands on every other, which is why the counts there agree with the published folding numbers and why nobody had to make the distinction before. On an unevenly creased one it is not empty at all.

So a state is the order restricted to the pairs that overlap, and the count of states is the count of distinct restrictions. Every move below lands in a state, every piece is a set of states, and a swap that changes the order without changing the state is counted as legal and not as a rearrangement.

That third clause is the one doing work. On the five-segment strip there are 200 swaps to try, 16 of them are legal and change what can be seen, and a handful more are legal and change nothing — and reporting the second kind as rearrangement would turn a strip with 34 isolated stackings into one with none.

The control, and it is the smallest case #

A move that is never legal is indistinguishable from a move that is not a move, and this site’s habit is that an assertion which has rejected nothing has proved nothing. So the claim above needs a case where the answer is yes, and the smallest strip supplies it.

At three segments — two creases, the strip everybody folds first — there are six stackings across four markings, two markings have a choice at all, and in both of them the stackings are joined by a legal swap. Four such swaps exist. A folder with a three-segment strip can genuinely rearrange it.

So the freeze arrives at four segments and does not go away. It is a fact about the paper rather than about the definition, and one more segment is what it costs.

The accordion, which has nothing to rearrange #

The other end of the range is worth a paragraph because it is the case a reader will picture.

An accordion — creases alternating mountain, valley, mountain — has exactly one legal stacking at any number of segments. There is nothing to rearrange, no move is legal, and no move is needed. The measurement reports one stacking and no moves rather than calling the set connected, because “connected” over a set of one is true and says nothing.

That is also why the table above counts markings with a choice separately. Most markings of a strip have their stacking forced, and a rule about rearranging them would be a rule about an empty collection.

Unevenly creased strips #

The strips above are all evenly creased, which is a special case and one whose specialness this site has been caught by before. So the same measurement is made on strips whose creases are at arbitrary positions.

A strip creased at 0.13, 0.31, 0.62 and 0.78 has eight markings that fold, four of which have a choice of stacking, and all four come apart — every stacking isolated. One creased at 0.08, 0.24, 0.28, 0.35 and 0.72 has sixteen markings that fold, eight with a choice, and all eight come apart.

And a strip creased at 0.20, 0.55 and 0.70 has four markings that fold and not one of them has a choice at all: every marking’s stacking is forced, so there is nothing to rearrange and nothing to report. That is the case the state count was invented for — where segments that never overlap were inflating an ordering count into a state count — and it behaves here exactly as it should.

Every piece is one stacking or two #

The table carries a stronger result than the one read off it, and the arithmetic that finds it is addition.

Set the isolated stackings beside the legal swaps. At three segments: two isolated, four swaps, six stackings. At five: thirty-four isolated, sixteen swaps, fifty stackings. At seven: 366 isolated, ninety-six swaps, 462 stackings. The isolated count plus the swap count is the stacking count, on every row.

So every stacking that is not isolated has exactly one legal swap out of it — no more. And a stacking with one way out, leading to a stacking that also has one way out, is a pair.

Check it against the piece counts. Fifty stackings in forty-two pieces: thirty-four singletons and eight pairs is forty-two pieces and fifty stackings. Four hundred and sixty-two in 414: 366 singletons and forty-eight pairs is 414 pieces and 462 stackings. Both close exactly.

Every connected piece is a single stacking or a pair of them. Not “rarely connected” — never connected past two.

What that does to the practical statement #

The essay’s conclusion is that a folder cannot rearrange their way from one state to another, and the piece structure says something more specific and more useful.

A folder who has folded a strip into a state with a move available has exactly one move, and having made it there is nothing further. The object does not have a neighbourhood to explore; it has at most one alternative, reachable in one step and reversible in one step.

That removes the last version of the hope the count invites. A count of fifty stackings suggests a space to search, and a set of thirty-four points and eight dominoes is not a space at all — there is no path of length three anywhere in the range measured, and no stacking is two moves from anything.

It also explains why the shares fall. Among the odd rows the share of stackings with any move at all runs two-thirds, then a third, then a fifth, while the number of pairs grows only from two to eight to forty-eight against a stacking count growing from six to fifty to 462. The pairs are being diluted rather than joined, which is what a longer strip does to whatever slack a shorter one had.

The one thing worth saying against over-reading it: seven segments is where the enumeration stops, and a piece of size three would be the first sign that the pairs are the small-case behaviour rather than the rule. Nothing in the argument for the parity predicts a maximum size of two, so that is a genuine open case rather than a settled one.

What it means at the table #

The practical statement is short and it is not the one a count suggests.

Two folded states of one marking are not variants. They are separated objects, and the separation is not a matter of difficulty — there is no sequence of small rearrangements between them, however patient the folder is. A model folded into the wrong one of its states has to be opened.

That is a sharper version of something the subject already half knows. The order is the half no notation records, and the field’s interchange format has a field for it that nobody fills in. The reason it matters is now measurable: the order is not merely unrecorded, it is unreachable — a reader given the crease pattern and the letters cannot recover the intended object by trying rearrangements, because the rearrangements do not exist.

How the pieces grow #

The count of pieces is worth reading beside the count of stackings, because the two grow at different rates and the gap between them is the only room a folder has.

At four segments there are sixteen stackings and sixteen pieces: every one is its own. At five there are fifty stackings and forty-two pieces, so eight of the fifty have somewhere to go. At six, 144 and 144 — nothing anywhere. At seven, 462 and 414.

Read as a share, the stackings that are joined to anything at all run 0 per cent, 32 per cent, 0 per cent, 21 per cent. There is no trend in that sequence and there is not meant to be one: it is the parity again, and the odd rows are the ones with any slack in them at all.

What there is a trend in is the absolute number of isolated stackings — 16, 34, 144, 366 — which is growing at very nearly the rate the stacking count itself grows. The proportion of a strip’s folded states that a hand can do anything about does not improve with size. A longer strip is not a more forgiving object; it is the same object with more of it.

Against the other space #

This site has now measured two spaces with the same question and got opposite answers, and putting them side by side is the point of having asked twice.

The letterings of a vertex are one connected piece under a change of any two creases, on every vertex measured, at every degree. A folder can walk from any folding of a vertex to any other.

The stackings of a marking are almost never connected under a change of two neighbouring layers. A folder can walk from almost none of them to anything.

The two are not the same kind of object and the moves are not the same kind of move, but the pairing says something about where this subject’s difficulty lives. The letters are the part the theorems reach and the part a hand can revise. The order is the part the theorems do not reach, and it is also the part a hand cannot revise. Whatever is hard about flat folding is hard in the same place twice.

What the freeze costs a designer #

Three consequences, and the last one is the reason this belongs in the design half of the subject rather than only in the theory.

A crease pattern plus a lettering is not a specification. It is a specification of a small set of objects, and the difference between them is not recoverable by any local operation. If a designer means one of them, the pattern does not say which.

A folding sequence is the specification. Publishing the pattern rather than the sequence has been the field’s practice since the 1980s and is usually defended on the grounds that a competent folder can reconstruct a sequence. That defence is about the letters, which a competent folder genuinely can reconstruct. It says nothing about the order, and the order is the half that cannot be corrected afterwards.

And a colour change is an order problem. Which panels can show the reverse of the paper is settled by the two-colouring, but which of them actually does show is settled by the stacking — and the stackings that show it are not reachable from the stackings that do not. A designer who wants the far side of the paper on top has to fold into that state, not fold and then arrange.

Where the measurement stops #

The enumeration has a ceiling and it is low: eight segments is 40,320 orderings to test against every swap, and the strip solver refuses past that rather than truncating. A truncated list of stackings would report a set with pieces missing as a set that has those pieces, which is the one failure mode this measurement cannot survive.

Nothing here extends to two dimensions either. A folded sheet’s layers are ordered over a region rather than along a line, and the move — lift this piece of paper over that one — is a statement about an overlap rather than about a height. That is the next rung and it needs a solver nobody here has written.