Deciding is not making

Four earlier essays here ask one question in four ways. Given a marked strip, is there any sequence of a given machine’s folds that flattens it? The machine that takes the whole pile is stopped by a strip with two creases in it; the machine that takes one layer is stopped by almost everything; the machine that crimps cannot fold an odd number of creases at all; and the machine that may choose its block is stopped by nothing in a sweep of a hundred and seventeen spacings.

That is the decision question, and it produces a lattice with one full column and three that have holes in different places.

The fifth of these essays asked something else of one machine. Not can it flatten this strip but how much of the strip’s repertoire can it produce — a strip has several folded states, and a machine that reaches one of them has not thereby reached the others. On an evenly creased strip the all-layers machine reaches every state there is, up to six stamps; on an unevenly creased one it reaches none.

This essay asks the same thing of the other two, and the answers are not a refinement of the decision table. They are a different table.

Four, and it keeps being four #

The one-layer machine is the striking one. It reaches exactly four folded states on a strip of three equal stamps, four on four, four on five, four on six — and four on every unevenly creased strip tried, including one with forty-eight states.

A number that does not move when its input grows by a factor of twenty-four is a number with a reason behind it.

Of the sixteen markings a five-stamp strip has, the one-layer machine reaches a state of exactly two: MVMV and VMVM. Those are the accordion, lettered from either face. Each of them has two folded states in the convention this site counts with — the stacking and the same stacking turned over — so two markings at two states each is four, and the four is not a property of the machine’s power but of how many accordions a strip has.

And on those two markings the machine is complete: it reaches every state they have. So the one-layer machine’s weakness is not that it reaches a fraction of each strip’s repertoire. It is that it reaches the whole of two markings and nothing whatever of the rest.

That sharpens the second of these essays’ finding rather than repeating it. The patient machine is the weak one reports that the one-layer machine fails everywhere except the accordion, which is a statement about the decision question. What the reach measurement adds is that there is no partial credit anywhere: the machine has no marking it can half-fold, no strip where it produces some states and misses others. Its whole repertoire is the accordion, and the mechanism is the one that essay identified — paper is joined, so a machine that declines to hold a layer also cannot move it, and the only sequence that never asks it to is the one that rolls a layer over at a time onto a pile that never has to travel.

What a repertoire of four is made of #

It is worth drawing the four, because “the accordion from either face” is a description and the thing itself is an argument.

A one-layer machine starts with a flat strip, which is one layer, so its first fold is any fold — there is nothing to hold back. After it, the pile is two layers, and the machine may hold the top one or the bottom one. Holding the top one and folding it at the next crease works, because nothing is attached to it beyond that crease except what is already on the moving side. Holding it and folding it at any earlier crease would have to drag the layer underneath, which is joined to it, and the move is refused.

So the machine has one move available at every step after the first: take the outermost layer and fold it at the next crease along. That is a roll, and a roll of alternating creases is the accordion. It can start from either end of the strip and it can start with the paper either way up, which is four sequences and four end states — and the reason the count does not grow with the strip is that the number of choices does not grow either. A longer strip is a longer roll.

That also says why the machine has no partial credit. Every state it reaches is the end of the only sequence available to it, so it either completes that sequence or it stops, and stopping produces no state at all. A machine with one move per step cannot reach part of a repertoire.

Everything, everywhere #

The some-layers machine reaches every folded state of every strip measured. Twelve of twelve at three stamps, thirty-two of thirty-two, a hundred of a hundred, two hundred and eighty-eight of two hundred and eighty-eight; and on the uneven strips eight of eight, sixteen of sixteen, twenty-four of twenty-four, twelve of twelve and forty-eight of forty-eight.

That is a stronger statement than the fourth of these essays’, and the difference is worth stating precisely. The machine that may choose measured the decision question over a hundred and seventeen spacings and found no strip that folds flat and that the machine cannot flatten. That is one state per strip. The measurement here is the whole set: on nine strips with a hundred and eight uneven states and four hundred and thirty-two even ones between them, the machine produces all of them.

And it does so on the five strips where the all-layers machine produces none. That is the part the ordering conceals. Reading down the table, the some-layers machine is at least as strong as the all-layers machine at every row, which is true by definition — an all-layers move is a some-layers move with the block set to the whole pile — and the interesting fact is not the ordering but the size of the step. On the uneven strips it is the whole of the repertoire against nothing at all.

Two columns that do not vary #

Set the three results side by side and the shape of the table is the finding.

Two of the three machines have no dependence on the strip at all. The some-layers machine reaches everything whatever it is handed; the one-layer machine reaches the accordion and nothing else whatever it is handed. Only the middle row varies, and it varies between the two extremes the other two rows are stuck at.

That is the opposite of what the decision table looks like. There, all three restricted machines have holes, the holes are in different places, and the interesting reading is the pattern of which strips each one loses. Here the pattern of which strips each one loses is empty for two of the three.

So the two questions are not the same question asked at different resolutions. The decision question is sensitive to the strip and the reach question is sensitive to the machine. A restricted machine either has a repertoire that is fixed and small, or a repertoire that is everything, and the one in between is the one whose restriction interacts with the spacing.

Why the all-layers machine is the one that varies #

The mechanism is the one the fifth of these essays identified, and it explains why that machine is the only one in the table with a column that reads differently for the two families.

An all-layers fold is legal only where every layer in the pile is severed at the line. On an evenly creased strip every crease still in play sits at the same place in the folded image, because every segment has the same length, so every line that severs one layer severs all of them and no fold is ever refused. On an unevenly creased strip that stops being true after the first fold, and the machine — which cannot decline to hold what it is holding — is stuck immediately.

The other two machines have no such coupling to the spacing. A one-layer machine holds one layer, so every legal fold line is available to it and the constraint is somewhere else entirely: it can sever anything and can move almost nothing, because paper is joined and moving a layer drags whatever is attached beyond the crease. That constraint is about connectivity, which does not care where the creases are. The some-layers machine can pick its block to contain exactly the layers that are severed, so the spacing is not a constraint on it either.

A machine’s answer depends on the spacing only when its restriction and the spacing are about the same thing. All-layers folding is a constraint on which lines are legal, and the spacing decides which lines are legal, so the two interact. One-layer folding is a constraint on what can move, and the spacing has nothing to say about that.

The ordering is real and nearly empty #

It is worth being careful with the word stronger here, because the measurements confirm an ordering and then make it useless.

The ordering is genuine: an all-layers move is a some-layers move with the block set to everything, and a one-layer move is a some-layers move with the block set to one, so the some-layers machine reaches at least what either of the others does, at every strip and every size. That is proved rather than measured, and the enumeration agrees with it everywhere.

What the enumeration adds is that the ordering is not a ranking. The all-layers machine is not between the other two: on the even strips it matches the strongest exactly, and on the uneven ones it falls below the weakest, which reaches four where the all-layers machine reaches none. Two machines neither of which dominates the other, both dominated by a third — and that pattern is invisible to any summary that reports one number per machine.

The same shape turns up wherever restricted models are compared on a family of instances, and it is the reason the fourth of these essays’ verdict — that the loss in these essays is due to being forced rather than to the atom — needed a machine with no restriction to state. A comparison between two restricted machines can only say which instances each loses, and it takes an unrestricted one to say whether the losses were necessary at all.

What the enumeration is #

The walk is exhaustive and the same for all three machines. From the flat strip, every legal move of the machine is tried; from each result, every legal move; a branch with no move left is examined, and if the strip is fully folded and the sequence has respected the marking, the state it finishes in is recorded.

Two things had to be repaired before the other two machines could be measured at all, and both are worth recording because the first had been latent since the walk was written.

A move that would tear the paper returns nothing, and the walk did not check. Folding a single layer at a crease drags everything attached beyond that crease along with it, and a machine holding only that layer is not holding the rest — so such a move is refused, and the refusal arrives as an empty result rather than as an error. The decision search has skipped those since it was written. The reach walk did not, and crashed on the first one. It never crashed for the all-layers machine because that machine cannot produce such a move: a fold that takes the whole pile cannot separate two pieces that are joined, so the failure was invisible for as long as only one machine was measured.

And the walk had no memory. A state reached twice by different sequences has the same leaves below it both times, so expanding it twice collects nothing new. The all-layers machine branches so narrowly that this never mattered. The one-layer machine has four moves at every line and the some-layers machine has twice the pile’s depth, and without the memo a five-crease strip does not finish. Adding it changes no answer and is the difference between a measurement and a hang.

What this does not measure #

Nine strips is nine strips. Four evenly creased and five unevenly creased, chosen to include a short segment, a near-coincidence and a spacing with no structure at all. The some-layers machine’s completeness is a measurement over those, not a theorem, and it is the same claim the fourth of these essays made over a hundred and seventeen spacings at one state each rather than all of them.

Nothing here is about cost. No sequence length is measured and no complexity class is named. A machine that reaches every state may need a great many folds to reach some of them, and a search’s cost lives somewhere other than its answer — the enumeration’s own expense is the reason the even table stops at six stamps.

The even strips’ state counts are the famous ones and the uneven strips’ are not. Twelve, thirty-two, a hundred and two hundred and eighty-eight are the sequence with no formula, doubled by the turning-over convention. The uneven strips’ counts — eight, sixteen, twenty-four, twelve, forty-eight — are in no table anywhere, because the object they count is not the one anybody has studied. That asymmetry is why the even column of this measurement can be checked against something and the uneven column cannot.

And the crimping machine is not in the table. It has a different atom — two creases at once — so its moves are not a restriction of the some-layers machine’s and the ordering does not contain it. Its reach is a separate measurement and nothing here bears on it.

The four is a count under a convention. The solver lists a marking’s stackings with the sheet one way up, so a stacking and its turned-over twin are counted as two, and the one-layer machine reaches both because it can start from either face. Counted as objects rather than as records the number is two, and the argument is unchanged.

Still open: how much of the pile is actually needed #

The one-layer machine and the some-layers machine are the two ends of one parameter, and the parameter has never been given a value between them.

A some-layers move takes a block of layers reaching the top or the bottom of the pile. Restrict the block to at most two layers, or three, and there is a machine nobody has defined — stronger than the patient one, weaker than the one that may choose, and with a number attached saying how much of the pile it is allowed to hold. The reach question can be asked of each of them, and the answer would say how much of the some-layers machine’s completeness is actually being used: whether it needs to reach deep into a pile of six, or whether two layers is already everything.

That is a sharper form of the fourth of these essays’ question. Was it the atom or the compulsion? has been answered — it was the compulsion — but “not compelled” is not one condition, it is a sequence of them, and where on that sequence the completeness arrives is what says how much freedom a machine actually needs. The measurement is the same enumeration with one number changed.

The second direction is the one this essay leaves standing. The all-layers machine’s completeness on even strips ends at seven stamps, where it misses twenty-eight of nine hundred and twenty-four states, spread as exactly two states each across fourteen of the sixty-four markings. Two per marking is a stacking and its turned-over twin, so it is fourteen states, one in each of fourteen markings, and what those fourteen have in common is not something this essay establishes. A characterisation of them would say what the even strip’s completeness actually rests on, rather than that it rests on being small.

Sideways from here, the some-layers machine’s completeness on uneven strips says something about folding a designed pattern that the all-layers result could not. The collapse of a tessellation is difficult precisely because taking hold of the whole sheet does not work, and this essay says the difficulty is not that the states are unreachable — every one of them is reachable, by a machine that may choose which layers to take. What the folder is doing when a collapse goes wrong is choosing the wrong block, which is a different kind of failure from being stopped.

The habit worth carrying is about what a comparison of models measures. Ask what each model can produce, not only what it can settle — the two questions have different sensitivities, and a family of instances that separates models sharply on one of them can be almost silent on the other. The decision table here is about the strips. The reach table is about the machines, and it took the same enumeration and a different question to see it.