The easiest strip needs the deepest reach

Deciding is not making leaves the three machines sorted into a table with one row that varies and two that do not. The machine that may take any block of layers reaching an edge of the pile reaches every folded state of every strip; the machine that may take only the outermost layer reaches four, whatever the strip is and however long; and between them sits the machine that takes the whole pile, complete on an evenly creased strip to six stamps and empty on an unevenly creased one.

Two of those three are the same machine with a number changed. A some-layers move takes a block of layers reaching the top or the bottom; set the block’s depth to one and it is the one-layer machine. Every value in between is a machine, and none of them has been defined, let alone measured.

The number is worth having because it turns a comparison of three models into a measurement of one quantity. How much of the pile does a machine actually have to be allowed to hold?

The machine, and the two ends it already had #

A move takes a run of layers starting at the top of the pile or at the bottom, reflects them about the fold line, and lifts the moving part clear. The run has a length, and that length is the parameter.

At one, the machine holds the outermost layer. It can sever anything and can move almost nothing, because paper is joined and folding a layer drags whatever is attached to it beyond the crease — so its only sequence is a roll from one end, and its whole repertoire is the accordion, four states counted with the convention that a stacking and its turned-over twin are two.

At the pile’s own depth, every block is available including the whole pile, so the machine is the some-layers machine and, as far as nine strips can say, it reaches everything.

Between them the machine is genuinely new. At a depth of two it may take the outermost layer or the outermost two; at three, up to three; and the interesting question is not whether the reach rises — it must, since a deeper machine has every shallower machine’s moves — but where it stops rising, because that is the depth the completeness actually needs.

Where it stops rising #

On the evenly creased strips the answer is uniform and it is one short of everything.

Four stamps is complete at three layers, five at four, six at five. A strip of $n$ equal segments needs a reach of $n-1$, and the column for $n$ — the whole pile — adds nothing at all.

That last clause is the sharper half. The some-layers machine’s completeness is usually explained by saying it may take everything, and on these strips it never has to: the move that completes the repertoire is always one layer short of the whole pile. A machine that was allowed every block except the full one would reach exactly the same states.

The rise on the way is not gentle. At six stamps the machine reaches 4 states of 288 at depth one, 60 at two, 172 at three, 236 at four and all 288 at five. Two layers instead of one multiplies the reach by fifteen; five layers instead of four adds the last eighteen per cent. The curve is steep at the bottom and the last step is the expensive one, which is the shape of a constraint that binds on a few awkward states rather than on the bulk.

And the uneven strips need less #

A strip with creases at .20, .55 and .70 has four segments and eight folded states, and the machine reaches all eight holding two layers. A strip with creases at .15, .40, .50 and .85 has five segments and sixteen states, and two layers is enough for those too. A strip with creases at .40, .50, .62 and .72 needs three; one with creases at .13, .31, .62 and .78 needs four; and the six-segment one needs five.

So the uneven strips run from two to $n-1$ and the even strips sit at $n-1$ every time. The evenly creased strip is the worst case for this machine, and it is the case the whole of this subject’s earlier work treats as the easy one.

What the last layer buys #

The tables carry a second reading that the completeness depth hides, and it is the one a designer of a machine would want.

At six equal stamps the machine reaches 236 of the 288 states at a depth of four and all 288 at five, so fifty-two states — eighteen per cent of the repertoire — exist only at the full depth. At five stamps it is twenty-four of a hundred, again the last step, again about a fifth. At four stamps the last step buys twelve of thirty-two.

So the deepest layer is not a formality that the enumeration happens to need. It is buying a substantial and consistent fraction of the repertoire, and that fraction does not shrink as the strip grows. A machine built to a depth of $n-2$ would not be nearly complete; it would be four-fifths complete, at every size measured.

The other end of the curve is more dramatic and points the same way. Depth one reaches four states, depth two reaches sixty of the six-stamp strip’s 288, and that single extra layer multiplies the repertoire by fifteen. The first layer and the last layer are both expensive and everything between them is cheap, which is a shape worth naming: the machine is not gaining reach smoothly as it is allowed more of the pile, it is passing two thresholds with a plateau between them.

The first threshold is the one the patient machine is stuck below. With one layer there is exactly one move available at every step past the first, so there is no branching at all and the machine has no repertoire in the ordinary sense — it has a procedure. Two layers gives it a choice, and a choice at every step of a sequence is what makes a set of outcomes rather than one.

The inversion, and what causes it #

Set the two facts together and they point opposite ways.

The all-layers machine is complete on the even strip and empty on the uneven ones. The depth-limited machine needs its deepest reach on the even strip and its shallowest on two of the uneven ones. The same property of the strip — that every segment has the same length — makes one machine’s life trivial and the other machine’s hardest.

The cause is the same property read twice, and it is worth writing out because the double reading is the result.

On an evenly creased strip every crease still in play sits at the same position in the folded image, because every segment is the same length. That is why the all-layers machine is never refused: a line that severs one layer severs all of them, so the restriction to taking everything costs nothing.

It is also why the pile is thick in exactly the same place. Every layer lies over every other layer along the whole of the folded image, so a fold line at any position cuts through all of them, and a machine that wants to move some of them and not others has to reach past the ones it is leaving. The more of the pile that lies together, the deeper the reach that is needed to get at a particular arrangement.

On an unevenly creased strip the layers land in different places. A fold line through one layer’s crease misses another layer entirely, so that layer is not in the way — and the machine can produce an arrangement by holding two layers rather than five, because there were never five layers over the line. What blocks the all-layers machine is exactly what saves the depth-limited one: a line that does not sever every layer is a line the greedy machine cannot use and a line the choosy machine does not have to reach past.

So the two results are not in tension. They are one fact about coincidence, of the kind a population of patterns drawn at random keeps producing: coincidences destroy distinctions, and a machine that needs distinctions is hurt by exactly what helps a machine that needs uniformity.

What the number means for the choice question #

The machine that may choose concludes that the loss in these essays is due to being forced rather than to the atom — three restricted machines with three unrelated failure modes, and one unrestricted machine with none. The verdict stands and this essay puts a size on it.

“Not forced” is not one condition. It is a sequence of conditions, and what the measurement says is that the essays near the bottom buy most of the freedom: one layer to two is the step that multiplies the reach by fifteen, and the steps after it are progressively smaller. The freedom the fourth of these essays’ machine needed was mostly the freedom to hold two layers instead of one.

That reframes the atom question rather than settling it. A machine’s atom is what it does in one move, and the some-layers machine’s atom was described as “any block reaching an edge” — one operation with a free parameter in it. Read as a family of machines, the parameter is the thing being spent, and the interesting statement is not that an unrestricted machine loses nothing but that a machine restricted to two layers loses far less than the restriction to one suggests.

There is also a warning in the even column. A parameter that always takes the value $n-1$ looks like a parameter that is really $n$, and it is not: the whole-pile column is empty on every row of both tables. A measurement that had stopped at “the some-layers machine is complete” would have carried the whole-pile move as part of the explanation, and it is never used.

What is enumerated #

The walk is the same exhaustive one, with the block depth passed through to the move generator. Every legal move of the restricted machine is tried from the flat strip, every legal move from each result, and a branch with no move left is recorded if the strip is fully folded and the sequence has respected the marking.

Two filters are load-bearing and were needed before any of the three machines could be measured. A branch that runs out of moves without finishing is the machine getting stuck, and recording it would report failure as coverage. A sequence that finishes flat having folded a mountain where the marking says valley has folded a different strip, and is discarded.

And a state reached twice by different sequences is expanded once. Without that the deeper machines do not finish: at depth $d$ the branching factor is $2d$ at every fold line, which on a six-segment strip at full depth is a tree nothing walks.

The depths reported are the smallest complete ones, found by sweeping from one upward rather than by predicting. The claim that no strip needs the whole pile is a claim about the last column of a table every cell of which was computed, and the figure is required to refuse any row whose completeness arrived at its own segment count.

What it does not show #

No property of an uneven strip measured here predicts its depth. The five run 2, 2, 3, 4 and 5, against segment counts of 4, 5, 5, 5 and 6 and state counts of 8, 16, 12, 24 and 48. The strip with sixteen states needs less depth than the strip with twelve, so the repertoire’s size is not it; two of the five have a repeated segment length and they need two and three; and the one with no two segments alike needs four. Whatever decides it is a property of how the segments land on one another when folded rather than of the list of lengths, which is the same distinction the counting of foldings runs into when it tries to read a strip’s difficulty off its markings.

Nine strips, and four of them evenly creased. The $n-1$ result for even strips holds at four, five and six segments, which is three data points of a pattern. Nothing here proves it continues, and a completeness that held at four sizes and stopped at the fifth is exactly what this subject has already been caught by once.

The depth is a depth, not a count of layers moved. A block of at most three layers may contain one piece or three, depending on how much of the pile the fold line actually crosses, so the parameter bounds what the machine may hold rather than what it moves. That is the right reading of a machine model — a hand can grip a certain thickness — and it is not a count of paper.

Nothing is measured about sequence length. A machine at depth two that reaches every state may need many more folds to reach some of them than a machine at depth five does, and the cost of a search is not the size of its answer. What is counted here is what is reachable.

The four states at depth one are four 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 two records of one object. Counted as objects the machine at depth one reaches two, and every number in these tables would halve without changing a comparison in them.

And the block still has to reach an edge of the pile. A machine allowed to take three layers out of the middle is a different machine again, and it is not one this model contains — the restriction to blocks that reach the top or the bottom is what makes a move a fold rather than a shuffle.

Still open: where the sequence length goes #

The reach is now measured against depth and the cost is not, and the two are likely to trade against each other in a way the reach table cannot show.

A shallower machine that reaches a state must reach it by a longer sequence, because each of its moves does less, and how much longer is a number this enumeration already walks past. The shortest sequence to each state, tabulated against the depth the machine is allowed, would say whether the depth-two machine’s completeness on an uneven strip is a practical completeness or one that takes a hundred folds to use. The measurement is a breadth-first walk instead of a depth-first one, on the same tree.

And the even strips’ $n-1$ wants an argument rather than three data points. The reason offered above — that on an even strip every layer lies over every other, so reaching a particular arrangement means reaching past the ones being left — predicts $n-1$ rather than merely something large, and a proof would say whether the last layer is unreachable for the reason given or for a different one. A strip of seven equal stamps would be the next data point and is within reach of the same walk.

Sideways from here, the parameter suggests one for the other machines too. The crimping machine folds two adjacent creases as a single motion, and two is as arbitrary a number as one is: a machine that folds $k$ adjacent creases at once is a second family with a dial, whose bottom essay is the machine that folds one line at a time. Whether its reach rises the same way — two thresholds and a plateau — or smoothly, is a measurement the same walk would make.

The habit worth carrying is about restrictions with a number in them. When two models differ by a restriction, look for the parameter the restriction is an extreme value of, and measure along it. Three machines compared pairwise give a lattice of who loses what; one machine with a dial gives a curve, and the curve says which part of the freedom was doing the work — which here is the first step of it, and never the last.