The tube a map makes

A map with both pairs of edges joined is a torus, which nobody can make out of paper and which has no bottom layer to count from. Joining one pair gives something else entirely: a tube, which anybody can make in ten seconds and which keeps most of what the enumeration needs.

It is the middle case, and it is the one the question is actually about for anything physical.

Making one #

Take a rectangle of paper and rule it into squares. Crease every line. Now roll it so that the left edge meets the right and tape them.

The result is a tube with creases running along it and round it. The creases that ran across the flat map now run round the tube; the ones that ran along it still run along it, from one open end to the other.

Push the two open ends towards each other. The tube collapses — sometimes.

The parity, in the hand #

Whether it collapses is decided by how many creases run along the tube.

A path drawn round the tube, at any height, crosses each of the lengthwise creases once. Crossing a crease exchanges which face of the paper is up. So the path returns the paper the other way up when the count is odd, and a sheet on which that happens has no flat folded state.

Two lengthwise creases flatten a tube, which is what happens when a drinking straw is pressed. Three do not: pushing two of them together forces a fourth crease to appear, and the fourth crease is the paper making the count even.

That is a condition the flat map does not have. Every rectangle of grid folds flat; half the tubes made from one do not.

What the tube keeps #

The reason the tube is the interesting middle is that it keeps the thing a torus loses.

Enumerating a map’s foldings works from the bottom: find the panel with nothing below it and build upward. The bottom panel lives at the rim, and a torus has no rim, so its layer order has no least element and the enumeration has nowhere to start.

A tube has two circles of rim — its two open ends — so it has a bottom panel, and the enumeration can start.

So the tube is a genuine map-folding problem: a sheet with a boundary, a well-defined layer order with a bottom, and an assignment to be chosen. It is a harder one, because the loop that cannot be shrunk adds a global condition and removes the free letters the rim was supplying.

Neither the strip nor the torus #

It is worth placing the tube against the two problems on either side of it.

The strip of stamps is one-dimensional: a sequence, folded onto itself, counted exactly to reasonable sizes.

The flat map is two-dimensional and open: the classical problem, unsolved, enumerated by brute force.

The tube is two-dimensional and closed in one direction: a parity condition, half the rim, a bottom layer, and no enumeration yet.

The torus is closed in both: two parity conditions, no rim, no bottom layer, and no enumeration that could be adapted.

Each step along that sequence removes something the previous problem’s methods relied on, and the tube removes the least.

What is measured #

The letter half, which is the half that transfers.

A two-period grid cell glued into a tube has ten free letters over six panels and settles in seven nodes of search. Cut out of the plane it has twelve letters over nine panels and settles in nine. Glued both ways it has eight over four and settles in six.

The counting half is not measured, because the enumeration this collection has starts from a bottom panel and counts a linear stack, and although a tube has a bottom panel, the arrangements it admits are constrained by the loop in a way the enumeration does not know about.

The experiment, in detail #

The parity is worth checking with paper, and the experiment takes four minutes and settles it.

Cut two rectangles of paper about twenty-four centimetres by twelve. On the first, rule three lines down the long direction, dividing the twenty-four into four equal strips, and crease them. On the second, rule two lines, dividing it into three, and crease those.

Roll each into a tube by bringing the two long edges together, and tape.

The first tube has four creases running along it: three ruled and one made by the tape. Push its ends together and it flattens, into a strip four layers thick, with two of the creases becoming the flattened edges.

The second has three creases along it: two ruled and the tape. Push and it does not flatten. It buckles somewhere, and the buckle is a fourth crease appearing where the arithmetic requires one.

Two things to notice. The tape counts — a seam is a line the flattened tube has to bend along whether or not anybody creased it, and leaving it out of the tally reverses every answer. And the failure is not local: every vertex of the three-crease tube is a perfectly ordinary vertex, and the refusal is a property of going all the way round.

Crossways creases, which are free #

The essay has been about the creases running along the tube, and the ones running round it deserve a sentence, because they behave entirely differently.

A crease running round the tube is a closed loop of crease. It does not affect the parity of any path round the tube, because such a path does not cross it — the two are parallel.

What it does is divide the tube into shorter tubes, stacked. Each of them has the same lengthwise crease count and therefore the same parity, so a tube that flattens flattens at every crossways crease and one that does not fails at all of them.

So the two families of crease on a tube play completely different roles: one is constrained by a parity and the other is not constrained at all. On a flat map they are symmetric, and gluing breaks the symmetry.

That asymmetry is the same one the Miura shows, where one direction has a parity condition and the other cannot have one at any size, and it is a general feature of gluing an anisotropic drawing.

What the letters cost #

The half that is measured is worth reporting properly, since it is the only number in the essay.

A two-period grid cell — four interior vertices, nine panels cut out — costs nine nodes of search cut out of the plane, seven glued into a tube either way, and six glued into a torus. Per panel that is one, one and a sixth, and one and a half.

A three-period cell costs sixteen nodes cut out and is refused in every glued form, since three is odd.

A four-period cell costs thirty-six cut out and eighteen glued both ways, at one node per panel and one and an eighth.

So the letters are cheap on all of them and the tube sits between the flat map and the torus in cost per panel, exactly as it does in every other count. That is the expected result and it is worth having because it says the letter half of the problem does not become interesting on a tube — whatever the tube’s difficulty is, it is in the ordering.

The tube nobody folds #

One honest note about scope. The tubes people manufacture are corrugated — Miura, Kresling, waterbomb — rather than gridded, and a grid tube is not a thing anybody makes.

The reason to use the grid here is that it is the map-folding problem’s own object. The classical question is about a rectangular grid, and asking it on a tube means gluing a rectangular grid, whatever anybody manufactures.

For the manufactured tubes the parity works the same way and the counts are different, and the Miura’s version has been measured: one crease per period across, so half the column counts refuse, and four per period along, so that direction never does.

A designer wanting the rule for their own pattern needs one number: how many creases a path round the tube crosses. Everything else follows.

Where the tube’s difficulty is #

Enumerating a tube’s foldings is harder than enumerating a flat map’s, and it is worth being specific about which part is harder.

The assignment is barely harder. There are fewer letters, since the gluing rejoins the creases the rim divided, and the vertex conditions are unchanged. The search costs about what it did.

The layer order is the problem. On a flat map, an ordering is a permutation of the panels along the stack, and the enumeration builds one from the bottom, adding panels and checking that no two pass through each other.

On a tube, the same enumeration builds an ordering — and then has to check that the ordering is consistent round the loop. A panel at one end of the row and its partner at the other are the same panel; their positions in the stack have to agree; and the enumeration has no way to know that until it has placed both.

So the constraint is not local and it is not checkable early. That is the classic shape of a hard enumeration: a global condition that can only be tested at the end of a long partial construction, which means the search does a great deal of work before discovering that it was wrong.

Why one loop is so much less trouble than two #

The tube and the torus differ by one identification and the difference in what they cost the enumeration is out of proportion to that.

A tube keeps two circles of rim, and a rim is where a stack has a bottom. So the enumeration’s basic move — find the panel with nothing below it — still works, and everything built on that move still works.

A torus keeps none, so the move fails at the first step, and every later step is built on it.

That is a discontinuity rather than a gradient. Half the rim buys a global constraint that makes the enumeration slower; the last half buys the loss of the concept the enumeration is written round.

The same discontinuity shows up in the search costs: the four sheets are within tens of per cent of each other below a certain size and a factor of twenty-six apart above it, and the two cylinders sit with the flat map rather than with the torus.

So the useful summary of the whole family is: a rim is what makes a stack countable, and having some rim is much closer to having all of it than to having none.

What a count would settle #

Supposing somebody wrote the enumeration, what would the number be worth?

The comparison against the flat map’s count at the same size would be the first thing this collection has that separates the map-folding problem’s difficulty into parts. If a tube of $n$ by $m$ has far fewer foldings than a flat map of the same size, the loop is doing a great deal of pruning, and the classical problem’s size is mostly slack. If it has comparably many, the loop is doing little and the difficulty is elsewhere.

Either answer is informative and neither is available. The flat map’s counts are known to about four by four; a tube’s are known for none.

That is a smaller gap than it sounds, since the enumeration is a matter of engineering rather than of ideas, and the parity halves the sizes worth attempting.

The sequence of objects #

It is worth listing the whole family once, since each member removes something from the last.

A strip of stamps. One dimension, two ends. Counted exactly, sequence known, no formula.

A ring of stamps. One dimension, no ends. A parity refuses the odd ones; the rest are counted up to a rotation of the layer stack, which is a different sequence.

A flat map. Two dimensions, four edges. The classical problem. Counted by brute force to small sizes.

A tube. Two dimensions, two edges. A parity refuses half the sizes; the rest have a bottom layer and a global constraint. Not counted.

A torus. Two dimensions, no edges. Two parities; no bottom layer at all. Not counted, and the enumeration would have to be rebuilt rather than extended.

Each step removes an edge and adds a condition, and the difficulty of the enumeration rises at each step for a reason that can be named. That is a more useful thing to have than any one of the counts.

Why the tube version is the useful one #

Three reasons, and they are practical rather than mathematical.

It exists. Folded tubes are manufactured, in quantity, for deployable structures and packaging. A parity condition on how many creases run along one is a design rule.

It is foldable in the hand. Everything above can be checked with a sheet of paper and a piece of tape in a couple of minutes, which is not true of a torus.

It is the smallest change from the classical problem. One pair of edges joined, one loop added, half the rim kept. If the classical problem’s difficulty is going to be located by varying the object, this is the first variation to try.

An old problem’s assumptions, listed #

Since the essay is an exercise in varying one of the classical problem’s assumptions, it is worth listing the others, because each is a variation somebody could make.

The sheet is a rectangle. Varied here, to a tube and a torus.

The creases are on a square grid. Could be varied to a triangular one, or to unequal spacings, and the counts would change.

Every line is creased. Could be varied by leaving some uncreased, which is a different problem again and is closer to what a real map has.

The folding is flat. Could be varied to partial folds, which is the rigid-folding question and has a completely different character.

The layers are unconstrained. Could be varied by requiring the folds to be made one at a time, which is simple foldability and is decidable where the general problem is not.

Of those five, the last two have substantial literatures, the third is barely studied, and the first two have not been varied at all before now.

That is a reasonable map of where the room is, and it says the sheet was not an obvious place to look — which is fair, since varying it needed machinery that had no reason to exist.

What a designer takes from it #

One rule, and it is not a rule anybody would guess.

A tube of corrugated or creased material collapses flat only if an even number of creases run along it. Not approximately, not usually: the odd ones have no flat state at all, and pushing them will produce an extra crease rather than a flat tube.

The rule is known in practice — people who make folded tubes arrive at even counts by trying — and it has not usually been connected to anything. It is the same rule that refuses a loop of paper with three creases and an odd cell of the grid on a torus, and all three are one statement about a path that cannot be shrunk.

What is left open #

The count. How many ways a tube of $n$ by $m$ squares folds flat is not known here and has not been attempted, and the obstacle is not conceptual.

A tube has a bottom panel, so the enumeration can start. What it lacks is a way to enforce the loop: an arrangement of layers that is perfectly valid locally can fail to close round the tube, and checking that is a global condition the enumeration would have to carry through every partial arrangement.

That is a real piece of work and it is the same piece of work the consistency test on a glued sheet needed — a relation that returns to its own panel one cell over is not a contradiction, and the enumeration would have to know it.

Recorded as owed, with the parity done and the letters measured.