Two stacks shuffled in glued pairs

A line in the columns, a binomial in the rows counted the folded states of Miura patches far past the size any listing could reach, by building each state from the pairs of panels that actually overlap and never ordering two that do not. The counts followed one formula on every patch reached. A patch $c$ columns by $r$ rows has $1 + (c - 2)(B_r - 1)$ folded states, where $B_r$ is the count at three columns, and the three-column counts were the central binomial coefficients: 2, 3, 6, 10, 20, 35, 70, 126 and 252 for one to nine rows.

Half of that was explained. Why the count grows by the same amount with every column could be watched happening, one state branching at the end of the strip each time. The binomial could not. The essay said what an explanation would have to look like — two stacks of $r$ panels merged into one, under some rule about where the merge may switch sides — and recorded that no such rule had been found in the stacks. The counts were offered as measured, from one row to nine, without exception and without a reason.

The rule is in the stacks, and it is simpler than the shape of the number suggested. It is a card shuffle in which some cards are glued together in pairs, and the glue has a location that can be drawn.

Every stack is the same three things #

Laid out side by side, the six states of the three-row patch have a structure that shows before any rule is stated. In every one, the middle column is at the bottom, its panels in row order with the first row lowest. Above it, every panel belongs to one of the two end columns. And each end column, read upward, takes its panels in reverse row order: the last row first, the first row last, as an accordion lies when it is closed from the far end.

So a state is not a free arrangement of nine panels. The middle column’s three are fixed. The left column’s three come in a fixed order and the right column’s three come in a fixed order. What varies between states is only how the two end columns are interleaved — which is exactly the operation of riffling two halves of a deck of cards together, where each half keeps its own order and the shuffle decides only which half each card comes from.

That much was guessed in the earlier essay and it holds in every state counted, from one row to nine: 524 states in all, each checked panel by panel. But it does not yet give the binomial. Two stacks of three cards can be riffled together in twenty ways, and the three-row patch has six states. Something forbids fourteen of the shuffles.

The block words under each state say what. In the three-row patch the left column’s first two panels — its first and second rows — are always adjacent in the stack, and so are the right column’s second and third. They behave like single cards. With two of the left column’s three panels glued and two of the right column’s three glued, each column is a deck of two cards, and two decks of two can be riffled together in exactly six ways. Every one of the six occurs, once.

A weld is a crease the other column lies across #

Two panels of a column are adjacent in the stack when nothing may come between them, and the subject has a name for the rule that says so. Which layer goes on top stated the conditions every folded stack must keep, and one of them is the taco-tortilla rule: if a crease joins two panels, and a third panel lies flat across the place where that crease is folded, the third panel cannot sit between the two panels the crease joins. It would have to pass through the fold to get there.

So the question of which pairs are glued is a question about where each column’s creases land in the folded patch, and which panels cover those places. The answer can be drawn.

Folded, every panel of the left column lies on one parallelogram, and every panel of the right column on another of the same shape, offset from the first along its length. The two overlap in a band. Each column is an accordion of its rows, folded back and forth along the straight creases between them, so its row creases land alternately along the parallelogram’s two long edges: the first along one edge, the second along the other, the third back on the first.

One long edge of each parallelogram lies inside the other column’s footprint, and one lies outside it. The left column’s upper edge crosses the band and lies over the right column’s paper; its lower edge lies beyond the right column altogether. The right column’s lower edge lies over the left column’s paper and its upper edge beyond. A crease that folds along an edge inside the other column’s footprint has the other column’s panels lying across its fold, and by the taco-tortilla rule none of them may come between the two panels it joins. Those two panels are welded. A crease along the outer edge has nothing of the other column across it, and the other column’s panels are free to slide between its two panels.

Both columns start their accordions on the same side. So the left column’s first, third and fifth creases fold along its inner edge and weld, and its second, fourth and sixth fold along its outer edge and do not. The right column is exactly out of step: its first crease folds along its outer edge, its second along its inner one. At every row crease, exactly one of the two end columns is welded, and which one alternates as the rows go up.

The arithmetic of glued cards #

That is the whole of the rule, and the count follows from it in two lines.

A patch $r$ rows high has $r - 1$ row creases, and each one welds exactly one pair of panels in exactly one end column. The two end columns hold $2r$ panels between them, and every weld turns two of them into one block, so the welds leave $2r - (r - 1) = r + 1$ blocks. The left column is welded at its odd creases and the right at its even ones, which divides the blocks as evenly as $r + 1$ can be divided: $\lfloor (r + 1)/2 \rfloor$ in one column and the rest in the other.

The blocks of each column keep their order, and the blocks of the two columns interleave freely. The number of ways to interleave two sequences of $k$ and $m$ items, each keeping its own order, is the number of ways to choose which $k$ of the $k + m$ positions the first sequence takes:

$$N(3, r) = \binom{r + 1}{\lfloor (r + 1)/2 \rfloor}.$$

The table sets the account beside the counts at every height that was measured. At one row there is no crease and nothing is welded: two single cards, two orders. At two rows the one crease welds the left column, which becomes a single block against the right column’s two, and there are three ways to place one block among three positions. At four rows the left column is welded at its first and third creases and the right at its second, two blocks against three, and there are ten interleavings; at nine rows, five against five, and 252. At every height the interleavings are the states, and the agreement is not only in number. Every state read panel by panel is the middle column followed by an interleaving of exactly these blocks, no interleaving occurs twice, and every interleaving occurs.

So the central binomial coefficient is not a coincidence of small numbers and not a hint at some deeper ballot problem. It is the arithmetic of a shuffle in which the welds take away one degree of freedom per row crease, alternately from each side.

Ten stacks at four rows #

The four-row patch is large enough for the welds to show their alternation, and small enough to see every state. The left column’s four panels come as two welded pairs — its first and second rows, its third and fourth — and the right column’s as a single panel, a welded pair and a single panel: its fourth row alone, its third and second together, its first alone. That is five blocks, two of one colour and three of the other, and $\binom{5}{2} = 10$.

Reading the stacks from the middle column upward, the welded pairs stay together through every state and the single panels go anywhere their column’s order allows. A right-column single can lie below every left-column panel, between the two left pairs, or above them both. The right column’s welded pair can do the same, but never splits around a left panel. Nothing else is decided. No state needs a second rule to explain it, which is what the count says and what the drawing shows.

At two rows the same structure is too small to recognise as a shuffle. The single crease welds the left column into one block against the right column’s two single panels, and the three states are the three places that one block can go. Every larger patch is the same picture with more creases, and the alternation of the welds is what keeps the two columns’ block counts within one of each other.

What the welds take away #

The shuffle also says how far the count sits from the two numbers it might have been.

If nothing welded the end columns, two decks of $r$ cards each could be riffled in $\binom{2r}{r}$ ways — 48,620 at nine rows. The welds remove $r - 1$ cards’ worth of freedom, and the count falls to 252. The earlier essay compared the binomial with $2^r$, the count a patch would have if each row’s letter fold chose which end lay on top independently; the binomial runs just below that, at 252 against 512, and the earlier essay attributed the difference to the rows not being independent.

The shuffle says what the dependence is. A row’s choice is not a choice between two ends at all: it is where in a shared sequence one block goes, and the sequence is shared by every row. The square root in the binomial’s growth, which keeps each added row from quite doubling the count, is the cost of the blocks having to keep their order — the same cost that separates riffling two decks from choosing a top card for each pair independently.

Where the shuffle ends #

The welds depend on the band. A row crease of the left column welds because its fold lies inside the right column’s footprint, and that requires the two footprints to overlap. The offset between them is the cell’s width times the sine of twice the shear, $W \sin 2a$, and they overlap only while that is less than the cell’s height $H$.

That is the condition the earlier essay found by sweeping the cells’ width, now with a reason attached. Past $W \sin 2a = H$ the band is gone. Neither column’s creases lie inside the other column’s footprint, so nothing is welded — and nothing is interleaved either, because two columns whose panels do not overlap are not stacked against each other in any sense a folded state records. The four-row patch drops from ten states to one at exactly the width where the band closes.

The shuffle therefore does not degrade gradually as the band narrows. Any band of positive depth carries the whole binomial, since the welds and the freedoms both depend only on which side of an edge a crease lies, not on how far inside. The band at the shear these essays draw is 0.356 of a cell height; at 43 degrees it is a few thousandths, and the count is still ten. The dial on the footprint figure shows the band closing, and the welded and free edges never trade places on the way.

The slant that stacks in one depth measured the same overlap from the other side, as thickness: the pile under a point of a folded Miura is more than two layers deep exactly when columns two apart overlap. The band in the footprint figure is that region. Two essays asking how thick a folded Miura is and how many ways it can be stacked were describing one parallelogram of shared paper, and this one is about what lies inside it.

The choice the first column makes #

The earlier essay reduced its whole formula to one question: why the first step, from two columns to three, makes $B_r - 1$ new states. A patch two columns wide has a single folded state at every height. Adding a third column adds $B_r - 1$ states, and every column after that adds the same number again.

The shuffle answers the question as asked. A two-column patch is a middle column and one end column, with nothing to interleave, so it has one state. The third column brings a second end column to lie against the first, and the two can be shuffled in $\binom{r+1}{\lfloor (r+1)/2 \rfloor}$ ways, one of which is the stack the two-column patch already had with the new column laid on top. So the first step makes exactly $B_r - 1$ new states, and the slope of every line in the earlier table is accounted for.

What the shuffle does not yet explain is the earlier essay’s other observation, that every later column adds the same number, which it measured by watching the branching states as the strip grew. A fourth column is not a third deck shuffled into the first two, since its panels sit over the third column’s, not over the middle one’s, and its own welds come from a different pair of footprints. The constant slope is still measured rather than derived; the binomial it is a slope of is now derived.

What the stacks cannot say #

Why the middle column lies at the bottom. In every state of every three-column patch counted, it does, and the shuffle takes that as given. It is a fact about how the two end columns are lettered against the middle one — both end columns fold toward the same side of it — and a patch lettered differently, with one end column folding under, would have a different structure that this account does not cover.

Whether any of the states can be reached. Every state counted is a legal stack and none of them is a sequence of folds. A collapse that takes the patch from flat to folded passes through intermediate positions in which the panels have not yet landed, and whether a folder’s ordinary collapse produces one particular shuffle, all of them, or only some, is a question about motion. What a dashed line can say is about the notation for those sequences, and nothing here connects them.

And four columns. The formula for wider patches still rests on the measured constant step. The shuffle reading has been checked on three-column patches only, and nothing here says what four columns look like as interleaved blocks.

Paper with no thickness and exact parallelograms #

The patches are ideal: paper of no thickness, panels that are exact parallelograms, creases that are lines, and a folded state that is an ordering of flat panels. A real folded Miura has thickness, and with thickness a weld is not a combinatorial fact but a geometric one: a panel can fail to pass a fold because it would have to bend round it, not because a rule forbids it. Whether a thick patch keeps every interleaving the ideal one has is the kind of question the field is empty where it would say nothing raised about recording a folded state at all.

The cells are also the standard ones these essays have used throughout: square, sheared by about twenty degrees. The shuffle holds at every shear tested on square cells and fails only where the band closes; nothing here was measured on cells of different heights in different rows.

Read state by state, not inferred from the count #

The shuffle could have been a story fitted to the binomial — the numbers agree, so the cards must be glued. It was checked the other way round.

Every state was read panel by panel: its middle column required at the bottom in row order, each end column required to appear in reverse row order, and every pair of panels the welds glue required to be adjacent. Every state’s block word was required to be distinct, and the number of distinct words to equal the number of interleavings, so that every shuffle occurs and occurs once. And the welds themselves were not read off the stacks: they were taken from the taco-tortilla rules the folded geometry generates, before any state was grown, and the footprint figure requires every crease that lies inside the other column’s footprint to be welded and every crease outside it not. All of this holds at every height from one row to nine.

The folded states are grown from the overlapping pairs, by the method the earlier essay checked against the exhaustive listing sign for sign on every patch the listing can finish. One choice with eleven answers proposed recording a folded state as a short list of free choices; for a three-column patch, the list is now one shuffle of $r + 1$ blocks.

Still open: the shuffle with a third deck #

A fourth column is the next object. Folded, it lies over the third column, and its own row creases land along the edges of its own parallelogram. Its welds come from where those edges fall against the third column’s footprint and against the first’s, and whether its states are a second shuffle on top of the first — an interleaving of its blocks into the sequence the end columns already form — or something that cannot be written as a shuffle at all, is exactly what would turn the measured constant step into a derived one. The states of a four-column patch four rows high number nineteen, few enough to read one by one.

The second direction is the record. A file has no paper and the half no notation records found that every format the subject has records the crease pattern and leaves the layer order empty. For a three-column Miura the layer order is now a word of $r + 1$ letters with a fixed number of each, and any shuffle can be written down in about $r$ bits. Whether the same compression exists for the printed six-by-four Miura’s 37 states — whether they too are a shuffle of welded blocks, in some arrangement of columns — would say whether the layer-order field can be filled with something a reader can read.

Sideways, the same structure is likely to sit inside every pattern whose folded state is a pile of accordions lying against each other. The map counted from the layers counts flat maps, where every panel overlaps every other and nothing is welded by a footprint; where the exponent comes from asks how fast their states multiply. A map is the case where the band covers everything, and the shuffle is the case where it covers a strip, and the difference between them may be most of the difference in how their counts grow.

The habit worth carrying is about counts that look like famous numbers. When a count matches a classical sequence, look for the objects the sequence counts before looking for a deeper reason. The central binomial coefficient has a dozen interpretations, several of them sophisticated; the one that fitted here was the plainest, two decks shuffled, and the only work was finding which cards were glued.