How many layers a draft can have

The integrity criterion returns a number rather than a verdict. One is an ordinary cloth, two is a double cloth, three is a triple cloth, and the whole point of asserting the number rather than asserting soundness is that two is a construction and not a mistake.

Which raises a question the site has been able to ask since its earliest essays and never has: how large can that number get? A draft is a grid of filled and empty squares, the criterion is a statement about a digraph, and there is nothing obviously stopping a small repeat from describing a great many cloths at once.

There is something stopping it, it is exact, and it is half the ends.

What the sweep says, counted #

The site has run one exhaustive enumeration since the beginning: every four-by-four draft in which each end and each pick interlaces at least once, which is 22,874 of the 65,536 there are. Every census here is a summary of it.

The integrity census has always reported one number off that sweep — that 144 drafts describe more than one cloth, rather under one per cent. What it has never reported is how many cloths those 144 describe, which is the question this essay needs and which nobody had asked.

Running it gives a shorter answer than expected. 22,730 drafts are one cloth. 144 are two. Not one is three.

The absence is the interesting half, and it is easy to under-read. A result of the form “nothing in the sample did such-and-such” is usually weak evidence: the sample may simply be too small, and this site has published exactly that caution about balanced drafts, where none of the ninety balanced four-by-fours separates and nothing here proves that it holds at larger sizes. The three-layer absence is not of that kind. It is not that three-layer drafts are rare at four ends; it is that a four-end repeat has nothing to make a third cloth out of.

Every cloth needs two ends and two picks #

The bound comes out of the criterion itself and takes a paragraph.

A component of the above-and-below digraph with more than one strand in it contains a cycle, because that is what strong connectivity means. A cycle in this digraph alternates ends and picks, since every edge runs between a warp end and a weft pick and never between two of a kind.

Can a cycle close in two steps? That would be an end above a pick and the same pick above the same end — one intersection asked to be both, which no draft can write. So the shortest cycle has four steps, and every component that is a cloth contains at least two ends and at least two picks.

A repeat of n ends and n picks has n of each to distribute among its cloths, no cloth may take fewer than two of either, and the cloths are disjoint. So there are at most ⌊n/2⌋ of them, and for a rectangular repeat the bound is whichever of ⌊ends/2⌋ and ⌊picks/2⌋ is smaller.

That is the whole proof. It has no free parameter, no material constant, no model of yarn in it, and it is the kind of statement this site exists to produce: a fact about cloth that follows from what cloth is rather than from what any particular cloth is made of.

One caution, because the criterion counts components and the bound is about cloths. A component of a single strand is not a cloth — it is a thread lying on the surface with nothing holding it down, which the sweep’s own filter removes. Counting those, a four-by-four matrix can have as many as eight components, all of them singletons, which is the all-warp-up matrix and is not a fabric at all. The bound is on the components that are fabrics.

A witness at four, at six, and at eight #

A bound is worth having only if something reaches it, and something does at every even size.

The construction generalises the double cloth with no new idea in it. Take k systems; give system number L the ends and the picks congruent to L; let each system interlace plainly with its own opposite number; and at every crossing between systems, let the upper one’s warp pass over the lower one’s weft. The result is k complete cloths in a repeat of 2k ends, and the criterion says k.

At four ends that is the double cloth. At six it is the triple cloth in the opening figure. At eight it is four cloths in a repeat a jacquard designer would think small.

Checked by exhaustion at four #

The proof is short enough to be trusted and short enough to be wrong in a way that is hard to see, so it is checked rather than believed.

Every four-by-four binary matrix — all 65,536, not merely the 22,874 the sweep keeps — is built and its components counted. The most cloths any of them holds is two, and ⌊4/2⌋ is two.

The check throws in an extra result nobody asked for. The matrices reaching the bound number exactly 144, and they are exactly the 144 the sweep already knew about: at this size a draft cannot describe two cloths and also have a thread lying loose. The filter that removes unwoven threads removes none of the two-cloth drafts, which is not obvious and is now asserted rather than assumed.

Six by six is not checked exhaustively and will not be. There are 2³⁶ matrices at that size, some 69 thousand million, and the argument does not need them: the proof is a proof, and the exhaustive check exists to catch an error in the proof rather than to establish it.

What was counted, and how #

Three separate enumerations produce the numbers on this page, and they are separate on purpose, because the failure mode of a count like this is a plausible wrong answer rather than a crash.

The census walks the sweep, builds each draft, runs the connectivity computation, and sorts each component by how many ends and how many picks it contains. Splitting the components that way is the part that is new here: every earlier summary of the sweep asked whether a draft separates, and this one asks into what.

The witness is built rather than searched for. A construction taking k systems and applying one rule at every crossing between them is written down, run through the same criterion every other figure on this site uses, and required to return k. A witness that returns anything else is refused while the figure is being drawn, which is the check that the drawing and the caption cannot come apart.

The exhaustive check is the third and is deliberately cruder than the other two: all 65,536 matrices, no filter, no cleverness, and a single number out of it. It is slow enough to be worth memoising and fast enough to run every time the figures are drawn, and its whole job is to disagree with the proof if the proof is wrong.

The three agree, and each of them would fail loudly on its own if it stopped agreeing. The counts in the figures are the output of running them while the pictures are drawn, not a recollection of an earlier run.

Three layers need six ends, whatever the fibre #

The statement worth carrying away is the contrapositive.

A weaver who wants three complete layers needs a repeat of at least six ends and six picks, and no yarn, sett, finish or ingenuity supplies a third layer on four. It is not a difficulty to be worked around; it is the same kind of impossibility as there being no satin on six ends, settled by counting before any thread is wound.

The number is small, which is exactly why it is worth stating carefully. Six ends is nothing — a plain-weave repeat is two and a five-end satin is five — so the ceiling almost never binds. A jacquard repeat of four hundred ends could in principle carry two hundred cloths, and nobody has ever wanted more than about five.

So the honest reading is: the ceiling is real, exact, and rarely the constraint. What is the constraint is in the same figures.

The float is what actually bites #

Stack k systems and the face warp passes over every pick of every layer beneath it. In the witness family the longest float is exactly 2k − 1: three at two layers, five at three, seven at four.

That is a much harder constraint than the ceiling, and it is one this site has already spent a ladder on. The float decides lustre, drape, snagging and abrasion; designing to a float limit is the constraint every jacquard designer works under; and a float of seven is a long float even in a satin meant to have them. The witness at eight ends reaches the ceiling and would make dreadful cloth.

Relieving it costs ends. Giving each layer a larger repeat of its own — three ends apiece rather than two — pushes the crossings apart but does not change the number of layers, and a six-end repeat carrying two layers of three sits comfortably below its own ceiling of three. The trade between how many layers and how good each one is happens well inside the bound, which is the ordinary relationship here between a decidable limit and a design decision: the arithmetic says what cannot exist and says nothing about what is worth making.

A float limit caps the layers, and the cap does not move #

The witness family has floats of 2k − 1, and it is tempting to read that as a property of the witness rather than of the problem — a crude construction that a cleverer designer would improve on. It is worth asking how far the cleverness can go, because the answer is another exact bound and it is the one that actually decides how many layers a fabric has.

Count the picks a face end passes over, round one whole repeat.

Let the face layer own r of the repeat’s picks. Every pick belonging to a lower layer passes under that end, by the definition of what makes the layers separate cloths — if a lower pick came over a face end anywhere, the two would be tied and the criterion would report one cloth rather than two. There are (k − 1)·r such picks in the repeat when the layers are equally weighted. Every one of them lies inside one of the face end’s r own-pick gaps.

So the face end’s floats, going round the repeat, are r runs containing (k − 1)·r foreign picks between them. Their average length is k − 1, and the longest is at least the average.

Every k-layer cloth has a float of at least k − 1 somewhere on its face, at every repeat size. Enlarging the repeat does not help: it adds own-picks and foreign picks in the same ratio, so the average gap is unchanged. That is the difference between this bound and the ⌊n/2⌋ one — the layer ceiling is bought off by making the repeat bigger, and the float floor is not.

Read the other way it is a design rule with no arithmetic in it. A float limit of F permits at most F + 1 layers. A furnishing jacquard worked to a limit of four floats to a repeat gets five layers and no more; an apparel cloth held to two gets three; and a filter fabric that will tolerate floats of nine could in principle carry ten cloths, which is roughly the number the industrial weavers of belting and preform actually build.

That is a much better explanation of the trade’s practice than the ceiling is. Nobody stops at five layers because six will not fit — six fits in twelve ends, which is a small repeat. They stop because the sixth layer puts a float of five on the face of a cloth whose face is the part anybody looks at.

Two things weaken the bound and both are visible in it. It assumes the layers are equally weighted; a backing cloth woven at half the face’s pick density contributes half as many foreign picks, so a light backing is cheaper in float than a full second cloth, which is exactly why backed constructions are commoner than true double cloths. And it counts the face end, which is the worst case: an end in the middle of a stack passes over the layers below it and under the layers above, so its runs are shorter. The bound bites on the surface and nowhere else — which is fortunate, because the surface is where a float costs anything.

Where the model stops #

The bound is on components, not on constructions. It says a repeat of n ends holds at most ⌊n/2⌋ separable cloths. It does not say those cloths are of equal weight, or usable, or that anybody could weave them: a shed with four systems in it is a mechanical problem the matrix has no opinion about.

Nothing here counts a spacer fabric correctly. Two cloths held apart by pile threads standing between them is a third thread system, and the quantity that makes it useful is the distance between the layers, which is not in the matrix at any repeat size. The criterion reports one cloth and is right and unhelpful.

Nothing here counts a preform’s thickness either. A three-dimensional woven preform has a top and a bottom, so it is not periodic through its thickness, and a bound on a repeat is a statement about a periodic object. The layer count of a preform is a specification rather than a ceiling.

And the census is of drafts, not of cloths. The 22,874 are writings, not fabrics; the sweep counts each asymmetric draft sixteen times, and the 144 will be a smaller number of cloths than 144. The ceiling itself is unaffected — it is a statement about a repeat of a given size and every writing of a cloth is a repeat of that size — but the proportions in the census are drafts and are labelled as such.

Who found it, and when #

The construction is old and the counting is not. Triple cloths are in the nineteenth-century pattern books, four- and five-layer constructions are standard in industrial fabrics for belting and for composite preforms, and nobody weaving them ever wondered whether a fourth layer might fit on four ends.

The formal apparatus arrived in the 1980s with the criterion itself, in Grünbaum and Shephard’s work on periodic fabrics — which supplied the observation that “how many cloths is this” is a well-posed question with a computable answer. Once it is a computable answer, “how large can it be” is an ordinary question about the computation, and the answer falls out of the shortest cycle in a bipartite digraph.

That the bound is half the ends is the sort of thing a subject only notices when it starts counting. The trade never needed it because the trade never approached it: five layers is a great many and a five-layer cloth has hundreds of ends in its repeat for entirely different reasons.

Where the ladder goes next #

This is the sixth rung of the integrity anchor and the first one that is a theorem about the criterion rather than an application of it. Below it are the base case, the braid and the nonwoven, where the criterion applies to structures it was not built for, the double cloth, where two is intended, and the boundary of the criterion itself.

The question this rung leaves open is the one the sweep cannot reach: whether the ceiling is attained at every size by good cloth rather than by the witness family, which reaches it with floats nobody wants. That needs an enumeration at six by six, which is 69 thousand million matrices and will need a smarter search than trying all of them.

The companion question belongs to the neighbouring anchor: given k layers, how few stitches join them and where those stitches may go — which has an exact answer, and it is one at any number of layers provided the stitch spans the outermost two.