A tone step does not need a satin

Every account of shading, in every design manual, builds the tone steps the same way. Take the satin on n ends, take its cosets — the satin shifted by one pick, by two picks, and so on — and add them one at a time. The rung this ladder starts from does exactly that, and it is right about everything that follows from it.

It is also wrong about what the construction needs, and the demonstration is a repeat that every weaver has used and no satin exists on.

There is no four-end satin and there is a four-end shading #

A satin on n ends is built from a move m, and the move has to be coprime to n — otherwise the line never visits every end — and has to be neither 1 nor n − 1, because those two give a twill. At four ends the numbers coprime to four are 1 and 3, and both are excluded. There is no four-end satin, and for the same reason there is no six-end satin either.

So the classical construction yields nothing at four ends and nothing at six. A designer who wanted four tones on four shafts, or six on six, would be told the repeat does not admit the thing the tones are made of.

The figure above is four tones on four shafts. Every step of it is one cloth, the tone at step k is exactly k/4, and no satin is anywhere in it.

What a tone step actually requires #

Go back to why the satin was there at all. A tone step has to have the same fraction of warp on the face everywhere, and “everywhere” is doing real work: if end three carries four marks and end four carries two, then the two ends are different tones and the cloth shows a warp stripe of contrast 1/n that nobody drew.

So the requirement is uniformity: every end carries exactly k marks and so does every pick. A satin coset satisfies it because a coset has exactly one mark in every end and every pick, and a union of k cosets therefore has exactly k. That is the whole of what the satin was contributing.

Anything with one mark in every end and every pick would do. A set like that is a permutation matrix, there are n! of them, and the satin’s cosets are n very particular ones.

The condition on a chain is stronger than that, though, and it is the condition that names the object. A shading has to be nested — each tone the previous one plus something — and the tone has to reach n − 1 over n, so the parts have to partition the whole repeat: n disjoint permutation matrices covering every intersection exactly once.

Number the parts and write each intersection’s number into the point paper, and the result is a Latin square: an n × n array in which every symbol appears once in every row and once in every column. That is the object a shading needs, and the satin’s cosets write one particular Latin square — the cyclic one, L[i][j] = i + j.

Which is König’s theorem read the other way #

There is a second route to the same statement and it says something the first does not.

Suppose only that a tone step is uniform: k marks in every end and every pick. Read the repeat as a bipartite graph — ends on one side, picks on the other, an edge wherever the warp is on the face — and uniformity says every vertex has degree exactly k. König’s edge-colouring theorem says a k-regular bipartite graph decomposes into k perfect matchings.

So a uniform tone step is automatically a union of k parts, each with one mark in every end and every pick, whether or not anybody built it that way. The decomposition is not a construction imposed on the problem; it is a consequence of the only requirement the problem has.

That is why the tone scale has exactly n − 1 steps and no others, and it is a much stronger statement than “the satin gives n − 1 of them”. There are no uniform tones between k and k + 1, not because nobody has found one, but because a degree is an integer.

How many there are, and at which quotient #

Counting them needs a decision about what counts as the same, and the decision is not the one the literature makes.

Renumbering the parts changes nothing at all. A part of a decomposition has no name, and a chain is an ordering of the parts, so a square and its symbol-relabelling admit exactly the same set of chains. That quotient is free.

Renumbering the picks is a different matter. Permuting the rows of a draft gives a different cloth with different floats, so the reduced Latin squares that the literature tabulates — first row and first column both in order — are one representative per row-ordering rather than one per decomposition. Reduced counts would undercount the shadings by a factor of (n − 1)!.

So the census here is over squares with the first row in order, which is L(n)/n!:

• four ends: 24 decompositions, from 576 Latin squares;
• five ends: 1,344, from 161,280;
• six ends: 1,128,960, from 812,851,200;
• eight ends, which is the commonest damask repeat: from 108,776,032,459,082,956,800.

The last of those is a twenty-digit number and it is quoted exactly because it can be: the totals come from the tabulated reduced counts, which are known up to eleven, and the arithmetic is done in exact integers rather than in doubles — a double stops being an integer at 9 × 10¹⁵ and returns a number ending in 960,000 for one that ends in 956,800.

Every one of them is cloth, which did not have to be true #

The site’s criterion is not a formality here. A union of parts has every end and every pick interlacing, which rules out a thread lying loose on one side — but interlacing everywhere is not the same as hanging together, and a draft can interlace at every thread and still fall into two cloths that slide apart.

Every tone step of every decomposition at four and five ends was built and run through the criterion: 336 steps at four ends, 40,320 at five. Not one of them separates.

That is a clean negative result and it is worth being explicit about. The criterion refuses the ends of the tone range — a tone of nought and a tone of one are not fabric at all, which is the bound on a shading’s contrast — and it refuses nothing in between, for any decomposition whatever. Uniformity is sufficient for integrity on a full repeat. Here the sufficiency is measured, at these orders; it holds at every order, by a double count of marks and gaps that the six-end census gives in full.

The four-end split, and the explanation that turns out to be wrong #

The twenty-four four-end decompositions do not behave alike, and the way they differ is the practical half of the whole business.

Four of them can put a plain weave at the midtone. Twenty cannot. The best any of the twenty reaches is a longest float of two at every intermediate tone; the four reach 3, 1, 3 — satin-like at the ends and plain weave in the middle, which is the shape the eight-end spread chain has and which the classical construction cannot produce at four ends because it cannot produce anything at four ends.

The obvious explanation is algebraic, and it is wrong. Four of the twenty-four squares are the multiplication tables of groups — the cyclic group of order four and the Klein four-group, in various labellings — and four reach a plain midtone, and they are not the same four. Two squares are both; two group tables reach no plain weave and two plain-reaching squares are not group tables at all.

So being a group’s table is not the criterion, and the property that decides it is not one this collection has found. What can be said is that the split exists, that it is four and twenty, and that a designer choosing a four-end shading is choosing between two genuinely different tone ramps with no name to ask for either by.

At five ends the satin’s own square is very good and not unique #

Five ends admits a satin — the move can be two or three — so the classical construction works there, and it can be asked how well.

The best profile any five-end decomposition reaches is 4, 2, 2, 4 — a longest float of four at the two extremes, which is forced, and two at both midtones. Fifty-four of the 1,344 reach it, and the cyclic square is one of them.

That is a real endorsement of the classical construction and a limited one. The satin’s decomposition is in the best class; it is not alone there, and it is one square in fifty-four rather than the unique answer the literature’s silence implies. Four hundred of the 1,344 cannot better 4, 3, 3, 4, so a designer picking a decomposition at random would be a full float length worse off nearly a third of the time.

No five-end chain reaches a plain weave at any tone, and the reason is not about decompositions: five is odd and a plain weave needs an even repeat.

Six ends, where the argument bites hardest #

Six is the case worth having, because six shafts is an ordinary loom and six-end shading is a thing a designer might actually want.

The enumeration here stops before six. There are 1,128,960 decompositions, and building each one’s sixty-two tone steps as drafts and walking its 720 orderings, with the whole list held at once, is tens of minutes and a gigabyte. So six-end squares are built by name rather than swept, and two of them make the point. A census that measures each square and lets it go does reach six ends, in seconds, and what it finds is that 2,816 squares hold every middle tone to a float of two and that none of them can do it through a plain weave.

The cyclic square — the one a 1/5 twill’s cosets write, since no satin will — reaches 5, 2, 2, 2, 5: a longest float of five at the two extremes, which the repeat forces, and two at all three midtones. It can also reach a plain weave at the midtone, at the cost of threes either side.

The table of the symmetric group on three letters reaches 5, 3, 2, 2, 5 and can reach no plain weave at all. It is a group table, exactly as the cyclic square is, so once again the algebra does not decide it.

So a six-end shading exists, and every account of shading says it does not. What it is built on is a twill rather than a satin, which means it carries a visible diagonal — and the diagonal is exactly what holds the tone ramp’s surface level, so the six-end shading nobody offers is the one with no relief in it.

What a four-end shading is actually for, which is not tone #

A construction that the manuals do not offer is worth being sceptical about, and the scepticism has an arithmetic form.

A shading on n ends reaches from 1/n to (n − 1)/n, because the ends of the tone range are not cloth. At four ends that is 0.25 to 0.75 — a contrast ratio of three, against an eight-end shading’s seven and a sixteen-end shading’s fifteen. And it is two intermediate tones between the extremes rather than six.

So a four-end shading is a poor greyscale and there is no arguing otherwise. Four tones spanning a contrast of three is not a halftone in any useful sense, and a designer who wants tonal modelling needs the repeat.

What four ends buys is the other axis. The four tones of the plain-reaching decompositions run from a 1/3 twill through a plain weave and back to a 3/1 twill, and those four cloths differ enormously in firmness, in how densely they can be set and in how they take a finish — while differing by only a factor of three in tone. That is a texture series with a tone gradient attached rather than a tone series, and it is exactly what a huckaback or a spot weave is doing.

It also costs four shafts and nothing else. Every tone step of a four-end shading needs four shafts, because a shaft is a distinct column and a four-end repeat has at most four of them, so the whole series runs on the plainest dobby there is. An eight-end shading needs eight, and a sixteen-end shading is a jacquard construction.

Which is the honest recommendation: at four ends the thing to design is the texture ramp, and the tone comes along with it whether it was wanted or not.

What was counted, and how #

The enumeration is a backtracking search with the first row fixed, and its output is checked against the tabulated counts rather than trusted: 24 at four ends, 1,344 at five, and the reduced counts 4 and 56 when the first column is fixed as well. An enumeration that agrees with an independently known count is an enumeration; one that does not is a bug.

Every tone step is built through the same guard, which counts the marks in every end and every pick of the result and refuses anything that is not uniform. That is the property the whole construction exists for, and checking it on the output rather than arguing it from the definition is what catches a decomposition that is not one.

The chain search is exhaustive. All n! orderings of the parts are walked against a table of every union, so “the best chain this square admits” is the best and not the best found. The tie-break — shortest worst float, then shortest total — is stated because it is a choice: at six ends the cyclic square’s best by that rule is 5, 2, 2, 2, 5 and it also admits 5, 3, 1, 3, 5, and which of those a designer wants is not a fact about the square.

Whether a square reaches a plain weave is asked of the table and not of the best chain. Every union of parts is the prefix of some ordering, so the question is whether any union is a plain weave — and an earlier version of this asked whether the chosen chain contained one, which reported the tie-break rule rather than the square.

And the guards are fed what they must refuse: an enumeration at six ends, a chain census at six ends, the table of the symmetric group asked for at five ends, a decomposition this collection cannot name, and a Latin square count on two ends. Each has to raise the collection’s own error type rather than merely raise something.

Where the model stops #

The census stops at five ends and the argument is about eight. Everything counted here is at four and five; six is two named squares here and a full census elsewhere, and eight is a number. The claim that the cyclic square is a good decomposition rather than the only one is therefore established where it can be established and assumed where it cannot, and an eight-end sweep is not going to happen — 2.7 × 10¹⁵ squares is not a census, it is a physical constant.

Nesting is imposed and not derived. A shading has to be a chain because a boundary between two tones would otherwise carry a line, and that argument is the rung below’s. Nothing here asks what a non-nested tone series would look like, and the answer is presumably that it looks like a design with lines in it.

Uniformity is treated as a hard requirement and it is a judgement. A tone step with one end carrying an extra mark is a stripe of contrast 1/n, which at eight ends is twelve and a half per cent and at thirty-two is three — so at a large enough repeat the requirement softens, and the whole Latin-square framing softens with it. Where that boundary is depends on the eye and is not computed.

And a decomposition is not a design. Choosing the square fixes what tones are available and how they can be ordered; it says nothing about which tones a figure uses, how large the regions are, or how the boundaries fall. A rectangular block is not half a rule is the same gap one level up.

Who found it, and when #

Latin squares are Euler’s, or at least his name is on them, and the decomposition of a regular bipartite graph into perfect matchings is König’s, from 1916. Neither had cloth in mind and both are in every combinatorics course.

The textile literature’s version is the coset construction, and it is old, correct and universally taught. What it does not say is that the coset is doing one job — uniformity — and that anything else doing that job would serve. The consequence is that four-end and six-end shadings are absent from the manuals, and absent as a construction rather than as a preference: there is no satin, so there is nothing to take cosets of, so the section stops.

The identification of the requirement with a Latin square, and the counts and the four-end split, are this collection’s. So is the negative result about groups, which is worth recording precisely because it is the explanation anybody would try first.

Where the ladder goes next #

Uniformity is what makes the tone scale exact, and it is expensive. A halftone screen of n × n dots gives n² + 1 greys and a weave of the same cell gives n − 1, which is a factor of n thrown away — and the whole of the loss is in this requirement rather than in anything about interlacing.