Every universe has one, or none

The previous census scanned all 349 shapes available in twelve semitones for four structural properties and found exactly one shape with all four. It is the diatonic set, and the result is genuine: one survivor out of 349 is a strong filter, and the essay that reported it was careful to say that a unique survivor is not by itself a reason to use it.

What that essay could not say is which half of the result belongs to the four properties and which half belongs to the twelve. The properties are stated in a way that never mentions the number twelve — a chain of one interval, two sizes of every step, as evenly spread as possible, every interval a different number of times — so there is nothing stopping the same census being run in a universe of any size at all. Nineteen, thirty-one and fifty-three are the universes this site has already had reason to build, and the machinery does not care whether anybody has ever played in them.

Twelve is not special. Twelve is a multiple of four.

What the sweep had to compute #

Enumerating every subset of a universe is hopeless past about nineteen notes: the number of shapes roughly doubles with each step, so a full census at 24 is eight million subsets and at 53 is a number with sixteen digits.

It does not have to be enumerated. One of the four properties is maximal evenness, and there are at most k maximally even shapes at each size k — the sets ⌊(ni + m)/k⌋ for m from 0 to k−1, most of which are rotations of one another. Any set with all four properties has that one, so the survivors are a subset of a list of about n²/2 candidates, and the other three properties need testing only on those.

That shortcut is worth exactly as much as the check on it. Run it at ten, twelve and fourteen against the full enumeration and the two must agree; at twelve it must return the one shape the earlier census reported, and nothing else. It does, and that is the negative test this figure rests on — a shortcut that is faster than the thing it replaces is only useful if it has been run against it and found identical.

Three cases, and twelve is in the middle one #

When n is a multiple of four there is exactly one survivor. It has n/2 + 1 notes and it is a chain of the interval n/2 − 1. Twelve gives seven notes generated by five semitones — the diatonic set, generated by the fourth, which is the fifth read the other way. Sixteen gives nine notes generated by seven. Twenty-four gives thirteen notes generated by eleven. Four gives three notes generated by one.

When n is odd there are two. They have (n−1)/2 and (n+1)/2 notes and both are chains of the interval 2 — a whole-tone chain, or as near to one as an odd universe allows. Nineteen gives a nine-note and a ten-note scale; fifty-three gives a twenty-six and a twenty-seven.

And when n is two more than a multiple of four there are none at all. Six, ten, fourteen, eighteen, twenty-two, twenty-six, thirty: not one shape of any size in any of them passes.

The near miss is always the same near miss #

An empty answer is only interesting if something came close, and something does — the same something every time.

The near miss at n = 10 is seven notes with the vector ⟨4, 4, 6, 5, 2⟩. At n = 14 it is nine notes with ⟨4, 5, 8, 4, 6, 7, 2⟩. At 18 it is eleven with ⟨4, 7, 8, 4, 10, 5, 6, 9, 2⟩. At 22, at 26, at 30: the same shape one size larger each time, generated, well formed, evenly spread — and failing deepness by a single repeated count, which is always the count four.

That is a regularity nobody was looking for, and half of it turns out to have a two-line proof.

The near miss always has (n+4)/2 notes, and the maximally even set of k notes in n steps is made of steps of one and two only. If there are a steps of one and b of two, then a + b = k and a + 2b = n, so a = 2k − n. Put k = (n+4)/2 and that is a = 4, in every universe, exactly.

So the near-miss set has exactly four unit steps, and its interval class 1 has a count of exactly four, because there is no other way to make a distance of one out of steps that are all one or two. The four is not a coincidence and it is not about parity: it is the number of small steps the set has, and it is four because the set is two notes larger than half the universe. Extending the computation to n = 42 confirms it, along with something the range this essay reports did not show — the other class that collides:

Writing n as 4m + 2, that class is 2⌈m/2⌉, rising by two every second universe. So the collision is between class 1 — which is pinned at four by the arithmetic above — and a class that walks steadily up the vector.

What remains open is only the second half: why some class always lands on four rather than stepping past it. That is now a much smaller question than the one this section started with, and it is the right shape for one — the claim is a computation over a range for one of the two facts and a theorem for the other, and the theorem is the one that looked most like a coincidence.

The same arithmetic says something about the survivors as well, which is worth one paragraph because it explains the three cases better than the sweep does. A maximally even set of k notes in n has 2k − n small steps, so a set with one small step needs k = (n+1)/2, which exists only when n is odd; a set with two needs k = (n+2)/2, which exists only when n is even. The two survivors in an odd universe and the one in a multiple of four are sets with one and two small steps respectively — the least irregularity a set of that size can have — and the empty universes are the ones where the size that would give the least irregularity is the size whose vector collides.

So all three cases in the sweep are statements about how many small steps the evenest set of a given size is allowed, which is one subtraction, and the parity of n decides which sizes are available. That is not a proof of the sweep’s pattern and it is the frame the pattern sits in, and it is the frame that makes twelve’s uniqueness a fact about twelve being a multiple of four rather than a fact about music.

What it does to “why seven” #

Nothing in the previous rung was wrong. In twelve semitones there is exactly one shape with all four properties and it is the diatonic set; that survives.

What changes is the interpretation, and it changes considerably.

The uniqueness is not evidence about music. It is the ordinary behaviour of this filter. The earlier rung was careful about this and said that a single survivor “does and does not” establish something; the sweep says which. It establishes that the four properties are jointly very restrictive, and it establishes nothing whatever about twelve. Every universe of the right shape has exactly one survivor, and there is nothing to explain about twelve having one — there would be something to explain if it did not.

The size of the survivor is forced. Seven of twelve is n/2 + 1 of n, so it is not seven because seven is a good number of notes; it is seven because twelve is twelve. Had European music settled on sixteen positions, this filter would have produced a nine-note scale, and had it settled on ten it would have produced nothing at all.

And the generator is forced with it. Five semitones of twelve is n/2 − 1 of n, which is the interval nearest to half the universe that is not half of it. That the fifth is both this arithmetic object and the second most consonant interval is the coincidence the whole subject rests on, and it is a real coincidence: consonance knows nothing about n and the filter knows nothing about the ear.

The pattern is arithmetic rather than musical, and it is complete: every universe has one survivor, two, or none, and which of the three is decided by n modulo four alone. Nineteen is odd and therefore has two; twelve is a multiple of four and has one; and the sizes with none are not near-misses but a residue class.

The two arguments for a division do not talk to each other #

Nineteen and thirty-one are proposed for a reason, and the reason is that their thirds are better: nineteen’s major third is 7.4 cents flat of the pure 5:4 against twelve’s 13.7 sharp, and thirty-one’s is within a cent. That is an argument about frequency ratios, made with an ear, and it has nothing to do with anything in this essay.

The census is an argument about counting, made with no ear at all. The two produce different shortlists and they do not overlap much.

It is worth being explicit about how little the two constrain each other, because the literature on alternative tunings tends to run them together. Twenty-four equal — quarter-tones, the division actually used in more repertoire than all the others put together — has thirds and fifths exactly as good as twelve’s and not one cent better, because it contains twelve. On the counting argument it has a single survivor of thirteen notes. Neither fact recommends it and it is used anyway, for a third reason: it is twelve with the gaps filled, and it can be played by two ordinary pianos.

A tuning system is chosen for at least three unrelated reasons and this ladder can only speak to one of them.

The generator series in another universe #

The previous rung showed that a chain of pure fifths has exactly two step sizes at 2, 3, 5, 7, 12, 17, 29 notes and no others. That machinery is not about twelve either, and running it inside nineteen shows what a familiar structure looks like when the universe changes underneath it.

That set is the diatonic scale, and it is worth seeing it arrive here as a survivor rather than as a starting point. Nothing in the census was told about Western music; it enumerated the subsets of twelve and applied four properties, and the thing that came out is the scale the properties were abstracted from — which is a check on the properties rather than a discovery about the scale.

So a nineteen-note universe still has a five-note and a seven-note scale generated by its own fifth, and they still have two sizes of every step. What it does not have is a deep one at those sizes, because deepness in nineteen needs nine notes rather than seven. The universe keeps the properties that are about the chain and loses the one that is about interval counts, and the diatonic-like object moves from seven notes to nine.

That is the cleanest statement of what the census is measuring. Two of the four properties travel with the generator and one travels with the size of the universe, and in twelve they happen to arrive at the same set.

Why deepness is doing all the work #

Three of the four properties are cheap and one is not.

Being a chain of one interval, having two sizes of every step, and being as evenly spread as possible are all satisfiable at almost any size in almost any universe. Deepness — every interval size occurring a different number of times — is the expensive one, and it carries a counting bound that decides everything above.

A universe of n has ⌊n/2⌋ interval sizes. Deepness requires ⌊n/2⌋ different non-negative counts, and the smallest that ⌊n/2⌋ different non-negative whole numbers can sum to is 0 + 1 + 2 + ⋯, which is ⌊n/2⌋(⌊n/2⌋ − 1)/2. A set of k notes has only k(k − 1)/2 intervals in total. So k must be at least about n/2 — a deep scale is necessarily about half its universe or more, whatever else is true of it.

That is why the survivors are all near half, why the twelve-note case gives seven rather than five, and why the pentatonic could never have qualified. The earlier essay said as much — five notes have ten intervals to spread over six sizes and 0+1+2+3+4+5 is fifteen — but stated it as a fact about the pentatonic. It is a fact about the property, it scales with n, and it means the filter can only ever select scales that are large fractions of their universe. In twenty-four it selects thirteen notes. In fifty-three it selects twenty-six.

It is also why the deep property was the last of the four to be found and the one with a name attached to a paper rather than to a tradition: it is the only one of the four that a musician could not have noticed by playing.

A filter that always returns about half the universe is not a filter for scales. It is a filter for something else, and it produced a scale at n = 12 because half of twelve is small enough to sing.

What the picture cannot show #

Nothing here is about counting onsets either, though it could have been. Maximal evenness is Bjorklund’s algorithm, and the sweep above is therefore also a statement about which cycle lengths admit an evenly spread rhythm with all four properties. Nobody has asked the rhythmic version of this question here and it costs nothing to ask it, which is the sort of thing that happens when two fields share a function.

Nothing here is about pitch. The universes are numbers of equally spaced positions, and whether the positions are 63 cents apart or 22 makes no difference to any count in this essay. A nineteen-note survivor of nine notes is a combinatorial object; whether anything in it is consonant is a question about ratios that this machinery never asks.

The sweep stops at thirty. The pattern is exact across twenty-seven universes and this essay does not claim it for all n, because it has not been proved here. Thirty-one and fifty-three were checked separately and behave as the rule predicts, which is corroboration and not proof.

The properties are a chosen four. They are the four the previous rung scanned, chosen because each is separately defensible and each is shared with some other shape. A different four would give a different sweep, and the regularity found here is a fact about this filter. What makes it worth reporting is that the filter was fixed before the generalisation was attempted rather than after.

And an equal division is an assumption. Every universe here is n equal steps. Traditions that do not divide anything equally — most of them, in the detail — have no n at all, and this entire ladder has nothing to say about them. Nor does it apply to a scale built on something other than the octave: the Bohlen–Pierce scale divides a 3:1 into thirteen, and thirteen is odd, so it has two survivors — of six and seven notes — and whether either is the scale anybody plays is a question about its spectrum rather than about its counts.

Whose music this is a claim about #

The nineteen- and thirty-one-note divisions were proposed by people who wanted better thirds, and they are used today by a small number of composers and instrument builders. This rung says something specific to them: an odd division has two of these objects and a division that is two more than a multiple of four has none, so a composer choosing between 19 and 22 is choosing between a universe with two structurally diatonic-like scales and one with none.

Whether that matters is not settled here and probably not settled anywhere. The four properties are what makes the diatonic set nameable and learnable, and there is a real argument that a universe without such a set is harder to write in. There is also a real argument that 22 is used precisely because its scales are not diatonic-like, which is a reason to want the empty answer.

What the census does is turn a preference into a checkable fact about a universe, before anybody has written a note in it.

The ladder from here #

Two rungs have now shown that the diatonic set’s structure is forced once the twelve is fixed. The next and last rung of this ladder asks what all of that structure cannot do, and the answer is sharp: not one of the four properties can tell which of the seven notes is home. Every one of them is invariant under rotation, so the census that distinguishes 349 shapes cannot distinguish a major scale from its own Aeolian mode — and everything a listener actually uses to find a tonic is outside it.