No term for an unequal beat

A bar of nine in Balkan practice is not nine of anything. It is four beats — three short and one long, 2+2+2+3 — and the essay that established that argued that the inequality sits at the beat level rather than inside it, which is a thing no single division of a bar produces.

This rung asks what the site’s own metre machinery does when handed one, and the answer is decided before any pattern is scored.

The candidate list is the problem #

The rules score a candidate metre by laying strong positions at a fixed period — every n steps from a starting point — and counting onsets on them, empty ones and strays.

That construction admits exactly the periods that divide the bar length. For a bar of nine those are 1, 3 and 9: nine beats, three beats of three, or one beat a bar. Four beats of unequal length is not a period, so it is not a candidate, so it cannot win and cannot lose.

The pattern given to that figure has onsets on exactly the four beats of the bar — steps 1, 3, 5 and 7 of nine. The model reads it as three beats of three, which puts a beat on step 7 where there is one and on step 4 where there is not, and calls the onsets at steps 3 and 5 strays.

Every part of that is defensible inside the model and every part of it is wrong about the music.

Move the long beat and it gets worse #

The other common ordering of the same bar is 2+2+3+2, with the long beat third rather than last. Nothing about the metre has changed; the beats are the same lengths in a different order.

The verdict flips from “three beats of three” to “nine equal beats” because moving one onset moved it off a strong position. So the model’s answer to what metre is this bar in is unstable under a rearrangement that every player of the repertoire regards as the same metre with a different accent pattern.

An unstable wrong answer is more informative than a stable one. It shows the model is not approximating the right answer badly; it is answering a different question.

Seven is prime, and that settles it #

The nine-step case at least gives the model something to choose between. A bar of seven does not.

Seven has no divisors between 1 and 7, so the candidate list has two entries and neither is a metre in any useful sense. One is the subdivision — calling every semiquaver a beat — and the other is the bar. A bar of 3+2+2 has three beats, and three is not available.

The same holds for five, eleven and thirteen, which between them cover most of the aksak repertoire. For a prime bar length the machinery has nothing to offer between the subdivision and the whole bar, and that is a fact about the integers rather than about the music.

It is worth noticing that the primes are where the repertoire lives. Five, seven, eleven and thirteen are the characteristic aksak bar lengths, and they are precisely the lengths at which an isochronous reading has nothing to say — a bar length with divisors gives the model somewhere to put a wrong answer, and a prime one leaves it with the subdivision. So the tradition and the model are picking out the same integers for opposite reasons: a prime length is what makes an additive metre necessary, since there is no equal division to fall back on, and it is what makes an isochronous model silent.

That is close to a definition and it is not one, because nine is the counterexample and it is this rung’s own headline case. Nine factorises perfectly well into three threes, the model offers that reading, and the tradition uses 2+2+2+3 anyway — so an available equal division is not taken up merely because it exists. What the primes establish is that an additive metre is forced at those lengths; nine shows it is also chosen at a length where it is not forced, which is a stronger claim about the tradition and a weaker one about the integers.

The nine case is therefore the one worth pointing the argument at. At seven the model has no right answer to offer and at nine it has one and offers the wrong one, and only the second is evidence about the rules rather than about arithmetic.

Both beats are beats, and the rate rule says so #

There is one piece of the machinery that survives this intact, and it is worth using because it settles a question that could otherwise be argued about indefinitely: whether an unequal beat is really a beat, or whether the short one is a subdivision and only the long one counts.

The rate rule has no opinion about periods and every opinion about durations, so it can be asked directly.

At a Bulgarian dance tempo of roughly 350 quavers a minute, the short beat is 342 milliseconds and the long is 513. The window runs from about 100 to about 2,000 and is preferred near 550. Both beats are inside it, and the long one is almost exactly at the preferred rate.

The whole bar, at 1,539 milliseconds, is near the slow edge and much less comfortable; the quaver, at 171, is fast and near the other. So of the four durations the bar contains, the two that the tradition calls beats are the two the rate rule likes best — which is a satisfying result and, since the rate rule was fitted to something else entirely, is not circular.

It also disposes of the alternative reading. If the short beat were a subdivision, its duration would be a subdivision’s duration, and 342 milliseconds is not: it is a beat, by the only quantitative criterion this site has for the word.

The weights fail for the same reason #

The trouble is not confined to induction. The syncopation measure needs a metrical weight for every position, and it gets one from a subdivision tree — a bar dividing into equal parts, each dividing again.

The trees available over nine steps are 3 × 3 and nine × 1. Neither is 2+2+2+3. So the weights that a syncopation count would be measured against are the weights of a metre the bar is not in.

Priced against the only tree of equal divisions that fits nine steps, two of the bar’s four beats — the ones at steps 3 and 5 — land on the weakest positions available, so the bar’s own beats read as syncopations against a metre it is not in. That is not a description of the music. It is a description of what happens when the only candidate on offer is the wrong shape.

That produces the worst kind of number: one that is computable, reproducible, and about the wrong object. A syncopation figure for an aksak bar computed this way says that the metre is heavily syncopated against itself, which is not a finding about Balkan rhythm but a finding about the tool.

What the model would need #

The repair is not small and it is worth saying what it is, because “the model can be extended” is the kind of sentence that hides the size of the extension.

A metre would have to be a list of beat lengths rather than a period — [2, 2, 2, 3] rather than “every 3” — with strong positions at the cumulative sums. The scoring would then work almost unchanged: count onsets on those positions, empty ones, strays.

That paragraph was written as a description of work to be done, and it is wrong about the work. A candidate in this machinery is not a number: it is a period and a list of strong positions, and the strong positions have never been required to be evenly spaced. An additive metre is period: 9 with the strong list [0, 2, 4, 6], which is the cumulative sums of 2, 2, 2 and 3. Nothing had to be built.

So the scoring was never the problem; the candidate list was. Every failure the sections above document is a failure to offer the model the right reading, not a failure to price it once offered, and that is a smaller and more embarrassing defect than the one this essay set out to describe. What it costs is the candidate list, and counting it rather than estimating it removes the objection. Isochronous candidates over a bar of n number the divisors of n. Unequal candidates number the compositions of n into parts of 2 and 3:

Five for nine, not nine, and thirty-seven for sixteen, not several hundred. The three-hundred figure is not reached until a bar of twenty-four, which no aksak metre is. At every bar length the repertoire actually uses the additive list is between two and sixteen entries — a shorter list than most of the candidate sets this ladder scores routinely — and it is exhaustively scoreable without any heuristic at all.

The seven-step case is the sharpest. Three additive candidates exist: 3+2+2, 2+3+2 and 2+2+3, which between them are the entire aksak seven repertoire. The right answer is one of three, and the machinery offers two of which neither is it. Extending the candidate list from two to three would have solved the case this rung is written about.

So the reason nobody did it is not the combinatorics, and it is worth saying what it is instead. The scoring would have to prefer among the additive candidates on some ground, and the preference rules have none: 2+2+2+3 and 3+2+2+2 place their strong positions differently and the rules can only count onsets on them, which is exactly the computation that flips its verdict when one onset moves. The extension makes the right answer available; it does not make the rules able to choose it, and a candidate list containing the truth alongside four near-misses that score similarly is a different failure from one that does not contain it.

That is a better account of the gap than a combinatorial one, and it points at the same repair the rest of this ladder points at — a model with a state, which keeps a metre once found rather than re-deriving it from each bar.

Counting is a composition, not a period #

There is a piece of evidence about what practitioners hold, and it is in what they say out loud.

The standard way of teaching these metres — in Bulgaria, in Turkey, and in every workshop that has ever imported them — is to count the groups: one-two, one-two, one-two, one-two-three. Not one to nine. Not one-two-three three times.

That is a list of beat lengths recited aloud, which is precisely the data structure the model does not have. What is being taught is a composition of nine into parts of two and three, and the fact that it can be said in a sentence is fairly strong evidence that it is what is being held.

The same distinction shows up one level higher, where this site has already met it. A four-bar group is a bar whose beats are bars, and hypermetre is routinely irregular — a five-bar phrase is three bars plus two, not five equal ones — which is an unequal composition at the level above rather than below. The machinery for hypermetre on this site has exactly the same limitation, for exactly the same reason, and nobody noticed because irregular hypermetre is usually described in prose.

What the notation does instead #

Western notation handles this by writing the bar length and leaving the grouping to a beaming convention, an added time signature such as 2+2+2+3, or nothing.

Drawn as a cycle the bar makes the point in one glance: its four onsets form an irregular quadrilateral rather than a square, so the inequality is visible immediately. In a time signature of 9/8 it is not written down at all — the signature counts the units and says nothing about how they group, which is the same gap the candidate list had and the notation has never closed.

“9/8” is the same signature used for three beats of three, and the two metres share nothing but a bar length. That the notation cannot distinguish them without an extra convention is the same shortfall as the model’s, arriving by a different route: both were built for a repertoire in which beats are equal, and both express an unequal bar by describing its subdivision and hoping.

What distinguishes an additive metre from a pattern of unequal spacing inside an equal one is nothing in the drawing of either — it is which of them a listener is counting, which is exactly the choice the candidate list either offers or withholds.

That last figure states the difficulty precisely. A tresillo and a 3+3+2 aksak bar have the same onsets — the even-distribution algorithm produces both, from the same two numbers, with no way of knowing which it has made. What differs is whether a listener counts four equal beats with the onsets falling awkwardly against them, or three unequal beats each of which is a beat. The onsets are identical and the metres are not, and no analysis of the onsets alone can tell them apart.

Whose music, and how much of it #

Additive metres are the norm across a wide band — Balkan, Turkish, Greek, much of the Middle East, parts of South Asia — and they are not exotic in the repertoire the rule set was fitted to either: a good deal of twentieth-century Western concert music uses them deliberately.

What varies is whether the unequal beats are felt as beats. In Bulgarian dance practice they are, unambiguously: the dancers’ steps are on them, the long beat carries a different step from the short ones, and a musician will name a rhythm by its grouping. In a Stravinsky bar of 5/8 the situation is less clear and is argued about.

So the claim here is narrow and it is enough. For at least one large repertoire, the object the model is supposed to find is not in the space the model searches. Whether some other repertoire’s unequal bars are really felt as unequal beats is a separate question that this rung takes no position on.

It is worth adding that this is the second time a rule set built for one repertoire has been found answering the wrong question about another. The bell-and-ensemble arrangement breaks it in a different direction — there the difficulty is that several groupings are in play at once and the model outputs one winner — and the two failures have a common root. A model that returns the metre of a passage is assuming that a passage has one, which is a property of a repertoire rather than of music.

What the picture cannot show #

It cannot show the long beat’s length. Every figure treats a bar of nine as nine equal steps with beats at 0, 2, 4 and 6. Measurements of Balkan performance find the long beat is not exactly 3/2 of the short one — the ratio drifts with tempo and with the ensemble, much as a swung pair does — so the even nine-step grid is itself an idealisation.

It cannot show the dance. The strongest evidence that these beats are beats is that people step on them, and a figure of onsets has no feet in it.

It cannot show what a listener from outside hears. Somebody raised on equal beats hearing an aksak bar for the first time is doing something, and it is probably closer to what the model does than to what a Bulgarian dancer does. That would make the model a decent account of an unfamiliar listener and a poor account of a competent one, which is an interesting thing for a model to be and is not tested here.

It cannot show the cycle’s other readings. Drawn as a circle a pattern has no beginning, and an aksak bar’s rotations are different metres with different names — 2+2+3+2 and 3+2+2+2 are distinct dances. Every figure here fixes a starting point.

And it cannot show that the model knows it has failed. Every score in every figure on this page is a number the model reports with the same confidence as any other. There is no output that means the right answer was not on the list, and there could not be, because the list is the model.

The ladder from here #

Two rungs have now found the same missing thing from two directions. The rules cannot keep a beat through a bar that does not mark it, because they have no state; and they cannot propose an unequal beat, because a candidate is a single number.

The first of those has a standard repair, and the last rung of this ladder is about it: a model whose metre is a period and a phase carried forward in time rather than a reading chosen afresh. It fixes the memory problem completely. It does not fix this one — an oscillator has one period too — which is worth knowing before it arrives.

Two things are left properly open. Whether an oscillator bank with two coupled periods, one for each beat length, would track an aksak bar is a real question with a real answer and it is not computed here. And whether the unequal beats are exactly 2:3 in performance is a measurement question that belongs with the milliseconds that make a groove rather than with a model of metre: the published measurements say they are not, by margins of the same order as swing, and every figure on this page has drawn them exact.