The form a first hearing cannot have

Four rungs of this ladder have now measured the same six chord schemes four ways, and all four measurements share an assumption so quiet that stating it is most of the work.

Each of them was computed from the whole matrix. Every bar against every other bar includes bar 3 against bar 34, which is a comparison nobody could make while bar 3 was sounding. The boundary operator reads a window centred on each bar, and half of that window is in the future. The lag profile averages a diagonal that runs the length of the piece.

A listener has none of that. A listener has a growing prefix — bar 1, then bars 1 to 2, then bars 1 to 3 — and every judgement is made from the corner of the matrix that exists so far. So the question this rung asks is simple and the answer has two halves that could hardly be less alike.

The operator, run over only what has been heard #

The natural expectation, and the one this essay was written to confirm, was that running Foote’s checkerboard causally would lose boundaries. It does not lose any.

The reason is in the kernel and takes one line to state. Foote’s window is a square running from $c-w$ to $c+w-1$ in both directions, so the last cell it needs is the one at bar $c+w-1$. By the time bar $c+w-1$ has been heard, every number the operator wants is available, and the value it computes is exactly — to the last decimal place — the value it will compute at the end of the piece.

The boundary operator is not blind to anything. It is late, by the reach of its own kernel, and by nothing else.

That is the opposite of the result this rung was slated to produce, and it is a better one, because it makes the delay a quantity rather than a mystery. At a kernel width of four bars the delay is three bars. At 108 beats a minute in common time, three bars is 6.7 seconds.

The delay is the width, and the width was already the question #

Rung 2 established that the kernel width selects which level of the hierarchy gets reported: a narrow window finds the four-bar groups inside a blues chorus, a wide one finds the chorus boundaries. That parameter now acquires a second meaning.

So the trade is exact and it is not a defect of the method. A boundary at the scale of a section cannot be reported sooner than a section-sized window can close over it. Anything that claims to have found an eight-bar-scale boundary within a bar of its occurrence has either used information from before the boundary alone — which is a different and weaker operation — or is guessing.

Seven bars at 108 beats a minute is 15.6 seconds. That is well outside the two-to-eight-second window inside which a series of events is heard as one thing, which means the operator’s verdict on a section boundary arrives after the material that produced it has stopped being present and become memory. Three bars, on the other hand, is 6.7 seconds, which is inside that window at its wide end.

The tempo at which the verdict stops being present #

That comparison is made once, at one tempo, and it has a boundary in it worth solving for. A delay of w − 1 bars of common time at B beats a minute is 240(w − 1)/B seconds, so for each kernel width there is a tempo above which the verdict lands inside the present and below which it does not.

A section-scale kernel never reports inside the present. Two hundred and ten beats a minute is faster than almost anything in the repertoires these six schemes come from, so at every ordinary tempo an eight-bar window’s verdict arrives after the material that produced it has left the present entirely — at 108 it is 15.6 seconds, which is twice the widest estimate of the window.

A four-bar kernel sits on the boundary: it needs 90 beats a minute to stay inside eight seconds and 206 to stay inside the typical three and a half, so at ordinary tempos its verdict arrives at the far edge of the present or just past it. Only the two-bar kernel is comfortably inside at any tempo anybody plays.

So the exchange rate the section above states — one bar of delay per bar of kernel — has a threshold in it. The operator can report a boundary while its material is still present only at the smallest scale it can see, and every boundary at the scale of a phrase or a section is a verdict about something the listener has already stopped holding.

Two smaller things were verified rather than asserted while that was computed. The claim that the causal curve equals the full curve shifted by w − 1 to the last decimal place is exactly true: over four schemes at three widths each, the largest difference between the causal value at bar t and the full value at bar t − (w − 1) is zero, not merely small. And the delay really is w − 1 at every width tried, which is the statement that the Gaussian taper the operator uses does not extend the kernel’s reach past the square it is applied to.

The window that delay has to be compared against does not move with the tempo, and the bar counts inside it do. A listener’s present is somewhere between three and a half and eight seconds long whatever is being played, so at 160 beats a minute in four a bar is a second and a half and five of them fit inside eight seconds; at 60 a bar is four seconds and two of them fill it. A delay measured in bars is therefore not a delay in the terms the listener works in at all. The same three-bar lag is comfortably inside the present at a fast tempo and well outside it at a slow one, and the operator has no way to know which case it is in, because its input is a list of bars with no clock attached. That mismatch — a structural quantity counted in bars against a perceptual one measured in seconds — is what this rung inherits and what the next one takes up.

The other half does not survive at all #

The lag profile is a different kind of object and it fails causally in a way the operator does not.

The lag profile is computed from a self-similarity matrix, and looking at where in that matrix the answer lives is enough to settle the causal question without running anything. For a verse-and-chorus plan the whole of the evidence for sixteen bars being the period sits in the upper-right region — bars 1 to 16 read against bars 17 to 32 — and not one cell of that region exists until bar 17 has been heard. A listener at bar 12 holds the leading twelve-by-twelve corner of the matrix, which contains none of it: every cell that would say “this is a repeat of something sixteen bars ago” is a cell whose second index has not arrived.

That is a stronger failure than the operator’s. The local operator’s verdict is late — it arrives, correct, some number of bars after the boundary it describes. The lag profile’s evidence is absent, and stays absent until the piece has supplied the second half of every comparison it needs.

The strongest lag is a mean over a diagonal that runs the length of the piece, and a listener at bar $t$ has only the leading $t \times t$ corner. Computing the profile on that corner is a legitimate operation — it is the same three lines of arithmetic — and what comes out is an answer that changes.

The hero figure runs that computation at every point in a verse-and-chorus plan. The final answer is sixteen bars. It is revised ten times on the way there, and it is not reached for the last time until bar 32, which is the last bar of the piece. A listener who has heard the whole thing once has the answer at the moment the piece ends and not a bar before.

Three schemes, three ways of being unavailable #

That is the sharpest of the three cases, because the wrong answer is held for so long. Over the thirteen bars from 16 to 28 the estimate is something other than four for ten of them — 22 seconds at 108 beats a minute — and for eight of those ten it is eight bars. It is wrong for a good reason: at bar 17 the piece really does look like a scheme with an eight-bar period, because two eight-bar A sections have gone by and the bridge has only begun to contradict them.

That is the second scheme to settle only at its final bar, and the pair matters because they are as unlike each other as the collection allows — a thirty-two-bar popular plan of four sections and a sixteen-bar classical theme of two. The property is not a quirk of one encoding.

The blues settles earlier, at bar 24 of 36, which is two full choruses — the point at which the twelve-bar diagonal first has enough pairs to outweigh everything else. The rondo settles at bar 32 of 40, when the third refrain confirms what the second suggested. The sixteen-bar period settles at bar 16, its last.

Across the five schemes with any structure at all, the answer arrives between 67 and 100 per cent of the way through, and it is revised between three and ten times before it does.

How many times a first hearing changes its mind #

The trajectories carry a second number that is easy to walk past, and it separates the schemes on something none of the earlier rungs could see.

The revision counts run from three for the sixteen-bar period to ten for the verse-and-chorus plan, and the ordering is not the ordering of any other measurement in this ladder. The verse-and-chorus plan has the sharpest single period of the six and the most unstable route to it, which sounds contradictory and is not: its material is four chords rotated, so almost every prefix of it supports some short period or other, and each new eight bars knocks the previous candidate over.

How hard a form is to find is a different quantity from how much repetition it contains, and this is the first figure in the ladder that can tell them apart. A scheme can be highly periodic and yet take the whole of itself to prove it.

There is a version of this that belongs to the listener rather than to the arithmetic, and the site has met it in three other fields. A metre is inferred rather than received, and the same onsets support several readings until one is committed to. A tonal hierarchy is built by counting what has been heard, so it is different at bar 4 and at bar 40. An expectation is a prediction that can be violated, which requires that it existed and was wrong. All three are quantities that a listener revises while the piece runs, and all three are usually drawn as though they were properties of the score.

The control, which knows itself immediately #

The control behaves the way a control should and its behaviour is worth reading carefully, because it inverts the essay’s headline.

Everywhere else in this ladder the ostinato has been the case that returns nothing: no blocks and no stripes, no boundaries at any kernel width, a completely flat lag profile. Here it is the case that returns its answer instantly and never changes it.

Both statements are the same statement. What makes a form unavailable to a first hearing is exactly the part of it that is worth having, and a piece with nothing to withhold withholds nothing. The four-bar cycle is knowable at bar 4 because there is nothing else it could turn out to be.

A boundary that becomes one afterwards #

There is a case that sits between the two halves of this essay and belongs to neither, and it is the one that explains why the slated version of this argument seemed obviously right.

Bar 9 of an AABA plan is a boundary, and no listener misses it. It is not a boundary because something changed there — nothing did. It is a boundary because bars 9 to 16 turn out to be a repeat of bars 1 to 8, and that fact is not established at bar 9. It is established somewhere around bar 11 or 12, when enough of the repeat has gone by to be sure it is one — which is the same four-bar threshold the next rung but two measures directly.

So there is a class of boundary that genuinely is retrospective, and this is it: the ones that are boundaries because something came back. The local operator never finds them, before or after. The lag profile finds that the material comes back but not where. Locating a return-boundary needs the stripe in the matrix, which is neither of the two methods this rung ran causally, and which needs the return to have happened.

The form of a piece is available on a first hearing in exactly one of its three parts. Changes of neighbourhood arrive a few bars late. Returns arrive when the return is far enough in to be recognised. The period arrives, if at all, near the end.

Whose music this is a claim about #

The six encodings are conventions with dates, and the delays computed above are delays for those conventions rather than for music in general.

The rondo plan is a classical one — the ABACA letter scheme is the shape of a great many finales from roughly 1770 to 1830, and the practice of putting the first episode in the dominant and the second in the relative minor is a convention of that repertoire and not a law. The thirty-two-bar AABA song form belongs to American popular song of roughly 1925 to 1955. The twelve-bar blues, as encoded, is the commonest of several variants in wide use across the twentieth century. The sixteen-bar period is a textbook object of the classical style, and the verse-and-chorus plan is a rough description of a great deal of popular song since about 1960.

That matters for one specific reason. A form that takes 90 per cent of itself to become identifiable is only a problem for music that gets one hearing, and almost none of this repertoire does. A blues chorus is the third of twelve; a song is heard on the radio for the fifth time; a rondo finale is preceded by a movement in the same key and by a listener who has heard fifty other rondos. The composers of every one of these plans were writing for an audience that already knew the plan, which is what a convention is for — and the key plan of a sonata-form movement is the same bargain at a larger scale.

So the honest form of the finding is not that listeners cannot hear form. It is that the information required to derive a form from the sound is not present until near the end, and listeners are not deriving it. They are recognising it, from a stock of plans learned elsewhere, which is a completely different operation and the one no figure in this ladder performs.

What this cannot show #

The prefix used here is the piece, and a listener’s prefix is almost never the piece. Somebody hearing the second chorus of a blues has heard the first, and somebody hearing a song for the fourth time has the whole matrix in memory and is not running any of this. What is modelled here is a genuinely first hearing of a piece that begins where the encoding begins — an idealisation, and the strongest one in the essay.

The model of memory is also the crudest possible: everything heard is available, perfectly, forever, and nothing else is. That is wrong in both directions and the next rung replaces it with something that at least has a time constant in it.

And the arithmetic here is arithmetic on a chord scheme. A real listener finding the form of a real piece uses register, orchestration, dynamics, lyrics and the fact that the singer has stopped, none of which is in a roman numeral. Every delay reported above is therefore an upper bound on the wrong quantity: the delay for this computation on this description, which a listener is not performing.

Where the ladder goes #

Two of this essay’s numbers were converted to seconds in passing and neither conversion was taken seriously. Three bars is 6.7 seconds at 108 beats a minute and 13.8 at 52; sixteen bars of holding a wrong period is 35 seconds or 74. The bar has been the unit of every measurement in this ladder and a bar is not a duration.

Converting the whole matrix to seconds and weighting it by a decay is the next rung, and it changes which of these schemes counts as the most repetitive — because a stripe at four bars’ lag and one at twenty-four are the same ink and are not the same experience.

Before that, one setting inside the encoding has been fixed at 0.18 in every figure above without being defended, and it decides whether a passage that comes back in another key counts as coming back at all. That parameter turns out not to have a good value, which is a more interesting result than a bad one.