A boundary costs the same wherever it is put

The number every ladder here has been quoting swept the one parameter the whole capacity model rests on and found the collection’s headline finding turns on it: at eleven cents of internal noise an octave holds about six nameable categories, and at six cents it holds twelve, which is the chromatic scale. Its closing sections listed the other things that computation holds fixed and could not price any of them.

The first item on that list is the one every figure in the ladder has quietly assumed since the boxes were first counted: the categories are all the same width. They divide the octave into $n$ equal bands of $1200/n$ cents, a stimulus falls anywhere in its own band, and the listener answers with the nearest centre.

No scale in the world is like that. The diatonic major has five tones and two semitones. The Turkish theory of Rast has steps of nine, eight, five, nine, nine, eight and five Holdrian commas — 204, 181 and 113 cents. If a narrow category is harder to name than a wide one, then the count is the wrong statistic and the narrowest step is the right one, and the traditions reorder.

They do not reorder #

Bhupali’s pentatonic — 200, 200, 300, 200, 300 — is named right on 96.343 per cent of trials at eleven cents of noise. Five exactly equal steps of 240 cents are named right on 96.343 per cent of trials. The measured slendro, at 231, 243, 243, 238 and 245, gives the same number again.

At seven degrees the same thing happens with a much wider spread of patterns. The tempered diatonic, with two 100-cent steps beside five 200-cent ones, gives 94.880 per cent. The Turkish Rast, whose narrowest step is 113 cents, gives 94.880. The Arabic Rast, which is 200, 150, 150, 200, 200, 150 and 150, gives 94.880. Seven equal steps of 171.4 cents give 94.880.

That is not a near miss dressed up. It is agreement to every digit the integration produces, on patterns whose narrowest step differs by a factor of three, and it holds at every value of the noise the ladder has ever drawn.

The obvious next move is to try to break it, since a null that has only been tested on scales anybody uses has only been tested on mild cases. Feed the model four categories at 0, 30, 600 and 900 cents — a thirty-cent sliver wedged between a five-hundred-and-seventy-cent slab and a three-hundred-cent one, which is not a scale and is not meant to be — and it returns 97.075 per cent, against 97.075 for four equal categories of three hundred. Squeeze the sliver to fifteen cents, which is narrower than the noise, and it returns 97.075 again.

The one arrangement that does move the number is three boundaries crowded together: 0, 10, 20 and 600 gives 97.256, which is better than equal. That is the first hint of where the argument’s limit is, and the last section returns to it. Everywhere a scale could plausibly live, the invariance is not approximate.

Why, and it is one line #

Take one category, with its centre at $c$ and its boundaries $a$ below and $b$ above. A stimulus at offset $x$ from the centre is named correctly when the noise keeps the estimate inside those boundaries, and averaging that over the band and multiplying by the band’s width — which is what averaging over the octave rather than over the categories amounts to — gives

$$\int_{-a}^{b}\!\left[\Phi\!\left(\tfrac{b-x}{\sigma}\right)-\Phi\!\left(\tfrac{-a-x}{\sigma}\right)\right]dx \;=\; \int_{0}^{a+b}\!\left[1-P(|X|>u)\right]du$$

which depends on $a$ and $b$ only through their sum. Summing over the categories, the octave’s total error is

$$\sum_i \int_{0}^{w_i} P(|X|>u)\,du \;\longrightarrow\; n\,\mathrm{E}|X| \;=\; n\,\sigma\sqrt{2/\pi}$$

as soon as every width $w_i$ is large compared with $\sigma$, because the integral has converged by then. Each boundary costs the mean absolute value of the noise and nothing else. There are $n$ of them however the octave is cut, so the error rate is $n\sigma\sqrt{2/\pi}$ over 1200 and the width pattern has dropped out of the arithmetic entirely.

The intuition is worth having as well as the algebra. Naming goes wrong only near a boundary, in a zone whose width is set by the noise; a wide category has the same two zones as a narrow one, and simply has more safe middle between them. Making one category narrower steals safe middle from itself and gives it to its neighbour. Nothing is lost, because the confusion zones have not moved.

What the search was computing #

The consequence for the ladder is larger than the null. Every capacity figure in it finds the nameable count by a loop — try every division of the octave, keep the largest that meets the criterion — and the loop has been evaluating a formula.

Setting the error rate equal to one minus the criterion and solving gives

$$n \;=\; \frac{(1-c)\,E}{\sigma\sqrt{2/\pi}} \;=\; \frac{1504\,(1-c)}{\sigma}\quad\text{for an octave}$$

with $E$ the equave in cents. At the laboratory noise of eleven cents and a criterion of ninety-five per cent that is 6.84, which floors to six — the collection’s headline, arrived at without integrating anything.

Two conventions that are one number #

The seventh rung noticed something it could state and not explain. Sweeping the noise moved the count, and so did moving the criterion, and it observed that “a criterion of ninety per cent at eleven cents and a criterion of ninety-five at six cents give thirteen and twelve, which are nearly the same answer from opposite corners.” It called the headline a surface over two conventions.

It is not a surface. The two conventions enter only through the ratio $(1-c)/\sigma$, exactly and everywhere, so the surface is a family of hyperbolae and every point on one of them is the same listener as far as this model is concerned. A listener with twice the noise and half the tolerance for error is not approximately the same as the original; the two are indistinguishable.

That collapses the audit’s uncertainty from two dimensions to one, and it makes the experiment the ladder is owed cheaper to specify. There is no need to fix a criterion before measuring a noise, because a measurement of either one at a stated other pins the ratio, and the ratio is the whole model.

It also says something uncomfortable about what has been quoted. Six categories per octave is not a fact about the ear; it is a fact about the ear divided by a convention somebody chose, and moving the convention from ninety-five per cent to ninety — which is a perfectly ordinary choice in an identification experiment — doubles the answer and takes the collection’s most-quoted finding from “fewer boxes than the notation has” to “one more than the notation has.”

What the equave does #

One term in the formula is not a property of the listener at all, and nothing in this ladder has ever varied it. The count is proportional to $E$, the interval being divided, because the confusion zones are a fixed number of cents and a larger interval simply has more room for them.

An octave gives 6.84 at the laboratory noise. A 3:1 — the equave of the scale built on a spectrum with no even partials — is 1902 cents, so it gives 10.84, and the same listener names ten of its degrees rather than six.

That is a prediction about an object this collection already carries. The Bohlen–Pierce scale divides its tritave into thirteen; the model says a listener with the laboratory noise can name ten of the thirteen at ninety-five per cent, against six of twelve in an octave. A tritave-based scale is not harder to learn than a chromatic one; it is easier per degree, and the reason is arithmetic rather than anything about thirds.

The other ceiling, which does depend on register #

There are two ceilings on how many notes an octave can hold and this ladder has always kept them apart. The one above is identification: how many categories can be named, which is what a scale degree requires. The other is resolution: how many differences can be told apart in a direct comparison, which is what the difference limen measures and which is a much larger number.

The contrast is worth drawing because it is the one place a register dependence enters. The limen moves by a factor of two and a half across the compass, so the resolution ceiling is genuinely different in the bass and the treble. The naming count is not: it has no frequency in it anywhere, only the noise, the criterion and the size of the equave.

That is a claim rather than an omission, and it can be wrong. If the identification noise inherits a floor from the discrimination limen, then $\sigma$ is register-dependent too, and the count would be smaller in the bass than in the treble. The arithmetic says the floor is nowhere near binding — eleven cents against a limen of 8.6 at the very bottom of the useful range and 3.4 in the middle of it — so the two would have to be within about thirty per cent of each other for the floor to show, and they are within thirty per cent only at the extreme bottom. A register dependence in naming is predicted by this model in the lowest octave of the piano and nowhere else, which is a small, specific and testable thing that follows from putting the two ceilings on one axis.

Where the closed form is not exact #

The derivation needs every category to be several times wider than the noise, and it is worth knowing what happens when they are not, because the answer is the opposite of the usual direction of error.

At a category width of four standard deviations the formula and the integration agree to three decimal places. At three, to two. At two they differ by 0.9 percentage points and at one by 16.6, and in both cases the formula is pessimistic: it charges a full confusion zone to every boundary, and when two boundaries are closer together than the noise a listener cannot independently fall over both of them.

The place that matters is where the ladder actually reads the formula, and there it is exact by a comfortable margin. Meeting a ninety-five per cent criterion means an error rate of five per cent, which means the categories are about sixteen standard deviations wide — a hundred and seventy-six cents at the laboratory noise. The correction is invisible at four widths and the criterion puts the answer at sixteen.

So the approximation bites only at accuracies nobody would call naming. At fifty-three equal divisions the formula says 61.2 per cent and the integration says 61.9, which is a difference between two ways of being unable to do the task.

What this leaves standing and what it takes away #

It takes away one candidate explanation the ladder had available. When a value is named as two different things or when a boundary refuses to move under a shifted prior, an appeal to unequal category widths was open — perhaps the Arabic Rast is harder to name than the diatonic scale because its 150-cent steps are narrower. The arithmetic says no: at seven degrees the two are the same task exactly, and every scale of the same cardinality is the same task.

It leaves the whole of the capacity result standing, in a stronger form. The count of nameable categories in an interval is $(1-c)E/(\sigma\sqrt{2/\pi})$ and depends on nothing else — not on where the degrees are, not on whether they are evenly spaced, not on whether the scale is a tradition’s or a random draw. Nineteen, thirty-one and fifty-three are all far beyond a trained listener’s naming capacity by the same margin they always were, and the reason temperament works is untouched, because that argument is about the width of one category rather than about how many there are.

And it sharpens what the ladder is owed. The experiment the seventh rung specified — an identification task run twice, in and out of a key — measures a difference of two noises. What the formula says is that the same experiment measures the whole model, because there is nothing else in it.

One more thing follows from the closed form being pessimistic rather than optimistic near the noise, and it is a small comfort. Every count the ladder has quoted is a floor. If two boundaries in a real scale are close enough for their confusion zones to overlap — which never happens in a scale but does happen in an ornament, a slide or a bend that passes through several degrees — the listener does better than the formula says, not worse. The model’s error is in the direction that does not flatter it.

Which computation produced the numbers #

Naming accuracy for an arbitrary set of centres puts the boundaries midway between adjacent centres with the octave wrapping, draws the stimulus uniformly within its own band, adds Gaussian noise of standard deviation $\sigma$, and answers with the nearest centre. Averaging over the octave rather than over the categories is what makes it reduce exactly to the equal-division function the ladder has been using, and the two are checked against each other on every equal division drawn.

The internal noise of eleven cents is the logistic scale this collection’s identification figures use, chosen so the drawn transition is the thirty cents identification studies of trained listeners report. It is the one number here taken from outside.

The traditions’ degrees are their own theory’s — cents for the Arabic convention, Holdrian commas of 22.64 cents for the Turkish one, and the measured slendro as measured.

The closed form uses $\mathrm{E}|X| = \sigma\sqrt{2/\pi}$ for a Gaussian, which is exact, and drops a correction term that is the tail integral beyond each category’s own width.

Where the model stops #

A single Gaussian is still a strong claim. Everything above inherits the model’s assumptions and only removes one of them. Real identification is not symmetric about a category centre, and the shifted-prior result is already evidence of something the model does not have.

Uniform presentation is doing work. Weighting the average by category width is the assumption that stimuli arrive uniformly across the octave. Present each category equally often instead and a narrow category’s high error rate is no longer discounted by its rarity, and the invariance goes. Which of the two an experiment implements is a design choice that has never mattered before and now decides whether the null holds.

One noise for a whole octave is the assumption that has not been touched. It is the subject of the last section and it is the one that would bring the step pattern back, so nothing here should be read as saying the pattern cannot matter — only that it cannot matter for this reason, under a noise that is the same at every boundary.

And the categories are assumed to tile. Every degree of the scale gets a category and every cent of the octave belongs to one, which is how an identification task with a forced choice works and is not how listening works. A listener hears an interval that is between two degrees as bad intonation rather than as one or the other.

Where this ladder goes next #

Eight rungs. The ear sorts into boxes; the boxes are wide enough that a tempered chord is still major; one value can be named as two things; an octave holds about seven of them; the boundaries barely move under a prior; all of that is a statement at one value of one parameter; the parameter is swept and the headline turns on it; and now the model’s remaining assumption is removed and the whole thing collapses to one line with one free number in it.

What is owed after this is the second dimension of the noise, and the closed form is what makes it askable. Sigma here is a single quantity applied to every interval in the octave, and there is no reason to expect that: an octave and a fifth are judged against references a listener holds far more securely than a tritone, so the confusion zones at those boundaries should be narrower than the zones elsewhere. The formula generalises immediately — the error rate becomes the sum over boundaries of each boundary’s own $\sigma\sqrt{2/\pi}$ rather than $n$ copies of one — and the moment the noise is allowed to differ by boundary, the step pattern matters again, because a scale can choose to put its boundaries where the ear is sharpest. That is the version of this question with musical content in it, it needs a per-interval noise this collection does not have, and it is one identification experiment away: the same task the seventh rung specified, scored per boundary rather than pooled.