Maximal Evenness Proofs

Maximally even scales were introduced by Clough and Douthett, 1991, and maybe somewhere earlier. They have some properties where either I haven't seen a proof, or I don't like it, so here are some of my own.

Works referenced can all be found in Manuel Op De Coul's bibliography


I'm using a notes in the chromatic scale and b notes in the diatonic, following Agmon. 0<b<=a and a and b are integers. The case where a=b is an equal temperament. The case where b=0 is a scale with no notes to the octave, so it's difficult to see what these properties would mean when applied to it. The maximally even scale is defined as

p(n) = floor(na/b)

Where p(n) is the nth pitch and floor is defined along with ceil and frac such that

x = floor(x) + frac(x)
x = ceil(x) - frac(x)

Where x is real, floor(x) and ceil(x) are integers and 0 <= frac(x) < 1. Some useful identities follow from this

floor(i+x) = i + floor(x) where i is an integer
floor(frac(x)) = 0
floor(x+y) = floor(x) + floor[frac(x) + y]
either floor(x) = ceil(x) or floor(x) = ceil(x) - 1
min[floor(x), ceil(x)] = floor(x)
max[floor(x), ceil(x)] = ceil(x)

The scale repeats about a period of b/gcd(a,b) diatonic steps and a/gcd(a,b) chromatic steps where gcd(a,b) is the greatest common divisor of a and b. This follows from na/b always being an integer when a is a multiple of b and floor(x) being the same if you add an arbitrary number of integers to x.

Agmon 1996, at any rate, only defines the scale in the range 0 <= p(n) < a and 0 <= n < b but it works for any integer n.

Myhill's Property

Myhill's property states that there should be two sizes of each diatonic interval class, except octaves. It's true for all maximally even scales where b is the period, that is gcd(a,b)=1. In general, any interval comes in two step sizes when the number of steps isn't a multiple of the period.

Define the number of steps in an interval of i steps as

dp(i) = p(n+i) - p(n)

for any n. From the maximal evenness formula it follows that.

dp(i) = p(n+i) - p(n)
dp(i) = floor[(n+i)a/b] - floor(na/b)
dp(i) = floor(na/b + ia/b) - floor(na/b)
dp(i) = floor(na/b) + floor(frac(na/b) + ia/b)) - floor(na/b)
dp(i) = floor(frac(na/b) + ia/b)

If ia/b is an integer, then

dp(i) = ia/b + floor(frac(na/b))
dp(i) = ia/b = floor(ia/b) = ceil(ia/b)

If ia/b is not an integer, then dp(i) is either floor(ia/b) or floor(ia/b)+1 = ceil(ia/b). So the general case is

dp(i) = {floor(ia/b), ceil(ia/b)}

This gives two different values provided ia/b is not an integer, that is (ia) mod b = 0 and i is not a multiple of the period. Hence Myhill's property holds whenever gcd(a,b)=1.


Propriety, or coherence to the Balzano school, is defined as

dp(i+j) >= dp(i) for all integer i, j where j>0

From the result above, this is

min[floor((i+j)a/b), ceil((i+j)a/b)] >= max[floor(ia/b), ceil(ia/b)]
floor[(i+j)a/b] >= ceil(ia/b)
floor(ia/b + ja/b) >= ceil(ia/b)
floor(ia/b) + floor(frac(ia/b) + ja/b)) >= ceil(ia/b)
floor(frac(ia/b) + ja/b) >= ceil(ia/b) - floor(ia/b)
floor(frac(ia/b) + ja/b) >= 1
ja/b >= 1

This is always the case, because a>=b and j>=1. Hence all maximally even scales are proper, in agreement with Agmon 1996. We can go one step further, and show that a scale is strictly proper (coherent with no ambiguities) when

floor[(i+j)a/b] > ceil(ia/b)
floor(frac(ia/b) + ja/b) > ceil(ia/b) - floor(ia/b)

When frac(ia/b)=0 this is trivially fulfilled, as the right hand size is zero and ja/b >= 1. So it remains to be proven for frac(ia/b) <> 0. In this case, the right hand side is 1 and

floor(frac(ia/b) + ja/b) > 1
floor(frac(ia/b) + ja/b) >= 2
frac(ia/b) + ja/b >= 2

The smallest non-zero value of frac(ia/b) is gcd(a,b)/b, where gcd(a,b) is the greatest common divisor of a and b. The smallest value of ja/b is 1 is j>0. That leaves us with the condition

gcd(a,b)/b + a/b >= 2
(gcd(a,b)+a)/b >= 2

So the maximally even b from a scale is strictly proper when (a+1) >= 2b provided a/b is in its lowest terms. The general rule a+gcd(a,b) >= 2b is both a necessary and sufficient condition for strict propriety, and covers scales like

1 2 2 2 2
1 2 2 2 2 1 2 2 2 2

Missing chromatic steps

For the case a >= 2b, there are no intervals of 1 chromatic step within the diatonic scale. This follows from the smallest diatonic step being floor(a/b) chromatic steps which is 2 when a=2b.

Hence no maximally even scale with a>=2b can fulfil Agmon's efficiency criterion, which may be why Agmon and Clough&Douthett both reject such scales as diatonics (see Agmon 1996). They prefer scales with this kind of efficiency (which is different to Rothenberg's) that are proper with no more than 1 ambiguous interval pair within the octave.

Such scales that fulfil Myhill's property with a chromatic notes to the octave must have floor(a/2)+1 diatonic notes to the octave. Agmon 1996 shows this, and apparently he proved it somewhere, but it's easy enough done:

Myhill's property states that if there are b notes to the octave, there must be 2b-1 distinct intervals between them, because there are two for every size except the octave. If the scale has no ambiguities, that means each of the 2b-1 intervals must be a distinct number of chromatic steps modulo the period. So the total number of chromatic steps must be a=2b-1. It also follows that a must be odd.

If you're allowed 1 ambiguity, then there will only be 2b-2 chromatic notes to the octave, and a=2b-2. And a must be even.

Turning these equations round, when a is even

b = (a+2)/2 = a/2 + 1

and when a is odd

b = (a+1)/2 = (a-1)/2 + 1

and floor(a/2)+1 covers both these cases. It's a necessary but not sufficient condition. Maximally even scales seem to work, and here are the simplest ones:

01 0
11 1
22*1 1
32 1 2
43 1 1 2
53 1 2 2
64*1 2 1 2
74 1 2 2 2
85 1 2 1 2 2
95 1 2 2 2 2
106*1 2 2 1 2 2
116 1 2 2 2 2 2
127 1 2 2 1 2 2 2
137 1 2 2 2 2 2 2
148*1 2 2 2 1 2 2 2
158 1 2 2 2 2 2 2 2
169 1 2 2 2 1 2 2 2 2
179 1 2 2 2 2 2 2 2 2
1810*1 2 2 2 2 1 2 2 2 2
1910 1 2 2 2 2 2 2 2 2 2
2011 1 2 2 2 2 1 2 2 2 2 2
2111 1 2 2 2 2 2 2 2 2 2 2
2212*1 2 2 2 2 2 1 2 2 2 2 2
2312 1 2 2 2 2 2 2 2 2 2 2 2

Those marked with a * in the degenerate column have a period smaller than the octave. So they don't strictly have Myhill's property but they do have the right number of ambiguities to the octave. It's up to you whether or not you include them. Agmon doesn't.

MOS Scales

These are also known as cyclic scales or well formed scales. Agmon 1996 showed that every maximally even scale is an MOS. Carey & Clampitt 1996 showed that every non degenerate MOS (MOS where the period equals the octave) has Myhill's property, which agrees with the above. I'm going to show that maximally even scales are all and only those that you'll find on the scale tree.

The Scale Tree is Erv Wilson's name for the Farey or Stern Brocot tree applied to musical scales. I think he mentioned it first in 1974. Each node can represent an MOS. The denominator is the number of notes in the scale, and the numerator the number of steps to the generator. Here, we'll take these to be chromatic steps.

Reconciling with the Scale Tree

To show an equivalence between the scale tree and maximally even scales, each node on the scale tree needs to be associated with exactly one MOS, and there needs to be a way to find that MOS for each maximally even scale.

First, the scale tree only contains ratios in their lowest terms. That means the period and octave must be identical. When this isn't the case, you can divide through by the gcd to get an equivalent scale. I'll redefine the numbers of chromatic and diatonic steps to the octave as a' and b' so that a and b can be a'/gcd(a',b') and b'/gcd(a',b'). A pitch on the maximally even scale then becomes

p = floor(n(a*gcd(a',b'))/(b*gcd(a',b')))
p = floor(na/b*(gcd(a',b')/gcd(a',b')))
p = floor(na/b)

So this doesn't change any of the notes, only the way the scale is defined.

Following Agmon, I'm labeling the generators as q/a and b/r where q <= 0 < a and b <= 0 < r. That means the scale tree is only taken between 0/1 and 1/1. You can take it from 0/1 to 1/0 if you like, but it means each scale will be found in more than one place.

If the ratio of a node is the chromatic generator q/a, the diatonic generator r/b is the ratio of one of the adjacent, simpler nodes. There are always two such, and there are always two different ways of specifing the generator. In general,

|ar-bq| = 1

So I'll make the arbitrary choice

ar - bq = 1

It's possible to find q and r for any a and b, as I have to do in my software. If you get q' and r' such that ar' - bq' = -1, replace them with q = a-q' and r = b-r'.

bq' - ar' = 1
b(a-q) - a(b-r) = 1
ba - bq - ab + ar = 1
ar - bq = 1

It also follows this relationship that

gcd(q,a) = gcd(r,b) = 1

As, if q', a' = iq, ia or r', b' = ir, ib

iar - ibq = i(ar-bq)

which can never be 1 or -1 if everything's an integer and i isn't -1, 0 or 1.

The MOS is defined as

n = kr mod b
p = kq mod a

From these, it followed that if you started with q' and r' where 0<=q'<a and 0<=r'<b isn't the case, you could reduce them to q = q' mod b and r = r' mod b so that it is. Hence, any MOS where |ar-bq| = gcd(a,b) maps to a maximally even scale, and we only need to consider the case ar-bq = 1.

Proof of equivalence

So far, I've shown that there's a mapping between maximally even scales and the scale tree. Now I'll show that the maximally even scale is identical to the MOS you get on the scale tree.

Starting from the definition of chromatic steps

p = kq mod a
pb = kqb mod ab

from ar-bq=1

pb = k(ar-1) mod ab
pb = [(kar mod ab) - k] mod ab
pb = [a(kr mod b) - k] mod ab

and we can substitute from the definition of n

pb = (an - k) mod ab

As n is only defined mod b, and there only needs to be one k for each n, I'll set

0 <= n < b
0 <= k < b
0 <= p < a

There's no loss of generality, because in both cases the scale is periodic with respect to a and b. An equivalence in one period means an equivalence in all. I'll now show that, with these restrictions,

pb = an - k

which follows from

0 <= an-k < ab when 0<=n<b and 0<=k<b

First, in the special case where k=0,

n = kr mod b
n = 0 mod b
n = 0
an-k = 0

In addition, this is the only case where n=0 and 0 <= k < b because gcd(b,r) = 1. Because k<b and b<=a that means k<a and an-k=0 is only true when k=0. So we still have to prove

0 < an-k < ab when 0 < n < b and 0 < k < b

The first part of this inequality is

0 < an-k
k < an

As an is always >0, the right hand side never gets smaller than a, giving.

k < a

Which is one of the restrictions. The other part if the inequality is

an-k < ab

Because k>0, this simplifies to

an < ab

and because a>0,

n < b

which is also on the list of restrictons, so we've proved that

0 <= an - k < ab


0 <= n < b, 0 < k < b

which means that

pb = (an - k) mod ab

is the same as

pb = an-k

for 0 <= p < a. And that's fairly obviously the same as

p = (an-k)/b
p = an/b - k/b
an/b = p + k/b

and you may recall

0 <= k < b
0 <= k/b < 1

From the original definition of p, it must be an integer. From the definition

x = floor(x) + frac(x)

where floor(x) is an integer and 0<=frac(x)<1, we can identify

k/b = frac(an/b)
p = floor(an/b)

The second one of those is the definition of a maximally even scale, and as we arrived at it from the definition of an MOS from the scale tree, it proves that the two are identical.