As there are a number of professional mathematicians on this list, I'd

like to direct a couple of questions still in need of an answer

directly at them...

(1) Given any arbitrary three-term scale (i.e., a scale made up of "a"

amount of small steps, "b" amount of medium steps, and "c" amount of

large steps), what global rule could govern their arrangement so as to

insure that the resulting scale is trivalent (i.e., has three sizes of

intervals in every unique interval class)?

(2) What role could the Tribonacci constant play in three-stepsize

Trihill scales that is analogous to the Fibonacci constant in

two-stepsize Myhill scales?

Any ideas here would be much appreciated.

thanks in advance,

--Dan Stearns

Even if I'm not mathematician, I hope to contribute answering the

unanswered Dan's questions. I shall write my answer in parts on distinct

posts from time to time. (I don't have much time to write).

I would like, first, to thank Paul Erlich who invited me, after "my last

message", to compare periodicity blocks with gammiers. I explored this link

particularly in january and found a great richness permitting to elucidate

questions about complexity dimension. I'm working with that to a theory of

convex bodies in primal lattices.

I would like also to thank Dan Stearns for his questions. I mentioned at

end of "my last message" the need for questions. I needed the Dan's

questions to organize simply, in a flash, a good part of my research. I

don't have time to write a paper now, so it's an opportunity to order

partially a profusion of discoveries.

The first part is like an introduction. Subsequently I shall write, by

example, about generalized Stern-Brocot tree in lattices. At the end of

this part it will be possile to give a better idea of the subjects to come.

Part I

I recall the first question of Dan Stearns in the message 18294 :

<<

(1) Given any arbitrary three-term scale (i.e., a scale made up of "a"

amount of small steps, "b" amount of medium steps, and "c" amount of

large steps), what global rule could govern their arrangement so as to

insure that the resulting scale is trivalent (i.e., has three sizes of

intervals in every unique interval class)?

>>

Let A < B < C be rational numbers corresponding to the steps such that

a log A + b log B + c log C = log 2

In rational systems these scales are subsets in abelian groups <2,p> or

<2,p,q> which contain all integer power combinations of the factors between

the <>. However I will neglect the case <2,p> for the complexity level in

<2,p> has to be too high to be musically pertinent. So the number of steps,

here, equals the base dimension. The numbers p and q are, in theory, simply

independant rational numbers but they are pratically small prime numbers

like 3, 5, 7, 11, 13, ...

We search here global relations independant of the prime numbers used,

therefore we search consistency in modular relations. By isomorphy, all

intervals are represented in the base [log 2 log p log q] by a unique

coordinate vector (x1,x2,x3). (This property results from equality

mentioned about steps and coordinates). I note also the coordinates by (x1

x2 x3)* where (*) indicate, as a transpose sign, that the vector is a

column matrix.

We know soon the number of classes (a + b + c), the degree of octave. How

are determined these classes? The classes are simply the quotient-group

<2,p,q>/<r,s> were r and s are the "commas" of Hellegouarch, or the "unison

vectors" of Fokker, or the "modal srutis" in my terminology. The group

<r,s> is the subgroup of <2,p,q> generated by the linear power combinations

of the vectors r and s.

By equality of dimensions (here) and for we know the order of the steps,

the modal srutis r and s are easily determined. It's simply B - A and C - B

represented by the vectors

(B1-A1 B2-A2 B3-A3)* = (r1 r2 r3)*

(C1-B1 C2-B2 C3-B3)* = (s1 s2 s3)*.

Let X = (x y z)* represent any interval. The image D of X (degree of X) by

the epimorphism applying intervals on Z corresponds to the determinant of

the matrix (X s r).

D = D(X) = det (X s r) =

(x s1 r1)

det (y s2 r2) = x det (s2 r2) + y det (s3 r3) + z det (s1 r1)

(z s3 r3) (s3 r3) (s1 r1) (s2 r2)

So D has the form x D1 + y D2 + z D3.

We could say also that D is the scalar product of X by (D1 D2 D3)* which is

the vectorial product of s and r.

If X = (1 0 0)* then D = D1, the degree of the octave (or the number of

classes modulo 2), for this X represents 2.

If X = (0 1 0)* then D = D2, the degree of p, for this X represents p.

If X = (0 0 1)* then D = D3, the degree of q, for this X represents q.

Examples:

Let A < B < C be 9/8 < 8/7 < 7/6 represented in [log 2 log 3 log 7] by

(-3 2 0)* < (3 0 -1)* < (-1 -1 1)*

Then r and s correspond to (6 -2 -1)* and (-4 -1 2)* which are the modal

srutis 64/63 and 49/48. The degree function is

(x -4 6)

D(X) = det (y -1 -2) = 5x + 8y + 14 z

(z 2 -1)

So in the slendro family, there are 5 degrees in which the harmonic 3

corresponds to degree 3 (for 8 mod 5 = 3) and the harmonic 7 to degree 4

(for 14 mod 5 = 4). And we can verify that A, B and C have degree 1 :

D(9/8) = D(-3,2,0) = (5)(-3) + (8)(2) + (14)(0) = 1

D(8/7) = D(3,0,-1) = (5)(3) + (8)(0) + (14)(-1) = 1

D(7/6) = D(-1,-1,1) = (5)(-1) + (8)(-1) + (14)(1) = 1

---

Let A < B < C be 16/15 < 10/9 < 9/8 represented in base log <2,3,5> by

(4 -1 -1)* < (1 -2 1)* < (-3 2 0)*

Then r and s correspond to (-3 -1 2)* and (-4 4 -1)*, which are the modal

srutis 25/24 and 81/80. Then D(x,y,z) is

(x -4 -3)

D(X) = det (y 4 -1) = 7x + 11y + 16 z

(z -1 2)

So in the Zarlino family, there are 7 degrees in which the harmonic 3

corresponds to degree 4 (for 11 mod 7 = 4) and the harmonic 5 to degree 2

(for 16 mod 7 = 2). Verifying tonal generator we find :

D(16/15) = D(4,-1,-1) = (7)(4) + (11)(-1) + (16)(-1) = 1

D(10/9) = D(1,-2,1) = (7)(1) + (11)(-2) + (16)(1) = 1

D(9/8) = D(-3,2,0) = (7)(-3) + (11)(2) + (16)(0) = 1

---

Returning to the question, the following relation express the independance

of the numbers 2, p and q. Let (A1 A2 A3)* (B1 B2 B3)* (C1 C2 C3)* be the

coordinates of A, B and C. Then

(A1 B1 C1) (a) (A1*a + B1*b + C1*c) (1)

(A2 B2 C2) (b) = (A2*a + B2*b + C2*c) = (0)

(A3 B3 C3) (c) (A3*a + B3*b + C3*c) (0)

To reach the octave, an equal amount of p (or q) have to be used in

numerator and denominator while an excedent 2 is required. This is the

basic constraint.

----------

So far it's only trite remarks. Now concerning

<< to insure that the resulting scale is trivalent (i.e., has three sizes

of intervals in every unique interval class) >>

I could say that we have just seen there exist an infinity of intervals for

each classes and that the question is to determine the three pertinent

intervals for each classes (except unison). I will show that they are

contained in a convex shape (easily calculated) around the unison.

We could question the necessity to use three values for each classes but

without using the gammier theory, this level is out of reach (I want here

to restrict my comments to Z-module properties).

This new subject is more sophisticated so I will begin with remarks to give

a better contextualization.

---

First we have to note that, independantly of prime numbers used (that we

suppose to be not too high), it subsists in lattice (or Z-module)

representation, a component of the global complexity that I will name, for

the time being, "modular complexity". For each interval represented by

(x,y,z) this modular complexity corresponds to the sum of the absolute

value of the coordinates.

MC = MC(X) = |x| + |y| + |z|

So MC is 0 for unison (0 0 0)* and grows up in any direction. Knowing prime

numbers, we could adjust relative rate of growing in different axis and

calculate complexity or sonance :

complexity = 2^|x| * p^|y| * q^|z|

sonance = |x|*log 2 + |y|*log p + |z|*log q

But independently of primal base used we can understand the importance to

use for each class of intervals those of minimal complexity which are

forcely located around the unison. We are not interested here in metric of

complexity but in topological invariance using MC in place of complexity.

( Beside, I suggest than N-limit is more pertinent with

complexity than with odds or primes. Convexity has really

sense with complexity in constrast with the two others.

Convexity means constraint to use all possibilities

within the limit. It would be easy to show, by some

examples, like slendro scales, the non-pertinence of

convexity for odds and primes, and I found that the

pertinent tonal structures are convex bodies.)

As alternative to MC, we can use MC2 = sqrt (xx + yy + zz), the usual

euclidian norm, in plunging our Z-module in R^3, to analyse ball containers

of intervals behind a limit of complexity. It's particularly helping in

dimension > 3 to figure a parallelotope having his vertex belonging to an

hyperquadric.

Knowing primal base we could represent by a spherical body (in R^3) the

domain containing the set of discrete intervals (of Z^3) having a maximal

distance limit from unison. This distance (MC2) is a represention of the

"sonance". (We have to use the log of complexity, the sonance, to be linear

with MC2).

This container spherical shape is not preserved by the representation in

Z^3 but we expect to obtain a container ellipsoidal shape with

1/log 2 : 1/log p : 1/log q

axis ratios and we expect that this shape be easily described by an

equation using a "diagonal" matrix M such that

(x y z) M (x y z)* = K

---

Another important remark concerns the effect of octave modularity on MC and

MC2. Independently of primal base, octave modularity means using <p,q>

rather than <2,p,q>. Thus we have to use only "reduced modular complexity"

with RMC = |y| + |z| and RMC2 = sqrt (yy + zz). So we can easily understand

that RMC and RMC2 are globaly more reduced relatively to MC and MC2 if

yz > 1 rather than yz < 1.

For example the intervals 15/8 and 5/3 have the same RMC = 2, but they have

respectively 5 and 2 for the MC ; the intervals 45/32 and 9/5 have also the

same RMC = 3, but respectively 8 and 3 for the MC.

By principle, if y and z in (y,z) have the same sign we can expect a great

value of |x| to obtain an interval of the first octave.

So a cube domain (having vertex on axis) with norm MC < K in <2,p,q> would

appear as a compressed hexagonal shape rotated counterclockwise in <p,q>.

We expect, with usual euclidian norm, that our ellipsoid container would

appear also like a counterclockwise rotated ellipse responding to the equation

(1 y z) M (1 y z)* = K

But the matrix M would be no more diagonal and would have the form

(1 0 0)

M = (0 a d)

(0 d c)

So

(1 0 0)

(1 y z) (0 a d) (1 y z)* = K

(0 d c)

axx + bxy + cyy = K (where b = 2d)

---

I will anticipate on arguments here to give now examples insuring that this

abstracted matter has sense in tonal structures.

By periodicity, the vectors s, r+s, r, -s, -r-s and -s determine 6 limit

points of complexity. (We will develop that later, and we will use the

convex hexagonal outline to study the algebraic properties). But we can

exhibit now the elliptic container of complexity determined by A, B and C

which satisfied to the six points. By symmetry, the s, s+r and r values are

sufficient to calculate this ellipse.

In the first example we had r = (6 -2 -1)* and s = (-4 -1 2)*. So the

points are (-1,2) (-3,1) (-2,-1) and we have to solve :

a - 2b + 4c = K

9a - 3b + c = K

4a + 2b + c = K

We find easily the quadratic form 3xx + 3xy + 7yy = K and we may verify

that the six points give K = 25 the square of classes.

In the second example we had r = (-3 -1 2)* and s = (-4 4 -1)*. So the

points are (4,-1) (3,1) (-1,2) and we have to solve :

16a - 4b + c = K

9a + 3b + c = K

a - 2b + 4c = K

We find easily the ellipses family 3xx + 3xy + 13yy = K and we may verify

that the six points give K = 49 the square of classes.

( We can note, in this case, that the elliptic shape

contains erroneously the points (4,0) and (-4,0)

with K = 48 which are 81/64 and 128/81. So the

approximative euclidian norm fail here. Strictly,

only the hexagonal shape inside, as we will see,

is truly periodical -- while ellipses may overlap --

and contains exclusively the three intervals of

minimal complexity for each classes. But this

hexagonal shape has his vertex located on the

calculated ellipse. And all that is generalizable

at N dimensions. )

By anticipation we presumed here that the set of intervals included in the

hexagonal shape corresponds strictly to the three periodical blocks of less

complexity. These blocks of one interval by class surround unison and have

only unison as intersection, so we have with N = a + b + c classes only the

3N-2 intervals searched. With gammier theory we could go deeper and

consider, with algebraic arguments, lesser numbers N of intervals included

in the first shape and see that these subsets are also convex bodies.

The two precedent examples permit to see how the ideal N-limit ball of

sonance (which has, by definition of sonance, a cubic form in R^3 with his

vertex on primal axis) is approximately approched by systems with

epimorphism to Z (i.e. having classes). The hexagonal shape correspond to

the sonance parallepiped defined by sonance limit of the modal srutis.

The following figures show an image of our slendro and zarlino examples. We

see the hexagonal shape defined by r and s vectors and the three periodical

blocks of 5 and 7 distinct classes surrounding unison. The intervals of

degree 1 and -1 are colored. In the second example, we distinct the 13

intervals of the Zarlino gammier among the complete 19 intervals gammier

(using harmonic 27 and 45). So we see how modulation at dominant or

sub-dominant keeps the confinement.

http://www.aei.ca/~plamothe/pix/r8.gif

http://www.aei.ca/~plamothe/pix/r7.gif

----------

I stop here the introduction. In the following parts I shall show how a

universal matrix transform successively a given periodical block in the

other surrounding blocks. In dimension N, this matrix appears as a root N+1

of the identity matrix. If we change negative signs in this matrix by

positive signs, the successive powers exibit N-bonacci series and we have

the analogy of Fibonacci in 2D. We can also construct an analog of Stern-

Brocot tree with N branches.

Beside, I found a simple method to represent 3D and 4D lattices using

Tribonacci series. It's so simple I imagine it's already known. If not

I shall show that.

I don't give other details. The matter grows each day and I did'nt take

time to organize.

Pierre Lamothe

Pierre Lamothe wrote,

>For each interval represented by

>(x,y,z) this modular complexity corresponds to the sum of the absolute

>value of the coordinates.

> MC = MC(X) = |x| + |y| + |z|

As you may know, I'm not a fan of this sort of complexity measure unless 2

is included as one of the coordinates, and the coordinates are lengthened by

the log of the corresponding prime (this then agrees with Tenney's Harmonic

Distance). In periodicity block work it usually isn't, and a metric more

suited to octave-invariant lattices is given by Kees van Prooijen at:

http://www.kees.cc/tuning/perbl.html

This sort of thinking leads naturally to triangular lattices like Erv

Wilson's at:

http://www.anaphoria.com/dal.PDF

(figure 6, 6c, 6d, 13, 14, etc.)

I only used a rectangular lattice in my "gentle introduction to periodicity

blocks" for simplicity's sake. If I were to begin looking for periodicity

blocks which best approximated a spherical region around a central point or

dyad in the lattice, I would certainly not use the rectangular lattice but a

triangular lattice instead. Of course, the interior the sphere, even if it

has an area of 31 or 41 or 72 or whatever, is not necessarily going to be a

periodicity block -- there could easily be two ratios in one equivalence

class and no ratios in another -- but a little rearragement could result in

a parallelopiped, hexagonal prism, or rhombic dodecahedron [or higher- (or

lower-) dimensional analogues] that is a periodicity block. This is because

sphere's don't tile space -- if you try to fill space with identical

spheres, there are spaces between the spheres, or they overlap, or both.

Anyway, Paul Hahn and I talked a bit about this very idea -- see the

archives.

>Beside, I found a simple method to represent 3D and 4D lattices using

>Tribonacci series. It's so simple I imagine it's already known. If not

>I shall show that.

It may be simple to you, but I'm sure most of us on this list can't even

begin to imagine what it is you mean, let alone how you would do it. Please

enlighten us, holding our hand as well as you can.

Cheers,

Paul

Hi Pierre,

Your post looks very interesting, and I've gone through the first part of it fairly

slowly.

Here is how I understand the main points - maybe it will help others

on the TL who are relatively new to the idea of a periodicity block, and maybe will help

give another perspective:

Notation:

lowercase for numbers of steps, upper case for step sizes.

a steps of size A, b of size B, and c of size C adding up to the octave.

<2,p,q> = all numbers of form 2^x1*p^x2*q^x3 where x1, x2 and x3 can be positive or negative

exponents.

Ex. For j.i. diatonic, p=3, q = 5, and <2,p,q> = all numbers of form 2^x1*3^x2*5^x3.

(for those completely new to this idea, note that this definition automatically givies

you only one representative for each ratio. E.g. 6/5 and 12/10 would both be

expressed as 2^1*3^1*5^-1.)

We can use vector notation:

j.i. diatonic

Notes:

1/1 3^2/2^3 5/2^2 2^2/3 3/2 5/3 3*5/2^3 (2/1)

=

(0,0,0), (-3, 2, 0), (-2, 0, 1), (2, -1, 0), (-1, 1, 0), (0, -1, 1), (-3, 1, 1), (1, 0, 0)

e.g. (0,-1,1) short for (3^-1)*(5^1), i.e. 5/3.

Now, let r = (r1,r2,r3) and s = (s1,s2,s3) be two fixed vectors, corresponding to the

Fokker unison vectors.

We need a third vector to make a parallelepiped in 3D, call this X = (x,y,z)

Then the volume of the parallelepiped with vertices (0, r, s, X, r*s, s*X , r*X, r*s*X)

is given by the determinant (this is a well known result)

> (x s1 r1)

> det (y s2 r2) = x det (s2 r2) + y det (s3 r3) + z det (s1 r1)

> (z s3 r3) (s3 r3) (s1 r1) (s2 r2)

since it is a unit lattice, each point occupies unit volume, so the volume of the parallelopiped

gives the number of lattice points in it, wherever it is located

in the lattice (with certain conventions on how one treats points on the boundaries

of it). (also a well known result, but not quite so much so as the volume of the parallelopiped).

Since the vectors that define the parallelopiped are also vectors joining lattice points, then in fact,

the determinant gives the number of distinct points in the lattice up to equivalence

by its three defining vectors.

So far, this is just a brief summary of the notion of a periodicity block.

Now, the interesting move, you want to make a scale with three step sizes

A, B and C, where A, B and C are expressible in terms of (2,p,q).

So you choose r = B-A, s = C-B.

where here we are talking about vector subtraction,

r= (B1-A1 B2-A2 B3-A3)* = (r1 r2 r3)*

s= (C1-B1 C2-B2 C3-B3)* = (s1 s2 s3)*.

As ratios, r=B/A, and s = C/B

In other words, you are treating lattice points as

equivalent if you can get from one to another

via B/A, or via C/B.

So for example if the steps are 9/8, 10/9 and 16/15

and if you start the scale with 10/9, then you have to leave

out 9/8 and 16/15, because they are equivalent to

10/9 under the unison vectors r and s.

Continuing with the diatonic j.i. steps

9/8, 10/9 and 16/15, and these values

of r, s, and using (x,y,z) = X as a third unison

vector, one can find the determinant from these.

Your determinant equation gives

D(X) = 7x + 11y + 16 z

so we can immediately read off from this the

number of notes you get using any other

type of equivalence.

So for octave equivalence, x=1, y=0, z=0,

so D(X)=7, so one has 7 notes.

For equivalence of 1/1 and 3/1

as in BP type scales, you will have

X = (0,1,0) (i.e. 2^0*3^1*5^0)

and so x=0, y=1, z=0.

So this time, we have 11 notes.

Similarly, equivalence under 5/1 gives a scale of 16 notes.

Here, there is one point I don't quite get.

You deduced that the 3/1 will be the fourth

degree of the diatonic scale, and I don't quite

see where you get that.

> So in the Zarlino family, there are 7 degrees in which the harmonic 3

> corresponds to degree 4 (for 11 mod 7 = 4) and the harmonic 5 to degree 2

> (for 16 mod 7 = 2). Verifying tonal generator we find :

I can see that 11 mod 7 = 4, but not the connection of this to

the fourth degree of the scale.

Another example:

for 9/8, 8/7 and 7/6,

treating 9/8

D(X) = 5x + 8y + 14 z

so one can read out that there are 5 notes up to octave equivalence.

(and 8 up to equivalence under 3/1, and 14 up to equivalence under

7/1)

One could also put x=-1, y=1, D(x)=3, so there are 3 notes up to

equivalence by 3/2, and so forth.

Actually finding what these three notes seems a bit tricky.

(this will sometimes give negative numbers for D(X), but that is

okay, as the volume is defined as abs(D(x)), and is always positive).

Are there any nice easy ways to find these three notes, or any of the scales,

without doing a lot of 3D geometry?

That's about as far as I've got, though I've looked ahead, and

see you haven't yet given the trivalence result, so look forward to see

how it works out.

I expect some on the TL may have immediately understood what you

are saying, but for others, this area may also be relatively new, or your

approach to it is new, or some of the notation and terminology,

so maybe what I've said here will help them to ask the right questions

to understand the rest of what you have said.

I look forward to what you have to say about trivalence. I can't at

present see how trivalence is going to emerge from this, and if you

can do that, I'm sure it will be fascinating!

Robert

I wrote,

<<(1) Given any arbitrary three-term scale (i.e., a scale made up of

"a" amount of small steps, "b" amount of medium steps, and "c" amount

of large steps), what global rule could govern their arrangement so as

to insure that the resulting scale is trivalent (i.e., has three sizes

of intervals in every unique interval class)?>>

I think that a "trivalent only" rule and a Tribonacci expansion have

to be separate conditions as an even numbered scale must result in a

sixth term that is even as well; [1,1,2] for example gives a

Tribonacci expansion of:

1,1,2,4,7,13,...

And a 4 note subset of 13 cannot be trivalent.

However, there I believe there is a very simple global rule that would

govern arbitrary three-term indexes so that their stepsize arrangement

results in a scale that is constructed of three stepsizes and

trivalent in all but what I'd term "unique" cases:

L-out-of-M-out-of-N

For a two-term index, [a,b], the global rule would lie in the

Fibonacci series where any four term segment corresponds to:

a,b,M,N,...

So let's say the arbitrary two-term index is [3,11], then you'd have:

3,11,14,25,...

And the L-out-of-M would be bbabbbbabbbabb:

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

Or:

0 2 4 5 7 9 11 13 14 16 18 20 21 23 25

For a three-term index, [a,b,c], the global rule would lie in the

Tribonacci series where any six term segment would correspond to:

a,b,c,L,M,N,...

Let's say the arbitrary three-term index is [1,4,3], then you'd have:

1,4,3,8,15,26,...

This would be an even numbered scale, but as the sixth term is even as

well, the results should be trivalent... let's see.

M-out-of-N gives:

0 2 3 5 7 9 10 12 14 16 17 19 21 23 24 26

And L-out-of-M gives either:

0 3 7 10 14 16 19 23 26

Or:

0 3 7 10 12 16 19 23 26

Both of these are a [1,4,3], and both are trivalent.

I believe that *THIS IS THE RULE*...

The exceptional cases I believe should occupy a different class, or

rather represent some distinct subgenre of these three-term scales.

(And though I have yet to check or much experiment with it, I'd

imagine that this same principle could be carried out to other n-term

scales as well with similar results.)

--Dan Stearns