Tribonacci and 3D lattices

[ To Paul Erlich ] (cf. 18650)

As I wrote, the method I found is so simple there might well be known.

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Using Tribonacci series for 3D lattices representation
------------------------------------------------------

We use normally square meshes for 2D lattices with horizontal and vertical
unit vectors.

QUESTION :

How to represent a third unit vector to have the nodes always distincts and
well distributed on the plane (even if there are many nodes on the third
axis)?

ANSWER :

To keep it simple, I will limit maths and give only recipes.

Let us suppose we have to draw on a squared paper a 3D lattice having less
than 7 nodes on the third axis. Using 7 x 7 squares for the unit mesh -- so
the end of the X-unit is (7,0) and the end of the Y-unit is (0,7) -- one of
the good solution for the end of the Z-unit is (5,4) :

Y . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . .(Z). .
. . . . . . . .
. . . . . . . .
. . . . . . . .
O . . . . . . X

Q. Why it is a good solution?

Let visualize in the same (modulo) mesh the 0-to-7 multiples of the Z-unit
vector. We observe a good distribution of the nodes in the plane suitable
for less than 7 nodes on Z-axis.

. . . . . . .(7)
. . . .(5). . .
.(3). . . . . .
. . . . .(1). .
. .(6). . . . .
. . . . . .(4).
. . .(2). . . .
(0). . . . . . .

We have to understand here that any lattice of whatever size in Z-axis will
have its nodes on the set of these 7 periodical possible positions. So if a
lattice has more than 6 nodes on the Z-axis then some nodes will be hidden.

In that case we could use another good solution with greater periodicity
such Z-unit = (9,7) when X-unit = (13,0) and Y-unit = (0,13) :

Y . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . .(Z). . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
O . . . . . . . . . . . . X

. . . . . . . . . . . . .(D)
. . . . . . . .(B). . . . .
. . .(9). . . . . . . . . .
. . . . . . . . . . .(7). .
. . . . . .(5). . . . . . .
.(3). . . . . . . . . . . .
. . . . . . . . .(1). . . .
. . . .(C). . . . . . . . .
. . . . . . . . . . . .(A).
. . . . . . .(8). . . . . .
. .(6). . . . . . . . . . .
. . . . . . . . . .(4). . .
. . . . .(2). . . . . . . .
(O). . . . . . . . . . . . .

-----

Q. How to find these good distributions?

Let 2 4 7 13 24 44 81 .. be the Tribonacci series (following 2) and a-b-c
be three successive numbers in that series (e.g. 2-4-7, 4-7-13, 7-13-24
..). With (c,0) and (0,c) for the end of X-unit and Y-unit vectors, our
precedent solutions for the end of Z-unit were (c-a,b).

Indeed (5,4) = (7-2,4) and (9,7) = (13-4,7).

-----

Q. Is that the only possibility?

The solution already calculated belongs to the first quadrant defined by X
and Y vectors. Naturally, by symmetry X | Y, the value (b,c-a) in the same
quadrant is also a good solution.

However the two solutions have a light flaw. The right-hand convention for
the direction of unit vectors is not respected (for we perceive the Z-axis
entering in the plane and the convention ask emerging from the plane). We
could simply permute Y and Z vectors to respect the convention.

We could also reverse the direction of Z taking (a-c,-b) in the third
quadrant to have a solution respecting the right-hand convention.

More, taking (a-c,b) and (c-a,-b) in the second and fourth quadrant we
would have yet equivalent solutions. In fact, there exist a total of 8
symmetrical solutions. Here is one of the third quadrant solution using
2-4-7 and the set of possible positions. The sign (-) indicates that the
right-hand convention is not respected so we have to permute two letters.
(That stands on not so obvious perception that Z-axis emerge in lower part,
for we normally look downwards).

. . . . . . . Y . . . . . . .
. . . . . . . . . . . . . . .
. . .(-). . . . . . .(-). . .
. .(-). . . . . . . . .-z . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
-x . . . . . . O . . . . . . X
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . Z . . . . . . . . .(+). .
. . .(+). . . . . . .(+). . .
. . . . . . . . . . . . . . .
. . . . . . .-y . . . . . . .

Let us note finally that (c,c) - (c-a,b) = (a,c-b) which gives the position
of multiple c-1 -- so we would have the reverse sequence c-1, c-2, c-3 ...
-- gives also a compressed Z-axis solution. Here are the 8 new
possibilities with the values 2-4-7 :

. . . . . . . Y . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . .(z). . .(z). . . . .
. . . .(z). . . . .(z). . . .
. . . . . . . . . . . . . . .
-x . . . . . . O . . . . . . X
. . . . . . . . . . . . . . .
. . . .(z). . . . .(z). . . .
. . . . .(z). . .(z). . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . .-y . . . . . . .

That corresponds to the same set of positions in reverse order. Here are
that positions with 4-7-13 :

Y . . . . . . . . . . . . .
. . . . . . . .( ). . . . .
. . .( ). . . . . . . . . .
. . . . . . . . . . .( ). .
. . . . . .( ). . . . . . .
.( ). . . . . . . . . . . .
. . . . . . . . .( ). . . .
. . . . Z . . . . . . . . .
. . . . . . . . . . . .( ).
. . . . . . .( ). . . . . .
. .( ). . . . . . . . . . .
. . . . . . . . . .( ). . .
. . . . .( ). . . . . . . .
O . . . . . . . . . . . . X

-----

Q. How to calculate with normal X-unit = (1,0) and Y-unit = (0,1)?

We have simply to divise the solution by c. The value for the Z-unit will
be then (1-a/c, b/c).

-----

Q. Finally how to have a universal solution with no periodicity so all
finite lattices have all their nodes always separated?

We know that the ratio, at the limit, of successive numbers in the
Tribonacci series is the Psi number. Using rt2 for square root and rt3 for
cubic root, we may calculate its value with this formula :

Psi = (1/3)(1 + rt3(19 + rt2(33)) + rt3(19 - rt2(33)))

= 1.839 286 755 ...

So if X-unit = (1,0) and Y-unit = (1,0), taking for a-b-c the successive

(1/Psi^2) (1/Psi) (1)

we obtain for (1-a/c, b/c) the expression (1-1/Psi^2, 1/Psi) whose values
approximated with 4 digits give

Z-unit = (0.7044, 0.5437)

With permutation and sign changes we obtain the 8 symmetrical solutions.

Pierre Lamothe

Pierre,

That is so beautiful! It gave me goosebumps.

Many list members will be familiar with the idea that phi or the
golden mean, with its Fibonacci approximations, results in the minimum
of near-coincidences in one dimension. For example between the
harmonic partials of two notes whose fundamantals have a frequency
ratio of phi.

Pierre has shown how to generalise this to give minimum coincidences
between regular 2D arrays.

The only problem is, I've forgotten what the intended application was
to tuning. Can someone remind us?

Regards,
-- Dave Keenan

Dave Keenan wrote:

> The only problem is, I've forgotten what the intended application
was
> to tuning. Can someone remind us?

What I'd like to do is automatically generate a family of scales that
approximate a given chord. Like now we can start with a perfect
fifths, and get mostly schismic scales out. What would happen if you
started with a major triad? Or a 5:6:7 diminished or 7:9:11 augmented
triad?

Of course we can get the right results by listing all ETs and taking
those with the best approximation. But a 3-D equivalent of the Scale
Tree would be sweet.

I haven't been following the mathematical detail of this at all. I
still don't know what a tribonacci series is. A friendly summary
would be much appreciated.

Graham

Dave Keenan wrote,

<<Pierre, That is so beautiful! It gave me goosebumps.>>

I'm glad to see someone enjoying Pierre's work here. Perhaps a summery
or paraphrasing of sorts by someone like yourself Dave (as I think you
have a knack for presenting potentially difficult concepts in a
patient and easily understandable way) might help others get a better
idea of what Pierre is doing here.

<<Pierre has shown how to generalise this to give minimum coincidences
between regular 2D arrays.>>

Again, could you elaborate on this a bit more? (Thanks.)

<<The only problem is, I've forgotten what the intended application
was to tuning. Can someone remind us?>>

If this is analogous to a three-term Stern-Brocot Tree, or three
fraction "adjacent fractions", then its applications would hopefully
run along similar lines. In other words think of Erv Wilson's scale
tree only it would pertain to three-term scales as opposed to two...
or "Tribonacci" scales as opposed to "Fibonacci" scales.

--Dan Stearns

Graham Breed wrote,

<<I haven't been following the mathematical detail of this at all. I
still don't know what a tribonacci series is. A friendly summary
would be much appreciated.>>

Though I've posted quite a bit on this over the last few months, I
also know that I'm definitely the wrong person to try and give a
(friendly) summary...

But I would like to bump this question back on up in the hopes that
(friendly) Robert Walker -- or anyone else who might want to give it a
(friendly) go -- is lurking!

--Dan Stearns