Symmetrical complexity for 5 and 7 limit temperaments

The symmetrical complexity for codimension one (5-limit linear,
7-limit planar) is the symmetrical lattice distance for the comma
defining it. For a 7-limit linear lattice, we have

Symcomp( <a b c d e f| ) = sqrt(3 (a^2+b^2+c^2) - 2(ab+ac+bc))

which is the symmetrical mapping lattice distance for the first three
components of the wedgie, considered as a note-class mapping. It might
be interesting to compare the lists of temperaments we get using this
definition of complexity with, L1 TOP, etc.

> The symmetrical complexity for codimension one (5-limit linear,
> 7-limit planar) is the symmetrical lattice distance for the
> comma defining it.

Is this the same as the "n" in your "interval count" message,
then? That was what I've been calling "taxicab distance to
the comma".

Between Tenney weighting, which gives log(n*d) for commas n/d
(is this what you're calling L1?), and no weighting (plain
"taxicab", your suggestion that 2401/2400 = 81/80) lies what
I've been calling "number of notes searched before finding
the comma" complexity. This is the volume of everything within
taxicab radius n of the origin, or n^pi(lim). In a sense,
weighting by limit. Clearly this is related to log-flat
badness...

(n-d)log(d)^e / d

...where e is pi(lim) or pi(lim)-1.

-Carl

--- In tuning-math@yahoogroups.com, "Carl Lumma" <ekin@l...> wrote:
> > The symmetrical complexity for codimension one (5-limit linear,
> > 7-limit planar) is the symmetrical lattice distance for the
> > comma defining it.
>
> Is this the same as the "n" in your "interval count" message,
> then? That was what I've been calling "taxicab distance to
> the comma".

No, it's the symmetrical Euclidean lattice distance.

>> > The symmetrical complexity for codimension one (5-limit linear,
>> > 7-limit planar) is the symmetrical lattice distance for the
>> > comma defining it.
>>
>> Is this the same as the "n" in your "interval count" message,
>> then? That was what I've been calling "taxicab distance to
>> the comma".
>
>No, it's the symmetrical Euclidean lattice distance.

Ok, let's take a look...

>To get to a note-class in what I called shell n^2 from the unison
>in 7-limit, you need at minimum n steps because a straight line
>path of 1-step intervals takes you only out to a distance n.

Oh yeah, here you're squaring n, even though the 7-limit is
3- or 4-dimensional.

So what the hell is a "step" in a "straight line path"?

>For 81/80 this is ceil(sqrt(13))=4 steps, and for
>2401/2400 it is ceil(sqrt(11))=4 steps also.

Where do 13 and 11 come from?

>In fact, both can be reached in four steps in only one way, up
>to commuitivity; we have
>
>81/80 = (6/5)(3/2)^3 (1/4)
>2401/2400 = (7/6)(7/5)^2(7/4) (1/4)

This sure looks like taxicab, but what are the "(1/4)" terms?

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> >For 81/80 this is ceil(sqrt(13))=4 steps, and for
> >2401/2400 it is ceil(sqrt(11))=4 steps also.
>
> Where do 13 and 11 come from?

81/80 is <-4 4 -1 0|, and 4^2+1^2+0^2+4*(-1)+4*0+(-1)*0=13, and
2401/2400 is <5 -1 -2 4|, from whence (-1)^+(-2)^2+(-4)^2+(-1)*(-2)+
(-1)*4+(-2)*4 = 11. Hence 2401/2400 is closer to the origin in the
symetrical 7-limit lattice we all know and love. It is a small
interval, but not really a complex one.

> >In fact, both can be reached in four steps in only one way, up
> >to commuitivity; we have
> >
> >81/80 = (6/5)(3/2)^3 (1/4)
> >2401/2400 = (7/6)(7/5)^2(7/4) (1/4)
>
> This sure looks like taxicab, but what are the "(1/4)" terms?

They make the math come out right. I don't really care about the
octaves, but those are them. You could write

49/48 = (7/6)(7/8), 49/50 = (7/5)(7/10), and then, look Ma!, no
octaves:

2401/2400 = (7/6)(7/8)(7/5)(7/10)

>> >For 81/80 this is ceil(sqrt(13))=4 steps, and for
>> >2401/2400 it is ceil(sqrt(11))=4 steps also.
>>
>> Where do 13 and 11 come from?
>
>81/80 is <-4 4 -1 0|, and 4^2+1^2+0^2+4*(-1)+4*0+(-1)*0=13,

So is this

|| <a b c d] || = b^2+c^2+d^2+bc+bd+cd

?

Wait, this looks familiar from a few mails back...

>The symmetrical complexity for codimension one (5-limit linear,
>7-limit planar) is the symmetrical lattice distance for the comma
>defining it. For a 7-limit linear lattice, we have
>
>Symcomp( <a b c d e f| ) = sqrt(3 (a^2+b^2+c^2) - 2(ab+ac+bc))

Ah, so this is new. It might be nice to surround new terms with
*asterisks* or something, to distinguish them from existing
definitions (such as you gracefully gave for Voronoi and Delaunay
cells in another post).

'symmetrical lattice distance' returns nil at Google and mathworld.

The difference between these two forms might be due to that you're
using wedgies for one and a comma-monzo for the other. But wait,
now I can check the "What the numbers mean" form!

I believe you meant "symcomp( <<a b c d e f || )". This means
there are a generators in 3/2, b in 5/2, c in 7/2, d in 5/3,
e in 7/3 and f in 7/5. Turning that into a 2&g map...

...whoops, looks like I'm hitting the same snag as Herman...

>>3&g: [<1 1 3 3|, <-6 0 -25 -20|] g ~ 7/72 ~ 11/114
>>
>>5&g: [<-2 -8 1 4|, <7 25 0 15|] g ~ 58/72 ~ 135/167
>>
>>7&g: [<1 7 -4 1|, <-2 -20 15 0|] g ~ 65/72 ~ 182/202
>
>That's useful to know. I can see where the second part of the maps
>come from, but how do you get the first part? It's clear that the
>element corresponding to the period is always 1 in this example,
>which makes sense, but is there any easy way to get the other three
>numbers other than trying a few until you find one that works? In
>other words, it's easy to determine that [<1 x y z|, <0 6 -7 -2|]
>is a possible mapping of miracle, as is [<x 1 y z|, <-6 0 -25 -20|],
>but I don't know how to get x, y, and z. I've been trying to find
>something like this in the archives, but I don't know where to look.

I don't see that this was ever answered. Did I miss it?

If I could turn a generic wedgie into a 2&g map, I might be able
to find a corresponding comma and see if b^2+c^2+d^2+bc+bd+cd is
the symcomp for a monzo.

>and
>2401/2400 is <5 -1 -2 4|, from whence (-1)^+(-2)^2+(-4)^2+(-1)*(-2)+
>(-1)*4+(-2)*4 = 11. Hence 2401/2400 is closer to the origin in the
>symetrical 7-limit lattice we all know and love. It is a small
>interval, but not really a complex one.

Yes, I gather you were saying that.

>> >In fact, both can be reached in four steps in only one way, up
>> >to commuitivity; we have
>> >
>> >81/80 = (6/5)(3/2)^3 (1/4)
>> >2401/2400 = (7/6)(7/5)^2(7/4) (1/4)
>>
>> This sure looks like taxicab, but what are the "(1/4)" terms?
>
>They make the math come out right. I don't really care about the
>octaves, but those are them. You could write
>
>49/48 = (7/6)(7/8), 49/50 = (7/5)(7/10), and then, look Ma!, no
>octaves:
>
>2401/2400 = (7/6)(7/8)(7/5)(7/10)

Ok. So I'm at a loss to describe how this is different from
taxicab distance.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> 'symmetrical lattice distance' returns nil at Google and mathworld.

People on tuning-math already knew the symmetrical 7-limit lattice of
note-classes when I got here. They didn't seem to know the formula for
calculating lattice distance, but clearly would have understood there
had to be one, so I don't regard this as a new topic. Anyway I've
talked about it endlessly in the last few years.

it's easy to determine that [<1 x y z|, <0 6 -7 -2|]
> >is a possible mapping of miracle, as is [<x 1 y z|, <-6 0 -25 -20|],
> >but I don't know how to get x, y, and z. I've been trying to find
> >something like this in the archives, but I don't know where to look.
>
> I don't see that this was ever answered. Did I miss it?

If you know the whole wedgie, finding x, y and z can be done by
solving a linear system. If you only know the period and generator
map, you first need to get the rest of the wedgie, which will be the
one which has a much lower badness than its competitors.

For instance, suppose I know the wedgie is <<1 4 10 4 13 12||. Then
I can set up the equations resulting from

<1 x y z| ^ <0 1 4 10| = <<1 4 10 4 13 12||

We have <1 x y z| ^ <0 1 4 10| = <<1 4 10 4x-y 10x-z 10y-4z||

Solving this gives us y=4x-4, z=10x-13; we can pick any integer for x
so we choose one giving us generators in a range we like. Since 3 is
represented by [x 1] in terms of octave x and generator, if we want
3/2 as a generator we pick x=1.

> Ok. So I'm at a loss to describe how this is different from
> taxicab distance.

It's clearly not taxicab Tenney distance, which is what Paul has been
calling that. It's taxicab distance with Fifth Element style flying
taxicabs, and routes which form an A3=D3 lattice.

>> 'symmetrical lattice distance' returns nil at Google and mathworld.
>
>People on tuning-math already knew the symmetrical 7-limit lattice of
>note-classes when I got here.

But they didn't call it that!

>They didn't seem to know the formula for
>calculating lattice distance, but clearly would have understood there
>had to be one, so I don't regard this as a new topic. Anyway I've
>talked about it endlessly in the last few years.

Have you ever called it "symmetrical lattice distance"?

>>>it's easy to determine that [<1 x y z|, <0 6 -7 -2|]
>>>is a possible mapping of miracle, as is [<x 1 y z|, <-6 0 -25 -20|],
>>>but I don't know how to get x, y, and z. I've been trying to find
>>>something like this in the archives, but I don't know where to look.
>>
>> I don't see that this was ever answered. Did I miss it?
>
>If you know the whole wedgie, finding x, y and z can be done by
>solving a linear system. If you only know the period and generator
>map, you first need to get the rest of the wedgie, which will be the
>one which has a much lower badness than its competitors.
>
>For instance, suppose I know the wedgie is <<1 4 10 4 13 12||. Then
>I can set up the equations resulting from
>
><1 x y z| ^ <0 1 4 10| = <<1 4 10 4 13 12||
>
>We have <1 x y z| ^ <0 1 4 10| = <<1 4 10 4x-y 10x-z 10y-4z||
>
>Solving this gives us y=4x-4, z=10x-13; we can pick any integer for x
>so we choose one giving us generators in a range we like. Since 3 is
>represented by [x 1] in terms of octave x and generator, if we want
>3/2 as a generator we pick x=1.

This is huge. Processing. . .

>> Ok. So I'm at a loss to describe how this is different from
>> taxicab distance.
>
>It's clearly not taxicab Tenney distance,

That's weighted. I'm saying it looks like unit/unweighted taxicab.

>It's taxicab distance with Fifth Element style flying
>taxicabs,

:)

>and routes which form an A3=D3 lattice.

Lost me. The only unweighted lattice types I'm aware of are
rectangular, where all the angles are 90deg, and triangular,
where I believe all the angles are 60deg, at least through FCC.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> Lost me. The only unweighted lattice types I'm aware of are
> rectangular, where all the angles are 90deg,

= cubic = Zn, Z3 in three dimensions.

and triangular,
> where I believe all the angles are 60deg,

= A2 in two dimensions, and An in n dimensions

at least through FCC.

FCC = A3 = D3

>> Lost me. The only unweighted lattice types I'm aware of are
>> rectangular, where all the angles are 90deg,
>
>= cubic = Zn, Z3 in three dimensions.

Thank you, thank you, thank you.

> and triangular,
>> where I believe all the angles are 60deg,
>
>= A2 in two dimensions, and An in n dimensions

Thank you.

> at least through FCC.
>
>FCC = A3 = D3

Then again, I think we're talking about the same thing!
You're counting the 'rungs' on the shortest path to the
target, no? (And where do An and Dn diverge?)

By the way, in 1999, Paul Hahn gave the following:

>Given a Fokker-style interval vector (I1, I2, . . . In):
>
>1. Go to the rightmost nonzero exponent; add the product of its
>absolute value with the log of its base to the total.
>
>2. Use that exponent to cancel out as many exponents of the opposite
>sign as possible, starting to its immediate left and working right;
>discard anything remaining of that exponent.
>
> Example: starting with, say, (4 2 -3), we would add 3 lg(7) to
> our total, then cancel the -3 against the 2, then the remaining
> -1 against the 4, leaving (3 0 0). OTOH, starting with
> (-2 3 5), we would add 5 lg(7) to our total, then cancel 2 of
> the 5 against the -2 and discard the remainder, leaving (0 3 0).
>
>3. If any nonzero exponents remain, go back to step one, otherwise
>stop.

This gives taxicab distance on an unweighted odd-limit An lattice,
IIRC.

-Carl

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> It's clearly not taxicab Tenney distance, which is what Paul has
been
> calling that.

I also call Paul Hahn's complexity in the symmetrical triangular
lattice 'taxicab'.

> It's taxicab distance with Fifth Element style flying
> taxicabs, and routes which form an A3=D3 lattice.

What do you mean by 'Element style flying taxicabs' here? I thought
the cabs drive only on the routes, as normal.

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> and triangular,
> where I believe all the angles are 60deg, at least through FCC.

That's the one!

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >FCC = A3 = D3

> Then again, I think we're talking about the same thing!
> You're counting the 'rungs' on the shortest path to the
> target, no? (And where do An and Dn diverge?)

They diverge everywhere except n=3.

> By the way, in 1999, Paul Hahn gave the following:

Thanks!