q for Gene

You showed a simple formula (two, actually) for symmetric
Euclidean distance with three factors. Is there a general
formula for any number of factors?

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
>
> You showed a simple formula (two, actually) for symmetric
> Euclidean distance with three factors. Is there a general
> formula for any number of factors?

It's easy to make all odd primes symmetrical and of the same length,
resulting in an An lattice. I can't see much point in this, because
the symmetries become unrealistic after the 7-limit. More reasonable
it seems to me is to break symmetry to some degree; for instance in
the 13-limit one might decide to make 7, 11, and 13 symmetrical, and 3
and 5 symmetrical, with 9 and 15 being the same as 7, 11, or 13.

>> You showed a simple formula (two, actually) for symmetric
>> Euclidean distance with three factors. Is there a general
>> formula for any number of factors?
>
>It's easy to make all odd primes symmetrical and of the same length,
>resulting in an An lattice.

Is there a formula like

||(a,b,c)||_p = ((|a|^p+|b|^p+|c|^p+|a+b+c|^p)/2)^(1/p)

for it?

>I can't see much point in this, because the symmetries become
>unrealistic after the 7-limit.

Does this have to do with nonprime odds? Can you give an
example of a bad result caused by them?

-Carl

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

> >It's easy to make all odd primes symmetrical and of the same length,
> >resulting in an An lattice.
>
> Is there a formula like
>
> ||(a,b,c)||_p = ((|a|^p+|b|^p+|c|^p+|a+b+c|^p)/2)^(1/p)
>
> for it?

That's the one; good for any prime limit if generalized in the obvious
way.

> >I can't see much point in this, because the symmetries become
> >unrealistic after the 7-limit.
>
> Does this have to do with nonprime odds? Can you give an
> example of a bad result caused by them?

If you have a 9-limit chord then applying the symmetry could net you a
121-limit chord, which is hardly a similar thing musically speaking.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >
> > You showed a simple formula (two, actually) for symmetric
> > Euclidean distance with three factors. Is there a general
> > formula for any number of factors?
>
> It's easy to make all odd primes symmetrical and of the same length,
> resulting in an An lattice. I can't see much point in this, because
> the symmetries become unrealistic after the 7-limit. More reasonable
> it seems to me is to break symmetry to some degree; for instance in
> the 13-limit one might decide to make 7, 11, and 13 symmetrical, and 3
> and 5 symmetrical, with 9 and 15 being the same as 7, 11, or 13.

Why not just use the 13-limit 'Kees' lattice?

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> Why not just use the 13-limit 'Kees' lattice?

Because it's not symmetrical.

>> >It's easy to make all odd primes symmetrical and of the same length,
>> >resulting in an An lattice.
>>
>> Is there a formula like
>>
>> ||(a,b,c)||_p = ((|a|^p+|b|^p+|c|^p+|a+b+c|^p)/2)^(1/p)
>>
>> for it?
>
>That's the one; good for any prime limit if generalized in the obvious
>way.

Just to be sure, is this right

sqrt( a^2 + b^2 ... + (a + b ...)^2 )

?

>> >I can't see much point in this, because the symmetries become
>> >unrealistic after the 7-limit.
>>
>> Does this have to do with nonprime odds? Can you give an
>> example of a bad result caused by them?
>
>If you have a 9-limit chord then applying the symmetry could net you a
>121-limit chord, which is hardly a similar thing musically speaking.

Hm. Does this mean 9 has the same length as 121?

-Carl

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

> Just to be sure, is this right
>
> sqrt( a^2 + b^2 ... + (a + b ...)^2 )

Yep.

> >If you have a 9-limit chord then applying the symmetry could net you a
> >121-limit chord, which is hardly a similar thing musically speaking.
>
> Hm. Does this mean 9 has the same length as 121?

If you make 3 symmetrical with 11, it will.

>> Just to be sure, is this right
>>
>> sqrt( a^2 + b^2 ... + (a + b ...)^2 )
>
>Yep.

Thanks!

>> >If you have a 9-limit chord then applying the symmetry could net you a
>> >121-limit chord, which is hardly a similar thing musically speaking.
>>
>> Hm. Does this mean 9 has the same length as 121?
>
>If you make 3 symmetrical with 11, it will.

Eh, my detergent causes me more trouble. :)

-Carl

At 05:36 PM 11/14/2005, you wrote:
>>> Just to be sure, is this right
>>>
>>> sqrt( a^2 + b^2 ... + (a + b ...)^2 )
>>
>>Yep.
>
>Thanks!

Drat, I think there's a /2 missing here.

Is *this* right:

sqrt( (a^2 + b^2 ... + (a + b ...)^2) / 2) )

?

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
>
> At 05:36 PM 11/14/2005, you wrote:
> >>> Just to be sure, is this right
> >>>
> >>> sqrt( a^2 + b^2 ... + (a + b ...)^2 )
> >>
> >>Yep.
> >
> >Thanks!
>
> Drat, I think there's a /2 missing here.
>
> Is *this* right:
>
> sqrt( (a^2 + b^2 ... + (a + b ...)^2) / 2) )

Even better, but it's just a constant factor. This way the distiance
to the tonality diamond notes from the unison is 1, not sqrt(2).

>> At 05:36 PM 11/14/2005, you wrote:
>> >>> Just to be sure, is this right
>> >>>
>> >>> sqrt( a^2 + b^2 ... + (a + b ...)^2 )
>> >>
>> >>Yep.
>> >
>> >Thanks!
>>
>> Drat, I think there's a /2 missing here.
>>
>> Is *this* right:
>>
>> sqrt( (a^2 + b^2 ... + (a + b ...)^2) / 2) )
>
>Even better, but it's just a constant factor. This way the distiance
>to the tonality diamond notes from the unison is 1, not sqrt(2).

I've got it working in scheme with the rest of my jazz.

-Carl

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > Why not just use the 13-limit 'Kees' lattice?
>
> Because it's not symmetrical.

What's so great about this symmmetry if you have to choose an arbitrary
place to break it (such as between 5 and 7 in your example) anyway? At
least the balls of "odd limit" are perfectly symmetrical in the 'Kees'
lattice (by which I mean the Euclidean embedding Carl and I have been
discussing).

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> What's so great about this symmmetry if you have to choose an arbitrary
> place to break it (such as between 5 and 7 in your example) anyway?

It allows you to play Smith-Miller type games with 13-limit JI, for
one thing.

At
> least the balls of "odd limit" are perfectly symmetrical in the 'Kees'
> lattice (by which I mean the Euclidean embedding Carl and I have been
> discussing).

I haven't been able to figure out what you've been disucssing, because
I don't know what either of you mean by "lattice", as you refuse to
use standard math terminology.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > What's so great about this symmmetry if you have to choose an
arbitrary
> > place to break it (such as between 5 and 7 in your example)
anyway?
>
> It allows you to play Smith-Miller type games with 13-limit JI, for
> one thing.

What are Smith-Miller type games.

> At
> > least the balls of "odd limit" are perfectly symmetrical in
the 'Kees'
> > lattice (by which I mean the Euclidean embedding Carl and I have
been
> > discussing).
>
> I haven't been able to figure out what you've been disucssing,

Why not interject with questions when you get lost/confused.

> because
> I don't know what either of you mean by "lattice",

Sure you do:

http://mathworld.wolfram.com/PointLattice.html

> as you refuse to
> use standard math terminology.

Unbelievable!

Take a look at this page again:

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

In particular, the third-to-last and second-to-last diagrams (and of
course the formulas and text). Do these make sense to you?

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> What are Smith-Miller type games.

As in "hexany phrase". You get a nontrivial permutation group on pitch
classes from isogenies of the lattice, and lift these to permutations
on JI notes.

> > because
> > I don't know what either of you mean by "lattice",
>
> Sure you do:
>
> http://mathworld.wolfram.com/PointLattice.html

That definition, which goes "Formally, a lattice is a discrete
subgroup of Euclidean space, assuming it contains the origin. That is,
a lattice is closed under addition and inverses, and every point has a
neighborhood in which it is the only lattice point" is precisely what
you've explicitly and emphatically rejected. For our purposes, a
generalization where instead of "Euclidean space" we say "real normed
vector space" or "finite-dimensional real Banach space" is what we
want; this *is* still standard terminology. But you don't want that
either, leaving me with no clue what you mean. This is *why*
mathematicians prefer precise definitions.

> > as you refuse to
> > use standard math terminology.
>
> Unbelievable!

If you did not intend to reject standard math terminology, then why
did you make a huge point of saying that's what you are doing?

> Take a look at this page again:
>
> http://www.kees.cc/tuning/lat_perbl.html

I've seen it before, but it makes little sense if you don't agree on
what a "lattice" even is.

> In particular, the third-to-last and second-to-last diagrams (and of
> course the formulas and text). Do these make sense to you?

Without definitions they are just pretty pictures.

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

> >> Is *this* right:
> >>
> >> sqrt( (a^2 + b^2 ... + (a + b ...)^2) / 2) )
> >
> >Even better, but it's just a constant factor. This way the distiance
> >to the tonality diamond notes from the unison is 1, not sqrt(2).
>
> I've got it working in scheme with the rest of my jazz.

As far as I can see, this is key to constructing a Euclidean lattice
which most closely resembles the Kees lattice. Set

u3 = log3(2)*a3 ... up = log2(p)*ap

where the ai are prime exponents/monzo coefficients. Then take

sqrt((u3^2 + u5^2 + ... + up^2)+(u3+u5+...+up)^2)/2)

as your ersatz Kees norm.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > What are Smith-Miller type games.
>
> As in "hexany phrase". You get a nontrivial permutation group on
pitch
> classes from isogenies of the lattice, and lift these to
permutations
> on JI notes.

Wouldn't quite a number of different lattices be useful for this?

> > > because
> > > I don't know what either of you mean by "lattice",
> >
> > Sure you do:
> >
> > http://mathworld.wolfram.com/PointLattice.html
>
> That definition, which goes "Formally, a lattice is a discrete
> subgroup of Euclidean space, assuming it contains the origin. That
is,
> a lattice is closed under addition and inverses, and every point
has a
> neighborhood in which it is the only lattice point" is precisely
what
> you've explicitly and emphatically rejected. For our purposes, a
> generalization where instead of "Euclidean space" we say "real
normed
> vector space" or "finite-dimensional real Banach space" is what we
> want; this *is* still standard terminology. But you don't want that
> either, leaving me with no clue what you mean. This is *why*
> mathematicians prefer precise definitions.

For the purposes of this thread, Euclidean space is where we want to
be. Again, I see no reason why different notions of "distance", such
as taxicab etc., can't happily coexist in a Euclidean lattice.

> > > as you refuse to
> > > use standard math terminology.
> >
> > Unbelievable!
>
> If you did not intend to reject standard math terminology, then why
> did you make a huge point of saying that's what you are doing?

I made a huge point os saying that I'm rejecting standard math
terminology? Where?

> > Take a look at this page again:
> >
> > http://www.kees.cc/tuning/lat_perbl.html
>
> I've seen it before, but it makes little sense if you don't agree on
> what a "lattice" even is.

Why not? Where does the sense/meaning come into question?

> > In particular, the third-to-last and second-to-last diagrams (and
of
> > course the formulas and text). Do these make sense to you?
>
> Without definitions they are just pretty pictures.

Incredible. Where, in your opinion, does this page stop being
meaningful, and what alternative meanings could it be said to have
under what alternative definitions of things?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
>
> > >> Is *this* right:
> > >>
> > >> sqrt( (a^2 + b^2 ... + (a + b ...)^2) / 2) )
> > >
> > >Even better, but it's just a constant factor. This way the
distiance
> > >to the tonality diamond notes from the unison is 1, not sqrt(2).
> >
> > I've got it working in scheme with the rest of my jazz.
>
> As far as I can see, this is key to constructing a Euclidean lattice
> which most closely resembles the Kees lattice. Set
>
> u3 = log3(2)*a3 ... up = log2(p)*ap
>
> where the ai are prime exponents/monzo coefficients. Then take
>
> sqrt((u3^2 + u5^2 + ... + up^2)+(u3+u5+...+up)^2)/2)
>
> as your ersatz Kees norm.

In the 5-limit case, does this agree with what Kees proposed and what
I latched onto, which is taking the square lattice and transforming
using the matrix:

[log3 log(5)/2]
[0 sqrt(3)log(5)/2]

?
Because what we want are for the expressibility balls to show up as
regular hexagons, right?

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> >
> > --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >
> > > >> Is *this* right:
> > > >>
> > > >> sqrt( (a^2 + b^2 ... + (a + b ...)^2) / 2) )
> > > >
> > > >Even better, but it's just a constant factor. This way the
> distiance
> > > >to the tonality diamond notes from the unison is 1, not sqrt
(2).
> > >
> > > I've got it working in scheme with the rest of my jazz.
> >
> > As far as I can see, this is key to constructing a Euclidean
lattice
> > which most closely resembles the Kees lattice. Set
> >
> > u3 = log3(2)*a3 ... up = log2(p)*ap
> >
> > where the ai are prime exponents/monzo coefficients. Then take
> >
> > sqrt((u3^2 + u5^2 + ... + up^2)+(u3+u5+...+up)^2)/2)
> >
> > as your ersatz Kees norm.
>
> In the 5-limit case, does this agree with what Kees proposed and
what
> I latched onto, which is taking the square lattice and transforming
> using the matrix:
>
> [log3 log(5)/2]
> [0 sqrt(3)log(5)/2]
>

Sorry; I meant log(3) but wrote log3 by mistake.

> ?
> Because what we want are for the expressibility balls to show up as
> regular hexagons, right?
>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> > u3 = log3(2)*a3 ... up = log2(p)*ap
> >
> > where the ai are prime exponents/monzo coefficients. Then take
> >
> > sqrt((u3^2 + u5^2 + ... + up^2)+(u3+u5+...+up)^2)/2)
> >
> > as your ersatz Kees norm.
>
> In the 5-limit case, does this agree with what Kees proposed and what
> I latched onto, which is taking the square lattice and transforming
> using the matrix:
>
> [log3 log(5)/2]
> [0 sqrt(3)log(5)/2]

Nope.

> Because what we want are for the expressibility balls to show up as
> regular hexagons, right?

I don't know what an expressibility ball is, but my ersatz Kees metric
gives the same result as Kees for certain intervals (5/4, 3/2, 7/4,
9/8, 128/125, etc) and is always greater than or equal to the Kees value.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> Incredible.

I don't think that word means what you think it means.

> Where, in your opinion, does this page stop being
> meaningful, and what alternative meanings could it be said to have
> under what alternative definitions of things?

The last lattice picture isn't even a picture of a lattice. Which
intervals do you want to place in a hexagon around the origin, and why
isn't this just the A3 lattice?

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

> I don't know what an expressibility ball is, but my ersatz Kees metric
> gives the same result as Kees for certain intervals (5/4, 3/2, 7/4,
> 9/8, 128/125, etc) and is always greater than or equal to the Kees
value.

Less than pr equal to.

>> At least the balls of "odd limit" are perfectly symmetrical in
>> the 'Kees' lattice (by which I mean the Euclidean embedding Carl
>> and I have been discussing).
>
>I haven't been able to figure out what you've been disucssing, because
>I don't know what either of you mean by "lattice", as you refuse to
>use standard math terminology.

What terminology should we use? As discussed before, it seems
"point lattice" isn't far off...

-Carl

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

> What terminology should we use? As discussed before, it seems
> "point lattice" isn't far off...

In math, it's usual to assume you mean "Euclidean lattice" unless you
say otherwise. For our purposes, "TOP lattice" and "Kees lattice"
would be non-Euclidean lattices, in TOP space and Kees space
respectively. "Point lattice" is something you might use if there is a
chance of confusing it with the partially ordered sets of "lattice
theory", which is another sense of the word.

If people have some other sense of lattice in mind, I wish they would
say so and give a definition. I've not been assuming Paul means
"lattice" in the sense of "point lattice", since he was so emphatic
about rejecting that, with a long lecture about how mathematicians
don't own the word. But if you don't use standard terminology you
should define your meaning.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > > u3 = log3(2)*a3 ... up = log2(p)*ap
> > >
> > > where the ai are prime exponents/monzo coefficients. Then take
> > >
> > > sqrt((u3^2 + u5^2 + ... + up^2)+(u3+u5+...+up)^2)/2)
> > >
> > > as your ersatz Kees norm.
> >
> > In the 5-limit case, does this agree with what Kees proposed and
what
> > I latched onto, which is taking the square lattice and
transforming
> > using the matrix:
> >
> > [log3 log(5)/2]
> > [0 sqrt(3)log(5)/2]
>
> Nope.

That's unfortunate.

> > Because what we want are for the expressibility balls to show up
as
> > regular hexagons, right?
>
> I don't know what an expressibility ball is,

Should I say "Kees ball"?

> but my ersatz Kees metric
> gives the same result as Kees for certain intervals (5/4, 3/2, 7/4,
> 9/8, 128/125, etc) and is always greater than or equal to the Kees
>value.

The transformation matrix I reproduced above should allow a regular
hexagon norm to give exactly the same result as Kees expressibility
for *all* intervals. Doesn't it?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > Incredible.
>
> I don't think that word means what you think it means.
>
> > Where, in your opinion, does this page stop being
> > meaningful, and what alternative meanings could it be said to have
> > under what alternative definitions of things?
>
> The last lattice picture isn't even a picture of a lattice.

Sir, I specifically was *not* talking about the last lattice picture.
Let me repeat: I don't agree with anything on that page after and
including the word "paradox". Did you pay attention when I told you
which parts of that page I was referring to?

> Which
> intervals do you want to place in a hexagon around the origin,

Those with Kees expressibility less than any given value.

> and why
> isn't this just the A3 lattice?

It seems the second-to-last lattice on Kees's page is the only one that
accomplishes this goal. Do you disagree?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
>
> > What terminology should we use? As discussed before, it seems
> > "point lattice" isn't far off...
>
> In math, it's usual to assume you mean "Euclidean lattice" unless
you
> say otherwise. For our purposes, "TOP lattice" and "Kees lattice"
> would be non-Euclidean lattices, in TOP space and Kees space
> respectively. "Point lattice" is something you might use if there
is a
> chance of confusing it with the partially ordered sets of "lattice
> theory", which is another sense of the word.
>
> If people have some other sense of lattice in mind, I wish they
would
> say so and give a definition. I've not been assuming Paul means
> "lattice" in the sense of "point lattice", since he was so emphatic
> about rejecting that, with a long lecture about how mathematicians
> don't own the word.

Gene, for the nth time, my "lecture" was in response to other people
who were giving the partially ordered set definition of lattice on
the tuning list:

http://mathworld.wolfram.com/Lattice.html

It's very frustrating that your misinterpretations of me stick in
your memory for years, despite my repeated attempts to correct them,
while you can't even remember for 1 day which parts of Kees's page I
told you to look at.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> Gene, for the nth time, my "lecture" was in response to other people
> who were giving the partially ordered set definition of lattice on
> the tuning list:

When you say lattice, what is it you mean--and do you consistently
stick to that?

> It's very frustrating that your misinterpretations of me stick in
> your memory for years, despite my repeated attempts to correct them,
> while you can't even remember for 1 day which parts of Kees's page I
> told you to look at.

Can't you simply explain what it is you want to do with out a lot of
damned pictures? This is like trying to read Erv Wilson.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> > Which
> > intervals do you want to place in a hexagon around the origin,
>
> Those with Kees expressibility less than any given value.

Is the plan to find a Euclidean lattice such that everything with
||v||_kees < M is inside a regular hexagon, for any M?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > Gene, for the nth time, my "lecture" was in response to other
people
> > who were giving the partially ordered set definition of lattice
on
> > the tuning list:
>
> When you say lattice, what is it you mean--and do you consistently
> stick to that?

Point lattice. Remember when I left the tuning list for a long while
after some very frustrating posts from you?

> > It's very frustrating that your misinterpretations of me stick in
> > your memory for years, despite my repeated attempts to correct
them,
> > while you can't even remember for 1 day which parts of Kees's
page I
> > told you to look at.
>
> Can't you simply explain what it is you want to do with out a lot of
> damned pictures? This is like trying to read Erv Wilson.

I find Kees's text perfectly straightforward. We want put 5-limit JI
on a lattice so that regular hexagons around the origin enclose all
the notes, and only those notes, below a certain expressibility. The
third lattice on the page (and only that lattice) accomplishes this,
does it not?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > > Which
> > > intervals do you want to place in a hexagon around the origin,
> >
> > Those with Kees expressibility less than any given value.
>
> Is the plan to find a Euclidean lattice such that everything with
> ||v||_kees < M is inside a regular hexagon, for any M?

It's not just a plan; it's a done deal. I find it amazing how glacially
slow communication on this list is (but I love you anyway). When did I
first bring this up?

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> I find Kees's text perfectly straightforward. We want put 5-limit JI
> on a lattice so that regular hexagons around the origin enclose all
> the notes, and only those notes, below a certain expressibility.

Then put 3/2 and 5/4 at 60 degrees from each other, and scale then so
that the length of 3/2 to 5/4 is in the proportion 1:log3(5).

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > I find Kees's text perfectly straightforward. We want put 5-limit
JI
> > on a lattice so that regular hexagons around the origin enclose all
> > the notes, and only those notes, below a certain expressibility.
>
> Then put 3/2 and 5/4 at 60 degrees from each other, and scale then so
> that the length of 3/2 to 5/4 is in the proportion 1:log3(5).

Looks about right by eye. Is that what you get when you start with the
square lattice and then transform using the matrix

[log(3) log(5)/2]
[0 sqrt(3)log(5)/2]

?

If so, you're agreeing with what Kees says (and depicts) on his page. I
don't see how Kees could have been any clearer about this there.

(http://www.kees.cc/tuning/lat_perbl.html)

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> Looks about right by eye. Is that what you get when you start with the
> square lattice and then transform using the matrix
>
> [log(3) log(5)/2]
> [0 sqrt(3)log(5)/2]
>
> ?

I was reading the rows before, but looking at the columns gives
a vector of length log(3) and a vector of length log(5), which is what
you want. The angle between them is 60 degrees, which is also correct.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> I don't see how Kees could have been any clearer about this there.
>
> (http://www.kees.cc/tuning/lat_perbl.html)

One way it would be clearer is if it said somewhere what the page was
about and what, exactly, these manipulations were intended to
accomplish. Another way it could be clear is if we were told the
*columns* of the second transformation matrix were supposed to be
generators of the 5-limit lattice of pitch classes, with the first
column being the "3" column and the second column being the "5" column.
Pointing out that these vectors are of length log(3) and log(5)
respectively, and that they are at an angle of 60 degrees from each
other, would not hurt. Finally, there is a lot of extraneous matter on
the page which serves no clear purpose I can see.

>> I find Kees's text perfectly straightforward. We want put 5-limit JI
>> on a lattice so that regular hexagons around the origin enclose all
>> the notes, and only those notes, below a certain expressibility.
>
>Then put 3/2 and 5/4 at 60 degrees from each other, and scale then so
>that the length of 3/2 to 5/4 is in the proportion 1:log3(5).

If you mean 1:log5(3), I think that's the "isosceles" measure we've
been discussing, which doesn't match up to expressibility, as Paul
shows...

http://kees.cc/tuning/erl_perbl.html

-Carl

>>> I find Kees's text perfectly straightforward. We want put 5-limit JI
>>> on a lattice so that regular hexagons around the origin enclose all
>>> the notes, and only those notes, below a certain expressibility.
>>
>>Then put 3/2 and 5/4 at 60 degrees from each other, and scale then so
>>that the length of 3/2 to 5/4 is in the proportion 1:log3(5).
>
>If you mean 1:log5(3),

Ack, the 1: means you were right.

>I think that's the "isosceles" measure we've been discussing,

Bzz, sorry.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
>
> >> I find Kees's text perfectly straightforward. We want put 5-limit JI
> >> on a lattice so that regular hexagons around the origin enclose all
> >> the notes, and only those notes, below a certain expressibility.
> >
> >Then put 3/2 and 5/4 at 60 degrees from each other, and scale then so
> >that the length of 3/2 to 5/4 is in the proportion 1:log3(5).
>
> If you mean 1:log5(3), I think that's the "isosceles" measure we've
> been discussing, which doesn't match up to expressibility, as Paul
> shows...

log5(3):1 amounts to the same thing as 1:log3(5). Why would you want
to make the third a smaller distance than the fifth?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > Looks about right by eye. Is that what you get when you start with
the
> > square lattice and then transform using the matrix
> >
> > [log(3) log(5)/2]
> > [0 sqrt(3)log(5)/2]
> >
> > ?
>
> I was reading the rows before, but looking at the columns gives
> a vector of length log(3) and a vector of length log(5), which is what
> you want.

So this *does* agree with the formula you proposed for Carl? When you
said earlier that it doesn't agree, I wrote, "that's unfortunate". Are
we now fortunate enough to be able to report agreement instead?

> The angle between them is 60 degrees, which is also correct.

Whew! And it's trivial to extend this to any prime limit, of any number
of dimensions, right?

P.S. I had earlier guessed that a ball in the 7-limit case would be a
rhombic dodecahedron but would it actually be a cuboctahedron?

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
>
> >> I find Kees's text perfectly straightforward. We want put 5-limit
JI
> >> on a lattice so that regular hexagons around the origin enclose
all
> >> the notes, and only those notes, below a certain expressibility.
> >
> >Then put 3/2 and 5/4 at 60 degrees from each other, and scale then so
> >that the length of 3/2 to 5/4 is in the proportion 1:log3(5).
>
> If you mean 1:log5(3), I think that's the "isosceles" measure we've
> been discussing,

Nope -- the angle between 3:2 and 5:4 is *not* 60 degrees in
the "isosceles" lattice.

> which doesn't match up to expressibility, as Paul
> shows...
>
> http://kees.cc/tuning/erl_perbl.html
>
> -Carl

Carl, you're so close and yet so far! Look again at this page:

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

The "isosceles" case is the second lattice on this page, while what
Gene proposes above is the third lattice on this page. Kees and Gene
both seem to be saying that the latter *does* have expressibility
boundaries agreeing with regular hexagons around the origin, which is
as close to circles (Euclidean balls) as we can get this to be on a
(Euclidean) lattice.

>> If you mean 1:log5(3), I think that's the "isosceles" measure we've
>> been discussing,
>
>Nope -- the angle between 3:2 and 5:4 is *not* 60 degrees in
>the "isosceles" lattice.
>
>> which doesn't match up to expressibility, as Paul
>> shows...
>>
>> http://kees.cc/tuning/erl_perbl.html
>>
>> -Carl
>
>Carl, you're so close and yet so far! Look again at this page:
>
>http://www.kees.cc/tuning/lat_perbl.html
>
>The "isosceles" case is the second lattice on this page, while what
>Gene proposes above is the third lattice on this page. Kees and Gene
>both seem to be saying that the latter *does* have expressibility
>boundaries agreeing with regular hexagons around the origin, which is
>as close to circles (Euclidean balls) as we can get this to be on a
>(Euclidean) lattice.

I subsequently noticed this.

-Carl

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> Whew! And it's trivial to extend this to any prime limit, of any number
> of dimensions, right?

Yes, the generalization is immediate.

> P.S. I had earlier guessed that a ball in the 7-limit case would be a
> rhombic dodecahedron but would it actually be a cuboctahedron?

The vertices of the lattice points of minimal distance from the
unison, the 7-limit tonality diamond lattice points, are on the
vertices of a cuboctahedron. The Voronoi cells are rhombic
dodecahedra, with two kinds of verticies corresponding to deep
(hexany) holes and shallow (tetrad) holes.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > Whew! And it's trivial to extend this to any prime limit, of any
number
> > of dimensions, right?
>
> Yes, the generalization is immediate.
>
> > P.S. I had earlier guessed that a ball in the 7-limit case would
be a
> > rhombic dodecahedron but would it actually be a cuboctahedron?
>
> The vertices of the lattice points of minimal distance from the
> unison, the 7-limit tonality diamond lattice points, are on the
> vertices of a cuboctahedron.

Not what I was asking, but anyway, now it sounds like you're talking
about the old equilateral triangular latice. In the 5-limit version
of the lattice we were actually talking about, the 5-limit tonality
diamond lattice points are *not* on the vertices of a regular
hexagon. So what you say about the 7-limit is impossible.

The question is what shape a ball of *any* expressibility value would
be in this 7-limit generalization of the 5-limit case where the
expressibility balls are regular hexagons (but no set of lattice
points actually lies on the vertices of a regular hexagon).

> The Voronoi cells are rhombic
> dodecahedra, with two kinds of verticies corresponding to deep
> (hexany) holes and shallow (tetrad) holes.

Not what I was asking.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> wrote:
> >
> > > Whew! And it's trivial to extend this to any prime limit, of
any
> number
> > > of dimensions, right?
> >
> > Yes, the generalization is immediate.
> >
> > > P.S. I had earlier guessed that a ball in the 7-limit case
would
> be a
> > > rhombic dodecahedron but would it actually be a cuboctahedron?
> >
> > The vertices of the lattice points of minimal distance from the
> > unison, the 7-limit tonality diamond lattice points, are on the
> > vertices of a cuboctahedron.
>
> Not what I was asking, but anyway, now it sounds like you're
talking
> about the old equilateral triangular latice. In the 5-limit version
> of the lattice we were actually talking about, the 5-limit tonality
> diamond lattice points are *not* on the vertices of a regular
> hexagon. So what you say about the 7-limit is impossible.

That is, if I interpret "cuboctahedron" to mean "semiregular
cuboctahedron". Which I probably sholdn't have done. So it may not
be 'impossible'. But note that these 'lattice points of minimal
distance from the unison' are at different distances from the unison -
- doesn't minimal distance imply they'd all be the same?

> The question is what shape a ball of *any* expressibility value
would
> be in this 7-limit generalization of the 5-limit case where the
> expressibility balls are regular hexagons (but no set of lattice
> points actually lies on the vertices of a regular hexagon).
>
> > The Voronoi cells are rhombic
> > dodecahedra, with two kinds of verticies corresponding to deep
> > (hexany) holes and shallow (tetrad) holes.
>
> Not what I was asking.

But there are more than two kinds of vertices since these Voronoi
cells wouldn't be 'regular' rhombic dodecahedra in the lattice we
were talking about.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> > The vertices of the lattice points of minimal distance from the
> > unison, the 7-limit tonality diamond lattice points, are on the
> > vertices of a cuboctahedron.
>
> Not what I was asking, but anyway, now it sounds like you're talking
> about the old equilateral triangular latice. In the 5-limit version
> of the lattice we were actually talking about, the 5-limit tonality
> diamond lattice points are *not* on the vertices of a regular
> hexagon. So what you say about the 7-limit is impossible.

I don't know what definition you use, or what's regarded as correct,
but what I mean by the 5-limit diamond is the hexagon of pitch classes
{6/5, 5/4, 4/3, 3/2, 8/5, 5/3} surrounding 1; hence, including 1, the
pitch classes {1, 6/5, 5/4, 4/3, 3/2, 8/5, 5/3}.

> The question is what shape a ball of *any* expressibility value would
> be in this 7-limit generalization of the 5-limit case where the
> expressibility balls are regular hexagons (but no set of lattice
> points actually lies on the vertices of a regular hexagon).

Your question makes no sense to me.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> That is, if I interpret "cuboctahedron" to mean "semiregular
> cuboctahedron". Which I probably sholdn't have done. So it may not
> be 'impossible'. But note that these 'lattice points of minimal
> distance from the unison' are at different distances from the unison -
> - doesn't minimal distance imply they'd all be the same?

Yes, it implies they are the same and no, they are not at different
distances. They are all at a distance of 1 from the unison.

> But there are more than two kinds of vertices since these Voronoi
> cells wouldn't be 'regular' rhombic dodecahedra in the lattice we
> were talking about.

I'm not at all clear what you are talking about, but it doesn't seem
to be what I mean by a symmetric lattice.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > > The vertices of the lattice points of minimal distance from the
> > > unison, the 7-limit tonality diamond lattice points, are on the
> > > vertices of a cuboctahedron.
> >
> > Not what I was asking, but anyway, now it sounds like you're
talking
> > about the old equilateral triangular latice. In the 5-limit
version
> > of the lattice we were actually talking about, the 5-limit
tonality
> > diamond lattice points are *not* on the vertices of a regular
> > hexagon. So what you say about the 7-limit is impossible.
>
> I don't know what definition you use, or what's regarded as correct,
> but what I mean by the 5-limit diamond is the hexagon of pitch
classes
> {6/5, 5/4, 4/3, 3/2, 8/5, 5/3} surrounding 1; hence, including 1,
the
> pitch classes {1, 6/5, 5/4, 4/3, 3/2, 8/5, 5/3}.

That's what I mean by it too, and what I thought you meant by it. In
the lattice we've been talking to each other about in this thread,
these points do *not* form a regular hexagon.

> > The question is what shape a ball of *any* expressibility value
would
> > be in this 7-limit generalization of the 5-limit case where the
> > expressibility balls are regular hexagons (but no set of lattice
> > points actually lies on the vertices of a regular hexagon).
>
> Your question makes no sense to me.

Where did you lose me? I thought we were on the same page here for a
bit. In the 5-limit case, a regular hexagon centered around the
origin in this lattice forms a ball for some expressibility value no
matter how big the hexagon is. You described the 60-degree angle
between, and the log(p) relative lengths of, the intervals octave-
equivalent to primes p in this lattice. I asked if this extended in
the obvious way to higher prime limits, and you said it did. That
means to me that there's some highly symmetrical (or "spherical") 3D
figure that functions as the regular hexagon did in the 2D case (or
did I make an incorrect inference there?). And I'm asking you what
that figure is.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > That is, if I interpret "cuboctahedron" to mean "semiregular
> > cuboctahedron". Which I probably sholdn't have done. So it may
not
> > be 'impossible'. But note that these 'lattice points of minimal
> > distance from the unison' are at different distances from the
unison -
> > - doesn't minimal distance imply they'd all be the same?
>
> Yes, it implies they are the same and no, they are not at different
> distances. They are all at a distance of 1 from the unison.

Clearly that's not the case.

> > But there are more than two kinds of vertices since these Voronoi
> > cells wouldn't be 'regular' rhombic dodecahedra in the lattice we
> > were talking about.
>
> I'm not at all clear what you are talking about, but it doesn't seem
> to be what I mean by a symmetric lattice.

If you scroll up (click on "up thread") in this thread, we were *not*
talking about the symmetric lattice here. I don't know where you
shifted gears, but I didn't.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> I asked if this extended in
> the obvious way to higher prime limits, and you said it did. That
> means to me that there's some highly symmetrical (or "spherical") 3D
> figure that functions as the regular hexagon did in the 2D case (or
> did I make an incorrect inference there?). And I'm asking you what
> that figure is.

The cubeoctahedron, of course.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> That's what I mean by it too, and what I thought you meant by it. In
> the lattice we've been talking to each other about in this thread,
> these points do *not* form a regular hexagon.

Since you were talking about lattice points on the verticies of a
cubeoctahedron, I assumed you could not be talking about the Kees norm .

> Where did you lose me? I thought we were on the same page here for a
> bit. In the 5-limit case, a regular hexagon centered around the
> origin in this lattice forms a ball for some expressibility value no
> matter how big the hexagon is.

Yes, but the verticies are not lattice points.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > I asked if this extended in
> > the obvious way to higher prime limits, and you said it did. That
> > means to me that there's some highly symmetrical (or "spherical")
3D
> > figure that functions as the regular hexagon did in the 2D case (or
> > did I make an incorrect inference there?). And I'm asking you what
> > that figure is.
>
> The cubeoctahedron, of course.

Thank you! So my initial statement that it was a rhombic dodecahedron
was wrong, and my subsequent guess that it was in fact a cuboctahedron
was right . . . A cool thing is that the 7-limit diamond and the 9-
limit diamond can each be enclosed in a centered-around-1/1 "regular"
cuboctahedron in the 7-limit version of this lattice. Also of interest
are the higher-dimensional analogues, though of course our power of
visualization rapidly decreases in these cases.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > That's what I mean by it too, and what I thought you meant by it.
In
> > the lattice we've been talking to each other about in this
thread,
> > these points do *not* form a regular hexagon.
>
> Since you were talking about lattice points on the verticies of a
> cubeoctahedron,

I was? Could you point to where? I don't think you're right. *You*
were the one who brought up lattice points on the vertices . .

> I assumed you could not be talking about the Kees norm .
>
> > Where did you lose me? I thought we were on the same page here
for a
> > bit. In the 5-limit case, a regular hexagon centered around the
> > origin in this lattice forms a ball for some expressibility value
no
> > matter how big the hexagon is.
>
> Yes, but the verticies are not lattice points.

True, and I never said they were.