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What the numbers mean

🔗Gene Ward Smith <gwsmith@svpal.org>

1/29/2004 3:20:37 AM

Here is a no doubt long overdue discussion of what the numbers in the
wedgie for a linear temperament mean.

The wedgie for miracle is <<6 -7 -2 -25 -20 15||. Normally we would
content ourselves with saying the mapping to primes is given by
[<1 1 3 3|, <0 6 -7 -2|], but that is a 2-centric way of putting it;
we have other maps for 3 and generator, 5 and generator, and 7 and
generator:

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

The reason for the second way of putting the generator is that this is
now in terms of 3, 5, or 7 of the <72 114 167 202| val. So, for
instance, we can consider miracle in terms of a generator which is 11
steps out of 114 representing 3, or 135 out of 167 steps representing
5, or 182 out of 202 steps representing 7.

If we look at the absolute values of the numbers in the miracle
wedgie, 6 is the number of generators to get to a 3 in a 2&g system,
or to get to a 2 in a 3&g system, or to get to 3/2 in either.
Similarly, 7 is the number of generator steps to 5/2, 2 is the number
of generator steps to 7/2, 25 is the number of generator steps to 5/3,
20 the number to 7/3 and 15 the number to 7/5, in the systems whose
period is one of the two primes in the ratio. So, the last three
numbers of the wedgie can be related to 5/3, 7/3, 7/5 just as the
first three can to 3/2, 5/2, 7/2.

If we take absolute values and normalize each by dividing by the log
base 2 of the two primes involved, we get

[6/p3 7/p5 2/p7 25/(p3p5) 20/(p3p7) 15/(p5p7)] =

[3.785578521, 3.014735907, .7124143742, 6.793166368, 4.494834256,
2.301151281]

The complexity is defined by the weighted value for 5/3, which is the
worst case (as, in fact, it clearly is.)

🔗Paul Erlich <perlich@aya.yale.edu>

1/29/2004 1:51:52 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> Here is a no doubt long overdue discussion of what the numbers in
the
> wedgie for a linear temperament mean.

I've noticed that the matrix of linear temperament commas with 0s
along the diagonal has both the upper triangle and lower triangle
very similar to the corresponding numbers in the wedgie (the latter
are divisible by the former, in fact).

> The wedgie for miracle is <<6 -7 -2 -25 -20 15||. Normally we would
> content ourselves with saying the mapping to primes is given by
> [<1 1 3 3|, <0 6 -7 -2|], but that is a 2-centric way of putting it;
> we have other maps for 3 and generator, 5 and generator, and 7 and
> generator:
>
> 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
>
> The reason for the second way of putting the generator is that this
is
> now in terms of 3, 5, or 7 of the <72 114 167 202| val. So, for
> instance, we can consider miracle in terms of a generator which is
11
> steps out of 114 representing 3, or 135 out of 167 steps
representing
> 5, or 182 out of 202 steps representing 7.
>
> If we look at the absolute values of the numbers in the miracle
> wedgie, 6 is the number of generators to get to a 3 in a 2&g system,
> or to get to a 2 in a 3&g system, or to get to 3/2 in either.
> Similarly, 7 is the number of generator steps to 5/2, 2 is the
number
> of generator steps to 7/2, 25 is the number of generator steps to
5/3,
> 20 the number to 7/3 and 15 the number to 7/5, in the systems whose
> period is one of the two primes in the ratio. So, the last three
> numbers of the wedgie can be related to 5/3, 7/3, 7/5 just as the
> first three can to 3/2, 5/2, 7/2.
>
> If we take absolute values and normalize each by dividing by the log
> base 2 of the two primes involved, we get
>
> [6/p3 7/p5 2/p7 25/(p3p5) 20/(p3p7) 15/(p5p7)] =
>
> [3.785578521, 3.014735907, .7124143742, 6.793166368, 4.494834256,
> 2.301151281]
>
> The complexity is defined by the weighted value for 5/3, which is
the
> worst case (as, in fact, it clearly is.)

Meaning what, exactly?

Can you show this process in action for the simpler, 3-limit and 5-
limit cases? And why do we take the worst case, instead of some sort
of product (which would appear to get you your spacial measure once
you've orthogonalized)?

Thanks for this post, though; I'll be referring to it in the future.

🔗Dave Keenan <d.keenan@bigpond.net.au>

1/29/2004 1:52:49 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> Here is a no doubt long overdue discussion of what the numbers in the
> wedgie for a linear temperament mean.
...

That was a good explanation. Thanks heaps Gene!

> If we take absolute values and normalize each by dividing by the log
> base 2 of the two primes involved, we get
>
> [6/p3 7/p5 2/p7 25/(p3p5) 20/(p3p7) 15/(p5p7)] =
>
> [3.785578521, 3.014735907, .7124143742, 6.793166368, 4.494834256,
> 2.301151281]
>
> The complexity is defined by the weighted value for 5/3, which is the
> worst case (as, in fact, it clearly is.)

You are dividing by the _product_ of the two logs, shouldn't you be
dividing by their _sum_ (the log of the product of the primes)?

🔗Paul Erlich <perlich@aya.yale.edu>

1/29/2004 1:57:51 PM

--- In tuning-math@yahoogroups.com, "Dave Keenan" <d.keenan@b...>
wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
> > Here is a no doubt long overdue discussion of what the numbers in
the
> > wedgie for a linear temperament mean.
> ...
>
> That was a good explanation. Thanks heaps Gene!
>
> > If we take absolute values and normalize each by dividing by the
log
> > base 2 of the two primes involved, we get
> >
> > [6/p3 7/p5 2/p7 25/(p3p5) 20/(p3p7) 15/(p5p7)] =
> >
> > [3.785578521, 3.014735907, .7124143742, 6.793166368, 4.494834256,
> > 2.301151281]
> >
> > The complexity is defined by the weighted value for 5/3, which is
the
> > worst case (as, in fact, it clearly is.)
>
> You are dividing by the _product_ of the two logs, shouldn't you be
> dividing by their _sum_ (the log of the product of the primes)?

Product makes a lot more sense to me than sum, in my vague intuitive
understanding of these things. You probably see that Gene's defining
p2 as 1. Now the wedgie represents a bivector in 4-dimensional space,
which means the relevant basis elements are *directed areas* in 4D,
of which there are of course 6 "multilinearly-independent" ones. Now,
the unit lengths in the lattice are scaled by (p2,) p3, p5, and p7,
so the unit areas would have to be scaled by the 6 possible
*products*, not *sums*, of these -- (p2)p3, (p2)p5, (p2)p7, p3p5,
p3p7, and p5p7.

🔗Gene Ward Smith <gwsmith@svpal.org>

1/29/2004 6:53:06 PM

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

> > The complexity is defined by the weighted value for 5/3, which is
> the
> > worst case (as, in fact, it clearly is.)
>
> Meaning what, exactly?

It's hard to find a reasonable take on septimal miracle which doesn't
have it that 5/3 is the most complex consonance.

> Can you show this process in action for the simpler, 3-limit and 5-
> limit cases? And why do we take the worst case, instead of some
sort
> of product (which would appear to get you your spacial measure once
> you've orthogonalized)?

"Orthogonalized" is one of those words which is holding up
communication, as I can only interpret that in terms of a L2 norm.
The vals, if we assume the Tenney metric, have an L_inf norm, and I
am regarding wedgies as multivals, hence the L_inf norm. It's logical
from a mathematical point of view, but not something we are forced to
beleive and adopt.

🔗Gene Ward Smith <gwsmith@svpal.org>

1/29/2004 6:55:59 PM

--- In tuning-math@yahoogroups.com, "Dave Keenan" <d.keenan@b...>
wrote:

> You are dividing by the _product_ of the two logs, shouldn't you be
> dividing by their _sum_ (the log of the product of the primes)?

No, not if we are using the dual norm induced on vals by the Tenney
norm. We can define that as normalizing the vals by dividing through
by log2(p); if we take products of pairs of these, which we do in
finding the wedgie, we end up dividing by the products of pairs of
primes.

🔗Herman Miller <hmiller@IO.COM>

1/29/2004 7:51:46 PM

On Thu, 29 Jan 2004 11:20:37 -0000, "Gene Ward Smith" <gwsmith@svpal.org>
wrote:

>Here is a no doubt long overdue discussion of what the numbers in the
>wedgie for a linear temperament mean.
>
>The wedgie for miracle is <<6 -7 -2 -25 -20 15||. Normally we would
>content ourselves with saying the mapping to primes is given by
>[<1 1 3 3|, <0 6 -7 -2|], but that is a 2-centric way of putting it;
>we have other maps for 3 and generator, 5 and generator, and 7 and
>generator:
>
>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.

--
see my music page ---> ---<http://www.io.com/~hmiller/music/index.html>--
hmiller (Herman Miller) "If all Printers were determin'd not to print any
@io.com email password: thing till they were sure it would offend no body,
\ "Subject: teamouse" / there would be very little printed." -Ben Franklin

🔗Dave Keenan <d.keenan@bigpond.net.au>

1/30/2004 12:56:35 AM

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:
> --- In tuning-math@yahoogroups.com, "Dave Keenan" <d.keenan@b...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> > wrote:
> > > Here is a no doubt long overdue discussion of what the numbers in
> the
> > > wedgie for a linear temperament mean.
> > ...
> >
> > That was a good explanation. Thanks heaps Gene!
> >
> > > If we take absolute values and normalize each by dividing by the
> log
> > > base 2 of the two primes involved, we get
> > >
> > > [6/p3 7/p5 2/p7 25/(p3p5) 20/(p3p7) 15/(p5p7)] =
> > >
> > > [3.785578521, 3.014735907, .7124143742, 6.793166368, 4.494834256,
> > > 2.301151281]
> > >
> > > The complexity is defined by the weighted value for 5/3, which is
> the
> > > worst case (as, in fact, it clearly is.)
> >
> > You are dividing by the _product_ of the two logs, shouldn't you be
> > dividing by their _sum_ (the log of the product of the primes)?
>
> Product makes a lot more sense to me than sum, in my vague intuitive
> understanding of these things. You probably see that Gene's defining
> p2 as 1. Now the wedgie represents a bivector in 4-dimensional space,
> which means the relevant basis elements are *directed areas* in 4D,
> of which there are of course 6 "multilinearly-independent" ones. Now,
> the unit lengths in the lattice are scaled by (p2,) p3, p5, and p7,
> so the unit areas would have to be scaled by the 6 possible
> *products*, not *sums*, of these -- (p2)p3, (p2)p5, (p2)p7, p3p5,
> p3p7, and p5p7.

Yes. I agree now. My confusion was caused by the omission of the
"p2"s, since this gave different dimensionality (in the physics sense
of "units") for the first three versus the last three.

🔗Paul Erlich <perlich@aya.yale.edu>

1/30/2004 1:52:42 PM

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

> > Can you show this process in action for the simpler, 3-limit and
5-
> > limit cases? And why do we take the worst case, instead of some
> sort
> > of product (which would appear to get you your spacial measure
once
> > you've orthogonalized)?
>
> "Orthogonalized" is one of those words which is holding up
> communication, as I can only interpret that in terms of a L2 norm.

Yes, but clearly there are *some* geometrical constructs that hold
even when you don't define angles, and this is, I believe,
what "affine geometry" refers to. The determinant gives you the
correct "area" (where there's one lattice node per unit area) of a
periodicity block regardless of the angles you assume.

I'll keep playing around on my own in an attempt to understand this.
If I construct a matrix of unison vectors (say, 3 of them for 12-
equal, or 4 of them for pajara), each of which has a 0 for a
different prime, so that I have 0s along the diagonal, the triangles
of numbers above and below the diagonal each look suspiciously like
the wedgie, and . . .

> The vals, if we assume the Tenney metric, have an L_inf norm, and I
> am regarding wedgies as multivals, hence the L_inf norm.

But didn't you say you could regard them as multimonzos instead? In
which case you would apply the L_1 norm? This confusion is why I
wanted you to provide the mathematical undergirding for my "Attn:
Gene 2" post, something I'm still strongly hoping you'll do.

🔗Paul Erlich <perlich@aya.yale.edu>

1/31/2004 2:28:05 AM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> wrote:
>
> > > The complexity is defined by the weighted value for 5/3, which
is
> > the
> > > worst case (as, in fact, it clearly is.)
> >
> > Meaning what, exactly?
>
> It's hard to find a reasonable take on septimal miracle which
doesn't
> have it that 5/3 is the most complex consonance.
>
> > Can you show this process in action for the simpler, 3-limit and
5-
> > limit cases? And why do we take the worst case, instead of some
> sort
> > of product (which would appear to get you your spacial measure
once
> > you've orthogonalized)?
>
> "Orthogonalized" is one of those words which is holding up
> communication, as I can only interpret that in terms of a L2 norm.

What if I just talk about "projections"? It seems the wedgie gives us
the size of the projection of the commas' bimonzo onto the
unit 'orthogonal' bimonzos of the lattice. If we use L_inf, two
temperaments with the same largest projection have the same
complexity. But clearly it requires more notes to fill the bimonzo if
its second-largest projection is nonzero. And in fact, in the taxicab
lattice, no matter how you shape it, you can't construct a bivector
with fewer notes than what you get by constructing the projections
and sticking them together like two walls and a floor in the 3D
case . . . So I see pretty clearly now that the L_1 norm of the
multimonzos is appropriate; unfortunately I think I just calculated
them wrong, so they might not actually disagree with:

> The vals, if we assume the Tenney metric, have an L_inf norm, and I
> am regarding wedgies as multivals,

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

2/20/2004 4:35:10 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> Here is a no doubt long overdue discussion of what the numbers in
the
> wedgie for a linear temperament mean.
>
> The wedgie for miracle is <<6 -7 -2 -25 -20 15||. Normally we would
> content ourselves with saying the mapping to primes is given by
> [<1 1 3 3|, <0 6 -7 -2|], but that is a 2-centric way of putting it;
> we have other maps for 3 and generator, 5 and generator, and 7 and
> generator:
>
> 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
>

Gene, is there a way to calculate the period values, or is it just
done "ad hoc?" (I had the same question regarding Graham's matrices,
I guess the answer would apply to both situations) Thanx

🔗Gene Ward Smith <gwsmith@svpal.org>

2/20/2004 5:32:31 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> Gene, is there a way to calculate the period values, or is it just
> done "ad hoc?" (I had the same question regarding Graham's matrices,
> I guess the answer would apply to both situations) Thanx

I don't know what you mean by the period values, but I'm sure there's
a way to calculate it. What, exactly, is a period value--the number
of periods to an octave in a linear temperament?

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

2/20/2004 6:38:04 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > Gene, is there a way to calculate the period values, or is it just
> > done "ad hoc?" (I had the same question regarding Graham's
matrices,
> > I guess the answer would apply to both situations) Thanx
>
> I don't know what you mean by the period values, but I'm sure
there's
> a way to calculate it. What, exactly, is a period value--the number
> of periods to an octave in a linear temperament?

I mean the like <1 1 3 3| below. So actually the number of periods
in a mapping (what I should have said). I this case of course they
are octaves.

[<1 1 3 3|, <0 6 -7 -2|],

- Paul

🔗Gene Ward Smith <gwsmith@svpal.org>

2/20/2004 7:20:33 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> I mean the like <1 1 3 3| below. So actually the number of periods
> in a mapping (what I should have said). I this case of course they
> are octaves.
>
> [<1 1 3 3|, <0 6 -7 -2|],

The first 1 of <1 1 3 3| is the number of periods per octave; in this
case the octave is the period. The others depend on precisely which
generator you choose--is it to be an approximate 15/14, or maybe a
28/15 or 15/7 instead? If it is a usual secor, s, then we must choose
a so that a+6s is an approximate 3, and so forth. Very recently I
posted how, if you know the wedgie, solving for x,y,z in
<1 x y z| ^ <0 6 -7 -2| = wedgie gives you the various possible
values, which is another approach. Finally, you can define a certain
set of vals directly from the wedgie, and Hermite-reduce a matrix of
these, and get one canonically-defined possibility.

So, it's not so much ad hoc as an embarrassent of riches.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

2/20/2004 7:28:36 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> The first 1 of <1 1 3 3| is the number of periods per octave; in
this
> case the octave is the period. The others depend on precisely which
> generator you choose--is it to be an approximate 15/14, or maybe a
> 28/15 or 15/7 instead? If it is a usual secor, s, then we must
choose
> a so that a+6s is an approximate 3, and so forth. Very recently I
> posted how, if you know the wedgie, solving for x,y,z in
> <1 x y z| ^ <0 6 -7 -2| = wedgie gives you the various possible
> values, which is another approach. Finally, you can define a
certain
> set of vals directly from the wedgie, and Hermite-reduce a matrix
of
> these, and get one canonically-defined possibility.
>
> So, it's not so much ad hoc as an embarrassent of riches.

Thanks. I still don't understand Hermite-reduction, (now I have two
reasons to learn it - this, and for extracting multiple commas, like
in 7-limit linear temperament). Proceeding backwards, I guess I'll
hunt for your post on solving for x,y,z, if it really was that recent.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

2/23/2004 5:32:05 AM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > I mean the like <1 1 3 3| below. So actually the number of periods
> > in a mapping (what I should have said). I this case of course
they
> > are octaves.
> >
> > [<1 1 3 3|, <0 6 -7 -2|],
>
> The first 1 of <1 1 3 3| is the number of periods per octave; in
this
> case the octave is the period. The others depend on precisely which
> generator you choose--is it to be an approximate 15/14, or maybe a
> 28/15 or 15/7 instead? If it is a usual secor, s, then we must
choose
> a so that a+6s is an approximate 3, and so forth. Very recently I
> posted how, if you know the wedgie, solving for x,y,z in
> <1 x y z| ^ <0 6 -7 -2| = wedgie gives you the various possible
> values, which is another approach. Finally, you can define a
certain
> set of vals directly from the wedgie, and Hermite-reduce a matrix
of
> these, and get one canonically-defined possibility.
>
> So, it's not so much ad hoc as an embarrassent of riches.

So, <1 x y z| ^ <0 6 -7 -2| = <<6 -7 -2 -25 -20 15||? That's all?

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

2/23/2004 10:38:27 AM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul.hjelmstad@u...> wrote:
> >
> > > I mean the like <1 1 3 3| below. So actually the number of
periods
> > > in a mapping (what I should have said). I this case of course
> they
> > > are octaves.
> > >
> > > [<1 1 3 3|, <0 6 -7 -2|],
> >
> > The first 1 of <1 1 3 3| is the number of periods per octave; in
> this
> > case the octave is the period. The others depend on precisely
which
> > generator you choose--is it to be an approximate 15/14, or maybe
a
> > 28/15 or 15/7 instead? If it is a usual secor, s, then we must
> choose
> > a so that a+6s is an approximate 3, and so forth. Very recently I
> > posted how, if you know the wedgie, solving for x,y,z in
> > <1 x y z| ^ <0 6 -7 -2| = wedgie gives you the various possible
> > values, which is another approach. Finally, you can define a
> certain
> > set of vals directly from the wedgie, and Hermite-reduce a matrix
> of
> > these, and get one canonically-defined possibility.
> >
> > So, it's not so much ad hoc as an embarrassent of riches.
>
> So, <1 x y z| ^ <0 6 -7 -2| = <<6 -7 -2 -25 -20 15||? That's all?
Yes, Paul, that's all. Does anyone have a recommendation on software
to use to handle linear algebra. Thanks!

🔗Gene Ward Smith <gwsmith@svpal.org>

1/12/2006 10:42:53 AM

We've had threads before on what the numbers in a wedgie "mean". Since
they do mulitple things, there are actually multiple answers. However,
I am not sure if I've pointed out the following before.

If we take a wedgie for a rank-two temperament where there are n odd
primes, then the first n numbers are the number of generator steps to
these primes, times the number of periods to the octave. In other words,
the complexity of the odd primes, in octave-equivalent terms. Signs
are such that a non-negative number of generator steps is taken to get
to "3".

What about the rest of the wedgie? Peel off the first n numbers, and
what is left is, when reduced, the corresponding Bohlen-Pierce wedgie,
where 2 is left out of the picture and 3 now takes the place of 2. You
can, of course, continue to peel the wedgie like an onion in this way.
"When reduced" means you'd do the usual wedgie reduction, taking out
any GCD and choosing the canonical sign.

Let's take miracle for an example. The 7-limit wedgie is

<<6 -7 -2 -25 -20 15||

Peel off the first three numbers, and you get <<-25 -20 15||, reduce
this to a wedige and you have <<5 4 -3||. Taking the conjugate of this
gives |-3 -4 5>, which is a no-octaves monzo for 3^(-3) 5^(-4) 7^5 =
16807/16875, so Bohlen-Pierce miracle tempers out this interval, which
is "mirkwai", or the smaller BP diesis.

The 11-limit wedgie is

<<6 -7 -2 15 -25 -20 3 15 59 49||

Peel off the first four numbers, which show 3, 5, 7 and 11 in terms of
secor steps, and you get <<-25 -20 3 15 59 49||; reduce that to a BP
wedgie and you have <<25 20 -3 15 59 48||. You can find the TM basis
in the usual way, getting {16875/16807, 7971615/7891499} as the basis
for BP miracle. Peeling the onion again, we get the triprime comma for
the primes 5, 7 and 11.