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Message: 7500 Date: Wed, 01 Oct 2003 16:15:53 Subject: Re: hey gene From: Carl Lumma

>i'm confused about that, because wouldn't b_2, b_2 + b_1, b_2 + >2*b_1, b_2 - b+1, etc., all have the same length in the orthogonal >complement of the space spanned by b_1?

I've heard of orthogonal and complement, but never "orthogonal complement". -Carl

Message: 7501 Date: Wed, 01 Oct 2003 23:38:13 Subject: Re: hey gene From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> >i'm confused about that, because wouldn't b_2, b_2 + b_1, b_2 + > >2*b_1, b_2 - b+1, etc., all have the same length in the orthogonal > >complement of the space spanned by b_1?

> > I've heard of orthogonal and complement, but never > "orthogonal complement". > > -Carl

you can always look it up! Orthogonal Complement -- from MathWorld * [with cont.]

Message: 7502 Date: Wed, 01 Oct 2003 01:34:09 Subject: [tuning] Re: Polyphonic notation From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> >no, scales with a period equal to a 1/N octave, where N is an

integer

> >greater than 1, are distributionally even but not MOS.

> > Oh, you're enforcing the 'new' definition of MOS. Who came up > with distrib. even? > > -Carl

john clough and nora englesbrshmegegel . . . i forget her last name. you can find the term in my 22 paper on your website.

Message: 7503 Date: Wed, 01 Oct 2003 23:54:08 Subject: Re: hey gene From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> What's this: >

> ># h2 scale blocks > > > >cm1 := [9/7, 6/5, 8/7]; > >c1 := [[-1, 0, 0], [-1, 0, 1], [-1, 1, 1], [0, 0, 0]]; > >s1 := [1, 15/14, 6/5, 5/4, 9/7, 3/2, 12/7, 7/4];

These are what I called "chord blocks", which are 7-limit scales analogous to Fokker blocks. This works because 7-limit tetrads, uniquely among prime limits, form a lattice. The resulting scales have the nice property of having a lot of chords to work with.

> And did anything ever become of this: > > FreeLists / tuning-math / [tuning-math] 38 lin... * [with cont.] (Wayb.)

This is it, I think.

> > Oh, and I don't see anything explaining TM reduction on > your website, or anywhere else for that matter.

Maybe it's time to add more stuff there.

Message: 7504 Date: Wed, 01 Oct 2003 23:58:22 Subject: Re: Canonical homomorphisms revisted From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> Gene, > > Any news on this front? Might it be useful for the notation > effort?

I was happy just to finally get it to work. You could base notation systems on it, I suppose, but there doesn't seem any clear connection.

Message: 7505 Date: Wed, 01 Oct 2003 02:00:44 Subject: [tuning] Re: Polyphonic notation From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> >>Who came up with distrib. even? > >> > >> -Carl

> > > >john clough and nora englesbrshmegegel . . . i forget her last

name.

> >you can find the term in my 22 paper on your website.

> > Wow, was this added in a later rev? All I remember is "maximal > evenness". > > -Carl

yes, that was changed. it was silly to keep using "maximal evenness" when () the definition clough and others use for this term reflects a philosophy i don't subscribe to () their definition didn't agree with the definition i gave () the definition i gave agreed with their term "distributional evenness" -- John Clough and Nora Engebretsen wrote a paper probably called _distributionally even scales_, which seems to be missing from the tuning and temperament bibliography.

Message: 7506 Date: Wed, 01 Oct 2003 04:42:44 Subject: [tuning] Re: Polyphonic notation From: Dave Keenan This is a followup to Yahoo groups: /tuning/message/47437 * [with cont.] Me (Dave):

> >Why would I want to? With the fifth as the generator of nominals, the > >natural number of nominals is 7. 6 is improper. You say "By doing X > >you're doing a bad thing". And I say "But I'm not doing X". And you > >say "Do X". This isn't making much sense to me. Sorry.

Carl:

> But you are (or were) advocating doing X, but trying to force 7 > nominals on other scales. It is, as you say, improper to do so. Which > is what I was trying to point out. It looks as though I've succeeded! >

Er Sorry. No. That was _Rothenberg_ improper. But of course it's worse than that, as Paul said, its not distributionally even. It's not MOS, or well-formed or whatever. When you introduce the Pythagorean-limma (or diatonic semitone) accidental needed to make it work, you find you have a redundant nominal. The analogy to your example, would be if I tried to force 7 nominals _in_a_chain_of_secors_ onto Blackjack. But that's not what I'm doing. I'm forcing 7 nominals in a chain of 2:3 approximations. George and I long ago conceded that there are advantages to notating a linear temperament with an appropriate number of nominals for that temperament. Our claim is only that, for those who are not willing to learn a totally new set of nominals every time they change tunings, 7 nominals in a chain of approximate fifths is by far the best general solution. And we basically have no idea how to notate ratios precisely if the generator of the nominals is not itself a _very_ simple ratio. As Paul kindly said, at least with fifths it's a manageable sort of mess. :-) And I would add: with many familiar landmarks, particularly in the harmony.

> >No. A little reflection allows me to explain that, as it stands now, > >the semantic foundations of Sagittal notation have absolutely nothing > >to do with any temperament.

> > I should have said, "good PBs" there. [I think of PBs as temperaments, > which always gets me into trouble.]

So what's a _good_ PB for notational purposes? That sounds even less likely to be agreed upon than a good linear temperament. How about we forget about this given our agreement below?

> >We quickly got beyond 72-ET, but for a very long time there was a > >constant tension between basing the notation on some equal temperament > >versus basing it on ratios and thereby keeping it open. The problem > >with a notation based on ratios is that, to keep everyone happy, you > >need a huge number of different accidentals,

> > With, say, 19-limit JI, I don't see a way around this.

Nor do I, unless you are willing to allow some symbols to be only approximations (which in many cases could be less than 0.5 c away). That's why we _are_ providing a huge number of symbols, for those who think they need them.

> With linear temperaments, you only need 1 accidental pair at a time, > as I've pointed out.

But Carl, that's like saying you only need 6 pairs of accidentals to notate 19-limit JI. One for each prime above 3. It becomes essentially unreadable once you go past 2 accidentals per note. For example, few people even want to refer to a note one degree above C in 31-ET, exclusively as Dbb for very long. They soon invent a new accidental for this and call it C^ or some such. The more you extend your chain of generators without closing, the more new accidentals you will want for these "enharmonics". Graham has already done this once for decimal. And even ignoring these "enharmonics", you need other accidentals when you have multiple parallel chains, i.e. when the period is not the whole octave.

> The particular comma involved will depend on > the limit and the number of notes in the base scale. This could be > handled two ways. The first way I suggested is to get a list of simple > 19-limit commas and assign accidental pairs to them.

OK. Well that's nearly done (to at least 23-limit), but not quite ready for publication. I keep letting myself get distracted by tuning list posts. :-)

> The same > accidentals could be used for planar temperaments, JI, whatever, with > more than one pair in use at a time.

Sure.

> If average use ("gimme 9 notes of such-and-such temperament in the > 13-limit") turns out to require more commas than can fit on a list,

I don't understand how average-use could require "more commas than can fit on a list". What could this mean except "an infinite number of commas"?

> you could try assigning (an) accidental(s) for each *temperament*, > with the understanding that it/they would take on TM-reduced value(s) > for the limit and scale cardinality being used.

Eek! So then we would have to learn not only new nominals for every temperament, but new accidentals too? There's definitely no need for this. We've got so many accidentals available in the sagittal system that if you can't find a simple enough one that fits, by using their primary comma interpretations, you can just choose the one whose primary comma has its untempered value nearest in cents to the untempered value of the comma you really wanted (so-called secondary interpretations of the symbols), and then you let it be known that, for that temperament only, the symbol exactly represents that secondary comma. We've already done that for some obscure multiples of 12-ET, notated using nominals in a chain of 12-ET-sized fifths. In this case we figure people don't care what ratio is being approximated and will just think of them as particular fractions of a tone or particular fractions of a sharp or flat.

> >We soon realised we could have our cake and eat it too - that every > >symbol could represent a single unique comma ratio but that users who > >want to notate rational tunings are free to choose larger or smaller > >sets of symbols to trade off economy-of-symbols against > >accuracy-of-representation (for those ratios which are not represented > >exactly by the chosen symbols).

> > Great. That's the master list idea.

OK. I though I made it clear long ago that's what we were doing. Sorry

> >We have also used a temperament to help decide on the actual symbols > >to be used for the comma ratios in the superset. This is an > >8-dimensional temperament

> > Representing how many harmonic dimensions?

There are primary commas on the list with primes up to 23 (maybe 29, it isn't finalised yet).

> >So the first part of my belief is that it is far better to have a > >notation system whose semantics are based on precise ratios and then > >use that to also notate temperaments, rather than trying to find the > >ultimate temperament and then using a notation based on that to notate > >both ratios and other temperaments.

> > Wow; this is exactly what I've been saying all along!!

Really? Then how have I managed to waste so much of my time answering this thread?

> >Then if that's accepted, the second part is that it is best if the > >simplest or most popular ratios have the simplest notations.

> > Right. And it's this aspect that makes the search more-or-less > equivalent to the search for good PBs.

Nope. You've lost me there.

> >I understand that you agree with this, and so it should be obvious > >that the simplest accidental is no accidental at all and so the > >simplest ratios should be represented by nominals alone. When we > >agree that powers of 2 will not be represented at all, or will be > >represented by an octave number, or by a distance of N staff positions > >or a clef, then surely you agree that the next simplest thing is to > >represent powers of three by the nominals.

> > Well, that's a weighted-complexity approach. But even with most > weighted-complexity lists I've seen, non-rational-generator > temperaments appear.

Huh? I thought you just agreed that we would first decide how to _precisely_ notate ratios? Therefore we don't care about weighted complexity, or any complexity (except at the 3-prime-limit), because we know we are going to represent ratios of the other primes as being _OFF_ the chain, by using accidentals. Whether we use rational or irrational generators we can only represent powers of _ONE_ ratio _EXACTLY_, _ON_ the chain, (modulo our interval of equivalence).

> >Well I think the fact that we have Graham proposing MIRACLE > >temperament with 10 nominals and Gene proposing ennealimmal > >temperament with 9 nominals should make it clear that there is > >unlikely to ever be agreement on which is the ultimate temperament > >for notating everything else including ratios.

> > It was the ultimate-temperament aspect of the project I objected > to since the beginning!

OK. Well I'm glad that's cleared up.

> >You seem to have been assuming that George and I were merely > >championing some other (fifth-generated) temperament as the > >ultimate for notating everything else. I hope I have explained > >why this is not so.

> > Ok, ok, I think we're more on the same page now.

Great!

> But certainly > the project didn't start out this way, and even in the last few > days I saw a blurb for George and/or you looking very confused > about non-heptatonic systems.

I think we're only confused about how a notation whose nominals are related by an irrational generator could be used notate ratios precisely.

Message: 7507 Date: Wed, 01 Oct 2003 05:11:54 Subject: [tuning] Re: Polyphonic notation From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...>

> So what's a _good_ PB for notational purposes? That sounds even less > likely to be agreed upon than a good linear temperament.

a sizable community, perhaps most ji composers, have agreed -- ben johnston's notation has the nominals on the so-called "major block". as you know, i prefer your choice, as do joe monzo and daniel wolf.

> > With linear temperaments, you only need 1 accidental pair at a

time,

> > as I've pointed out.

> > For example, few > people even want to refer to a note one degree above C in 31-ET, > exclusively as Dbb for very long. They soon invent a new accidental > for this and call it C^ or some such.

anyway, it's a beautiful note to play over a G major chord before resolving to C major.

> And even ignoring these "enharmonics", you need other accidentals

when

> you have multiple parallel chains, i.e. when the period is not the > whole octave.

that's where it's nice to have new nominals. in particular, i like a half-octave from G to be written as an upside-down G, and perhaps notated with the notehead *between* the positions for C and D, with a slash through the notehead just so no one mistakes it for a C or a D . . .

> > you could try assigning (an) accidental(s) for each *temperament*, > > with the understanding that it/they would take on TM-reduced value

(s)

> > for the limit and scale cardinality being used.

> > Eek! So then we would have to learn not only new nominals for every > temperament, but new accidentals too?

i think carl meant this as a way of deciding which of the accidentals to use in a particular scenario, not as a way of introducing *additional* accidentals . . . in other words, it means that for both meantone diatonic and 10-tone pajara you'd use the symbol for 25:24 to mean the single accidental of the temperament, while if blackjack qualified you might use 36:35 . . .

> you can just choose the one whose primary comma has its untempered > value nearest in cents to the untempered value of the comma you

really

> wanted

no, carl's solution is better . . .

Message: 7508 Date: Wed, 01 Oct 2003 05:41:18 Subject: [tuning] Re: Polyphonic notation From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:

> --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> >

> > So what's a _good_ PB for notational purposes? That sounds even less > > likely to be agreed upon than a good linear temperament.

> > a sizable community, perhaps most ji composers, have agreed -- ben > johnston's notation has the nominals on the so-called "major block". > as you know, i prefer your choice, as do joe monzo and daniel wolf.

I actually meant, what are the _criteria_ that make a PB _good_ in this application (as a universal set of nominals).

> > And even ignoring these "enharmonics", you need other accidentals

> when

> > you have multiple parallel chains, i.e. when the period is not the > > whole octave.

> > that's where it's nice to have new nominals. in particular, i like a > half-octave from G to be written as an upside-down G, and perhaps > notated with the notehead *between* the positions for C and D, with a > slash through the notehead just so no one mistakes it for a C or a > D . . .

Good point. Additional chains can use additional nominals rather than additional accidentals.

> > you can just choose the one whose primary comma has its untempered > > value nearest in cents to the untempered value of the comma you

> really

> > wanted

> > no, carl's solution is better . . .

Which was .... ? Maybe if _you_ explain it ...

Message: 7509 Date: Wed, 01 Oct 2003 05:54:56 Subject: [tuning] Re: Polyphonic notation From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:

> > > you can just choose the one whose primary comma has its

untempered

> > > value nearest in cents to the untempered value of the comma you

> > really

> > > wanted

> > > > no, carl's solution is better . . .

> > Which was .... ? > > Maybe if _you_ explain it ...

you want the chromatic alteration symbol to correspond to the simplest ratio it actually represents in the temperament. so for the diatonic/meantone case, your choice is between { . . . ,2187:2048, 135:128, 25:24, 250:243, . . .}, and the simplest is 25:24, so you use the symbol for 25:24 . . .

Message: 7510 Date: Wed, 01 Oct 2003 06:44:43 Subject: [tuning] Re: Polyphonic notation From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:

> --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> > wrote: >

> > > > you can just choose the one whose primary comma has its

> untempered

> > > > value nearest in cents to the untempered value of the comma you

> > > really

> > > > wanted

> > > > > > no, carl's solution is better . . .

> > > > Which was .... ? > > > > Maybe if _you_ explain it ...

> > you want the chromatic alteration symbol to correspond to the > simplest ratio it actually represents in the temperament. so for the > diatonic/meantone case, your choice is between { . . . ,2187:2048, > 135:128, 25:24, 250:243, . . .}, and the simplest is 25:24, so you > use the symbol for 25:24 . . .

There's no conflict here with what I wrote above. You're saying that "the comma you really want" is 25:24. So first we look to see if we have a symbol for this (i.e. with 24:25 as its primary (exact) interpretation). It turns out that we do, namely )||( , so the part you've quoted above does not apply. But imagine if the chromatic comma you needed an accidental for was (for some bizarre reason) 5569:5801, we would not find a symbol for that, so rather than invent a new symbol, we would calculate the untempered size of this to be about 70.66 cents and find that the closest symbolised comma has an untempered size of about 70.67 cents, namely 24:25, and so we would use the symbol for that. We would say that we are using a secondary interpretation of the )||( symbol.

Message: 7511 Date: Thu, 02 Oct 2003 01:01:18 Subject: Re: [tuning] Re: Polyphonic notation From: Carl Lumma

>> Oh, you're enforcing the 'new' definition of MOS.

> >Not on me, I hope. I don't like it and there are too many terms >floating about as it is. We could just stick to Myhill's property.

Yes, I must admit I don't think we should change our usage of MOS, because of something Kraig said. I don't think Erv would mind. He reminds me somewhat of Derrida in that he considers names a necessary evil. He has said repeatedly, to me, and in a public lecture, that he's simply struggling to communicate what he's doing, and that he hopes only that someone will come along and find something useful, and improve the names if possible. -Carl

Message: 7512 Date: Thu, 02 Oct 2003 23:37:46 Subject: Re: hey gene From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> Doesn't go to a chord? Aren't you connecting the centers of the > triangles? Then I get Cm->CM->C#m. If you connect the roots, or > any of the vertices, I get Cm->CM-Am. In fact, since there's so > much symmetry in this thing, I can't imagine ending up anywhere > other than on the corresponding part of some chord.

I'm connecting the centers of the triangles with a line whenever there is a common line between two of the triangles. This gives hexagons, where you have a line from Cm to CM, and lines *in different directions* from CM to Am and CM to Em. If you head in the *same* direction, you end up in the center of a hexagon, which does not correspond to either a major or a minor triad. I could do this algebraically instead if it would help.

Message: 7513 Date: Thu, 02 Oct 2003 08:31:45 Subject: Re: hey gene From: monz hi paul and Carl, --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:

> --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote: >

> > I wonder if this has anything to do with straightness.

> > yes, a reduced basis will have good straightness, because > the set of basis vectors is, in some sense, as short as > possible. and, as we discussed before, shortness implies > straightness. the "block" always has the same "area", so > if the vectors are close to parallel, they'll have to be > long to compensate. remember that whole confusing > discussion? >

> > Oh, and I don't follow this... > >

> > > That's astoundingly simple! Wouldn't it be quite > > > reasonable to further require that the only ratio t/u > > > in the set {(p/q)^i (r/s)^j} such that T(t/u) < T(r/s), > > > is p/q itself? The idea would be that otherwise, the > > > two unison vectors are "mismatched".

> > > > -Carl

> > that would mean that otherwise, r/s is much longer than p/q, > and the basis itself is an odd one to choose because it > necessarily involves two unison vectors of such different > proportions.

i understand this, in a nutshell, to mean that the reduction process places the bounding vectors of the periodicity-block as close as possible to the center/origin, and as short as possible, thus ensuring that the entire block is compacted as much as possible towards the center/origin. yes? -monz

Message: 7514 Date: Thu, 02 Oct 2003 19:02:08 Subject: Re: hey gene From: Carl Lumma

>> Doesn't go to a chord? Aren't you connecting the centers of the >> triangles? Then I get Cm->CM->C#m. If you connect the roots, or >> any of the vertices, I get Cm->CM-Am. In fact, since there's so >> much symmetry in this thing, I can't imagine ending up anywhere >> other than on the corresponding part of some chord.

> >I'm connecting the centers of the triangles with a line whenever >there is a common line between two of the triangles.

So CM and C#m aren't connected then? Why wouldn't you connect them?

>This gives hexagons,

So does connecting the centers of all the triangles.

>where you have a line from Cm to CM, and lines *in >different directions* from CM to Am and CM to Em. If you head in the >*same* direction, you end up in the center of a hexagon, which does >not correspond to either a major or a minor triad.

If you continue in the same direction and distance as from Cm -> CM, you wind up at C#m. Is there some sort of reasoning behind not including it because it only shares one pitch (instead of two) with CM? -Carl

Message: 7515 Date: Thu, 02 Oct 2003 08:32:25 Subject: Re: hey gene From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> >> What's this: > >>

> >> ># h2 scale blocks > >> > > >> >cm1 := [9/7, 6/5, 8/7]; > >> >c1 := [[-1, 0, 0], [-1, 0, 1], [-1, 1, 1], [0, 0, 0]]; > >> >s1 := [1, 15/14, 6/5, 5/4, 9/7, 3/2, 12/7, 7/4];

> > What are cm1, c1, and s1?

s1 is the scale. If I recall correctly, cm1 is the comma basis, and c1 is something obtained from cm1 and used to calculate s1.

> I remember this stuff. But I don't remember the bit about the > 7-limit being unique in this. What is it that makes, say, the > 5-limit triads not form a "lattice"?

5-limit triads can be represented as the vertices of a hexagonal tiling; this isn't a lattice in the sense of the word I use since it isn't a group. If you extend to the group it generates, you get a triangular lattice by adding the centers of the hexagons. These centers don't represent major or minor triads, but the unique note in common to all of the chords of a hexagon. We can decree that they represent something--for instance the augmented triad q--5/4 q--8/5 q where "q" is the central note. However, this is not symmetrical and rather artificial.

Message: 7516 Date: Thu, 02 Oct 2003 01:36:42 Subject: Re: hey gene From: Carl Lumma

>5-limit triads can be represented as the vertices of a hexagonal >tiling; this isn't a lattice in the sense of the word I use since it >isn't a group.

Why isn't it a group? -Carl

Message: 7517 Date: Thu, 02 Oct 2003 08:43:03 Subject: Re: TM reduction From: monz thanks, Gene!!!! --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:

> --- In tuning-math@xxxxxxxxxxx.xxxx "monz" <monz@a...> wrote: >

> > i asked Gene for a good definition for TM reduction a > > long time ago ... and Gene, if you gave it to me and i > > lost it in the shuffle, i apologize. can you send it again?

> > First we need to define Tenney height: if p/q is a positive

rational

> number in reduced form, then the Tenney height is TH(p/q) = p q. > > Now suppose {q1, ..., qn} are n multiplicatively linearly

independent

> positive rational numbers. Linear independence can be equated, for > instance, with the condition that rank of the matrix whose rows are > the monzos for qi is n. Then {q1, ..., qn} is a basis for a lattice > L, consisting of every positive rational number of the form

q1^e1 ...

> qn^en where the ei are integers and where the log of the Tenney > height defines a norm. Let t1>1 be the shortest (in terms of Tenney > height) rational number in L greater than 1. Define ti>1

inductively

> as the shortest number in L independent of {t1, ... t_{i-1}} and

such

> that {t1, ..., ti} can be extended to be a basis for L. In this way > we obtain {t1, ..., tn}, the TM reduced basis of L.

Message: 7518 Date: Thu, 02 Oct 2003 08:44:35 Subject: Re: hey gene From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> >5-limit triads can be represented as the vertices of a hexagonal > >tiling; this isn't a lattice in the sense of the word I use since

it

> >isn't a group.

> > Why isn't it a group?

One step takes a C minor chord to a C major chord. Where does the next step go? It doesn't go to a chord at all--we don't have a group, since we don't have closure under addition.

Message: 7519 Date: Thu, 02 Oct 2003 03:29:51 Subject: Re: hey gene From: Carl Lumma

>>>5-limit triads can be represented as the vertices of a hexagonal >>>tiling; this isn't a lattice in the sense of the word I use since >>>it isn't a group.

>> >> Why isn't it a group?

> >One step takes a C minor chord to a C major chord. Where does the >next step go? It doesn't go to a chord at all--we don't have a group, >since we don't have closure under addition.

Doesn't go to a chord? Aren't you connecting the centers of the triangles? Then I get Cm->CM->C#m. If you connect the roots, or any of the vertices, I get Cm->CM-Am. In fact, since there's so much symmetry in this thing, I can't imagine ending up anywhere other than on the corresponding part of some chord. -Carl

Message: 7520 Date: Thu, 02 Oct 2003 00:01:07 Subject: [tuning] Re: Polyphonic notation From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> >no, scales with a period equal to a 1/N octave, where N is an

integer

> >greater than 1, are distributionally even but not MOS.

> > Oh, you're enforcing the 'new' definition of MOS.

Not on me, I hope. I don't like it and there are too many terms floating about as it is. We could just stick to Myhill's property.

Message: 7521 Date: Thu, 02 Oct 2003 00:12:06 Subject: [tuning] Re: Polyphonic notation From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:

> But imagine if the chromatic comma you needed an accidental for was > (for some bizarre reason) 5569:5801, we would not find a symbol for > that, so rather than invent a new symbol, we would calculate the > untempered size of this to be about 70.66 cents and find that the > closest symbolised comma has an untempered size of about 70.67

cents,

> namely 24:25, and so we would use the symbol for that.

Not necessarily bizarre--I was proposing 6561/6250 = (4374/4375)*(21/20) for 5-limit ennealimmal notation.

Message: 7522 Date: Thu, 02 Oct 2003 00:34:50 Subject: TM reduction From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "monz" <monz@a...> wrote:

> i asked Gene for a good definition for TM reduction a > long time ago ... and Gene, if you gave it to me and i > lost it in the shuffle, i apologize. can you send it again?

First we need to define Tenney height: if p/q is a positive rational number in reduced form, then the Tenney height is TH(p/q) = p q. Now suppose {q1, ..., qn} are n multiplicatively linearly independent positive rational numbers. Linear independence can be equated, for instance, with the condition that rank of the matrix whose rows are the monzos for qi is n. Then {q1, ..., qn} is a basis for a lattice L, consisting of every positive rational number of the form q1^e1 ... qn^en where the ei are integers and where the log of the Tenney height defines a norm. Let t1>1 be the shortest (in terms of Tenney height) rational number in L greater than 1. Define ti>1 inductively as the shortest number in L independent of {t1, ... t_{i-1}} and such that {t1, ..., ti} can be extended to be a basis for L. In this way we obtain {t1, ..., tn}, the TM reduced basis of L.

Message: 7523 Date: Thu, 02 Oct 2003 00:39:10 Subject: Re: TM reduction From: Paul Erlich what's the point of defining tenney height as p*q if you're only going to use the log anyway, and tenney harmonic distance is already log(p*q)? --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:

> --- In tuning-math@xxxxxxxxxxx.xxxx "monz" <monz@a...> wrote: >

> > i asked Gene for a good definition for TM reduction a > > long time ago ... and Gene, if you gave it to me and i > > lost it in the shuffle, i apologize. can you send it again?

> > First we need to define Tenney height: if p/q is a positive

rational

> number in reduced form, then the Tenney height is TH(p/q) = p q. > > Now suppose {q1, ..., qn} are n multiplicatively linearly

independent

> positive rational numbers. Linear independence can be equated, for > instance, with the condition that rank of the matrix whose rows are > the monzos for qi is n. Then {q1, ..., qn} is a basis for a lattice > L, consisting of every positive rational number of the form

q1^e1 ...

> qn^en where the ei are integers and where the log of the Tenney > height defines a norm. Let t1>1 be the shortest (in terms of Tenney > height) rational number in L greater than 1. Define ti>1

inductively

> as the shortest number in L independent of {t1, ... t_{i-1}} and

such

> that {t1, ..., ti} can be extended to be a basis for L. In this way > we obtain {t1, ..., tn}, the TM reduced basis of L.

Message: 7524 Date: Thu, 02 Oct 2003 00:40:44 Subject: Re: hey gene From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> Yes, but I meant does the difference between KZ and M have to do > with straightness?

One difference is that KZ requires a Euclidean norm, which Tenney doesn't give us.

7000
7050
7100
7150
7200
7250
7300
7350
7400
7450
**7500**
7550
7600
7650
7700
7750
7800
7850
7900
7950

**7500 -**
7525 -