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Message: 8575 - Contents - Hide Contents Date: Wed, 26 Nov 2003 21:53:11 Subject: Re: Finding Generators to Primes etc From: Graham Breed Paul Erlich wrote:> Isn't he *assuming* that the fifth is the generator here? Sorry, i'm > having trouble following his reasoning . . .Yes, but that page is out of date, and wrong anyway for octave equivalent vectors. But it's the best that's currently "published". I've explained the modern way several times on this list, but as you've obviously forgotten I'll try again. You form a matrix with the octave (1 0 0 ...) at the top, then a chromatic unison vector (it doesn't matter which) and below them the commas. Take the adjoint (the inverse multiplied by the determinant). The left hand column is an equal temperament mapping -- describing the periodicity block corresponding to the chromatic UV. The gcd of the next column is the number of periods to the octave, and when you divide through by that GCD you get the generator mapping. The wedge product version I pretty much showed in a recent bra/ket post. Graham

Message: 8576 - Contents - Hide Contents Date: Wed, 26 Nov 2003 21:54:40 Subject: Re: Finding Generators to Primes etc From: Graham Breed

Message: 8577 - Contents - Hide Contents Date: Wed, 26 Nov 2003 22:09:14 Subject: Re: Finding Generators to Primes etc From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> Paul Erlich wrote: >>> Isn't he *assuming* that the fifth is the generator here? Sorry, i'm >> having trouble following his reasoning . . . >> Yes, but that page is out of date, and wrong anyway for octave > equivalent vectors. But it's the best that's currently "published". > I've explained the modern way several times on this list, but as you've > obviously forgotten I'll try again.If I don't have a direct understanding of how something works, I won't retain it. Sorry.> You form a matrix with the octave (1 0 0 ...) at the top, then a > chromatic unison vector (it doesn't matter which)Is this one of those cases where you're saying chromatic unison vector but don't really mean it? Anyway, thanks, and I hope you'll update your pages.

Message: 8578 - Contents - Hide Contents Date: Wed, 26 Nov 2003 15:52:54 Subject: Re: Finding the wedge product? From: Carl Lumma>Now we list them in alphabetical (also numerical) order of index >inside the correct number of brackets."Lexigraphic order", no? More later; I can't think in this hick country! (Montana) -Carl

Message: 8579 - Contents - Hide Contents Date: Wed, 26 Nov 2003 09:06:27 Subject: Re: Finding Generators to Primes etc From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul G Hjelmstad" > <paul.hjelmstad@u...> wrote: > As I was saying, for each prime, use the mapping common to both 5/12 > and 8/19, which is > > prime 2 = 1 period > prime 3 = 2 periods - 1 generator > prime 5 = 4 periods - 4 generators >>> And then is rms applied after >> that? >> yes, you want to 'solve' the above system of equations for the > generator, to minimize your desired error function. Paul H,I hope you've installed the optional Solver Add-in for Excel.

Message: 8580 - Contents - Hide Contents Date: Wed, 26 Nov 2003 09:28:58 Subject: Re: Finding the wedge product? From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:> So, can we get a version of your Gospel with this rolled in?Lets first take the simplest case worth considering. The wedge product of two 3-limit (2D) vectors. [a1 a2> ^ [b1 b2> The procedure is to first list every product of a coefficient from A with a coefficient from B, i.e. their ordinary scalar products. So with 2 coefficients in each there will be 2x2 = 4 products to consider, a1*b1, a1*b2, a2*b1, a2*b2. As you calculate each product, combine the indices of the two coefficients to make a compound index for it. It is important to keep the indices in their original order at this stage. So we have product index a1*b1 11 a1*b2 12 a2*b1 21 a2*b2 22 There are certain rules about what to to with each product now, depending on its compound index. There are 3 possibilities: 1. If the indexes have a digit in common then ignore it. Just throw the product away. So we throw away a1*b1 and a2*b2. 2. Otherwise if the digits in the compound index are already in alphabetical order, do nothing. So a1*b2 is just fine as it is. 3. Otherwise if they are not in alphabetical order, then put them in alphabetical order. But first, look at each digit of the compound index in turn, and count how many larger digits are to the left of it. Add up all these left-and-larger counts as you go, or just keep counting so their counts accumulate. If the result is an odd number then negate the product, otherwise leave it as it was. Consider the index 21. There are zero larger digits to the left of the 2 (because there are _no_ digits to the left of it), and there is one larger digit to the left of the 1, namely the 2. So the total of the left-and-larger counts is 1, an odd number. So a2*b1 becomes -a2*b1. We now have product index a1*b2 12 -a2*b1 21 Now find any products that have the same index and add them together. So we have only product index a1*b2 - a2*b1 12. Now list all these sums in alphabetical order of their indices, inside as many brackets as the sum of the number of brackets in the two arguments, and pointing in the same direction. The wedge product is only defined for values having their brackets pointing the same way. So our answer is [[a1*b2-a2*b1>> Now lets try something more messy. A 7-limit (4D) vector wedged with a 7-limit bivector. This might represent combining a third comma with two that have already been combined, as an intermediate result on the way to finding the ET mapping where these all vanish. [a1 a2 a3 a4> ^ [[b12 b13 b14 b23 b24 b34>> We first make the list of products of all pairs, with their compound indices. product index a1*b12 112 a1*b13 113 a1*b14 114 a1*b23 123 a1*b24 124 a1*b34 134 a2*b12 212 a2*b13 213 a2*b14 214 a2*b23 223 a2*b24 224 a2*b34 234 a3*b12 312 a3*b13 313 a3*b14 314 a3*b23 323 a3*b24 324 a3*b34 334 a4*b12 412 a4*b13 413 a4*b14 414 a4*b23 423 a4*b24 424 a4*b34 434 Now we get rid of all those with two digits the same. Of course once you've got the idea, you wouldn't even bother writing them down in the first place. This leaves. product index left-and-larger count a1*b23 123 a1*b24 124 a1*b34 134 a2*b13 213 1 a2*b14 214 1 a2*b34 234 a3*b12 312 2 a3*b14 314 1 a3*b24 324 1 a4*b12 412 2 a4*b13 413 2 a4*b23 423 2 And we do the left-and-larger counts on the indices that aren't already in alphabetical order (shown above), and negate the product if this is odd. And we end up with: product index a1*b23 123 a1*b24 124 a1*b34 134 -a2*b13 123 -a2*b14 124 a2*b34 234 a3*b12 123 -a3*b14 134 -a3*b24 234 a4*b12 124 a4*b13 134 a4*b23 234 Now we sum the products having the same index. product index a1*b23 + a3*b12 - a2*b13 123 a1*b24 - a2*b14 + a4*b12 124 a1*b34 - a3*b14 + a4*b13 134 a2*b34 - a3*b24 + a4*b23 234 Now we list them in alphabetical (also numerical) order of index inside the correct number of brackets. [[[a1*b23+a3*b12-a2*b13 a1*b24-a2*b14+a4*b12 a1*b34-a3*b14+a4*b13 a2*b34-a3*b24+a4*b23>>> Voila!

Message: 8581 - Contents - Hide Contents Date: Wed, 26 Nov 2003 10:52:18 Subject: Re: Finding Generators to Primes etc From: monz --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul G Hjelmstad" > <paul.hjelmstad@u...> wrote: >>>> Hmm. Sorry, let me approach it this way. in 12&19, >> you have 5/12 for the one and 8/19 for the other. >> How do you come up with one raw generator-to-prime mapping. >> As I was saying, for each prime, use the mapping > common to both 5/12 and 8/19, which is > > prime 2 = 1 period > prime 3 = 2 periods - 1 generator > prime 5 = 4 periods - 4 generatorsi'm not understanding a lot of this ... but i am curious about this: why are you overshooting the prime with the periods and then subtracting generators (instead of coming as close under the prime as you can with the periods, then adding generators)? the latter is the way i've always thought of prime-mapping. is there some special reason to do it "backwards" like this? -monz

Message: 8582 - Contents - Hide Contents Date: Thu, 27 Nov 2003 01:37:22 Subject: Re: Finding the wedge product? From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:>> Now we list them in alphabetical (also numerical) order of index >> inside the correct number of brackets. >> "Lexigraphic order", no?Yes. Although I think it's "lexicographic".

Message: 8583 - Contents - Hide Contents Date: Thu, 27 Nov 2003 01:45:54 Subject: Re: Finding Generators to Primes etc From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> > wrote: >>> Now *I* need to figure out how to get generators from commas! >> My personal approach is to turn everything into wedgies, and then > derive everything *from* wedgies.OK. Now it's time for me to wrestle with the term "wedgie". It this a synonym for "multivector" (both contravariant and covariant)?

Message: 8584 - Contents - Hide Contents Date: Thu, 27 Nov 2003 02:05:17 Subject: Re: Finding Generators to Primes etc From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> > wrote:>> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> >> wrote: >>>>> Now *I* need to figure out how to get generators from commas! >>>> My personal approach is to turn everything into wedgies, and then >> derive everything *from* wedgies. >> if there's only one comma, is that the wedgie?It is by my usage; if you were going to treat the 5-limit the same as all the other limits you'd need to take the complement and make the first nonzero coefficent positive. I see no point in that excess of consistency.

Message: 8585 - Contents - Hide Contents Date: Thu, 27 Nov 2003 02:14:21 Subject: Re: Finding Generators to Primes etc From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> OK. Now it's time for me to wrestle with the term "wedgie". It this a > synonym for "multivector" (both contravariant and covariant)?No; these correspond to regular temperaments. An (m-1)-D regular temperament is determined by m vals. Take the wedge product of the vals and divide through by the gcd of the coefficients. If the first nonzero coefficient is negative, change signs. The result is the wedgie. You may also start from (n-m) commas, which you then wedge and take the complement, followed by the above procedure. The wedgie for a regular temperament is unique, and even if their are torsion problems in the vals or commas you start from there are none in the wedgie they define.

Message: 8586 - Contents - Hide Contents Date: Thu, 27 Nov 2003 02:28:24 Subject: Re: Finding Generators to Primes etc From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> You form a matrix with the octave (1 0 0 ...) at the top, then a > chromatic unison vector (it doesn't matter which) and below them the > commas. Take the adjoint (the inverse multiplied by the determinant).This proceedure only works if you have a linear temperament. Something else you might try is finding a basis for the nullspace of the matrix formed from the commas alone, without your additions, and using this to obtain a reduced set of vals (which could involve some extra work.) From there, one can put the vals in the form you like; I am partial to Hermite reduction unless we are dealing with linear temperaments, in which case we do period-generator and make the the generator as small, greater than one, as possible, to get a standard reduced form. Sometimes it suffices to simply find all standard vals which make all of the commas zero and use this to start with. Finding the wedgie from the commas, and the matrix from the wedgie, will also work; that is how I would do this but I use Maple's Hermite reduction function for it.

Message: 8589 - Contents - Hide Contents Date: Mon, 01 Dec 2003 05:46:44 Subject: Enumerating pitch class sets algebraically From: Gene Ward Smith I took all 4095 nonempty subsets of 12 notes, and for each of these I formed a corresponding sum of 12-th roots of unity, mod 13^8, of the form sum z^i, i in I, where "I" is the subset of notes, and z is -288746898, which is a primitive 12th root of unity mod 13^8. In order to make this independent of circular permutations, I then raised all of these to the twelvth power. This resulted in 361 different pitch class sets. Pitch class sets so counted are invariant under transposition. It turns out we need only go as far as 13^3 = 2197, with z = 418, in order to insure all of the hash values for the pitch class sets are distinct. Does anyone know if 361 is what other people have gotten for this?

Message: 8591 - Contents - Hide Contents Date: Mon, 01 Dec 2003 06:22:45 Subject: Re: Enumerating pitch class sets algebraically From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> Does anyone know if 361 is what other people have gotten for this?It occurred to me that if this method was going to work, it should distinguish all 4095 pitch class sets to start out with, and since the primitive 12th root of unity is algebraic of degee four (with polynomial x^4 - x^2 + 1) it was by no means obvious this is true; in fact, it isn't. Oh well.

Message: 8594 - Contents - Hide Contents Date: Mon, 01 Dec 2003 22:02:29 Subject: Re: Enumerating pitch class sets algebraically From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx jon wild <wild@f...> wrote:> > Gene - pc-sets are usually enumerated using Polya's method. IIRC there are > 351 in 12-tet.Thanks, I'll research that. I don't see how there can be 351; 361 should be a minimum. Anyway, my method isn't certain of working unless the et is prime, which 12 is a long way from being. Robert Walker made some good posts illustrating this a> couple of years ago on the main list, I *think* the subject had the number > 351 in it, if you want to search for the thread. I started to write a FAQ > entry about enumerating pitch-class sets (there's still an 11k email on > this in my postponed messages folder), but like so many other things, it > died (as did the whole FAQ project, I think).I'd be interested in what you can dig out; if you are willing to forward the email that would be nice. I've also generated all> prime-form set-classes (and their interval vectors) for ets up to 31, and > have them sitting zipped up on another computer.Is prime-form a reduced form?> Just recently a very good article appeared in "Music Theory Online", the > most recent issue, by Rick Cohn. He has come up with a very simple formula > for enumerating tetrachordal set-classes in even ETs. It would be great if > you could see how to generalise that formula to n-chords!I had in mind exploring how to do such things, but I didn't know what had been done, since googling didn't turn anything up.

Message: 8595 - Contents - Hide Contents Date: Mon, 01 Dec 2003 22:31:31 Subject: Re: Enumerating pitch class sets algebraically From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx jon wild <wild@f...> wrote: >>>> Gene - pc-sets are usually enumerated using Polya's method. IIRC > there are>> 351 in 12-tet. >> Thanks, I'll research that. I don't see how there can be 351; 361 > should be a minimum.352 is the number I remember -- maybe that includes the empty set. This got discussed on the music theory yahoogroup (from which I've since unsubscribed due to anti-semitic and anti-german postings).

Message: 8596 - Contents - Hide Contents Date: Mon, 01 Dec 2003 22:44:16 Subject: Re: Enumerating pitch class sets algebraically From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx jon wild <wild@f...> wrote:> > Gene - pc-sets are usually enumerated using Polya's method.Remarkably, there seems to be an entire book on this: "From Polychords to Polya", by Michael Keith. I suppose if I was a combinatorialist, it would have been the first thing I tried, but I still would like to do it using number theory. Alas, not only is not to be found at San Jose State, it isn't anywhere in the interlibrary loan system we have either. I've put it on my Christmas wish list, but if I don't get it I may buy it.

Message: 8597 - Contents - Hide Contents Date: Mon, 01 Dec 2003 22:45:22 Subject: Re: Enumerating pitch class sets algebraically From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:> 352 is the number I remember -- maybe that includes the empty set. > This got discussed on the music theory yahoogroup (from which I've > since unsubscribed due to anti-semitic and anti-german postings).Flavell acting up?

Message: 8599 - Contents - Hide Contents Date: Mon, 01 Dec 2003 22:51:49 Subject: Re: Enumerating pitch class sets algebraically From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> > wrote: >>> 352 is the number I remember -- maybe that includes the empty set. >> This got discussed on the music theory yahoogroup (from which I've >> since unsubscribed due to anti-semitic and anti-german postings). >> Flavell acting up?Yeah, that was the name. Are you familiar with it from elsewhere?

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