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Message: 8575 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 Yes, but that page is out of date, and wrong anyway for octave equivalent vectors.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 Date: Wed, 26 Nov 2003 21:54:40 Subject: Re: Finding Generators to Primes etc From: Graham Breed > if there's only one comma, is that the wedgie? Yes.
Message: 8577 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 Yes, but that page is out of date, and wrong anyway for octave 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 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 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 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 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 generators i'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 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 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: 8595 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: 8599 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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