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Message: 6001 Date: Thu, 16 Jan 2003 11:58:16 Subject: Re: Nonoctave scales and linear temperaments From: Graham Breed Gene Ward Smith wrote: > I think plusing is the same as taking the wedge product. If you want > n-dimensional temperaments (where linear is 2, planar 3, etc.) then > you can wedge n et maps. You may also wedge pi(p)-n commas together for the same result, where p is the prime limit and pi(x) is the number theory function counting primes less than or equal to x. The dual/complement of the wedge product. But the wedge product can't distinguish torsion from contorsion. I'm using the & operator for combining equal temperaments to get a linear temperament. The + operator can then be for adding equal temperaments to get another equal temperament. So h19&h31=meantone, but h19+h31=h50. Graham
Message: 6002 Date: Thu, 16 Jan 2003 22:48:41 Subject: Re: Nonoctave scales and linear temperaments From: Carl Lumma > > Hmm. How bad are such cases? Could we take them out at the > > end? > > i have no idea what you're asking. bad? Can you find them at the end of a map-space search, and take them out? >>That's what I call taxicab complexity, I think. > > not quite. for one thing, read this: > > lattice orientation * > > including the link to my observations. // > > >if there's more than one comma being tempered out, we need > > >a notion of the "angle" between the commas . . . > > > > Please explain. > > search for "straightess" in these archives . . . Working... > > You assume there was an untweakable one in the 5-limit case? > > Bah! > > are you saying the octave should never be assumed to be > exactly 1200 cents? Yes, yes, and... yes. >>And in the 'now it's planar' case, you would no longer have an >>untweakable one. Something's got to give! > > i'm not following you. If you insist on there being an untweakable generator, you won't have it if you take a linear temperament, tweak the octave, and call it planar. > > Thanks. There is also, I gather, such a thing as a basis of > > commas? TM (TM stands for?) > > tenney-minkowski. tenney is the metric being minimized, and > minkowski provided a basis-reduction algorithm applicable to > such a case. > > > reduction applies to commas only, > > right? > > right. Thanks again. So if reduction is necc., it means that a temperament can be described by two different lists of commas, right? This means we'll have the same problem searching comma space as we did map space. So wedgies are our last hope. > > > no. you use a badness cutoff simply to define the list of > > > temperaments in the first place. > > > > That's the same as taking the 20 "best" temperaments. > > well, if your badness cutoff, extreme error cutoff, and extreme > complexity cutoff leave you with 20 inside, and if such a clunky > tripartite criterion is what you define as "best". Are you saying a badness cutoff is not sufficient to give a finite list of temperaments? -Carl
Message: 6004 Date: Thu, 16 Jan 2003 23:03:24 Subject: Re: Nonoctave scales and linear temperaments From: Graham Breed > > >Yes, it's the volume of a hypersphere. So it climbs dramatically with >the number of dimensions, which is one less the number of primes (or odd >numbers if you're being masochistic). For 8 dimensions, you still get >O((r**7)**7) = O(r**49) candidate wedgies. > Oops! That should be O((r**7)**6) or O(r**42) and 6 from 42 only gives 5 million or so combinations. I made the same mistake in the earlier post: there are n-2 commas needed to define a linear temperament using n primes. Graham
Message: 6005 Date: Thu, 16 Jan 2003 13:48:33 Subject: Re: Nonoctave scales and linear temperaments From: Graham Breed Carl Lumma wrote: > If the above is true about commas, then complexity should be > defined in terms of commas, and we could search all sets of > simple commas... What sets of simple commas do you propose to take? The files I have here show equivalences between second-order tonality diamonds. If you have p odd numbers in your ratios, the tonality diamond will have of the order of p**2 (p squared) ratios. The second order diamond is made by combining these, so that gives O(p**4) ratios. You're then setting pairs of these ratios to be equivalent, giving O(p**8) commas. You then need to take combinations of pi(p)-1 commas. (That is one minus the number of primes in the ratios.) If you have many more commas than primes, that will go as the pi(p)-1st power. So the total number of wedgies you have to consider is O((p**8)**(pi(p)-1)) Lets simplify this by setting pi(p)~p. To get an understimate, I'll count it as pi(p) but still call it p. The complexity of finding p prime linear temperaments is then O(p**8(p-1)) In the 7-limit, there are only 4 primes, so the calculation is O(4**(8*3)) = O(4**24) = O(2.8e14) candidates (not that many, but that doesn't matter because we don't know how long each one will take). In the 19-limit, there are 8 primes. So we need O(8**(8*7)) = O(8**56) = O(3.7e50) If that really is 3.7e50 candidates, it's impossible. But even in comparison to the 7-limit case, it's huge. And (without much optimisation) I couldn't even do the 7-limit calculation! For display purposes, I'm only showing commas between the first n*3 ratios from the second-order diamond, where n is the size of the second order diamond. Using that would mean only O(p**4(p-1)) candidates. So only O(8**28) in the 19-limit, or O(2e25). But this isn't good enough -- most of my top 19-limit temperaments don't have 7 such equivalences. Whereas combing equal temperaments only gives O(n**2) calculations, where n is the number of ETs you consider. I find n=20 works well, requiring O(400) candidates. This is true in the 5-limit and also the 21-limit. I haven't heard of anybody doing a similar search with unison vectors in the 21-limit, or even suggesting ways to reduce the complexity. Graham
Message: 6013 Date: Thu, 16 Jan 2003 20:38:26 Subject: Re: Ultimate 5-limit again From: Carl Lumma > look at this: > > Yahoo groups: /tuning/database? * > method=reportRows&tbl=10&sortBy=5&sortDir=up This's great! I don't see the heuristic on it, though. -C.
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