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Message: 10575 - Contents - Hide Contents Date: Mon, 08 Mar 2004 22:49:31 Subject: Re: Dual L1 norm deep hole scales From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: >>> P.S. Why aren't they simply the "L1 deep hole" results? Why dual? I >> tend to think of the monzo space or lattice of notes itself as >> the 'standard' space, while the space of linear functionals on it >> (breeds) as its dual. Or am I misunderstanding? >> Presumably L1 would mean > > ||3^a 5^b 7^c|| = |a|+|b|+|c| > > This has a different geometry from that, certainly.If that's your concern, I would say "Triangular L1 norm" or something similar, for this.> "Dual" comes in > because if you use an L1 norm when measuring errors of tunings,What do errors have to do with any of this? These are simply symmetrical JI scales . . .> dual > to that is the "Dual L1 norm" on note-classes.Since the dual of L1 is L-inf, is this actually a "Triangular L-inf norm" on note-classes?

Message: 10576 - Contents - Hide Contents Date: Mon, 08 Mar 2004 23:11:42 Subject: Re: Dual L1 norm deep hole scales 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:>>> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" > <gwsmith@s...> >>> wrote: >>>>>>> Ball 2 radius 6 14 notes >>>> [1, 21/20, 7/6, 6/5, 49/40, 5/4, 21/16, 7/5, 3/2, 8/5, 42/25, >>>> 12/7, 7/4,9/5] >>>>>> Just as I suspected. We've finally constructed the Stellated > Hexany, >>> aka Mandala! >>>> Eh, I think that's turned up already. :) > > Did it?Sure. It came up as the scale of the unit cube in connection with chord-lattice scales, and again as the union of the first two Euclidean shells around a deep hole.

Message: 10577 - Contents - Hide Contents Date: Mon, 08 Mar 2004 23:15:12 Subject: Re: Dual L1 norm deep hole scales From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:>> "Dual" comes in >> because if you use an L1 norm when measuring errors of tunings, >> What do errors have to do with any of this? These are simply > symmetrical JI scales . . .Quite a lot, as these correspond to the ways we measure error. The Hahn norm corresponds to the minimax error, the Euclidean norm to the rms error, and this "dual L1" gadget to the L1 error measurement which you were advocating.>> dual >> to that is the "Dual L1 norm" on note-classes. >> Since the dual of L1 is L-inf, is this actually a "Triangular L-inf > norm" on note-classes?No, because "L-inf" is an L-inf on errors of consonances, and hence has the symmetry properties deriving from that.

Message: 10578 - Contents - Hide Contents Date: Mon, 08 Mar 2004 23:17:01 Subject: Re: Dual L1 norm deep hole scales From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- 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:>>>> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" >>>>> wrote: >>>> >>>>> Ball 2 radius 6 14 notes >>>>> [1, 21/20, 7/6, 6/5, 49/40, 5/4, 21/16, 7/5, 3/2, 8/5, 42/25, >>>>> 12/7, 7/4,9/5] >>>>>>>> Just as I suspected. We've finally constructed the Stellated >> Hexany, >>>> aka Mandala! >>>>>> Eh, I think that's turned up already. :) >> >> Did it? >> Sure. It came up as the scale of the unit cube in connection with > chord-lattice scales, and again as the union of the first two > Euclidean shells around a deep hole.Oops -- missed that.

Message: 10579 - Contents - Hide Contents Date: Mon, 08 Mar 2004 23:23:44 Subject: Re: Dual L1 norm deep hole scales From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: >>>> "Dual" comes in >>> because if you use an L1 norm when measuring errors of tunings, >>>> What do errors have to do with any of this? These are simply >> symmetrical JI scales . . . >> Quite a lot, as these correspond to the ways we measure error.The raison d'etre for these scales would seem to be if someone's looking for a good JI scale to use, without regard to its melodic structure. What does he or she care how we measure error?> The > Hahn norm corresponds to the minimax error,I thought the "Hahn scales" we were coming up with were minimax in terms of note-classes. So wouldn't the dual of that be L1 in terms of error?> the Euclidean norm to the > rms error, and this "dual L1" gadget to the L1 error measurement which > you were advocating.Don't get it.>>> dual >>> to that is the "Dual L1 norm" on note-classes. >>>> Since the dual of L1 is L-inf, is this actually a "Triangular L- inf >> norm" on note-classes? >> No, because "L-inf" is an L-inf on errors of consonances, and hence > has the symmetry properties deriving from that.Don't get it. But either way, dual or not, wouldn't "L1" indicate a rectangular, rather than triangular, lattice or metric? You were the one to bring this up but I don't see how duality ameliorates it.

Message: 10580 - Contents - Hide Contents Date: Tue, 09 Mar 2004 08:37:28 Subject: Re: Octave equivalent calculations (Was: Hanzos From: Graham Breed Gene Ward Smith wrote:> If you wedge <21 33 49 59| with <41 65 95 115| you get > <12 -14 -4 -50 -40 30|; dividing this through by two gives you the > wedgie for 7-limit miracle, <6 -7 -2 -25 -20 15|. If you wedge the > monzo for 16805/16087 with that for 1029/1024 and take the compliment, > once again you get twice the wedgie for miracle, and hence miracle. > These are the results you should be getting by any correct method, but > don't seem to be.Oh, right. Yes, I get a different result in the first case if I don't use wedge products. I think the non-wedge version is correct. As there's no way of knowing where a wedge product comes from I have to treat contorsion the same way as torsion. If I wanted to remove the contorsion, that'd be easy enough by looking at the mapping, and much easier than using wedge products. So, despite your arrogance in telling me that I'm incorrect (still with no reason) this can't be a reason for using wedge products. I can't factorize 16805/16087, and I still wouldn't be able to factorize it if I weren't using wedge products. I get both 3361 and 16087 to be primes. Graham

Message: 10581 - Contents - Hide Contents Date: Tue, 09 Mar 2004 20:39:02 Subject: Re: Graham on contorsion From: Paul Erlich My rambling thoughts: If the wedge product of the monzos (i.e., bimonzo) of two commas lead to torsion, they lead to torsion; one should make a big deal about this fact but not sweep it under the rug. Similarly, if the wedge product of the breeds (i.e., cross-breed) of two ETs lead to contorsion, they lead to contorsion; one should make a big deal about this fact but not sweep it under the rug. It's fine to then define the "wedgie" (why don't we call this the smith) as either of these wedge products with the (con)torsion removed by dividing through by gcd, and then insist that true temperaments correspond to a wedgie/smith.

Message: 10582 - Contents - Hide Contents Date: Tue, 09 Mar 2004 10:01:08 Subject: Re: Octave equivalent calculations (Was: Hanzos From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> Gene Ward Smith wrote: >>> If you wedge <21 33 49 59| with <41 65 95 115| you get >> <12 -14 -4 -50 -40 30|; dividing this through by two gives you the >> wedgie for 7-limit miracle, <6 -7 -2 -25 -20 15|. If you wedge the >> monzo for 16805/16087 with that for 1029/1024 and take the compliment, >> once again you get twice the wedgie for miracle, and hence miracle. >> These are the results you should be getting by any correct method, but >> don't seem to be. >> Oh, right. Yes, I get a different result in the first case if I don't > use wedge products. I think the non-wedge version is correct.Why is it correct? Is there a reason you want to give, or is this what we call in the States a faith-based initiative? As> there's no way of knowing where a wedge product comes from I have to > treat contorsion the same way as torsion.That there is no way to tell where a wedge product comes from seems to be to be a major advantage. If I wanted to remove the> contorsion, that'd be easy enough by looking at the mapping, and much > easier than using wedge products. So, despite your arrogance in telling > me that I'm incorrect (still with no reason) this can't be a reason for > using wedge products.I don't know why it is arrogant to assume we remove contorsion; this is what we've been doing all along. If you are not doing so, I'd suggest people take both your results and your temperament finder with a grain of salt.> > I can't factorize 16805/16087, and I still wouldn't be able to factorize > it if I weren't using wedge products. I get both 3361 and 16087 to be > primes.It should have been clear that this was a typo for 16875/16807, as that is a well-enough known seven-limit comma. In any case, that is the one you could try.

Message: 10583 - Contents - Hide Contents Date: Tue, 09 Mar 2004 21:29:42 Subject: Re: Octave equivalent calculations (Was: Hanzos From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> Gene Ward Smith wrote: >>> Why is it correct? Is there a reason you want to give, or is this what >> we call in the States a faith-based initiative? >> An equal scale with 21 notes should combine with an equal scale of 41 > notes to give an equal scale of 62 notes with maximally even subsets of > 21 and 41 notes. Alternatively, an MOS of 62 notes with a maximally > even sub-MOS of either 21 or 41 notes. All the scales I give for 21&41 > comply with this, but miracle does not.Since the question concerns temperaments, not scales, this is hardly relevant. I agree that programs which take two ets and produce scales out of them are a good thing, but they aren't temperament programs.>> That there is no way to tell where a wedge product comes from seems to >> be to be a major advantage. >> Why on earth would that be? It loses information.Why should you care if the temperament comes from putting together vals or putting together monzos? As for the information lost in reduction of the wedge product to a wedgie, that is deliberately eliminated. The sign involves ordering of the product, and I would be interested to hear what use this precious information is. The rest could be used to define quasi-temperaments with contorsion if you wanted to, but this strikes me as a pointless proceedure. If you want that, just take a temperament and contort it in all the various ways that suit you.

Message: 10584 - Contents - Hide Contents Date: Tue, 09 Mar 2004 10:05:55 Subject: Graham on contorsion From: Gene Ward Smith This is what Graham says on his web site: "Whether temperaments with contorsion should even be thought of as temperaments is a matter of debate. They're really a way of constructing a scale with a simpler temperament. One problem is that there's always more than one qualitatively different generator for a given linear temperament with contorsion. So the temperament isn't uniquely determined by the mapping by period and generator." Is there some reason why this is no longer applicable?

Message: 10585 - Contents - Hide Contents Date: Tue, 09 Mar 2004 21:31:33 Subject: Re: Graham on contorsion From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> which is exactly the same as the 5/24 scale above. So given a wedge > product of <<-2 -8 -8]], how do you know if it should give the 5/24 > family or the 7/24 family?The question is meaningless. It should give both if you want contorsion, and neither if (like me) you don't.

Message: 10586 - Contents - Hide Contents Date: Tue, 09 Mar 2004 10:31:01 Subject: Re: Graham on contorsion From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> This is what Graham says on his web site: > > "Whether temperaments with contorsion should even be thought of as > temperaments is a matter of debate. They're really a way of > constructing a scale with a simpler temperament. One problem is that > there's always more than one qualitatively different generator for a > given linear temperament with contorsion. So the temperament isn't > uniquely determined by the mapping by period and generator." > > Is there some reason why this is no longer applicable?Another way of looking at contorsion is that it is sometimes trying to be a higher-limit temperament, so to speak. 68 et in the 3-limit is <68 108|, divisible by 4, and in the 5-limit is <68 108 158|, divisible by 2. So far it is a miserable mess, but in the 7-limit we get <68 108 158 191| and suddenly there is a point to it all. Graham gives the example of 5&19 for contorsion, but really 5&19 makes more sense if you up the ante to the 7-limit and then you have hemifourths. On the other hand, there is no guarantee that there is a decent higher limit temperament lurking out there, if only it can be found.

Message: 10587 - Contents - Hide Contents Date: Tue, 09 Mar 2004 21:35:46 Subject: Re: Graham on contorsion From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:> My rambling thoughts: > > If the wedge product of the monzos (i.e., bimonzo) of two commas lead > to torsion, they lead to torsion; one should make a big deal about > this fact but not sweep it under the rug. Similarly, if the wedge > product of the breeds (i.e., cross-breed) of two ETs lead to > contorsion, they lead to contorsion; one should make a big deal about > this fact but not sweep it under the rug. It's fine to then define > the "wedgie" (why don't we call this the smith) as either of these > wedge products with the (con)torsion removed by dividing through by > gcd, and then insist that true temperaments correspond to a > wedgie/smith.Up to a change in language (let's call that erlich) this is my point of view. The fact that the gcd is not one signals torsion or contorsion; note that fact and move on. If you want contorsion, you can always produce it. No biggie.

Message: 10588 - Contents - Hide Contents Date: Tue, 09 Mar 2004 10:52:07 Subject: Re: Graham on contorsion From: Graham Breed Gene Ward Smith wrote:> This is what Graham says on his web site: > > "Whether temperaments with contorsion should even be thought of as > temperaments is a matter of debate. They're really a way of > constructing a scale with a simpler temperament. One problem is that > there's always more than one qualitatively different generator for a > given linear temperament with contorsion. So the temperament isn't > uniquely determined by the mapping by period and generator." > > Is there some reason why this is no longer applicable?It looks good to me, and shows why wedgies are inadequate for generating such scales. Graham

Message: 10589 - Contents - Hide Contents Date: Tue, 09 Mar 2004 22:34:38 Subject: White keys and black keys From: Gene Ward Smith I thought I'd mention again my approach to this, which does not involve either matricies or wedge products, and does not get into any unfortunate confusion of scales with linear temperaments. Linear temperaments do arise, but in terms of a generator for them in a particular equal temperament. Let's take 5 and 19. The convergents of 5/19 are 1/3, 1/4, 5/19; so we take 1/4 and consider 1/5 and 4/19. Here 1/5 < 4/19 are adjacent in the Farey sequence, and the mediant is 5/24. Hence, *in terms of scales*, we get a 5/24 generator for the 19 white key, 5 black key scale partition of 24 equal. The generator 5/24 is not associated to any particular prime limit, and hence not with any particular linear temperament. We can, however, associate it to various linear temperaments, but we *should not* confuse it with a linear temperament as it stands. In the 5-limit, we may regard it as a contorted meantone. In the 7-limit, more reasonably, as hemifourths, with TM basis 49/48 and 81/80. In the 11-limit, we get 11-limit hemifourths, with the comma 56/55 added to the list, and in the 13-limit we can add 91/90 to that. The choice of temperament is not unique; it's simply a matter of picking the most reasonable alternative. Here's the mapping for 13-limit hemifourths: [<1 2 4 3 2 6|, <0 -2 -8 -1 7 -11|] The complexity is low enough that the 19 note DE is getting the benefit.

Message: 10590 - Contents - Hide Contents Date: Tue, 09 Mar 2004 10:52:31 Subject: Re: Graham on contorsion From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> Gene Ward Smith wrote:>> This is what Graham says on his web site: >> >> "Whether temperaments with contorsion should even be thought of as >> temperaments is a matter of debate. They're really a way of >> constructing a scale with a simpler temperament. One problem is that >> there's always more than one qualitatively different generator for a >> given linear temperament with contorsion. So the temperament isn't >> uniquely determined by the mapping by period and generator." >> >> Is there some reason why this is no longer applicable? >> It looks good to me, and shows why wedgies are inadequate for generating > such scales.It doesn't mention wedgies, so it can hardly do that. A wedgie is a wedge product which has deliberately had the common factor taken out if the gcd is greater than one; if you wanted to leave it in and proceed on that basis you could, of course.

Message: 10591 - Contents - Hide Contents Date: Tue, 09 Mar 2004 23:20:45 Subject: Re: White keys and black keys From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> Let's take 5 and 19. The convergents of 5/19 are 1/3, 1/4, 5/19; so we > take 1/4 and consider 1/5 and 4/19.Incidentally, though I didn't do this before we *could* allow semiconvergents into the mix, and I think this shows 5/24 is not really so unique. The semiconvergents from 1/4 to 5/19 are 2/7, 3/11 and 4/15, and these lead to generators of 9/24, 14/24 and 19/24 respectively. If we derive temperaments from these generators, from 9/24=3/8 one gets augmented in the 5-limit, with a cheesy extension adding 1323/1250 as a comma in the 7-limit. 14/24=7/12 gives us 648/625 and 1323/1280 as commas and [<8 13 19 22|, <0 -1 -1 1|] as a mapping, and 19/24 = 1-5/24, so this is the same as 5/24. ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ Yahoo! Groups Links <*> To visit your group on the web, go to: Yahoo groups: /tuning-math/ * [with cont.] <*> To unsubscribe from this group, send an email to: tuning-math-unsubscribe@xxxxxxxxxxx.xxx <*> Your use of Yahoo! Groups is subject to: Yahoo! Terms of Service * [with cont.] (Wayb.)

Message: 10592 - Contents - Hide Contents Date: Tue, 09 Mar 2004 11:07:16 Subject: Re: Octave equivalent calculations (Was: Hanzos From: Graham Breed Gene Ward Smith wrote:> Why is it correct? Is there a reason you want to give, or is this what > we call in the States a faith-based initiative?An equal scale with 21 notes should combine with an equal scale of 41 notes to give an equal scale of 62 notes with maximally even subsets of 21 and 41 notes. Alternatively, an MOS of 62 notes with a maximally even sub-MOS of either 21 or 41 notes. All the scales I give for 21&41 comply with this, but miracle does not.> That there is no way to tell where a wedge product comes from seems to > be to be a major advantage.Why on earth would that be? It loses information.> I don't know why it is arrogant to assume we remove contorsion; this > is what we've been doing all along. If you are not doing so, I'd > suggest people take both your results and your temperament finder with > a grain of salt.It's not what I've been doing, and I've been not doing it for longer than you've been here.> It should have been clear that this was a typo for 16875/16807, as > that is a well-enough known seven-limit comma. In any case, that is > the one you could try.Then it's the one I tried on Sunday, and it gives the same result today -- miracle, but with a warning about the period not being what was expected (indicating some kind of torsion) for the matrix method. Graham

Message: 10593 - Contents - Hide Contents Date: Tue, 09 Mar 2004 11:30:30 Subject: Re: Graham on contorsion From: Graham Breed Gene Ward Smith wrote:> It doesn't mention wedgies, so it can hardly do that. A wedgie is a > wedge product which has deliberately had the common factor taken out > if the gcd is greater than one; if you wanted to leave it in and > proceed on that basis you could, of course.It doesn't mention wedgies, but does mention what wedgies are used to derive -- the mapping by (period and) generator. But if you want wedge products, you can have wedge products. the 5&19 example comes from these two vals: <5 8 12] <19 30 44] whose wedge product is <<-2 -8 -8]] The "scale with tempering" (if you don't want to call it a temperament) is: 5/24, 251.7 cent generator basis: (1.0, 0.20975898813907931) mapping by period and generator: [(1, 0), (2, -2), (4, -8)] mapping by steps: [(19, 5), (30, 8), (44, 12)] highest interval width: 8 complexity measure: 8 (9 for smallest MOS) highest error: 0.004480 (5.377 cents) unique You get a different MOS from the same family if you do 19&24 or 24&43. A very similar scale can be found by combining 24 and 31: 16/55, 348.3 cent generator basis: (1.0, 0.29024101186092066) mapping by period and generator: [(1, 0), (1, 2), (0, 8)] mapping by steps: [(31, 24), (49, 38), (72, 56)] highest interval width: 8 complexity measure: 8 (10 for smallest MOS) highest error: 0.004480 (5.377 cents) unique Count back and this is the same as 17&7 with best 5-limit approximations: 7/24, 348.3 cent generator basis: (1.0, 0.29024101186092066) mapping by period and generator: [(1, 0), (1, 2), (0, 8)] mapping by steps: [(17, 7), (27, 11), (40, 16)] highest interval width: 8 complexity measure: 8 (10 for smallest MOS) highest error: 0.004480 (5.377 cents) unique That's generator of 7/24, which is different to the 5/24 of the previous scale. The original vals are <24 38 56] <31 49 72] and their wedge product is <<-2 -8 -8]] which is exactly the same as the 5/24 scale above. So given a wedge product of <<-2 -8 -8]], how do you know if it should give the 5/24 family or the 7/24 family? Graham ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ Yahoo! Groups Links <*> To visit your group on the web, go to: Yahoo groups: /tuning-math/ * [with cont.] <*> To unsubscribe from this group, send an email to: tuning-math-unsubscribe@xxxxxxxxxxx.xxx <*> Your use of Yahoo! Groups is subject to: Yahoo! Terms of Service * [with cont.] (Wayb.)

Message: 10594 - Contents - Hide Contents Date: Tue, 09 Mar 2004 00:33:02 Subject: Re: Dual L1 norm deep hole scales From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:> The raison d'etre for these scales would seem to be if someone's > looking for a good JI scale to use, without regard to its melodic > structure. What does he or she care how we measure error?It seems to me the duality provides another and convincing raison d'etre.>> The >> Hahn norm corresponds to the minimax error, >> I thought the "Hahn scales" we were coming up with were minimax in > terms of note-classes. So wouldn't the dual of that be L1 in terms of > error?The situation is confusing, because we've got three different (at minimum) norms to contend with. We have a norm which is Linf (minimax) or Euclidean (rms.) This leads to *another* norm by applying it to the consonances. Starting with the Euclidean norm (norm #1) (x^2+y^2+z^2)^(1/2), we apply it to 3,5,7,5/3,7/3,7/5 and get the following: x^2+y^2+z^2+(y-x)^2+(z-x)^2+(z-y)^2 = 3(x^2+y^2+z^2)-2(xy+xz+yz) Taking the square root of this is norm #2. Now form the symmetric matrix for the above, and take the inverse: [[3,-1,-1],[-1,3,-1],[-1,-1,3]]^(-1) = [[1/2,1/4,1/4],[1/4,1/2,1/4],[1/4,1/4,1/2]] The quadratic form for this last is (a^2+b^2+c^2+ab+ac+bc)/2, which gives us norm #2; of course we can rescale by multiplying by 2. The same situation, three *different* norms, we find if we start with norm #1 being the L_inf norm. Then norm #2 has a unit ball which is the convex hull of the twelve consonances--ie, a cuboctahedron. From the 14 faces of this we get norm #3.

Message: 10595 - Contents - Hide Contents Date: Tue, 09 Mar 2004 00:34:18 Subject: Re: inspiration for "jumping jacks"? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote: I'm trying to recall if I was thinking about this and can't, but I had heard it.

Message: 10596 - Contents - Hide Contents Date: Tue, 09 Mar 2004 01:01:13 Subject: Re: Dual L1 norm deep hole scales From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: >>> The raison d'etre for these scales would seem to be if someone's >> looking for a good JI scale to use, without regard to its melodic >> structure. What does he or she care how we measure error? >> It seems to me the duality provides another and convincing raison d'etre.Can you outline this raison d'etre, please? I have no idea what it could be.>>> The >>> Hahn norm corresponds to the minimax error, >>>> I thought the "Hahn scales" we were coming up with were minimax in >> terms of note-classes. So wouldn't the dual of that be L1 in terms of >> error? >> The situation is confusing, because we've got three different (at > minimum) norms to contend with. We have a norm which is Linf (minimax) > or Euclidean (rms.) This leads to *another* norm by applying it to the > consonances. > > Starting with the Euclidean norm (norm #1) (x^2+y^2+z^2)^(1/2),How is this used below?> we > apply it to 3,5,7,5/3,7/3,7/5 and get the following: > > x^2+y^2+z^2+(y-x)^2+(z-x)^2+(z-y)^2 = 3(x^2+y^2+z^2)-2(xy+xz+yz) > > Taking the square root of this is norm #2.This is the rms error criterion (or rms 'loss function'), right?> Now form the symmetric > matrix for the above, and take the inverse: > > [[3,-1,-1],[-1,3,-1],[-1,-1,3]]^(-1) = > [[1/2,1/4,1/4],[1/4,1/2,1/4],[1/4,1/4,1/2]] > > The quadratic form for this last is > > (a^2+b^2+c^2+ab+ac+bc)/2, which gives us norm #2; of course we can > rescale by multiplying by 2.So you're saying the dual of rms error is euclidean norm in the symmetric oct-tet lattice, yes?> The same situation, three *different* norms, we find if we start with > norm #1 being the L_inf norm. Then norm #2 has a unit ball which is > the convex hull of the twelve consonances--ie, a cuboctahedron. From > the 14 faces of this we get norm #3.I guess I must have been wrong above. What's the difference between this latter #2 and #3? And what about duality and the fact that on your Tenney page, you say that the dual of L1 is L_inf, but today you seem to be saying something different (is it the triangularity of the lattice that alters the situation)? Gene, it takes a huge amount of guesswork to navigate your posts. Being concise is all well and good, but taking things step-by-step will reap many rewards in human understanding, I predict. ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ Yahoo! Groups Links <*> To visit your group on the web, go to: Yahoo groups: /tuning-math/ * [with cont.] <*> To unsubscribe from this group, send an email to: tuning-math-unsubscribe@xxxxxxxxxxx.xxx <*> Your use of Yahoo! Groups is subject to: Yahoo! Terms of Service * [with cont.] (Wayb.)

Message: 10597 - Contents - Hide Contents Date: Wed, 10 Mar 2004 15:31:03 Subject: Re: Octave equivalent calculations (Was: Hanzos From: Graham Breed Gene:>>> Why is it correct? Is there a reason you want to give, or is this what >>> we call in the States a faith-based initiative? Me:>> An equal scale with 21 notes should combine with an equal scale of 41 >> notes to give an equal scale of 62 notes with maximally even subsets of >> 21 and 41 notes. Alternatively, an MOS of 62 notes with a maximally >> even sub-MOS of either 21 or 41 notes. All the scales I give for 21&41 >> comply with this, but miracle does not. Gene:> Since the question concerns temperaments, not scales, this is hardly > relevant. I agree that programs which take two ets and produce scales > out of them are a good thing, but they aren't temperament programs.If the question concerns temperaments, why didn't the question contain come hint as to this fact? Like maybe the word "temperament" along with a pointer to a definition so that we can all see what you're taking "temperament" to mean today. I was certainly talking about the results of my program, which include a definition of a temperament (most of the time, depending on how you define "temperament") but also (at least implicitly) a generic MOS (excluding the precise tuning, although this was supplied for the results I showed here) with a particular number of notes to the octave. And, definitely by implication this time, a family of MOS scales consistent with the mapping from JI. Well, nice to see that you now think my program is "a good thing". But how does that suddenly stop it from being a temperament program? It produces, as far as I can tell, a homomorphic mapping from what may be rational numbers to a tone group with a smaller rank, in line with your definition here: Regular Temperaments * [with cont.] (Wayb.) That's somewhat vague as to whether contorsion is allowed. First it says that an "icon" has to be epimorphic. But then "given an icon" we have to "find a homomorphic mapping". Well, you can certainly find the mapping for Vicentino's 7&31 system using the icon for meantone, so I suppose it must be a temperament. Still if you've now decided that temperaments don't have contorsion, that's fine, we'll find another name for the things like temperaments that may have contorsion. Gene:>>> That there is no way to tell where a wedge product comes from seems to >>> be to be a major advantage. Me:>> Why on earth would that be? It loses information. Gene:> Why should you care if the temperament comes from putting together > vals or putting together monzos? As for the information lost in > reduction of the wedge product to a wedgie, that is deliberately > eliminated. The sign involves ordering of the product, and I would be > interested to hear what use this precious information is. The rest > could be used to define quasi-temperaments with contorsion if you > wanted to, but this strikes me as a pointless proceedure. If you want > that, just take a temperament and contort it in all the various ways > that suit you.If you don't care, then it can hardly be a "major advantage" not to be told. If not being told is a "major advantage" then you must deeply care about not knowing, which is bizarre. At the minimum, reasonable people wouldn't care (not about whether the result came from vals or monzos, which is a straw man, but about which vals it came from) in which case it doesn't matter how they do the calculation. If people do care about getting the correct answer to the question they asked, then a method that throws away some of the information required for the answer and then tries to reconstruct it looks to be at a disadvantage compared to a method that goes straight to the correct answer. The information may be deliberately lost, depending on who is doing the deliberating. I don't think the sign carries any information. Why did you mention it? Yes, producing contorted quasi-temperaments from wedge products may be pointless. That could be why nobody, except you, has ever advocated doing so. Anyway, those who want contorsion-free temperaments will be pleased to learn that the module at # Temperament finding library -- definitions * [with cont.] (Wayb.) has been updated with a method to remove any contorsion from a "LinearTemperament" object. So, to get a contorsion-free temperament consistent with the best 5-limit mappings of 19- and 5-equal, you can do:>>> temper.Temperament(19,5,temper.limit5).uncontort()5/12, 503.4 cent generator basis: (1.0, 0.41951797627815951) mapping by period and generator: [(1, 0), (2, -1), (4, -4)] mapping by steps: [(7, 5), (11, 8), (16, 12)] highest interval width: 4 complexity measure: 4 (5 for smallest MOS) highest error: 0.004480 (5.377 cents) unique Graham

Message: 10598 - Contents - Hide Contents Date: Wed, 10 Mar 2004 15:40:40 Subject: Re: Graham on contorsion From: Graham Breed Me:>> which is exactly the same as the 5/24 scale above. So given a wedge >> product of <<-2 -8 -8]], how do you know if it should give the 5/24 >> family or the 7/24 family? Gene:> The question is meaningless. It should give both if you want > contorsion, and neither if (like me) you don't.The question is perfectly meaningful, and if you don't understand it you can't have been paying attention. Go back and re-read the message in which I asked it. I give a particular result for a pair of vals, and a different result for a different pair of vals. It isn't a mattor of "want contorsion" or "don't want contorsion" but wanting a *particular scale* with tempering, which happens to contain contorsion. Being given a list of scales that may be correct isn't good enough. If you don't want contorsion, then you shouldn't be following this thread, which is about contorsion (and clearly says so in the subject line). Graham

Message: 10599 - Contents - Hide Contents Date: Wed, 10 Mar 2004 16:32:48 Subject: Re: Graham on contorsion From: Graham Breed Paul Erlich wrote:> My rambling thoughts: > > If the wedge product of the monzos (i.e., bimonzo) of two commas lead > to torsion, they lead to torsion; one should make a big deal about > this fact but not sweep it under the rug. Similarly, if the wedge > product of the breeds (i.e., cross-breed) of two ETs lead to > contorsion, they lead to contorsion; one should make a big deal about > this fact but not sweep it under the rug. It's fine to then define > the "wedgie" (why don't we call this the smith) as either of these > wedge products with the (con)torsion removed by dividing through by > gcd, and then insist that true temperaments correspond to a > wedgie/smith.But the two cases are different. There is at least one historically important instance of contorsion: Vicentino's enharmonic of 1555. Although he gives higher limit ratios (sometimes incorrect) the harmony is all 5-limit, but with an equal division of the chromatic semitone. I expect other "quartertone" composers work on similar principles. My program will give this system of any two 7, 24 and 31-equal (7 and 31 are 5-limit consistent, 24 is twice 12). These numbers are important because: - 7 is the number of notes to the octave in staff notation. The alternative 5-limit contorted-meantone can't be written in staff notation with "quartertones". - 24 is the number of notes to a quartertone scale. If you think in terms of a 12 note chromatic, this is the simplest subdivision that gives you quartertones, or neutral thirds, or whatever. Vicentino never invokes such a scale, but never gives examples that lie outside it. Also Example 48.3 of Book III (p.205 of the Maniates translation) includes a fourth descending through all quartertones and Example 46.1 (p.199) includes a note that is incorrect given his definition of the enharmonic, but belongs to the correct "quartertone" scale. - 31 is the number of notes in Vicentino's tuning. If you asked for 7&24 or 24&31 than you should expect to get Vicentino's system. If the program automatically removed contorsion, you would be rightly surprised to get a result that wasn't consistent with 24-equal. If you asked for 7&31 in some other context, you may be surprised to get the quartertones coming back. The program can't really be sure which question you meant to ask. It could give you an uncontorted scale if neither of the inputs were contorted, but that means it won't give the right answer if you did want a microtonal system with 7 nominals consistent with 31-equal. So it currently returns the contorted result and leaves you to explicitly remove the contorsion (explicit is better than implicit). Torsion of unison vectors is a different matter. I don't know of any cases, theoretical or otherwise, in which torsion is desired in a tempered MOS. I'm not even sure what it would mean. A periodicity block with torsion certainly doesn't correspond to an MOS with contorsion. Where you don't supply a chromatic unison vector, it's even more likely that you simply wanted a torsion free scale that tempers out all the commatic unison vectors, and that's what you get. Maybe if you supplied the chromatic unison vector, you would want to be warned of torsion. But you aren't. So there. Graham

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