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Message: 2376 - Contents - Hide Contents Date: Mon, 10 Dec 2001 20:21:56 Subject: Re: Systems of Diophantine equations From: genewardsmith --- In tuning-math@y..., graham@m... wrote:> I found a reference for this: > > <Solving a System of Diophantine Equations with... * [with cont.] (Wayb.)> > > It says they do lattice base reduction as a step to solving the equations. > Whereas we're trying to do a twist on base reduction. Here's a paper > about that:That's a fancy new method for an old classic problem, and presumably becomes interesting mostly when the number of simultaneous Diophantine equations are high. Can you solve a system of linear equations over the rationals in Python?
Message: 2377 - Contents - Hide Contents Date: Mon, 10 Dec 2001 20:35:55 Subject: Re: More lists From: paulerlich --- In tuning-math@y..., "genewardsmith" <genewardsmith@j...> wrote:> --- In tuning-math@y..., "dkeenanuqnetau" <d.keenan@u...> wrote: >>> Whoa there! It's arguable that one could omit 3:9 since it is a >> duplicate of 1:3 (although Paul and I agree it should stay > duplicated>> because the interval really does occur at two different places in > the>> complete chord) but 1:9 isn't a duplicate of anything. >> Since Graham checked with me, he must be including 9, and counting 3 > once (on the grounds that there is only one 3 among all the numbers.) > Since we already double up on 3 a lot by having 3,9,5/3,9/5,7/3,9/7, > 11/3,11/9 we are giving it quite a lot of weight as it is. I don't > think there are any pragmatic arguments either way, but counting it > twice seems a little random to me.Sorry, I have to disagree. Graham is specifically considering the harmonic entity that consists of the first N odd numbers, in a chord. If a particular interval occurs twice, then we _have_ to weight it twice. And this is to say nothing of all the other saturated chords Graham is not using!
Message: 2378 - Contents - Hide Contents Date: Mon, 10 Dec 2001 00:52:18 Subject: hi guys . . . From: paulerlich just taking a little vacation . . . be back with you shortly . . .
Message: 2379 - Contents - Hide Contents Date: Mon, 10 Dec 2001 20:37:57 Subject: Re: More lists From: paulerlich --- In tuning-math@y..., graham@m... wrote:> What about 3/3, 5/5, 7/7, 9/9 and 11/11? Should they be included in the > average?It wouldn't affect the result.> How about 11/8, 16/11, 32/11, 11/4, 11/2, etc?If you included an equal number of octave-equivalents to each ratio, again, the result wouldn't be affected.
Message: 2380 - Contents - Hide Contents Date: Mon, 10 Dec 2001 02:14:21 Subject: Wedgies from ets and generators From: genewardsmith It's easy to compute a wedgie from a pair (n,g) where n is an equal temperament and g is a generator, and this could be added to the list of ways of cooking up candidate wedgies. We first reduce n to m, where gcd(m,g)=1, by dividing out common factors. We then form one val vector, the generator map, by taking [0, hn(3)/g mod m, hn(5)/g mod m, hn(7)/g mod m], where the modulo is taken in a range -m/2 < i <= m/2. The other val vector is simply the n et map: [n, hn(3), hn(5), hn(7)]. We take the wedge product of these two, and reduce it to the standard form wedgie, and there it is. Of course this is a little arbitary in borderline cases, but those are the least interesting anyway, and if desired a modulo in the range -m/2 <= i < m/2 could also be tried.
Message: 2381 - Contents - Hide Contents Date: Mon, 10 Dec 2001 20:40:43 Subject: Re: Wedge products From: genewardsmith --- In tuning-math@y..., graham@m... wrote:> In-Reply-To: <9v0ruo+hdll@e...> > So it isn't the octave-equivalent part of the wedge product of the two ETs > at all. But you've been ignoring this distinction. Do you get it to work > in the 5-limit?The 5-limit wedge product of two ets is the corresponding comma, and of two commas the corresponding et. We've been doing this all along; you can consider it to be a cross-product, or a matrix determinant. The wedgie of a 5-limit linear temperament would reduce the cross- product until it was not a power (getting rid of any torsion) and standardize it to be greater than one.>> You'd certainly can't ignore a basis element with a coefficient of >> zero unless it is beyond the range of dimensions you are working in. >> But before you said the definition of wedge products was > > ei^ej = -ej^ei > > nothing about keeping zero elements.That's the defintion, but it's an element in a vector space. You can't wish way a basis element ei^ej simply because it has a coefficient of zero, that isn't the way linear algebra works. A zero vector is not the same as the number zero.>>> If I could enumerate over all pairs, I could fix that. But that >> still>>> leave the general problem of all combinations of N items taken from >> a set.>>> I'd prefer to get rid of zero elements altogether.Why not simply order a list of size n choose m, and if one entry has the value zero, so be it? A function which goes from combinations of the first n integers, takrn m at a time, to unique integers in the range from 1 to n choose m might help.
Message: 2382 - Contents - Hide Contents Date: Mon, 10 Dec 2001 03:19:17 Subject: Re: The grooviest linear temperaments for 7-limit music From: paulerlich --- In tuning-math@y..., "dkeenanuqnetau" <d.keenan@u...> wrote:> --- In tuning-math@y..., "paulerlich" <paul@s...> wrote:>> Dave, if you don't have a cutoff, you'd have an infinite number of >> ETs better than 1547. Of course there has to be a cutoff. >> Yes. This just shows that this isn't a very good badness metric. > A decent badness metric would not need a cutoff in anything but > badness in order to arrive at a finite list. >>> I mean that only Gene's measure tells you exactly _how much_ better > a>> system is than the systems in their vicinity, >> How do you know it does that? "Exactly"?Sure, in a limit-probability sense. How many digits did Gene report? Anyhow, I'll have to refer you to Gene on the details of how it does that. I'd just like this paper to have some very simple systems with large errors, where a combined adaptive-tuning & adaptive-timbre approach would be needed, as well as systems to satisfy people like Rami Vitale, for whom even the _melodic_ distinctions of 225:224 cannot be tempered out.
Message: 2383 - Contents - Hide Contents Date: Mon, 10 Dec 2001 20:42:42 Subject: Re: More lists From: genewardsmith --- In tuning-math@y..., "paulerlich" <paul@s...> wrote:> Sorry, I have to disagree. Graham is specifically considering the > harmonic entity that consists of the first N odd numbers, in a chord.That may be what Graham was doing, but it wasn't what I was doing; I seldom go beyond four parts.
Message: 2384 - Contents - Hide Contents Date: Mon, 10 Dec 2001 03:26:35 Subject: Re: The grooviest linear temperaments for 7-limit music From: paulerlich --- In tuning-math@y..., "dkeenanuqnetau" <d.keenan@u...> wrote:> But why ever do you think the size of the wiggles should be flat? I > think it is quite expected that the size of the wiggles in badness > around 1-tET to 9-tET are _much_ bigger than the wiggles around 60- tET > to 69-tET.The two ranges would gave to be the same size logarithmically, for example 1-tET to 9-tET and 10-tET to 90-tET.
Message: 2385 - Contents - Hide Contents Date: Mon, 10 Dec 2001 21:41:50 Subject: Re: Wedgies from ets and generators From: genewardsmith --- In tuning-math@y..., graham@m... wrote:>> It's easy to compute a wedgie from a pair (n,g) where n is an equal >> temperament and g is a generator, and this could be added to the list >> of ways of cooking up candidate wedgies. >> This is a pair of numbers?Correct, as for instance (31,13)> Those two vals are what I get by going from unison vectors (or a wedgie) > to linear temperaments. So hn(i) is the approximation to the ratio i/1 in > the linear temperament.If "i" is a rational number, hn(i) is how many steps in the n-et are used to represent it.> hn(3)/g mod m must be equivalent to Carey and Clampitt's inverse modulo.I don't know about Carey and Clampitt, but this is standard mathematical notation. If hn(3)/g is not divisable by m then we may reduce mod m. If you want to get fancy it's a matter of "localization"; if hn(3)/g is in the localization of the integers Zm at m, meaning reduced fractions whose numerator may be any integer but whose denomimator is relatively prime to m, then we may reduce Zm mod m by taking Zm/mZm just as we do Z/mZ.> Yes, there are additional pieces of information you need, beyond the ET > and the generator. You need the list of prime intervals. You need a rule > for approximating them by integers. And you need a rule for choosing the > best of the mod n mappings. You won't get the simplest result for > 11-limit Meantone31 by your rule. The 11-direction will come out as 13 > fourths rather than 18 fifths.It comes out as 18 when I use an alternative program which centers around the mean of the 7-limit generators. By some measures, this will be the> simplest 11-limit mapping of 31, so it's a big one to miss.If you really don't want to miss anything you could take all the different equivalence possibilities up to your generator limit, and test all of them. If I use the g^(5/3) cents measure, the two come out almost the same: the 18 version as 313.6, and the -13 version as 311.4; presumably we want both.
Message: 2386 - Contents - Hide Contents Date: Mon, 10 Dec 2001 03:35:32 Subject: Re: More lists From: dkeenanuqnetau --- In tuning-math@y..., graham@m... wrote:> Yes, won't change the position. But, looking carefully at your previous > mail, I see you're including 1/3, 9/3 and 9/1, so that'll be it. I remove > the duplicates.Whoa there! It's arguable that one could omit 3:9 since it is a duplicate of 1:3 (although Paul and I agree it should stay duplicated because the interval really does occur at two different places in the complete chord) but 1:9 isn't a duplicate of anything. You can't ignore the 1:9 error. It really exists. When you listen to a bare 4:9 you don't hear two 2:3 errors.
Message: 2387 - Contents - Hide Contents Date: Mon, 10 Dec 2001 22:49:25 Subject: Re: More lists From: paulerlich --- In tuning-math@y..., "genewardsmith" <genewardsmith@j...> wrote:> --- In tuning-math@y..., "paulerlich" <paul@s...> wrote: >>> Sorry, I have to disagree. Graham is specifically considering the >> harmonic entity that consists of the first N odd numbers, in a > chord. >> That may be what Graham was doing, but it wasn't what I was doing; I > seldom go beyond four parts.Even if you don't, don't you think chords like 1:3:5:9 1:3:7:9 1:3:9:11 10:12:15:18 12:14:18:21 18:22:24:33 which contain only 11-limit consonant intervals, would be important to your music?
Message: 2388 - Contents - Hide Contents Date: Mon, 10 Dec 2001 03:42:49 Subject: Re: What's so Super about Superparticularity? From: paulerlich --- In tuning-math@y..., "unidala" <JGill99@i...> wrote:> --- In tuning-math@y..., "genewardsmith" <genewardsmith@j...> wrote: >>> Actually, superparticular ratios are associated with each branch of >> the Stern-Brocot tree, and not confined to any sub-branch. Simply >> take the ratio between the node at level n and a branch node at > level>> n+1, and label the branch connecting them with this superparticular >> ratio. > > Gene, >> I see what you mean above (where the branches in the Stern-Brocot > tree, as opposed to the nodes, are concerned). I was addressing the > ratios of the nodes only in my statement (portion of S-B Tree below): > > > 3/2 > / \ > / \ > / \ > / \ > / \ > / \ > / \ > / \ > / \ > / \ > 4/3 5/3 > / \ / \ > / \ / \ > / \ / \ > / \ / \ > 5/4 7/5 8/5 7/4 > / \ / \ / \ / \ > 6/5 9/7 11/8 10/7 11/7 13/8 12/7 9/5 > / \ / \ / \ / \ > > JG: While all nodes branching from a given node are "Farey adjacent" > in terms of their rational values, only some of the nodes values are, > themselves, superparticular in value.Similarly, while the "Greek" scales don't have only superparticular _pitch ratios_, the size of the steps between adjacent ratios are all superparticular. I think it's a fetish, but a benign one.> > JG: And, it follows, that the percentage of superparticular ratios > (of nodes)"pops up" to a maximal degree in the low-numbered ratio > values. Perhaps, then, it is their attributes of appearing at such > low-numbered ratios (and the corresponding increased coincidence of > overtones of such scale interval ratios), where by "scale interval > ratios", I am describing the chosen tones of a given scaleIn that sense, I really haven't seen a tendency to use superparticular ratios, either in ancient Greek theory or today.>(rather > than the step-sizes between the tones of the scale),That's where you tend to see superparticulars pop up. Not only in scales, but in ordered surveys of ratios within a given distance from 1/1 -- and hence also in choices for unison vectors. (See below.)>>> GWS: If they had done >> the same with superparticulars with square or triangular or fourth >> power, etc. numerators it would have been more to the point, if so. >> JG: Or would it? Can anyone demonstrate an implicit advantage of > utilizing superparticular scale interval ratios with large valued > integers existing in the numerators and/or denominators of such scale > interval ratios? Or am I missing something regarding Gene's points > made regarding "square or triangular or fourth power" numerators?Gene is primarily talking about the great propensity for superparticulars among choices for unison vectors.> The "magic", then, if not arising from low-numbered integers existing > in certain superparticular ratios, might exist only as a beneficial > characteristic of a "rational" (bad pun) choice of the subdivision of > the octave which follows from progressively applying the Stern- Brocot > tree?Sounds like part of the answer.
Message: 2389 - Contents - Hide Contents Date: Mon, 10 Dec 2001 18:40:58 Subject: Formatting test (please ignore) From: David C Keenan test 4 8 16 end test -- Dave Keenan Brisbane, Australia Dave Keenan's Home Page * [with cont.] (Wayb.) -- A country which has dangled the sword of nuclear holocaust over the world for half a century and claims that someone else invented terrorism is a country out of touch with reality. --John K. Stoner
Message: 2390 - Contents - Hide Contents Date: Mon, 10 Dec 2001 03:47:28 Subject: Re: More lists From: paulerlich --- In tuning-math@y..., graham@m... wrote:> Me:>>> It may depend on whether or not you include the zero error for 1/1 >> in the >>> mean. > > Dave:>> I don't. Seems like a silly idea. And that wouldn't change _where_ the >> minimum occurs. >> Yes, won't change the position. But, looking carefully at your previous > mail, I see you're including 1/3, 9/3 and 9/1, so that'll be it. I remove > the duplicates.9/3 is not a duplicate of 3/1 -- all saturated chords in the 11-limit contain not one, but two of the intervals that this is equal to. And 9/1 is a duplicate of what?
Message: 2391 - Contents - Hide Contents Date: Mon, 10 Dec 2001 18:41:46 Subject: Formatting test (please ignore) From: David C Keenan test 4 8 16 end test -- Dave Keenan Brisbane, Australia Dave Keenan's Home Page * [with cont.] (Wayb.) -- A country which has dangled the sword of nuclear holocaust over the world for half a century and claims that someone else invented terrorism is a country out of touch with reality. --John K. Stoner
Message: 2392 - Contents - Hide Contents Date: Mon, 10 Dec 2001 05:07:46 Subject: Re: More lists From: genewardsmith --- In tuning-math@y..., "dkeenanuqnetau" <d.keenan@u...> wrote:> Whoa there! It's arguable that one could omit 3:9 since it is a > duplicate of 1:3 (although Paul and I agree it should stay duplicated > because the interval really does occur at two different places in the > complete chord) but 1:9 isn't a duplicate of anything.Since Graham checked with me, he must be including 9, and counting 3 once (on the grounds that there is only one 3 among all the numbers.) Since we already double up on 3 a lot by having 3,9,5/3,9/5,7/3,9/7, 11/3,11/9 we are giving it quite a lot of weight as it is. I don't think there are any pragmatic arguments either way, but counting it twice seems a little random to me.
Message: 2394 - Contents - Hide Contents Date: Mon, 10 Dec 2001 13:06 +0 Subject: Re: More lists From: graham@xxxxxxxxxx.xx.xx In-Reply-To: <9v1b8g+f9bd@xxxxxxx.xxx> Me:>> Yes, won't change the position. But, looking carefully at your > previous>> mail, I see you're including 1/3, 9/3 and 9/1, so that'll be it. I > remove >> the duplicates. Paul:> 9/3 is not a duplicate of 3/1 -- all saturated chords in the 11-limit > contain not one, but two of the intervals that this is equal to. And > 9/1 is a duplicate of what?9/3 and 3/1 are duplicates. The both simplify to be the same. 9/1 isn't a duplicate, so I don't remove it. What about 3/3, 5/5, 7/7, 9/9 and 11/11? Should they be included in the average? How about 11/8, 16/11, 32/11, 11/4, 11/2, etc? Graham
Message: 2395 - Contents - Hide Contents Date: Mon, 10 Dec 2001 13:06 +0 Subject: Re: Wedgies from ets and generators From: graham@xxxxxxxxxx.xx.xx In-Reply-To: <9v15pt+7bem@xxxxxxx.xxx> Gene wrote:> It's easy to compute a wedgie from a pair (n,g) where n is an equal > temperament and g is a generator, and this could be added to the list > of ways of cooking up candidate wedgies.This is a pair of numbers?> We first reduce n to m, where gcd(m,g)=1, by dividing out common > factors. We then form one val vector, the generator map, by taking > [0, hn(3)/g mod m, hn(5)/g mod m, hn(7)/g mod m], where the modulo is > taken in a range -m/2 < i <= m/2. The other val vector is simply the > n et map: [n, hn(3), hn(5), hn(7)].Those two vals are what I get by going from unison vectors (or a wedgie) to linear temperaments. So hn(i) is the approximation to the ratio i/1 in the linear temperament. What you're showing is the inverse process. hn(3)/g mod m must be equivalent to Carey and Clampitt's inverse modulo. We need integers p(i) such that (g*p(i)) mod m = hn(i) mod m. That's the same as (g*p(i) - hn(i)) mod m = 0. That's how I understand it. If you prefer to write it as p(i) = hn(i)/g mod m, all well and good.> We take the wedge product of > these two, and reduce it to the standard form wedgie, and there it is.In my case "reduce it to the standard form wedgie" means getting the complement() and simplify()ing it.> Of course this is a little arbitary in borderline cases, but those > are the least interesting anyway, and if desired a modulo in the range > -m/2 <= i < m/2 could also be tried.Yes, there are additional pieces of information you need, beyond the ET and the generator. You need the list of prime intervals. You need a rule for approximating them by integers. And you need a rule for choosing the best of the mod n mappings. You won't get the simplest result for 11-limit Meantone31 by your rule. The 11-direction will come out as 13 fourths rather than 18 fifths. By some measures, this will be the simplest 11-limit mapping of 31, so it's a big one to miss. It is indeed easy when you ignore all the complications. Graham
Message: 2396 - Contents - Hide Contents Date: Mon, 10 Dec 2001 13:06 +0 Subject: Re: Wedge products From: graham@xxxxxxxxxx.xx.xx In-Reply-To: <9v0ruo+hdll@xxxxxxx.xxx> Me:>> Is your matrix of vals my mapping by steps? [(41, 31), (65, 49), > (95,>> 72), (115, 87), (142, 107)] for Miracle. If so, I'm with you until > you>> get to the Diophantine equations. I think it's solving systems of > linear>> Diophantine equations that I need to know how to do. Gene:> No, but if you have an easy way to get your two vals, and if they > produce the correct wedgie, then they will work also.Yes, I can get the wedgie no problem. It's going from that to the simplest set of unison vectors that's the problem. It now looks like it's systems of Diophantine equations I need to look at, and wedge products are incidental.>>>>> wedgie = reduce(temper.wedgeProduct, >> map(temper.WedgableRatio, >> [(225,224),(385,384),(243,242)])) >> I get 225/224^385/384^243^242 = h31^h41 = > [6,-7,-2,15,-25,-20,3,15,59,49] in the ordering I'm using now; this > has the correct number of dimensions, ten. If you want to mess around > with wedging equivalence classes (but what's the point?) then they > should come out in six dimensions. The equivalence class wedgies are > just subsets of the full wedgie, but they don't correspond any more, > and don't get rid of torsion, and so don't seem very useful.What? The octave-equivalent part of the unison-vector wedgie is the mapping by generators. It uniquely defines the temperament, and is the most important thing we're trying to find. The other equivalence class wedgies give us an ET mapping, if sometimes a silly one. These are the two pieces of information we need to construct the temperament. So they look like the most useful things the wedgie gives us.>>>>> wedgie.octaveEquivalent().flatten()>> (0, -6, 7, 2, -15) >> >> but with the wedge of the temperaments >> >>>>> (h31^h41).octaveEquivalent().flatten()>> (-25, -20, 3, 15, 59, 49) >> Both of these are only part of the correct wedgie, so naturally they > are not in correspondence.Yes, but in the bit you cut off Me:>> One good thing is that the generator mapping (ignoring the period > mapping)>> which I'm using as my invariant key, is simply the octave- > equivalent part>> of the wedge product of the commatic unison vectors! Gene:> Or of the wedge product of two ets.So it isn't the octave-equivalent part of the wedge product of the two ETs at all. But you've been ignoring this distinction. Do you get it to work in the 5-limit?>> Yes, for matrices you need to have consistent dimensions, but you > can get>> away without them for wedge products. At least the way I've > implemented >> them. >> That may be your problem. You could do this by assuming infinite > dimensions, and ignoring things after the dimension becomes larger > than your inputs, where all coefficients become zero, but the normal > way is to stick with a certain number of dimensions.Yes, but then 5:4 can't be both a 5-limit and 7-limit interval at the same time. I'm okay until we get to the flattening, which doesn't seem to be important.>>> But some zero elements aren't always present. Either I can>>>> get rid of them, which might mean that different products have > the >>> same>>>> invariant, or enumerate the missing bases when I calculate the >>> invariant. >> You'd certainly can't ignore a basis element with a coefficient of > zero unless it is beyond the range of dimensions you are working in.But before you said the definition of wedge products was ei^ej = -ej^ei nothing about keeping zero elements. And it doesn't make any difference so far. (May sometimes make the complements wrong.)>> Take the multiple-29 wedgie: >>>>>>> h29 = temper.PrimeET(29,temper.primes[:5]) >>>>> h58 = temper.PrimeET(58,temper.primes[:5]) >>>>> (h29^h58).invariant()>> (0, 29, 29, 29, 29, 46, 46, 46, 46, -14, -33, -40, -19, -26, -7) >> There are fifteen of these, so presumably it is 13-limit, but you > don't say. For a 13-limit wedgie of these two, I get your result, so > that seems to be what it is.I do say.>>>>> h29 = temper.PrimeET(29,temper.primes[:5]) ^5 octave-equivalent dimensions, so 13-limit.>>>>> wedgie.invariant()>> (29, 29, 29, 29, 46, 46, 46, 46, -14, -33, -40, -19, -26, -7) >> This has dimension 14, which is wrong, and there is your missing zero.I know there's a missing zero. I was showing you why it went missing.>> If I could enumerate over all pairs, I could fix that. But that > still>> leave the general problem of all combinations of N items taken from > a set.>> I'd prefer to get rid of zero elements altogether. >> For programming purposes? I think the program should follow the math, > and not vice-versa; otherwise you are asking for trouble.Either way, the program's following the math you originally gave.>> Right? Some pairs have torsion as well: >> My algorithm gets rid of the torsion, that's really the point of it > all.That's the thing with isolve? Graham
Message: 2397 - Contents - Hide Contents
Date: Mon, 10 Dec 2001 14:38 +0
Subject: Systems of Diophantine equations
From: graham@xxxxxxxxxx.xx.xx
I found a reference for this:
<Solving a System of Diophantine Equations with... * [with cont.] (Wayb.)>
It says they do lattice base reduction as a step to solving the equations.
Whereas we're trying to do a twist on base reduction. Here's a paper
about that:
<Lattice basis reduction and integer programmin... * [with cont.] (Wayb.)>
It says "The Gram-Schmidt vectors, however, do not necessarily belong to
the lattice, but they do span the same real vector space as b1, ..., bi,
so they are used as a "reference" for the basis reduction algorithm."
Also defines a lattice: "The lattice L spanned by b1, ..., bi is the set
of vectors that can be obtained by taking integer linear combinations of
the vectors b1, ..., bi."
It looks like what we want is between the Gram-Schmidt space (integer
combinations) and lattice (real combinations) because it has rational
combinations. Or is that the same as Gram-Schmidt?
I don't know, it's starting to get complicated. I think I'll ignore it
for a while.
Graham
Message: 2398 - Contents - Hide Contents
Date: Tue, 11 Dec 2001 20:04:11
Subject: Badness with gentle rolloff
From: David C Keenan
I now understand that Gene's logarithmically flat distribution is a very important starting point.
But
even Gene sees the need to apply cutoffs to it for (effectively if not
directly) very large errors and very high numbers of steps or
generators.
I agree that none of the alternatives I have suggested so far have
been very good. But I spent some more time on it and I thought "How
could I make a single badness function with a parameter that lets me
vary it continuously between Gene's log-flat-dist measure and
something more like what I've been groping towards." Well I found it,
and it gives the effect of a gentle rolloff as opposed to a sharp
cutoff for those properties mentioned above.
steps^(4/3) * exp((cents/k)^r)
The parameter r gives the rate of rolloff. From my spreadsheet
expeiments I'd guess that putting r at about 0.5 is going to make the
most people happy, but maybe it could go a bit lower to make Paul and
Gene happy. Note that as r gets closer to 0 this approaches:
steps^(4/3) * exp(ln(cents/k))
= step^(4/3) * cents/k
Which gives the log-flat distribution for 7-limit (If I've remembered
correctly). The parameter k becomes a mere scale factor here and so
becomes irrelevant to the ranking.
In the more general rolled-off form, the parameter k indirectly (and
in conjunction with r) determines where the middle of the band is for
rolloff. My favoured way of calibrating this (for 7-limit ETs) is to
make 9-tET come out with only slightly more badness than 72-tET, thus
agreeing with the log-flat-dist in this regard. When r is 0.5 that
makes k about 2.1 cents. The formula can probably be massaged into a
form where k doesn't have to be readjusted whenever you change r.
[To see the columns line up in the tables below, when reading from
Yahoo's web interface, you can choose "Message Index" then "Expand
Messages".]
Here's a top-20 7-limit ET ranking according to
steps^(4/3) * exp(sqrt(cents/2.1))
Steps RMS Badness
error
(cents)
-------------------------
31 4.0 390
22 8.6 466
27 7.9 562
41 4.2 583
19 12.7 595
26 10.3 709
46 4.5 714
15 18.5 720
53 3.5 722
72 1.8 747
9 29.1 776
10 27.4 799
68 2.4 815
37 7.6 824
12 24.5 836
50 5.0 858
99 0.9 876
36 8.6 896
58 4.0 899
62 4.0 982
Here for comparison is the log-flat
steps^(4/3) * cents
with a cutoff above 612-tET.
Steps RMS Badness
error
-------------------------
171 0.3 254
270 0.2 384
31 4.0 393
99 0.9 405
4 64.6 410
5 52.2 446
441 0.1 457
1 478.8 479
72 1.8 526
22 8.6 530
9 29.1 545
10 27.4 591
41 4.2 596
27 7.9 638
342 0.3 641
19 12.7 645
12 24.5 673
2 268.0 675
68 2.4 676
15 18.5 684
Regards,
-- Dave Keenan
Brisbane, Australia
Dave Keenan's Home Page * [with cont.] (Wayb.)
Message: 2399 - Contents - Hide Contents Date: Tue, 11 Dec 2001 06:49:47 Subject: Re: More lists From: genewardsmith --- In tuning-math@y..., "paulerlich" <paul@s...> wrote:> 1:3:5:9 > 1:3:7:9 > 1:3:9:11 > 10:12:15:18 > 12:14:18:21 > 18:22:24:33 > > which contain only 11-limit consonant intervals, would be important > to your music?Indeed they are, but they are taken care of.
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