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Message: 1500 - Contents - Hide Contents Date: Wed, 29 Aug 2001 03:23:23 Subject: Re: Defining CS and propriety for newbies (was: Now I think "the hypothesis" is tru) From: Carl Lumma>> >Interval class" is just a bad way to say "scale step". >> Eek! No it isn't. I remember you had this problem before. > > A scale step is the distance between two _consecutive_ scale > degrees. For example, in western diatonic scales we have > whole-tone steps and half-tone steps. We don't have minor third > steps or perfect fifth steps.You're right. I usually use "scale interval" here. I think my past problem was "scale degrees", which of course are pitches, not intervals.>> "Every interval" is just a bad way to say "every acoustic >> interval". Does that help? >> I don't think it helps. "Acoustic" simply means "relating to > sound". Everything we deal with here is acoustic.Not really. Propriety has only to do with the relative sizes of a scale's intervals, not with the actual sizes -- in fact, Rothenberg discards the interval matrix in favor of the rank- order matrix very early on. Now, "acoustic" may not be the best way to get this across, I'll agree. How would you say it?> //propriety stuff// Great job! > A more meaningful definition for CS would be of the form: > > A scale is CS if all intervals in the same range of sizes (in > cents), (with all ranges defined so as to be disjoint), span > the same number of scale steps.Now you've got me confused.> Notice that this is almost equivalent to strict-propriety, written > conversely. However a scale which is not strictly proper (i.e. it > has number-of-step ranges that meet or overlap) might be able to > have these ranges split into sub-ranges in such a way thay they no > longer overlap and it is thereby CS. > > 4. CS? > Number of steps in interval 4 4 > 1 2 3 <-> 3 <---> > Ranges <--------> <--------> <---> <-> > | | | | | | | | > 0 100 200 300 400 500 600 700 > > Interval size (cents) > > But clearly, this division into non-overlapping sub-ranges must be > musically meaningful and in particular the sub-ranges must not be > allowed to be too narrow, or else we are back to the trivial case > where every subrange can consist of a single size.There are many scales which have overlapping intervals subtending the same number of scale steps, which also lack ambiguous intervals. Therefore they are CS but not proper. These cases don't represent a failure of either concept, though. Propriety is based on the idea that listeners order scale intervals by size. CS is based on the idea that listeners can recognize particular intervals. Actually, both properties can result in what I'd call "convenience items", such as transpositional coherence (cough!), and handy things for PBs. I think these are primarily what motivated Wilson with the concept. But on perceptual grounds, I'd say CS doesn't have much of a leg to stand on (Paul's consonance-only CS may be another matter), but if it does, it would still have the leg for improper CS's. -Carl
Message: 1501 - Contents - Hide Contents
Date: Wed, 29 Aug 2001 06:02:29
Subject: Some definitions
From: genewardsmith@xxxx.xxx
DEFINITIONS
(1) A *tone group* T is a finitely generated subgroup of the group
(R+, *) of the positive real numbers under multiplication. A *tone*
is any element of (R+, *), that is, any positive real number
considered multiplicatively.
Hence, T is generated by n tones {t1, ..., tn}, which have the
property that
t1^e1 * ... * tn^en != 1
so long as the integer exponents e1...en are not all zero. The number
n is the *rank* of T, and any element t \in T is a tone of T. The
tones {t1, ..., tn} are the *generators* of T. The canonical examples
are the p-limit groups, so we also define the *p-limit group* T_p to
be the tone group generated by the primes less than or equal to the
prime p.
(2) A *note group* N is a finitely generated free abelian group which
we may identify with row vectors with integer entries. It is a sort
of generalized musical notation, since the canonical example makes
[a, b] represents "a" octaves and "b" fifths in a meantone system. If
we equate A 440 to [0,0], then any note in ordinary musical notation
may be translated into or out of this system--ordinary musical
notation can be thought of as representing this note group. Here of
course we do *not* equate B# with C, etc. The elements of N we call
notes, or N-notes if we need to be specific.
(3) A *tuning map* or *tuning* for the note group N is a
homomorphism "tune" of N into
(R+, *); it is defined by its values tune([1, 0, .. , 0]) = t1, tune
([0, 1, 0, ...,0]) = t2, ... tune([0, 0, ..., 1]) = tn. The image T =
tune(N) under this map is the *corresponding tone group*; if T is
also of rank n then tune is a *tuning isomorphism* and {t1, ..., tn}
are generators for T.
(4) If there are n primes less than or equal to a prime p, we define
the note group N_p to be the rank n free group, and the just tuning
map to be the tuning
just([1, 0, ..., 0]) = 2, just([0, 1, ... , 0]) = 3, ..., just([0,
0, ..., 1]) = p.
"Just" is therefore a tuning isomorphism from N_p to T_p.
(5) The dual N` to a note group N is the group N` = Hom(N, Z) of
homomorphisms from N into the integers. The elements of N` we call
ets, or N-ets if we need to be specific. If we take N concretely as
consisting of row vectors with integer entries, then we may take N`
to be column vectors with integer entries. If h and g are any two ets
and v is a note, then h+g is an et defined by "h+g"(v) = h(v)+g(v);
the 0-division et "0" which sends all notes v to 0 is the identity;
this defines the group structure on N`. Concretely, it is represented
by adding the two column vectors, just as the group N is defined by
adding row vectors.
If n is any positive integer, we define the et h_n in the dual to the
p-limit group N_p` to be the column vector with entries round(n log_2
(p_i)), so that h_n =
[ n ]
[round(n log_2(3)]
[round(n log_2(5)]
.
.
.
[round(n log_2(p)].
Here "round(x)" is the real number x rounded to the nearest integer,
so that round(x)-1/2<x<=round(x)+1/2 We should note that it is *not*
always the case that h_n + h_m = h_{n+m}.
(6) If M is any subgroup of the note group N, then null(M) is the
subgroup of N` consisting of all elements h \in N such that h(m) = 0
for every m \in M.
(7) If M is any subgroup of the dual to a note group N`, then null(M)
is the subgroup of N consisting of all v \in N such that h(v) = 0 for
every h \in M.
(8) If h \in N` is any nonzero et, then kernel(h) or null(h), the
*kernel* of h is the subgroup of rank n-1 of N defined as null(H),
where H is the group generated by h. In other words, the kernel is
the set of all notes v such that h(v)=0.
(9) A set of n-1 notes {u1, ..., u_{n-1}}in kernel(h) is a
*generating set* if any element u of the kernel is a Z-linear sum u =
j_1 u_1 + ... + j_{n-1}, where the j_i are integers.
(10) A nonzero et h is *reduced* if the coordinates of h are
relatively prime; that is, if no integer greater than one divides all
the v_i where v_i is the number in the ith row of v.
(11) If {u_i} is a generating set for h, then we may divide it
into "a" *commas* and "b" *chromas*, where a+b = n-1. The subset of
kernel(h) generated by the commas is the *commatic kernel* K, and the
quotient group N/K is the *commatic note group*. If all the
generators are commas, the commatic note group is simply Z and if h
is reduced we may identify it with the homomorphism from N to N/K. If
all of the generators are chromas, the commatic kernel is {0} and the
commatic note group is N. In general, the commatic note group is of
rank b.
Message: 1502 - Contents - Hide Contents Date: Thu, 30 Aug 2001 19:19:17 Subject: Re: Finding MOS the EZ way From: Paul Erlich Can you use the same argument to tighten up my "proof"?
Message: 1503 - Contents - Hide Contents Date: Thu, 30 Aug 2001 07:05:28 Subject: Finding MOS the EZ way From: genewardsmith@xxxx.xxx Let m and n be integers, and suppose we are seeking a MOS of m scale steps within an n-et. We may first reduce to the case where m and n are relatively prime, by setting d=gcd(m,n) and M=m/d, N=n/d. The scale will then repeat an interval of repetition N d times. Then we have the following theorem: (Theorem EZ) Let 0<a<M, 0<b<N be integers such that |a/b – M/N| < 1/(bM) Then b/N is a generator for a MOS with M scale steps in an N-et. Proof: If we multiply the above inequality on both sides by b/M, we obtain |a/M – b/N| < 1/M^2 which entails a/M is a semiconvergent to b/N, and hence we have a MOS. Corollary: If a/b is the penultimate convergent to M/N (i.e., the closest rational approximation with denominator less than N) then b/N is the generator for a MOS. Proof: The condition implies |a/b – M/N| = 1/(bN), and since N>M this implies |a/b – M/N| < 1/(bM). If b is small compared to M, the above becomes both sufficient *and* necessary. If b<=M, then 1/(bM)<=1/b^2 and a/b must be a semiconvergent to M/N. If b<=2M, then 1/(bM)<=1/(2b^2) and a/b is a convergent to M/N.
Message: 1504 - Contents - Hide Contents Date: Thu, 30 Aug 2001 19:22:21 Subject: Re: Defining CS and propriety for newbies (was: Now I think "the hypothesis" is tru) From: Paul Erlich Guys, CS is useful in the context in which it was intended -- strict JI scales that are connected in the lattice. Forget about random scales or ETs. Now, it seems that if a PB, in untempered from, is CS, then tempering all but one of the unison vectors will lead to an MOS. I'm hereby betting that this is true. Can we prove it?
Message: 1505 - Contents - Hide Contents Date: Thu, 30 Aug 2001 19:57:08 Subject: Re: Finding MOS the EZ way From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:> Can you use the same argument to tighten up my "proof"?It's quite possible EZ could help with that; I'll think about it.
Message: 1506 - Contents - Hide Contents Date: Thu, 30 Aug 2001 22:28:48 Subject: Re: Finding MOS the EZ way From: Dave Keenan --- In tuning-math@y..., genewardsmith@j... wrote:> Let m and n be integers, and suppose we are seeking a MOS of m scale > steps within an n-et.I hope you realise that MOS exist which are not in any ET?> We may first reduce to the case where m and n > are relatively prime, by setting d=gcd(m,n) and M=m/d, N=n/d. The > scale will then repeat an interval of repetition N d times. > > Then we have the following theorem:Very nice. Now can you give us the _algorithm_ to find all (generator, period) pairs (say in ET steps) for a MOS of m steps within an n-ET? On second thoughts, much more useful would be an algorithm to give us the generator and period (in cents) for a tempering of any octave-equivalent periodicity block where all but one unison vector is tempered out. Or even a proof that such is possible. Regards, -- Dave Keenan
Message: 1507 - Contents - Hide Contents Date: Thu, 30 Aug 2001 23:43:19 Subject: Re: Finding MOS the EZ way From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:> I hope you realise that MOS exist which are not in any ET?Yes, but the theorem won't apply to them. :)> Very nice. Now can you give us the _algorithm_ to find all (generator, > period) pairs (say in ET steps) for a MOS of m steps within an n-ET?The Euclidean algorithm will always give one such pair by the corollary to EZ. It is also easy to find semiconvergents for M/N and test them. In general we only need to look at the range a <= M/2, b <= N/2, which makes the above quite a bit more powerful. Of course, this doesn't quite answer your question, but it's a start.
Message: 1508 - Contents - Hide Contents Date: Fri, 31 Aug 2001 19:27:31 Subject: Re: Defining CS and propriety for newbies (was: Now I think "the hypothesis" is tru) From: Paul Erlich --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:>> Guys, CS is useful in the context in which it was intended -- strict >> JI scales that are connected in the lattice. Forget about random >> scales or ETs. >> The silly definition of CS works fine for subsets of ETs-with-less- > than-100-notes. In fact it works best for these. Rational scales have > the same problem as random scales. > > Consider a connected rational scale that has, for example, a 5:7 > spanning 8 steps while a 343:480 (still 7-prime-limit and only 0.7c > different) spans 9 steps. That would be CS according to the silly > definition. But everyone would hear the 343:480 as a 5:7 and so this > definition is useless.It may become useful in the context of proving the hypothesis. In the case above, 2400:2401 would not be a commatic unison vector in the scale and thus would probably be enlarged when the commatic unison vectors are tempered out. I'd prefer to think of it as an abstract, "group-theoretical" property of the set of ratios at this stage, rather than a perceptual characterization of the scale.
Message: 1509 - Contents - Hide Contents Date: Fri, 31 Aug 2001 19:29:40 Subject: Re: Defining CS and propriety for newbies (was: Now I think "the hypothesis" is tru) From: Paul Erlich --- In tuning-math@y..., "Carl Lumma" <carl@l...> wrote:>> Consider a connected rational scale that has, for example, a 5:7 >> spanning 8 steps while a 343:480 (still 7-prime-limit and only 0.7c >> different) spans 9 steps. That would be CS according to the silly >> definition. But everyone would hear the 343:480 as a 5:7 and so >> this definition is useless. >> For purposes of issue (2), PBs containing a step 2401:2400 would > be odd -- and unless all their uv's were smaller still, they > wouldn't be CS by Wilson's definition anyway. In short, I think > Wilson's definition is splendid for (2).Great minds think alike, Carl. ;^)
Message: 1510 - Contents - Hide Contents Date: Fri, 31 Aug 2001 00:17:23 Subject: Re: Defining CS and propriety for newbies (was: Now I think "the hypothesis" is tru) From: Dave Keenan --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:> Guys, CS is useful in the context in which it was intended -- strict > JI scales that are connected in the lattice. Forget about random > scales or ETs.The silly definition of CS works fine for subsets of ETs-with-less- than-100-notes. In fact it works best for these. Rational scales have the same problem as random scales. Consider a connected rational scale that has, for example, a 5:7 spanning 8 steps while a 343:480 (still 7-prime-limit and only 0.7c different) spans 9 steps. That would be CS according to the silly definition. But everyone would hear the 343:480 as a 5:7 and so this definition is useless. I agree with Carl, that a useful definition of CS would recognise some intervals as the same, even if they differ by a few cents. Paul, I understand your consonant-CS satifies this requirement. My current thinking is still that CS needs to be qualified with a cent value, as in "this scale has a CS-level of 33 cents" which means "this scale is CS provided you don't conflate any intervals differing by more than 33 cents". Bigger numbers are better. In that case, the above hypothetical rational scale only has a CS-level of 0.7 cents, which everyone should realise is useless. CS-level is meaningful for any kind of scale and doesn't require us to agree on which intervals are consonant, or what deviations are still recognisable as the same interval, or on deviations from what. To determine the CS-level of a scale: 1. Consider the set of pairs (interval-size-in-cents, number-of-scale-steps) over all the scale's intervals. List these pairs in order of interval-size-in-cents. (If two or more have the same size, their order relative to each other doesn't matter, and we know immediately that the scale has a CS-level of zero.) 2. Partition this list so that each sub-list is as large as possible with all its members having the same number-of-scale-steps. 3. For each pair of adjacent sub-lists, consider the difference in size between the intervals on either side of the partitioning boundary between them. The minimum of all such differences is the CS-level. -- Dave Keenan
Message: 1511 - Contents - Hide Contents
Date: Fri, 31 Aug 2001 03:17:00
Subject: More definitions
From: genewardsmith@xxxx.xxx
Let N be a note group. The dual group to N, N` = Hom(N, Z) had
elements I called "ets". I think a better name would be *vals*, since
these are generalized valuations. An et is a val which is close to
being the real valuation, or valuation at infinity, on some p-limit
group N_p. The dual group of a note group is its *val group*, it is a
finitely generated free group isomorphic (non-canonically) to its
dual. The dual of a val group is its note group. If N_p is the
p-limit note group we will call the dual group to this, N_p`, the
p-limit val group.
Let p be a prime and N be a note group representing (exactly or
approximately) notes in the p-limit, and let N_p` be the p-limit val
group. If the number of primes up to p is n, then N has rank k <= n.
Suppose {g_1, g_2, ... g_k} are k vals in N_p` which generate a
vector space of dimension k and such that any Q-linear combination,
(that is, any e1*g1 + ... + ek*gk where the coefficients are rational
numbers) which belongs to N_p` is a Z-linear combination (that is,
the coefficients are integers.) Then we define the k-tuple of vals
[g_1, g_2, ..., g_k] to be a *notation* for N. The elements of N are
thereby given musical meaning in a different way than by giving a
tuning for N; we can give N a tuning, a notation, or both in order to
make it musically meaningful.
If q is a rational number which when factored into primes has p^e as
the factor corresponding to the prime p, then a standard definition
of number theory defines the p-adic valuation v_p by
v_p(q) = e.
If we apply this to N_p, where 2^e2 * 3^e3 * 5^e5 ... corresponds to
[e2, e3, e5, ...] then v_2([e2, e3, e5, ...] = e2,
v_3([e2, e3, e5, ...] = e3 and so forth. These v_p are therefore
vals, and it seems reasonable to use this same standard notation. We
therefore obtain a notation for N_p which consists of
[v_2, v_3, ..., v_p], which as a matrix is the n by n identity
matrix. This notation is the identity notation. In general, if the
notation consists of k vals in an N_p` of rank n (meaning there are n
primes <= p) then it may be written explicitly as a matrix with n
rows and k columns.
To take a more interesting example, suppose we define vals g_2, g_3
in N_5` by
g_2([a, b, c]) = a-4c,
g_3([a, b, c]) = b+4c.
If u is a note in N_p, then g_2(u) defines how many octaves and g_3
(u) how many twelfths in a meantone system we require to get to the
note associated to u; [g_2, g_3] is therefore a notation for the
meantone note group. As a matrix, this notation is
[ 1 0]
[ 0 1]
[-4 4].
We get a different notation for the same group from [h_7, h_12],
which tells us how many steps respectively in the 7-et and in the
12-et the note u is represented by; as a matrix this notation is
[ 7 12]
[11 19]
[16 28].
The matrix
[ 7 12]
T = [11 19]
transforms a note in the first notation to the same note in the
second notation; the fact that the matrix T is unimodular (that is,
that |det(T)| = 1) shows that T is inevitable and the two notations
define the same meantone note group. The inverse matrix
[ 19 -12]
T^(-1) = [-11 7]
transforms something in the second notation back to the first. Hence,
for instance if [1, 2] is one step in the 7-et and two steps in the
12-et, [1, 2] T^(-1) = [-3, 2], which corresponds to 9/8 in a
meantone system (so 10/9 and 9/8 are represented by the same note)--
in other words, a full tone. A tone (in this sense!) is, as we might
expect, [1, 2] in the [h_7, h_12] notation.
Message: 1512 - Contents - Hide Contents Date: Fri, 31 Aug 2001 03:56:33 Subject: Re: Defining CS and propriety for newbies (was: Now I think "the hypothesis" is tru) From: Carl Lumma>> >uys, CS is useful in the context in which it was intended -- >> strict JI scales that are connected in the lattice. Forget >> about random scales or ETs. >>The silly definition of CS works fine for subsets of ETs-with-less- >than-100-notes. In fact it works best for these. Rational scales >have the same problem as random scales.There are really two issues here: (1) How does CS fit, if at all, into a model of melodic perception? (2) What happens to a PB when it is CS? Items belonging to (2) I referred to in my last message as "convenience items". I did mention that Wilson probably had these in mind when he coined "CS". It is for issue (1) that my criticisms of the measure, and yours Dave, are important. And (1) is also the category which Paul's consonance-CS addresses (or so it seems to me).>Consider a connected rational scale that has, for example, a 5:7 >spanning 8 steps while a 343:480 (still 7-prime-limit and only 0.7c >different) spans 9 steps. That would be CS according to the silly >definition. But everyone would hear the 343:480 as a 5:7 and so >this definition is useless.For purposes of issue (2), PBs containing a step 2401:2400 would be odd -- and unless all their uv's were smaller still, they wouldn't be CS by Wilson's definition anyway. In short, I think Wilson's definition is splendid for (2).>I agree with Carl, that a useful definition of CS would recognise >some intervals as the same, even if they differ by a few cents. >Paul, I understand your consonant-CS satifies this requirement.As I understood it, Paul's consonance-CS (as I've called it) is the same as Wilson's, except it only cares about ambiguous intervals if they are consonances in the scale. So this doesn't address near-collisions. Once again, my interp. of a possible application of CS for (1) is that it speeds scale interval tracking if listeners can recognize exact intervals. I don't think they can, with the possible exception of strong consonances. For melodic intervals in general, I agree with Rothenberg that listeners simply order melodic intervals by size. But you're right Dave -- for either Rothenberg or CS(1), near collisions are a problem! R. addresses this somewhat by replacing propriety with stability. I've addressed it for propriety by creating what is called in Scala, "Lumma stability" and "Lumma impropriety".> My current thinking is still that CS needs to be qualified with > a cent value, as in "this scale has a CS-level of 33 cents" which > means "this scale is CS provided you don't conflate any intervals > differing by more than 33 cents". Bigger numbers are better.That's a good idea. -Carl
Message: 1513 - Contents - Hide Contents Date: Sat, 01 Sep 2001 09:15:12 Subject: EZ converse From: genewardsmith@xxxx.xxx It turns out that the converse theorems to the ones I stated before about convergents and semiconvergents are a/M a convergent to b/N ==> |a/M - b/N| < 1/M^2 a/M a semiconvergent to b/N ==> |a/M - b/N| < 2/M^2 From this we can get (EZ converse) If b/N is a generator for a MOS with M scale steps out of an N-rt, then there exists an a/b such that |a/b - M/N| < 2/(bM) Proof: Multiply both sides of |a/M - b/N| < 2/M^2 by M/b. If b<M/2, we can replace the above condition by |a/b - M/N| < 1/b^2 and so we need look only at semiconvergents to M/N. If b<M/4 then we can replace the above condition by |a/b - M/N| < 1/2b^2 and we only need the convergents to M/N. However, we have a nice, general algorithm: (EZ integers) If b/N is a generator for an M-in-N system, then |aN - bM| < 2N/M. Proof: Multiply both sides of |a/M - b/N| < 2/M^2 by MN. We then need only check a finite list of values of e up to +-floor(M/N) in the equation aN - bM = e. This equation is easily solved, since modulo N it reduces to b = -e/M (mod N) and mod M it reduces to a = e/N (mod M), where the congruences are solvable since M and N are relatively prime. The usual way of solving such congruences is again via the Euclidean algorithm/continued fractions, the inverse up to sign can be found from |pN - qM| = 1, where p/q is the penultimate convergent to M/N. This means that if you want find all the MOS where M and N are both very large it's actually not that hard so long as N is not a great deal larger than M. (What's hard is figuring out the practical utility of doing so.)
Message: 1514 - Contents - Hide Contents Date: Sat, 01 Sep 2001 09:21:23 Subject: Re: EZ converse From: genewardsmith@xxxx.xxx --- In tuning-math@y..., genewardsmith@j... wrote:> We then need only check a finite list of values of e up to > +-floor(M/N) in the equation aN - bM = e.I meant +-floor(2M/N), of course.
Message: 1515 - Contents - Hide Contents Date: Sat, 01 Sep 2001 09:25:11 Subject: EZ converse From: genewardsmith@xxxx.xxx It turns out that the converse theorems to the ones I stated before about convergents and semiconvergents are a/M a convergent to b/N ==> |a/M - b/N| < 1/M^2 a/M a semiconvergent to b/N ==> |a/M - b/N| < 2/M^2 From this we can get (EZ converse) If b/N is a generator for a MOS with M scale steps out of an N-rt, then there exists an a/b such that |a/b - M/N| < 2/(bM) Proof: Multiply both sides of |a/M - b/N| < 2/M^2 by M/b. If b<M/2, we can replace the above condition by |a/b - M/N| < 1/b^2 and so we need look only at semiconvergents to M/N. If b<M/4 then we can replace the above condition by |a/b - M/N| < 1/2b^2 and we only need the convergents to M/N. However, we have a nice, general algorithm: (EZ integers) If b/N is a generator for an M-in-N system, then |aN - bM| < 2N/M. Proof: Multiply both sides of |a/M - b/N| < 2/M^2 by MN. We then need only check a finite list of values of e up to +-floor(2N/M) in the equation aN - bM = e. This equation is easily solved, since modulo N it reduces to b = -e/M (mod N) and mod M it reduces to a = e/N (mod M), where the congruences are solvable since M and N are relatively prime. The usual way of solving such congruences is again via the Euclidean algorithm/continued fractions, the inverse up to sign can be found from |pN - qM| = 1, where p/q is the penultimate convergent to M/N. This means that if you want find all the MOS where M and N are both very large it's actually not that hard so long as N is not a great deal larger than M. (What's hard is figuring out the practical utility of doing so.)
Message: 1516 - Contents - Hide Contents Date: Sun, 02 Sep 2001 12:04:22 Subject: Question for Gene From: Paul Erlich Gene, could you please comment on this: lattice orientation * [with cont.] (Wayb.) In particular, I'm assuming a city-block or taxicab metric. Is Kees observing that in his final lattice? It looks like he isn't. What else can you say?
Message: 1517 - Contents - Hide Contents Date: Sun, 02 Sep 2001 18:02:30 Subject: Re: Question for Gene From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:> What else can you say?Before I can say much else, I need to read up on what you two are doing, but I can make one comment now: if you want to put a symmetrical metric onto the note-classes defined as octave equivalents of 3^a 5^b, the way to do it is with the quadratic form q(3^a 5^b) = a^2 + ab + b^2. This and the bilinear form associated to it will make 5/4 and 5/3 the same size, both being one step away from 1. This is quite interesting when looking at automorphisms of the 5- limit which preserve octave equivalence and also the 3-et; these can be thought of as generalizations of the major/minor transformation. In the same way, putting the quadratic form q(3^a 5^b 7^c) = a^2 + b^2 + c^2 + ab + ac + bc onto the octave-equivalence classes of the 7-limit gives a lattice structure containing tetrahedrons and octahedrons, which is the only semiregular 3D honeycomb; around each vertex are eight tetrahedra and six octahedra in a pattern like the triangles and squares of a cubeoctahedron. If we look for automorphism groups which preserve this metric and the 4-et we get another interesting group of musical transformations--the octahedral group applied to 7-limit music.
Message: 1518 - Contents - Hide Contents Date: Sun, 02 Sep 2001 22:09:55 Subject: Symmetric 5-limit and 7-limit distance measures From: genewardsmith@xxxx.xxx Suppose we look at octave equivalence classes in the 5-limit, which we may represent by vectors [a, b] (representing the class containing element 3^a 5^b.) Then we define the quadratic form q5 by setting q5([a, b]) = a^2 + ab + b^2. The square root of q5 is an ordinary Euclidean distance to the origin [0, 0], and we may define distance between any two classes v1, v2 by d(v1, v2) = ||v1 - v2|| = sqrt(q5(v1, v2)). The 5-limit note classes now defines the regular tessellation of the plane by equilateral triangles, each triangle representing a triad. (The dual tessellation by hexagons represents the 5-limit harmonic relationships between triads.) We may define interesting, highly symmetric scales by picking a center point m and taking everything in a sphere of radius r around m. The simplest example is a triad, where we can take q5(v - [1/3, 1/3]) <= 1/3, giving 1 - 5/4 - 3/2. If we take everything in a sphere of radius 1 around [0, 0] we get 1 - 6/5 - 5/4 - 4/3 - 3/2 - 5/3 - (2), while q5(v - [1/3, 1/3]) <= 4/3 gives 1 - 6/5 - 5/4 - 3/2 - 5/3 - 15/8 - (2). Similarly, we may define q7([a, b, c]) = a^2 + b^2 + c^2 + ab + ac + bc. If we take q7(v - [1/4, 1/4, 1/4]) <= 1/4 we get 1 - 5/4 - 3/2 - 7/4. More interesting is what happens when we center at [1/2, 1/2, -1/2]: q7(v - [1/2, 1/2, -1/2]) <= 1/2 gives us 1 - 15/14 - 5/4 - 10/7 - 3/2 - 12/7 - (2). This harmonic octahedron is very well supplied with harmony--while it has only six notes, corresponding to the six verticies of an octahedron, it has eight chords, corresponding to the eight faces of an octahedron. (If you've seen D&D dice, the 8-sided dice are octahedra.) Concentrating on the octahedra seems to be the best way of visualizing 7-limit harmony in this symmetrical form. We get figures like 21/20 ----- 3/2 ----- 15/14 | | 7/4 5/4 | (6/5) (12/7) | 7/5 ----- 1 ------ 10/7 Paul or someone could draw something much better, but the 1, 3/2, 15/14, 10/7, 7/5 and 21/20 are all in a plane, the 7/4 and 5/4 are above, forming the top verticies of two octahedra, and the 6/5 and 12/7 are on the bottom, completing the octahedra. You can also see the tetrahedron 1 - 5/4 - 3/2 - 7/4 in there if you squint hard. This may all have something to do with what Monzo is working on.
Message: 1520 - Contents - Hide Contents Date: Mon, 03 Sep 2001 20:59:43 Subject: Re: Symmetric 5-limit and 7-limit distance measures From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> Suppose we look at octave equivalence classes in the 5-limit, which > we may represent by vectors [a, b] (representing the class containing > element 3^a 5^b.) Then we define the quadratic form q5 by setting > > q5([a, b]) = a^2 + ab + b^2. > > The square root of q5 is an ordinary Euclidean distance to the origin > [0, 0], and we may define distance between any two classes v1, v2 by > > d(v1, v2) = ||v1 - v2|| = sqrt(q5(v1, v2)). > > The 5-limit note classes now defines the regular tessellation of the > plane by equilateral triangles, each triangle representing a triad. > (The dual tessellation by hexagons represents the 5-limit harmonic > relationships between triads.) > > We may define interesting, highly symmetric scales by picking a > center point m and taking everything in a sphere of radius r around > m. The simplest example is a triad, where we can take > > q5(v - [1/3, 1/3]) <= 1/3, > > giving 1 - 5/4 - 3/2. > > If we take everything in a sphere of radius 1 around [0, 0] we get > > 1 - 6/5 - 5/4 - 4/3 - 3/2 - 5/3 - (2),What happened to 8/5? This _should_ be the 5-limit Tonality Diamond.> More interesting is what happens when we > center at [1/2, 1/2, -1/2]: > > q7(v - [1/2, 1/2, -1/2]) <= 1/2 > > gives us > > 1 - 15/14 - 5/4 - 10/7 - 3/2 - 12/7 - (2).That's the hexany. You should familiarize yourself with CPS (Combination Product Set) scales . . . the hexany is the 2)4 (1,3,5,7) hexany, meaning it's the set of numbers you get when you take products of 2 numbers at a time out of {1,3,5,7}. The 3)6 (1,3,5,7,9,11) and 3)6 (1,3,7,9,11,15) are called Eikosanies, and are similarly symmetrical in the equilateral-triangular lattice. So is the 2)5 (1,3,5,7,9) dekany, when 9 gets its own axis -- Dave Keenan created a splendid rotating dekany that actually plays the notes as the dekany rotates in 4-dimensional space. We went through a lot of interesting mathematics on the tuning list -- for example, we got in touch with George Olshevsky who has a vast knowledge of polychora in 4-, 5-, and 6-dimensional space.> > Concentrating on the octahedra seems to be the best way of > visualizing 7-limit harmony in this symmetrical form. We get figures > like > > 21/20 ----- 3/2 ----- 15/14 > | > | 7/4 5/4 > | (6/5) (12/7) > | > 7/5 ----- 1 ------ 10/7 > > Paul or someone could draw something much better, but the 1, 3/2, > 15/14, 10/7, 7/5 and 21/20 are all in a plane, the 7/4 and 5/4 are > above, forming the top verticies of two octahedra, and the 6/5 and > 12/7 are on the bottom, completing the octahedra. You can also see > the tetrahedron 1 - 5/4 - 3/2 - 7/4 in there if you squint hard.As well as the tetrahedron 1 - 6/5 - 3/2 - 12/7, the so-called "minor tetrad".> This > may all have something to do with what Monzo is working on.Hmm . . . the triangular lattice is the way _I_ do things . . . Monz uses a different lattice definition (with different angles), and only shows the connections corresponding to _primes_ -- so intervals like 5:3 don't get a connector.
Message: 1521 - Contents - Hide Contents Date: Mon, 03 Sep 2001 05:29:42 Subject: Re: Question for Gene From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:> Gene, could you please comment on this:It looks to me like he's trying to do something which doesn't quite work, which is to analyze a taxicab metric in Euclidean terms.
Message: 1522 - Contents - Hide Contents Date: Mon, 03 Sep 2001 21:00:55 Subject: Re: Question for Gene From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote: >>> Gene, could you please comment on this: >> It looks to me like he's trying to do something which doesn't quite > work, which is to analyze a taxicab metric in Euclidean terms.What if we forget about the Euclidean part -- can you shed any light on the "paradox" in pure taxicab terms?
Message: 1523 - Contents - Hide Contents
Date: Mon, 03 Sep 2001 05:49:55
Subject: Distance measures cut to order
From: genewardsmith@xxxx.xxx
Rather than screwing around trying to transform taxicab metrics, I
would suggest starting off with a Euclidean metric which works the
way you want it to work. For instance, you can give 3, 5, and 5/3 any
values you like, and this determines the metric.
If we want them all to be the same, as I did for my symmetric metric,
then if q([x, y]) = ax^2 + bxy + cy^2, we may substitute for
q([1,0])=1, q([0,1])=1, and q([-1,1])=1, getting the system of three
equations in three unknowns, {a=1, c=1, a-b+c=1}. Solving this gives
us a=b=c=1. If by some voodoo I determine 3 should have a distance of
1 from the unison, and 5 and 5/3 a distance of sqrt(2), I solve
another set of three equations and get x^2+y^2...and so forth.
In the 7-limit, we can solve six equations for six unknowns, by
specifying distances for 3,5,7,5/3,7/3, and 7/5. Etc.
Message: 1524 - Contents - Hide Contents Date: Mon, 03 Sep 2001 21:01:03 Subject: Re: Question for Gene From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote: >>> Gene, could you please comment on this: >> It looks to me like he's trying to do something which doesn't quite > work, which is to analyze a taxicab metric in Euclidean terms.What if we forget about the Euclidean part -- can you shed any light on the "paradox" in pure taxicab terms?
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