Thursday, April 10, 2014

Partition Puzzle 5: Feldman Analysis of Z-rels mod 31

Yesterday David Feldman left a comment on a previous E&EN blog entry. I felt it ought to be brought forward as a post of its own so it doesn't get overlooked. He has agreed to let me post it here as a Preliminary Report.

[Jon Wild wrote:]

"There is a quintuplet of Z-related hexachords in 31-tone equal
temperament whose interval vectors consist solely of 1s--they are all interval hexachords, expressing 15 interval classes in just six pitches. This set of 5 hexachords can be packed into the 31-tone aggregate leaving only one pitch uncovered. The odds that this is by chance are astronomical-there is some other principle at work, related to the observations Stephen has started to make, but I don't know what that principle is."

Here is a description of [this] example which may remove some of the mystery.

Modulo 31, the five sets can be written as[1]         

{3^0 , 3^10, 3^20, 3^3,  3^13, 3^23}         
{3^2 , 3^12, 3^22, 3^5,  3^15, 3^25}         
{3^4 , 3^14, 3^24, 3^7,  3^17, 3^27}         
{3^6 , 3^16, 3^26, 3^9,  3^19, 3^29}      
{3^8 , 3^18, 3^28, 3^11, 3^21, 3^31}

In the language of abstract algebra,
              {3^0 , 3^10, 3^20}
constitutes a subgroup of the multiplicative group of the field with 31elements and
              {3^3,  3^13, 3^23}

one of it's cosets.
              {3^0 , 3^6, 3^12, 3^18, 3^24}
constitutes a disjoint subgroup with 5 elements.

The five sets come from the first by multiplication, as one see quickly if one keeps in mind the congruence of 3^30 with 1 modulo 31.  Multiplication permutes intervals,but preserves all interval hexachords.

One could still argue that {3^0 , 3^10, 3^20, 3^3,  3^13, 3^23} giving an all-interval hexachord seems miraculous, but one surely gets all the intervals if no interval occurs twice. So write
            A = {3^0 , 3^10, 3^20}  B = {3^3 ,  3^13 , 3^23}
Obviously distinct pairs in A give distinct intervals (or the interval would have to be preserved by multiplication by 3^10); similarly distinct pairs in B; similarly a pair in A compared with a pair in B similarly two pairs each crossing A to B. 
Thus the only not obvious case would be a pair in A, say, and a pair crossing from A to B.  One can also rule out some of these cases a priori, and reduce the total number by symmetry, but in any case one doesn't have so many cases that the non-occurrence of an equality seems miraculous anymore.
– David Victor Feldman
9 April 2014 


[1] For a direct relationship to these sets in more recognizable integer notation, it may be helpful to use the following distribution of 3^n as an overlay:
1 25 5 27 24 11
9 8 14 26 30 6
19 10 2 17 22 23
16 28 18 29 12 21
20 4 7 13 15 3
If X is any one of these five sets,
V(X) = [6111111111111111]
V(X, complX) = [0 a a a a a a a a a a a a a a a]  (a=10)
remembering that complX always includes 0, the odd man out.

Sunday, April 6, 2014

Partition Puzzle 4: Knight's Gambit*

King Knight's Gambit

There appear to be two knotty issues in attempting to find a general solution to the homometric partition puzzle. First is the obvious problem posed by n-tuples. It is well known that z-related sets don't only come in pairs; there are z-related triples, quadruples, etc. This appears to be the most difficult problem. I'll put it on hold for now.

Second is a more subtle conjunct-disjunct problem. If two z-related sets S and T can be rotated or reflected (transposed or inverted) so they are in positions disjunct from one another, then we can always find a partition solution by the basic partition theorem given in the previous post. However, there are many cases for which two z-related sets cannot be positioned such that they are disjunct.

The easy case is when S and T have cardinalities larger than half the size of their K-space, but they are z-related because their smaller complements are z-related. A corollary to the basic partition theorem takes care of this type of case:

An example can be taken from K12. S={0,1,4,5,7} and T={3,6,9,10,11} are z-related. Q={2,8} and V(S,Q)=V(T,Q)=[0,2,2,2,2,2,0]. Their z-related complements are S'={2,3,6,8,9,10,11} & T'={0,1,2,4,5,7,8}, and Q'={0,1,3,4,5,6,7,9,10,11}. V(S',Q')=V(T',Q')=[5,12,12,12,12,12,5]. (Remember the first element of the spanning vector is the number of 0's (harmonic/melodic unisons, rhythmic simultaneities).)

When we get to K13 we run into a situation that at first appears to defy any attempt to apply any variation of the approach taken by the HPT. There are just two pairs of z-related hexads in K13.  {0,1,2,4,5,8} & {0,3,4,5,6,9} share the ICV [6323322] and {0,1,3,5,7,8} & {0,1,2,4,7,9} share ICV [6232233]. There are several notable things here. K13 is the first K-space with an odd number of nodes in which z-related sets appear. And 13 is not only odd, it's prime. Also, note the peculiar relationship between the vectors: their non-zero entries appear to be related by exchanging the ic cardinalities 2 and 3. Even though you may know why these two pairs are related this way, please stick with me through the following exposition which will lead off on a path that, at least within music theory (to my knowledge), is untrodden.i

Recall that in K12 there is a sort of sub-canonical atonal operation, usually denoted M5, which is simply multiplication by 5 mod 12. One of the interesting properties of this operation is that by modding back to 12 it exchanges interval classes 1 and 5, so if the ICV of some random set X is [aBcdeFg], the ICV of 5X will be [aFcdeBg], no matter how the operation may warp the shape of the set. And if B=F, X and 5X will also be Tn, TnI, or z-related. In K12 this operation only "works" (i.e., yields a full chromatic cycle) for multiplication by 1, 5, 7, and 11 because these are the only integers available that are co-prime with the modulus 12.

But since 13 is a prime number, in K13 all available integers 1 through 13 are co-prime with 13. For example, if X={0,1,2,4,5,8}, then 2X(mod13)={0,2,4,8,10,3}, so that V(X)=[6323322] and V(2X)=[6232233] (Figure 1).
Figure 1
So there isn't really an exchange of components between the vectors of these two pairs of sets, but a shuffle permutation that, in this example, gives a first-blush impression of such an exchange.[1] All of this is familiar territory to seasoned music theorists, even those who have never considered this particular case in K13 before.

Now comes the break-out question that might help us generalize the basic partition theorem to any pair of homometric sets, whether or not they are disjunct, and whether or not they appear in an even or odd K-space. Here's the question:

What if we don't mod out after multiplication?

In the example at hand, such a shift in procedure, rather than keeping us in K13, maps us into the subset consisting of only the even-numbered elements of K2X13 (i.e., K26). Furthermore, this shift retains the original ICV on the even-numbered components of the ICV of the transformed set(s) (Figure 2).
Figure 2
To reiterate & make the next move more clear, switch to a pictorial description. Look at the inscribed hexagons X={0,1,2,4,5,8} (the red hexagon) and Y={0,3,4,5,6,9} (blue hexagon) situated in K13 (Figure 3).
Figure 3
To apply the HPT we would like to rotate or reflect one of these hexagons such that X and Y are disjunct, leaving a singleton set for Q. Unfortunately, there is no way to do this – unless we think outside the circle, so to speak. Multiplying by 2 (without modding back into K13) yields the situation shown in Figure 4.
Figure 4
This gives us X'=2X={0,2,4,8,10,16} and Y'=2Y={0,6,8,10,12,18}.
The shapes of the hexagons in both cases are identical (invariant under a linear homothetic transformation). What has changed is that now they are situated only on the even numbered nodes, leaving us with an additional 13 nodes (the odd ones) to play with (Figure 5).
Figure 5
Rotating or reflecting either hexagon to odd-numbered nodes will guarantee that the two are disjunct. So let's move the blue hexagon by one click clockwise (Figure 6)
Figure 6
and ...

(1) we now have two hexagons with the same shape[2] as those we started with in K13;
(2) the K13 homometric relation is carried over into K26;
(3) the initial interval-class vector shared by X and Y in K13 is retained in the even components of the K26 vector for X' and Y' as well as any transposition of Y' (Figure 2);
and, finally, what is gained by this transformation,
(4) X' & Y' are disjunct and the basic partition theorem HPT can be applied. And ... voilĂ !

For the situation shown in Figure 6, X'={0,2,4,8,10,16}, Y"=T1(Y')={1,7,9,11,13,19}, = compl(X'UY") = {3,5,6,12,14,15,17,18,20,21,22,23,24,25}, and by the HPT:
V(X',Q') = V(Y",Q') = [06658866658785].

Obviously, this can also work in reverse, given the right initial conditions. If(!) two homometrically related figures can be rotated so that they are both placed on the even nodes of a K2n-space, whether that rotation makes them disjunct or conjunct, after the entire space is divided by two, in the resulting Kn-space the original figures will appear retaining their same shape and homometric relationship, as well as the same vector (with the zeros removed).

☛ Note to self & others: Lack of formality is still making this whole edifice unstable. It's still in a sense a CIP,  a conjecture-in-process.


* If the metaphorical connection of the title to the post content is unclear after reading, the answer will be provided in the next a future post.

[1] In passing, it's no secret that modded-out multiplication of pitch classes and interval classes can be alternatively expressed as permutations. Multiplication of ic's by 6 (mod 13) is summarized by the permutation (0)(615243), a spiral permutation in the same category as the sestina poetry form and the Klondike card shuffle. I'll be returning later to a few unusual but significant 20-21st century examples of permutations as compositional tools.

[2] A note from Charles Ives re escaping Diatony's gravitational pull: It may be helpful at first to understand "shape" as "chord." Example: If you have an augmented triad {0,4,8} in 12tET it can be described by its intervals expressed in semitones as (4,4,4). If you multiply by 2 mod 12, you get the same augmented triad {0,8,4}. Now, if you multiply by 2 without modding back into 12tET, you effectively still end up with a triad that sounds exactly the same as the initial {0,4,8} triad, but now "spelled" {0,8,16} in 24tET and described as an interval string using quarter tones, (8,8,8). If multiply-by-2 is all you do, you gain nothing - the two triads sound the same whether you claim to be in 12tET or 24tET - until a second augmented triad comes along that occupies 24tET's odd-numbered nodes, say it's spelled {9,17,1}.  At that point you have gained a trichord relationship with the voice-leading (spanning) vector [0300000303000] - and a hexachord ({0,1,8,9,16,17} w/string (1,7,1,7,1,7)) - not possible in 12tET, even though the generating "shapes" (augmented triads), considered individually, are indistinguishable. This is just the general idea. I used much more complex relationships in an experiment a while back.

Thursday, April 3, 2014

Partition Puzzle 3: A Homometric Partition Theorem

The following theorem is a corollary of the more intuitive ICV decomposition theorem for any three disjunct sets A,B,C which states
V(AUBUC) = V(A) + V(B) + V(C) + V(A,B) + V(A,C) + V(B,C).

Adding the stipulation of vector equivalence between A & B and requiring their cardinalities be less than half the modulus (so the cardinality of C is not empty) creates a tri-partition of any K-space, and the following theorem emerges.

The following proof comes from David Victor Feldman[1]. Using David Lewin's interval function, we are looking for IFUNC(A,C)=IFUNC(B,C). For clarity abbreviate IFUNC(X,Y)(i) as #XY, i.e., the number of intervals (i) up (clockwise) from an element in X to an element in Y. Display all the spanning possibilities for A,B,C as follows:
#AA  #AB  #AC
#BA  #BB  #BC
#CA  #CB  #CC
The sum of all nine entries in this matrix for any i, will be the corresponding i-entry in V(AUBUC); but since A and B have the same cardinality, we have four equal numbers: the sums of the first two rows and the sums of the first two columns. (For example, the sum of the first row must give the number of elements in A because the interval i up from an element in A must land in A, B, or C.) So in particular we have #AA + #AB + #AC = #AB + #BB + #CB. By elimination, #AA + #AC = #BB + #CB. Since we have stipulated that V(A)=V(B), we also have #AA=#BB, so we are left with #AC + #CB. Likewise, working from the equal sums of the first column and the second row, we get #CA = #BC. Thus #AC + #CA = #BC + #CB. Or, summarily for all values of i, V(A,C) = V(B,C).∎

Note that the theorem does not require A and B to be z-related (i.e., the relationship between A and B is not exclusively non-trivially homometric.) In a music theoretic context, A and B may be related by transposition and/or inversion (see Example 1 and 2 below) as well as by the z-relation (Example 3). But there are still surprises (Example 2). So we can identify the theorem as defining partitions that involve homometric pairs generally rather than being limited to the non-trivial case of the z-relation.

Example 1. In K12/12tET, A={C,E,G}, B={F#,A#,C#}. B=T6A so V(A)=V(B); C={D,D#,F,G#,A,B} and V(A,C) = V(B,C) = [0442440] as expected. But if instead B={G,B,D} and C={C#,D#,F,F#,G#,A,A#}, while V(A)=V(B), A & B are not disjunct & so A,B,C is not a partition of K. Calculating all values of i, we come out with V(A,C)=[0544323] and V(B,C)=[0543423] – close but no cigar.[2]

Example 2. Staying in K12/12tET, again let A={C,E,G}, but now B={F#,A,C#}. Now B=I1A so still V(A)=V(B); C={D,D#,F,G#,A#,B} and V(A,C)=V(B,C)=[0442431]. If B={E,G,B}, then C={C#,D,D#,F,F#,G#,A,A#}; V(A,C)=[ 0564333]=V(B,C) even though A and B are not disjunct! The theorem doesn't say this can't happen outside the theorem – which will become important as we go deeper. In this case, C is reflectively symmetric and therefore C's complement, the union of A & B, is also reflectively symmetric, and the symmetry is carried over in the spanning vectors, leading to future speculations about a role for neo-Riemannian transformations where L, P and R all result in reflectively symmetric tetrachords, as well as maximally even pitch structures (including  their twin "Euclidean rhythms"), and reflectively symmetric sets generally.

Example 3. Now, still in K12, let A={0,1,3,7}, B={10,11,2,4}, two AITs (z-related) whose union is not the octatonic this time. C={5,6,8,9} and V(A,C)=V(B,C)=[0232342]. But if instead B={0,1,4,6} so C={2,5,8,9,10,11}, again V(A)=V(B) but A and B are not disjunct and V(A,C)=[0463551] n.e. V(B,C)=[0453651].

The point of including the contrast between disjunct and conjunct A & B in the examples is to emphasize that the theorem identifies homometricity of pairs as a relationship of two sets to a third "reference" set, and not just a relationship between two sets. HPT won't work on just any random partition where  cardA=cardB.

Next in this thread, a transformation KmKn extending the HPT to virtually any homometric pair for any modulus.

[1] Private correspondence 19 March 2014.
[2] In comments to me just before I put this entry on line, David Feldman also noted the following:
All I can get now without disjointness is: V(A,C)-V(A,A∩B) =V(B,C)-V(B,A∩B)In Example 1, this explains the "close" in "close but no cigar" since a small A∩B will make for small V(A,A∩B) and V(B,A∩B).In Example 2, V({C,E,G},{E,G})=V({E,G,B},{E,G}) because the whole story on the left inverts to the story on the right. So V(A,A∩B) =V(B,A∩B)and the theorem (in the form V(A,C)-V(A,A∩B) =V(B,C)-V(B,A∩B) ) really *does* explain (and not merely not preclude) V(A,C) =V(B,C) in this instance.