Saturday, February 22, 2014

Thursday, February 13, 2014

Observations from Jon Wild on the Partition Puzzle

Many thanks to Jon Wild for the following e-mail. It was also posted on the SMT list on 2.10.14, but I wanted to copy it here in situ for reference & so it wouldn't get lost or forgotten.

* * *
Stephen Soderberg posted to propose the following puzzle regarding Z-related tetrachords in higher equal temperaments:

No one responded (at least not on-list, and there is no comment section on the blog) so I have dusted off some Z-sets from an old hard drive, and can offer the following (without even the hint of a claim to musical relevance):

Stephen investigated up to 24-tone equal temperament. Going higher, in 28-tone equal temperament we find there are three pairs of Z-related tetrachords. We can take one representative from each of the six set-classes, along with the maximally even set-class [0,7,14,21], to form the 28-tone aggregate--in fact we can do so in 416 distinct ways--so your hypothesis holds, in that case.

I don't have the data for 32-tone ET, but we can test other generalisations of your hypothesis, involving pc-sets larger than tetrachords:

Mod 24, there are 9 triples of Z-related heptachords. Of the 9, two can be packed into the aggregate alongside the maximally even set [0,8,16]. (Each of the two triples that works can be rearranged around [0,8,16] in eight inequivalent ways, so there are 16 solutions in all.) These have very much the same flavour as the partitionings on your blog. One such partitioning: {0,1,3,5,7,8,16} + {6,14,17,19,21,22,23} + {4,9,11,12,13,15,20} + {2,10,18}.

Mod 18, there is a triple of Z-related pentachords: [0,1,3,6,10], [0,1,4,7,9] and [0,2,3,6,11]. It would be nice if we could partition the 18-tone aggregate into one each of these, leaving just the maximally even set [0,6,12]. But we can't! There are 6 ways we can pack the three pentachords into the aggregate; each way, though, leaves an [0,3,6] uncovered. This remainder of [0,3,6] is suggestive in connection with the following case:

Mod 30, there are 1090 triples of Z-related nonachords. Of these 1090 triples, only 19 can be packed into the aggregate at all (i.e. regardless of what trichordal set they leave uncovered). Of these 19, none can be packed into the aggregate in a way that leaves uncovered a set whose primeform is [0,10,20]. But: seven triples can be packed into the aggregate leaving the set [0,5,10].

(Interpolating between the last two cases--18-tET leaving [0,3,6] and 30-tET leaving [0,5,10]--we could predict that in the mod 24 case above, it should also have been possible to pack Z-related heptachord triples leaving behind the set [0,4,8]. And sure enough this is possible, in plenty of ways. I bet there is a family of such solutions that exists for each ET cardinality of the form 6n (greater than 12), using three Z-related (2n-1)-chords and leaving a remainder of [0,n,2n].)

Here are some other case studies for you:

Mod 27, there are three Z-related triples of hexachords. Each triple can be packed into the maximally even set {0,1,3,4,6,7,9,10 ... 24,25} leaving behind the set {2,5,8,11,14,17,20,23}.

For example, the triple of prime forms {0,1,6,10,12,19}, {0,1,7,9,13,18}, and {0,2,8,9,14,18} can be combined as follows: {0,1,6,10,12,19} + {4,9,13,15,21,22} + {3,7,16,18,24,25}. (This is one of two ways--the other triples also have two or three ways each of being combined to form the same maximally even 18-out-of-27 set.)

Mod 28, there are 81 Z-related triples of octachords. Of these 81, only 12 can be packed into the aggregate at all, covering 24 of its 28 pitches. And none of the 12 leave an uncovered set whose primeform is the maximally even [0,7,14,21].

This is my favourite example: 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. I suspect that Z-related sets with flat or near-flat distribution of intervals--like these 31-tone hexachords and the familiar all-interval tetrachords--will generally work better in such tiling problems than those Z-related sets whose interval vectors are very unbalanced.

And now back to some lovely Schubert for my class tomorrow...

--Jon Wild, McGill University

Wednesday, February 5, 2014

Partition Puzzle

'No wonder kids grow up crazy. A cat's cradle is nothing but
a bunch of X's between somebody's hands, and little kids look
and look and look at all those X's ...'
'No damn cat, and no damn cradle.'
– Kurt Vonnegut, Cat's Cradle

To explain this partition puzzle (I know how to find the cats, but can I find cradles for all of them?), it's helpful to begin with a motivating example or two. First take a look at Table 1.

Table 1

This list of z-related[1] pairs of tetrads was extracted from the list I posted recently of the 1,591 tetrads from K4 through K24. Musicians can relate to the complete tetrad list in practical terms as either the complete tetrachord content in all harmonic spaces from the single tetrachord in the equal 4-division of the octave (4tET) through the 256 tetrachords in quartertone space (24tET), or all 4-onset rhythms possible in a single measure of 4 beats through a measure of 24 beats (disallowing rhythmic tuplets). I tend to think of it as an uninterpreted "event space" leaving it open to other applications both inside and outside the music domain.

Next take a look at this fairly simple diagram (Example 1) which represents the entry at the top of Table 1 as two tetragons inscribed in K8-space[2]:

Example 1

This is the first appearance of z-related sets in any K-space[3]. The box to the lower left indicates two "descriptions" – shape and content – of the figures inscribed in the K8-space. On the left are the interval strings for the two figures, (1214⤸ and (1331⤸ which describe the unique shapes of the inscribed polygons. All intervals in a string are measured clockwise. On the right in the box is the interval-class vector [2121] common to both strings. An interval class (ic) is defined as the shortest distance between two nodes, whether that is measured clockwise or counterclockwise. And the interval-class vector of any inscribed polygon describes the total ic content of the polygon. Examples 2a & 2b demonstrate the ic-vector equivalence of the two figures.

Example 2a: (1214⤸                                    Example 2b: (1331⤸
(ic content in both figures is [2121]:
two ic1's,one ic2, two ic3's & one ic4)

Example 1 is also the first appearance of z-related sets that can be explained by the Babbitt "hexachord theorem." Of course, the theorem does not only apply in K12-space where it was first observed. It  should be thought of as a set-complement theorem that applies in any K2n-space (i.e., any K-space with an even number of nodes). An informal generalized statement of the theorem might be:
Babbitt Complement Theorem. When you choose any n nodes from any K2n-space that define a structure S, their complement – the n remaining nodes – will define a structure T that has the same interval-class vector as S.
The two figures described by the two complementary sets of n nodes in K2n-space might be transformationally related by rotation (musical transposition) or reflection (musical inversion)[5]; but if they are not, they are said to be z-related. Example 3a shows (1124⤸↔(4211⤸ by reflection; Example 2 shows (2123⤸↔(2123⤸ by rotation (as well as reflection). (1133⤸ and (1214⤸ in Example 3c are not related by any (known) transformation, but because they are complements, by the theorem their ic-vectors are nevertheless identical – z-related.

Example 3a                         Example 3b                        Example 3c 

So far I have presented nothing new. It's been no more than a review, a generalization – to point to basic concepts beneath applications. Now, with one very slight adjustment in nomenclature, I can present the puzzle. The generalized hexachord theorem above is based on the idea of complementation which assumes a bifurcation – placing all the elements of a set into one or the other of  two baskets. The Babbitt Theorem for pairs of tetrads only applies in K8. But could it be that the Babbitt Complement Theorem is a special case of a more general partition theorem such that z-related tetrads appearing in K-spaces other than K8 can be arranged as a partition of the space in which they are found?

First, the simplest case. What happens if we double the size of the space in the previous example from K8 to K16? Obviously, any figure in K8 will appear as a congruence (simply multiplied by 2) in K16, and any z-relation in K8 will be carried over to K16. So: (1214⤸ → (2428⤸; (1331⤸ → (2662⤸; and (2428⤸ & (2662⤸ will be z-related (but only indirectly because of the Babbitt Theorem), sharing the same ic vector [02010201]. But looking back at Table 1, we see that there is also a new pair of z-related figures, (1357⤸ and (3418⤸, not congruent to any figure in K8. Inverting (3418⤸ to (1438⤸ and experimenting with the placement of both z-pairs in  K16, we discover (empirically) that we can partition K16 with its z-related tetrads (Example 4).[6]
Example 4

Now this is getting interesting. Tripling K8 to K24 and looking back at Table 1 again, we find K24 has exactly three pairs of z-related figures, and all of these taken together can partition K24 (Example 5).
Example 5

(363C⤸ & (3993⤸ are the result of tripling from K8, and (157B⤸ & (165C⤸ are new, but this partition brings in another player, the familiar pair of all-interval tetrads found in K12 appear doubled in K24: (4215⤸ → (842A⤸ & (2316⤸ → (462C⤸.[7]

But this is as far as you can go using the tetrad relationships in Table 1. For one thing, the only z-related tetrads in K12, the all-interval tetrad pair, can't partition K12. But this glitch actually opens up to an even more interesting possibility. As it happens, the well-known octatonic string in K12, (12121212⤸, acts just like the entire space in which it is embedded with respect to the Babbitt Complement Theorem (see "Z-Related Sets as Dual Inversions," PNM, 39.1 (1995), §3.18). Choose any 4-element substring in the octatonic, and its complement with respect to the octatonic will have the same ic vector. This includes the all-interval pair:

Example 6

The complement of the octatonic string is (3333⤸ – in 12tET, the diminished 7th chord – indicated in Example 6 as a gray square. So K12 can be partitioned by two z-related all-interval tetrads and the maximally even tetrad (3333⤸. This suggests a remote possibility that any K-space might be partitioned by combinations of z-related and maximally even sets. It becomes surprisingly less bizarre when we note how all the tetrads possible in K20 can be situated around a dilation of that same maximally even square:

Example 7

Actually, this might work if you consider the previous Examples 1, 4 and 5 to be z-sets situated around a maximally even set of order zero. Ah, but what about that one z-related structure in Table 1 that we've ignored – the pair in K13. Well, it works with a different max even set (32323⤸. Here it is:

And that accounts for all the z-related tetrad pairs listed in Table 1. Now it gets complicated.

[Correction & remarks coming soon]

To be continued.........


[1] While it is tempting here to point to the recent important (and surprising) music–crystallography connection noted by others in the music theory literature by replacing "z-related" with the much more descriptive "homometric" used in crystallography, graph theory & elsewhere, unfortunately "homometric" comes with its own set of baggage, not all of which do I understand adequately enough to reference with confidence. But the real problem that Forte, evidently unwittingly, set up by calling this phenomenon the "Z-relation" is that it immediately becomes confused with Zn, the set of all congruence classes of the integers Z for the relation congrunce mod n. This forces tortured locutions such as "the set of all Z-related sets in the set Zn." My solution here (which I remain unhappy with) is to keep the "z-relation" designation in its current meaning in music theory (but in an expanded sense and use) and refer to pitch–/rhythm– (event–) space as K–space or more specifically as Kn–space.  Kn–space can be represented visually as the 1-dimensional space defined by n nodes equally spaced around the circumference of a circle.   The nodes in Kn–space may be labelled, as in the venerable circle of fifths (C, G, D, ..., Bb, F, (C)), or unlabelled, as it is throughout this blog entry.

[2] vid. note [1]. For the curious: I'm using "K" for two personal reasons. First, a reference I once read (I believe it was by Nick Collins) referred to these as "Krenek circles" but I rather doubt Krenek was the first to use them in a music theory context. The other, more whimsical, reference is to the Greek κύκλος (kyklos) from Archimedes' last words to the Roman soldier who was about to kill him: "μή μου τοὺς κύκλους τάραττε" – Do not disturb my circles! (Probably apocryphal, but I like to believe it.)

[3] First used by Arthur Lindo Patterson to illustrate homometric structures (our z-related sets) in the phase problem in X-ray crystallography. [Cite: ______]

[4] For those who need an immediate "real" application in music, the diagram can be interpreted rhythmically this way:
Where the symbols "(" and "⤸" indicate that the measure can be "recycled" to begin on any of the eight notes, just as the same symbols were previously used to indicate the equivalence class of cyclically related permutations.

[5] – or dilation mod 2n (along with rotation to achieve a complement relationship), a kind of quasi-homothetic transformation that warps the K–space into itself. (Music's special case for dilation in 12tET is multiplication by 5 or 7 mod 12). But there is no need to introduce this complication here.

[6] NB: For all of these cases there is likely more than one way to partition any space with z-sets. In this first case, for example, the entire partition can be mirrored.

[7] A=10, B=11, C=12, etc.