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.

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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

1 comment:

David Feldman said...

You 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 your example which may remove some of the mystery.
Modulo 31, the five sets can be written as
{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 31 elements 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 case that the non-occurrence of an equality seems miraculous anymore.