[Smt-talk] Partition Puzzle

[Jon Wild]

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

> http://essaysandendnotes.blogspot.com/2014/02/partition-puzzle_5.html

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