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:

**K**12.

*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

**K**

**13.**{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 5*X*will be [aFcdeBg], no matter how the operation may warp the shape of the set. And if B=F,*X*and 5*X*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 2*X*(mod13)={0,2,4,8,10,3}, so that V(*X*)=[6323322] and V(2*X*)=[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.,**K****26**). Furthermore, this shift retains the original ICV on the even-numbered components of the ICV of the transformed set(s) (Figure 2).Figure 2 |

*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'*=2*X*={0,2,4,8,10,16} and*Y'*=2*Y*={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 |

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"*=**T**1(*Y'*)={1,7,9,11,13,19},*Q*= compl(*X'*U*Y"*) = {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

[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.