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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.


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[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.