Saturday, March 15, 2014

Partition Puzzle 2: Spanning Vectors

You will have to brace yourselves for this –
not because it is difficult to understand,
but because it is absolutely ridiculous:
All we do is draw little arrows
on a piece of paper –
that's all!
~ Richard Feynman
(QED: The Strange Theory of Light and Matter, 1985, p. 24)

~ David Lewin
(Generalized Musical Intervals and Transformations, 1987, Figure 0.1)

Now that I finally have my computer back from the shop and have restored or recreated most files from the defunct hard drive, I can jump back in to the partition puzzle.

A good place to start is with a thank you to Jeremiah Goyette for pointing out to me the mistake I made in the final example in my first partition puzzle blog entry. While repositioning a (2146) tetragon inscribed in K13, it came out as a (2155) tetragon, displacing one of the vertices by one click.  My enthusiasm for this neatly-fitting result convinced me there was no need to even consider the possibility of a clumsy error. But the blunder has turned out to be a reminder to me to look past a search for superficial symmetries into deeper sub-surface symmetries.

The main thing that came to mind was the basis for an article, "The T-hex Constellation" (JMT, Fall 1998, 42.2, pp. 207-16). The relevant page is reprinted here:



The idea that grabbed me after re-visiting this passage and, as usual, reading some Lewin[1] was a "ghost set" conjecture that has remained undeveloped in the back of my mind for years: Given two disjoint sets S and T that are related by V(S) = V(T), there exists some set Q such that V(S,Q) = V(T,Q).  And/or vice versa: if V(S,Q) = V(T,Q) then necessarily V(S) = V(T). And if a strong form of this conjecture were not true, then, given it does work in many specific cases, what are the further conditions that make it true in those cases?

This might be a useful alternative way of defining homometricity. Instead of only looking for identical ic counts within two separate objects S and T, it would be primarily a search for a "ghost" object Q under conditions C that defines a spanning  relationship (S-to-Q) ↔ (T-to-Q).

To clarify by example:

Musically, this would be like finding that the total voice-leading possibilities between chord S and chord Q are identical to those between chords Q and T. In fact, there are examples of this (albeit hypothetical – at this point I know of nowhere in the literature that a composer has actually used these directly). We already know that the all-interval tetrachords C-C#-D#-G and E-F#-A-A# are z-related. And since together these particular transpositions form the octatonic scale, the ghost set Q here ought to be the complement of the octatonic, the diminished-7th chord G#-B-D-F. And that's exactly the case:
Figure 1
Between S and Q there are four each ic1, ic2, ic4, ic5, and no instances of ic0, ic3 or ic6. And the same between T and Q. Summarizing this as an interval-class spanning vector relationship:

V(S,Q) = V(T,Q) = [0440440].
[NB! Unless noted otherwise, spanning vectors will include the cardinality of ic0. So in the usual 12tET chromatic, vectors will have seven places rather than six. Thus in the current example of AITs, not only V(S,Q) = [0440440] but also V(S) = V(T) = [4111111] (see article above). In this case S and T are disjunct (i.e., card(ic0) spanning S and T is 0), so, calculating V(S,T) = [0226222][1], we have one of the many possible "decompositions" of the octatonic set: V(octatonic) = V(SUT) = V(S)+V(T)+V(S,T) = 2[4111111]+[0226222] = [8448444]. A formal algebra for spanning vectors will be interesting to work out, but intuition with examples is enough to get by for now.]

Extending to any selection of four pitches from the octatonic, the partition principle for the above example results in the same ic spanning vector [0440440] ; for example, Figure 2 shows {dom7, dim7, halfdim7}, a quasi-tonal cover of the chromatic.
Figure 2
Again, V(S,Q) = V(T,Q) = [0440440]. This tidy factoid will not always be the case.

Returning to K13, Figure 3 shows one (now correct) version of the tri-partition of K13 by the two z-related sets S = {5,7,8,12} and T = {9,11,1,2} and the "ghost" set Q = K–(SUT) = {10,0,3,4,6}.

Figure 3

Figure 4 separates the three figures to show the n-gon "shape" and complete interval-class structure of each set.
Figure 4a                               Figure 4b                               Figure 4c

Correcting S meant losing a ghost set that was the maximally even 5-in-13 symmetry. Playing with various rotations (transpositions) and reflections (inversions) of S and T that leave them disjunct, it is now apparent that all possible shapes for Q are asymmetric. But this makes the possible partition solutions here much more tantalizing because a deeper symmetry begins to reveal itself in the spanning vectors. Taking the case in hand from Figure 4, we get
V(S,Q) = V(T,Q) = [0442352].
Holding T constant and rotating (transposing) S by –2, we get T = {3,5,6,10} resulting in Q = {0,4,7,8,12} with the string (14143). V(S) still equals V(T), but now
V(S,Q) = V(T,Q) = [0354224].
The "answer"to V(X,Q) has changed, but the relationship remains the same, suggesting that the equivalence per se is an invariant under certain (we know not yet what) conditions. One more test. What if we hold T constant again and this time reflect S (invert by I2) to {3,7,8,10} resulting in Q = {0,4,5,6,12} (string=(14116))? Of course V(S) = V(T) as usual, but this time
V(S,Q) = V(T,Q) = [0245432].
Again, the values of these spanning vectors have changed, but the relationship between the two is invariant.

Obviously, there's still a lot more ground to cover. What happens when S∩T≄∅? What about homometric triples, quadruples? Etc. Will the "ghost" turn out to be just another ignis fatuus?


[1] E.g., "Going even further, we may ask under what conditions among the four sets X1, Y1, X2, and Y2 we will have the relation IFUNC(X1,Y1) = IFUNC(X2,Y2).... This is all a vast open ground for mathematical and musical inquiry, even in atonal set-theory." (GMIT, p.103) The reader will note many of Lewin's ideas running through the threads here – way too many to sort out at this point.

[2] Calculating spanning vectors by hand is cumbersome and distracting. I've devised a simple spreadsheet "calculator" for quickly determining spanning vectors for any two sets in any K-space. I'll send this via email to anyone requesting it (at no charge), but you must have Microsoft Excel to use it. Send your request to: essaysandendnotes at icloud dot com.