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In a traditional 12-note tone row, it seems to me that the first six notes will always either be in the same set class as the last six notes, or will be z-related, but I'm having trouble confirming this. For example:

Schoenberg's Fourth String Quartet: 2 1 9 T 5 3 | 4 0 8 7 6 E

Both discrete hexachords are (014568)

Berg's Violin Concerto: 7 T 2 6 9 0 | 4 8 E 1 3 5

The first hexachord is (013468) and the last is (012469), both have interval vector <233331> and are thus z-related.

Are there exceptions to this? Alternatively, is there a proof of this fact without having to try all 12!=479,001,600 possibilities?

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I've got Allen Forte's book on my desk at the moment, as it happens; but I must confess I haven't read all of it! I presume, by your question, you're familiar with this book… (+1 BTW…) –  Bob Broadley Jun 18 at 22:44
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@BobBroadley Yes, although my primary set-class/serial study involved the Straus Post-Tonal Theory textbook. I read both long ago, although I frequently still refer to the Straus. It's possible that one or both answer the question, but I'm having difficulty locating an answer specifically. Seems like the kind of thing more likely to be an offhand reference than a specific topic of discussion... –  Pat Muchmore Jun 18 at 22:47
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This question would be great to look into - I'm unfortunately away from my library :(. Are you looking for specific examples of pieces / textbooks that refute the claim? Have you read any Babbitt? –  jjmusicnotes Jun 19 at 0:11
    
@jjmusicnotes I'm looking either for a specific example or textbook that refutes the claim, or a paper or a priori proof that proves the claim. –  Pat Muchmore Jun 19 at 0:47
    
Or, I suppose, a list of all 479 million possible rows showing that it's true. :) But seriously, I'm all but certain that it must be the case, just looking for some kind of confirmation. –  Pat Muchmore Jun 19 at 0:54

2 Answers 2

Yes, this is true. We refer to it, in the business, as The Hexachordal Combinatoriality Theorem. It can be mathematically proven, but I can't (without reliving some traumatic graduate school experiences) do it for you here. :)

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Fair enough! Any references as to where I could find this discussed though? A google search for Hexachordal Combinatoriality Theorem in quotes yields no results. As a general rule, the texts I've read use Combinatoriality to refer to the combination of two different row forms such that the first hexachord of each form and the last hexachord of each form complete the aggregate, not to the two discrete hexachords of a single row. –  Pat Muchmore Jun 19 at 0:45
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This looks like a proof: maa.org/sites/default/files/269122711809.pdf –  Micah Jun 19 at 5:26
    
Yep, looks like that covers it, thank you! Apparently it's just called the Hexachordal Theorem. –  Pat Muchmore Jun 19 at 13:22
up vote 4 down vote accepted

For the sake of future readers, I'd like to synthesize a full answer to the question so that it isn't only buried in a link inside a comment.


1.

As Robert Fink's answer suggests, the TL;DR answer to the question is yes, the final hexachord of any tone row will either be the same as the first hexachord, or will be its z-partner.


2.

A slightly longer answer suggested by a non-SE friend is as follows: The complement of any set-class of any size (that is, the set class of the notes remaining out of all 12 possibilities in the traditional Western system once you've removed the notes of the original set) will always be the same regardless of the specific notes. For example, the notes D, D#, F#, G#—a member of the all-interval tetrachord (0146)—has the complement C, C#, E, F, G, A, A#, B, a member of (01234689). A completely different member of (0146) would be G, A, C, C#; but its complement—D, D#, E, F, F#, G#, A#, B—is still of member of (01234689).

Obviously, this will still hold true for hexachords. Once you have the first hexachord of a traditional 12-tone series, the remaining hexachord is, by definition, the complement of the first. All that remains to prove the original hypothesis is the fact that the complement of any hexachord is either itself or its z-partner, as can be verified by looking at any hexachord list like the one in the Appendix of Straus' Introduction to Post-Tonal Theory.


3.

A somewhat more formal proof by Steven K. Blau was provided in a link by Micah, the full paper is here: http://www.maa.org/sites/default/files/269122711809.pdf

I'll provide a brief summary. If we pick an arbitrary hexachord, let's call it A, then it implies a complementary hexachord (B) that marks its completion in a tone row. We can envision these two hexachords in a clockface diagram like so:

     B   A   A
  B              B
 A                A
  B              A
      B       B
          A

Now we imagine switching one of the A notes with one of the B notes, I'll do a switch of the 2 o'clock and 3 o'clock positions.

     B   A   A
  B              A
 A              B
  B              A
      B       B
          A

The only intervallic changes in A will be matched by identical intervallic changes in B. For example, the A note that used to be at 3 o'clock was 3 semitones away from the A note at 12 o'clock, but it is now only 2 away. This changes the A hexachord's constituent interval content but we've simultaneously changed the B hexachord in that the B note that used to be at 2 o'clock was also 3 semitones away from the B note at 5 o'clock and is now only 2 away. In other words, at the same time that we changed one of the intervals in A from IC3 to IC2, we also changed one of the intervals in B in precisely the same way. The article at that link covers all possibilities (although I think the author accidentally reverses n+1 and n at one point)

Therefore, the interval content of the two hexachords will always be the same no matter what, which is just another way of saying that the discrete hexachords of any row will either be the same set class or z-related. QED

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Thanks @Pat, this explains a fundamental idea in an easily understandable way. I must confess I followed Micah's link, but didn't read the whole paper - now that you've given me this information, I'm likely to return to that paper. –  Bob Broadley Jun 19 at 22:36

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