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. . . The all-trichordal
set is of great value to many
composers because it contains ten different trichords, each one of which
can generate a set and thus get you from one area into another. There are
ten because we drop out the diminished triad, which can’t generate a set,
and the augmented triad, which is ambiguous under the operations of the
system. We could get them, too, by going around the corner, linking end
to beginning, but who needs them? It’s not hard to write an
all-trichordal
twelve-tone set. Here is the one from Images, my saxophone piece.
It has all kinds of excellent properties. [C, G, E-flat, D, E, F, B, C-sharp,
G-sharp, F-sharp, B-flat and A.] This one, by the way, does not have a
3 or a 9, which might already surprise you. It also happens to be all-combinatorial.
Here’s another one if you want to compare notes. [C, D-flat, B, E-flat,
A-flat, B-flat, E, A, G, F-sharp, D and F.] This is the one that I used
for my Indiana piece, Ars Combinatoria, because I was extremely
generous with them—I gave them one that has everything.
Now here’s what defies intuition totally: it is absolutely impossible to construct an all-trichordal set which is also all-interval. Why should that be? You’ve got all this intervallic variety—ten different trichords. It seems to me that the two properties should very well conjoin, but they can’t. This is where order begins to impose itself upon collection. The ordering necessary to get an all-trichordal set defeats the possibility of an all-interval set. (By the way if there’s anybody in this room who can prove this by any so-called mathematical methods, he can beat anybody I know in the field of combinatorial mathematics.) Intuitively, it just doesn’t seem right. It shows how deeply these properties of order and collection begin to contradict each other.
Milton Babbitt
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