In compiling an interval-class vector (where an interval class is an ascending interval less-than-or-equal-to one tritone), it seems that tritones should be counted (ascending) from both ends of that interval. For example, given the following PC set, [C E G Bb], we should count the ICs as follows:

C ascending to E: 4 semitones [No further ascending intervals from C <= 6 semitones]

E ascending to G: 3 semitones

E ascending to Bb: 6 semitones [No further ascending intervals from E <= 6 semitones]

G ascending to Bb: 3 semitones

G ascending to C: 5 semitones [No further ascending intervals from G <= 6 semitones]

Bb ascending to C: 2 semitones

Bb ascending to E: 6 semitones (Note: this is not a duplicate of E ascending to Bb). [No further ascending intervals from Bb <= 6 semitones]

Adding up the ICs, we have:

number of single semitones = 0

number of 2 semitones = 1

number of 3 semitones = 2

number of 4 semitones = 1

number of 5 semitones = 1

number of 6 semitones = 2 [emphasis added]

This IC vector should therefore be listed as: < 0 1 2 1 1 2 > [Emphasis added]

Whereas the "standard" list would be: < 0 1 2 1 1 1 >.

Among other things, this means that common-tone theorems of transposition do not have to make exceptions for tritones. Also, the difference in IC vectors between triads and nonads (I'm using Howard Hanson's terminology) is then < 6 6 6 6 6 6 >, not < 6 6 6 6 6 3 >. And between tetrads and octads, it's now < 4 4 4 4 4 4 >, rather than < 4 4 4 4 4 2 >. And between pentads and heptads, it's now < 2 2 2 2 2 2 >, rather than < 2 2 2 2 2 1 >.

Yes, I understand that, for interval classes, "don't count an interval and its inverse" is the conventional wisdom. For interval classes, we restrict the intervals to be less than or equal to one tritone. So interval larger than a tritone don't get counted. The point I'm trying to make is that if you count the (minor-third) interval from E ascending to G, you don't also count the interval from G descending to E. So when you count the (tritone) interval from E ascending to Bb, you (of course) don't count the interval descending back down to E. But, when counting intervals ascending from Bb (less than or equal to one tritone), we still have the ascending interval to E to count—and, of course, we don't count that interval descending back down to Bb.

Alternatively, if you count all intervals less than one octave, always ascending from every PC in the set, you will accumulate as many major-sevenths as minor-seconds, minor-sevenths as major-seconds, and so on. And as many tritones from one end of that interval as there are from the other end(s)—twice as many as are usually tallied with interval classes. If you then catalogue just the interval classes (less than or equal to one tritone), you get to keep all those extra tritones!

  • 1
    I'll leave this behind: consider the fact that you're now double-counting only 6-semitone intervals.
    – Dekkadeci
    Mar 24 at 17:41

1 Answer 1


Each pair of pitches receives one interval. C-E receives 4, not 4 and 8; Bb-c receives 2, not 2 and 10. Likewise, the pair of pitches Bb-E is counted only once.

Another way to think of it is that one does not count both an interval and that interval's inverse. It just happens that with a tritone, that interval is its own inverse. Nevertheless, we don't count both.

It's misleading to think in terms of ascending intervals or intervals smaller than an octave — these are not what the count represents. Each pair of notes is represented by a single interval between 1 and 6.

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