Are there really only 4 sets of notes that satisfy this Persichetti requirement?

Question

In 20th Century Harmony, Persichetti classifies major 2nds and minor 7ths as mild dissonances and minor 2nds and major 7ths as sharp dissonances.

He gives the following examples for his classification system:

In a later exercise, he assigns the reader to write a passage using three-note chords that use exactly one consonance, one sharp dissonance, and one mild dissonance.

I am only able to find 4 discrete chords that fit this (ignoring octave displacement for now). If the top note is a G, the consonance must be a minor third or major sixth away. This allows the final note to be either a 7th or 9th away, and if it is carefully chosen it forms a mild dissonance with one note and a sharp one with the other.

Is there any other set of notes besides these that would also count? I deliberately didn't include the tritone because he doesn't classify it as a dissonance; the fourth can be dissonant but isn't described as either mild or sharp.

Answer 1

TL;DR

By eliminating chords that can be transformed into each other by changing voicing and/or transposition, there are only two unique solutions: C-Db-Eb and C-D-Eb. This is because there is only one collection of intervals that can uniquely satisfy the requirements of the original problem: a chord containing a minor second, a major second, and a minor third.

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Discussion

Interval classes

Breaking the interval qualities into interval classes (i.e., 0 = unison, 1 = half step, 2 = whole step, ..., 12 = octave; where 0 = 12 = 24 = ..., 1 = 13 = 25 = ....), we have:

• consonances: 3, 4, 5, 7, 8, 9
• mild dissonances: 2, [5?], 10
• sharp dissonances: 1, [5?], 11

Exclude 5 (perfect fourth), since it's not clear what type of dissonance Perischetti considers it.

Chords and pitch classes

Since there's no need to consider chords that differ only by transposition, we'll base all of our chords on C. Thus, with pitch classes, we have 0 = C, 1 = C#, 2 = D, ..., 11 = B, and subtraction mod 12 of two pitches will tell us the interval between them. (E.g., C–E = pc 0–pc 4, 4 – 0 = 4 = major third.)

Additional details

• Interval qualities are invariant under inversion. That is, inverting a consonant interval results in a consonant interval; inverting a mild dissonance gives a mild dissonance; and inverting a sharp dissonance gives a sharp dissonance. Because of this, we can limit our chords to those that span less than an octave. Any chord spanning more than an octave can be reduced by octave displacement of one or more notes, resulting in either an inverted interval or an equivalent non-compound interval.

Example: Consider the first chord in Perischetti's Ex. 1-14, C-G-E (0, 7, 16). This chord contains the following interval classes [7–_mod120=7, 16–_mod127=9, 16–_mod120=16=4]. If we drop the E one octave, we have (0, 3, 7), resulting in the same interval qualities (i.e., consonances). Alternatively, we could raise the C an octave, resulting in (4, 7, 12), containing interval classes [3, 5, 8]. Either way, the resulting chord contains only consonances.

• The two smaller intervals must sum to the largest interval.

Proof: In a 0-based triad, we have pitches (0, x, y), where x < y. (We can require x < y, because in a triad x <> y [otherwise we'd have a dyad], and x > y can be resolved by octave displacement and/or inversion without changing the interval qualities.) This means 0 < x < y.

The chord (0, x, y) contains the intervals [x–0=x, y–x, y–0=y], with y the largest. (y > x is given, and y > y – x is true since y > x > 0. But if we combine the two smallest intervals, we have x+(y–x)=y.

Restating the problem

Create triads containing one of each of the three interval qualities, such that the largest interval is 11 (i.e., spanning less than an octave).

Solution

• WLOG, we can eliminate interval 11, because any chord using interval 1 can be inverted to have interval 11.
• WLOG, we can eliminate interval 10, because any chord using interval 2 can be inverted to have interval 10.
• Therefore, every chord must include intervals 1 and 2, to satisfy the requirement of one interval of each dissonance type.
• By the same logic as above, intervals 7, 8, and 9 can be eliminated in favor of 5, 4, and 3, respectively.
• This leaves only three possible solutions in intervalic content: [1, 2, 3]; [1, 2, 4], and [1, 2, 5].
• Since the two smaller intervals must sum to the larger interval, we have only one possible solution: [1, 2, 3]. In pitch terms, there are only two triads with these intervals: (0, 1, 3) and (0, 2, 3) [(C, Db, Eb) and (C, D, Eb)].

Showing the equivalence of the four original solutions.

The original question offers four solutions: (F#, E, G); (F, E, G); (A, Bb, G); and (Ab, Bb, G).

Reconfigure these to stay within the octave: (E, F#, G); (E, F, G); (G, A, Bb); and (G, Ab, Bb).

Now transpose to C: (C, D, Eb); (C, Db, Eb); (C, D, Eb); and (C, Db, Eb).

We see immediately that there are two pairs of equivalent solutions.

• The intervallic content of (C, D, Eb) is [Eb–D=1, D–C=2, Eb–C=3].
• The intervallic content of (C, Db, Eb) is [Db–C=1, Eb–Db=2, Eb–C=3].

Therefore, there are only two fundamentally unique solutions. All other solutions are different voicings and/or transpositions of these.