Please note: this is an archived version of the textbook. Visit IntMus for up to date content!

23a Examples - Basic PC Set Manipulation

In Unit 22, we introduced integer notation for pitch classes. Allen Forte first formalized this system in his seminal work, The Structure of Atonal Music, creating a broad analytical framework to compensate for tonal analytical methods inadequacy when music leaves tonality. His set theory is a specialized form of analysis that looks for intervallic patterns equally across all twelve chromatic pitches, rather than focusing on the relationship between a central pitch (i.e. tonic) and others.

Scratching the surface

Set theory is a broad field of theoretical study, and we will only introduce a few basic concepts in this unit. If you would like to explore this topic further–and if you intend to go beyond the undergraduate level of music study, you should–I highly recommend that you read some of the many fine books written on the subject. My favorites include:

  • The Structure of Atonal Music - Allen Forte
  • Introduction to Post-tonal Theory - Joseph Straus
  • Basic Atonal Theory - John Rahn

There have been numerous refinements and alterations to this method since the first publication of Forte’s text in 1973, so this unit will attempt to distill the basics into a foundation that will make this sort of analysis more accessible.

Notating pitch-class sets

When we introduced pc sets (pcs) in Unit 22, we did not formalize their notation, however, because we will study various levels of organization within pc sets, it is important to establish a standard form for writing an unordered pcs–a pitch class set that you have not specifically organized according to a predetermined, formalized sorting method. For unordered pcs, you should always notate this using parentheses with a comma between each pitch class. The following are all examples of unordered pc sets:

  • (0,1,5,t)
  • (B-flat,C,D,G)
  • (2,4)
  • (0,1,3,5,6,7,8,9,e)
  • (B,C,D,E,F,G,A)

You will notice that it doesn’t matter if you use pitch names or integer notation; a pitch class set is any collection of pitch classes.

Modulo 12 arithmetic

Before we begin using integer notation to study pitch classes and pitch-class sets, we must establish the mathematical method necessary for manipulating pitches. Because music uses twelve equally-spaced pitches but our standard counting system is based around tens, we must create a specific equivalency around the number 12. For this, we use modulo 12 (mod12) arithmetic. You already use modulo 12 arithmetic every time you count time in hours, assuming that you use a 12-hour system. For example, if it is 11:00 and you have a meeting in three hours, what time is your meeting? From experience, you understand that when you reach 12, you must reset your counting to find the meeting’s start time at 2:00–not 14:00. (Again, this assumes that you are not using a 12-hour clock cycle, not 24-hour.)

This is exactly how pitch-class integer notation works, although the system begins numbering on 0 instead of 12. In this system, each time you pass 11, you begin again at zero, therefore making every multiple of 12 equal to zero. Complete the following chart to show the first two equivalencies for every number in this system. Can you create a method to quickly find any level of equivalency? What would happen if you crossed below zero?

Pitch-class integer First equivalency Second equivalency
0 12 24
1 13 25
2 14
3
4
5
6
7 31
8
9
t (10)
e (11) 23

Conclusion

While it is easy to complete this chart by simply counting sequentially downward within each column, it is still useful in that it demonstrates the continuum that is mod12 arithmetic. All whole integers can be represented by an integer between 0 and 11, giving us exactly one integer for every possible pitch class in the chromatic system.

Pitch-class integer First equivalency Second equivalency
0 12 24
1 13 25
2 14 26
3 15 27
4 16 28
5 17 29
6 18 30
7 19 31
8 20 32
9 21 33
t (10) 22 34
e (11) 23 35

This also works in moving backward. Using basic arithmetic, it is easy to understand that 12 - 5 = 7. But because 12 is equivalent to 0 in mod12, this also means that 0 - 5 = 7. This is correct, but it skips an step in demonstrating how.

  • 0 - 5 = -5
  • You must then convert -5 into an integer between 0 and 11 using multiples of 12. Therefore, -5 + 12 = 7.

This shows that any whole integer, whether positive or negative, can be converted to an integer between 0 and 11. You could create a similar chart to the one above for negative integers.

Pitch-class integer First negative equivalency Second negative equivalency
0 -12 -24
1 -11 -23
2 -10 -22
3 -9 -21
4 -8 -20
5 -7 -19
6 -6 -18
7 -5 -17
8 -4 -16
9 -3 -15
t (10) -2 -14
e (11) -1 -13

Transposition using integer notation

Fixed-zero integer notation assigns the following integers to pitch-classes:

Pitch-class integer Pitch 1 Pitch 2
0 C B-sharp
1 C-sharp D-flat
2 D
3 D-sharp E-flat
4 E F-flat
5 F E-sharp
6 F-sharp G-flat
7 G
8 G-sharp A-flat
9 A
t (10) A-sharp B-flat
e (11) B C-flat

This is relatively simple to memorize, but it is more important to consider what these numbers truly represent. Consider the following pitch-class set:

(E, F, G, B-flat)

First, transpose these pitches up a P4 using our traditional intervallic method for transposition, and as you do this, pay careful attention to the process that you use for transposing. Do you think of each pitch as part of a key? If so, what scale degree do you use for each pitch? Do you count half-steps? Do you find the first pitch and then as the base for finding the rest of the intervals?

Next:

  • Convert this pitch-class set into integer notation
  • Then add 5 to each number
  • For any resulting number greater that 11, convert that number to its lowest possible integer using mod12 from above.
    • For example, if you were to get 13, you would need to subtract 12 to get 1.
  • Convert this new set of integers into pitch letter names.

Did you get the same result as when you transposed the pitches using the traditional method for pitch transposition? If so, why does this work? What do the numbers for integer notation actually represent? (Hint: it is not the assigned pitches.)

Notating transposition

Hopefully, you came to the same result using both methods of transposition. The pc set becomes (4,5,7,t), and then when you add 5 to each integer it becomes (9,t,12,15). All numbers greater than 11 must be reduced by mod12, so the set becomes (9,t,0,3). When these numbers are converted back into pitch letter names, you get (A,B-flat,C,E-flat) which is exactly a P4 above the original pc set.

This works, of course, because the numbers in integer notation actually represent the number of half-steps above a given pitch. When using fixed zero, each integer represents that many half-steps above C. From this, we derive the following chart, although if you have already memorized the integers for fixed-zero notation, you can always “reverse engineer” this chart using your knowledge of intervals within the C major scale.

Pitch-class integer Interval 1 Interval 2 Interval 3
0 unison d2 A7
1 m2 (augmented unison)
2 M2 d3
3 m3 A2
4 M3 d4
5 P4 A3
6 A4 d5
7 P5 d6
8 m6 A5
9 M6 d7
t (10) m7 A6
e (11) M7 (diminished unison)

When notating transposition, we write Tn where n = the interval by which you will transpose. The example above would be notated:

T5(E, F, G, B-flat) = (A,B-flat,C,E-flat)

OR using integer notation

T5(4,5,7,t) = (9,t,0,3)

Transposing downward

Any positive integer implies a transposition upward, so a negative integer is used to transpose downward. For example, perform the following transposition:

T-3(4,5,7,t)

This example is simple enough because each of the integers is easily reduced by -3 to create (1,2,4,7). However, try the following:

T-5(4,5,7,t)

In this example, 4 - 5 = -1. What does -1 equal? Because we use mod12, you can simply add 12 to this number to find its simplest equivalency.

-1 + 12 = 11

Just as you must reduce any number greater that 11 to an integer between 0 and 11, you must also increase any number less than 0. Therefore the set above is:

T-5(4,5,7,t) = (e,0,2,5)

Normal form (normal order)

In order to make comparing pc sets more useful, we need to use standardized orders to create consistent comparisons. You can imagine how difficult it would be to make sense of multiple pc sets if each set was in a random order. For this, we use normal form.

Normal form is similar to the way in which we analyze standard tonal harmony. For example, how would you condense the following open voicing of a seventh chord in its simplest form while retaining its inversion?

Most people will reduce this to a closed-position chord such as the following, because it shows every pitch and the inversion in the smallest space possible. This makes it easy to see the intervals and construction of the chord:

Normal form (or normal order) applies the same principles for organizing pitches to pc sets as you did when rearranging this seventh chord. Normal form is an ascending arrangement of a pc set in which the outside interval is smallest. For example, take the following pc set and arrange it so it fills the smallest space possible.

(1,7,t)

More precisely, find the ascending arrangement that has the smallest interval between the outside pitch classes. You may use any method that works, but common methods include using mod12, notating the pitches on a staff, or even drawing the pitches on a clock face. As you work through this, consider the speed and efficacy of each method that you try.

Conclusions

For this trichord, it is probably easiest to translate the pitches into standard pitch letters, and then just treat it as a triad. In this case, the pitches D-flat, G, and B-flat form a G diminished triad, so you can quickly tell that the normal form for this should be:

[7,t,1]

Notice that when notating a pc set in normal form, the parentheses are replaced by brackets. (This notation method is outlined by Joseph Straus in his Introduction to Post-Tonal Theory.)

But what happens when you have a more complicated and ambiguous pc set? Try putting the next pentachord into normal form. What difficulties do you find? Which method seems to get the quickest result? Which seems the most consistent?

(1,3,7,t)

Breaking “ties” for ordering pcs

With any pcs, it is possible that there are multiple arrangements that have the same interval between the outside pitches. Try putting the following trichord into normal form.

(0,5,7)

If you arrange this on a staff in all possible ascending orders, you get:

Conclusion

After looking at this on the staff, it is obvious that you can eliminate the last arrangement, but how do you choose between the first two which both have an outer interval of 7 (P5)? In pcs theory, the first tie-breaker is looking at the distance between the lowest pitch and the second highest pitch. In the above example, the second interval for the first arrangement is 5 (P4), but the second interval for the second arrangement is only 2 (M2). In this case the normal form this trichord is [0,2,7].

Developing a method for normal form

Let’s try to create a step-by-step process for finding normal form by using a more difficult example. The following pentachord has multiple ties and will require you to work through multiple levels. Try to develop this method without using a staff or clock faces. While working through this, how many permutations in ascending order are there for a this set? How does that number relate to the cardinality for any set? Remember that each tie is broken by measuring the first pitch against the next closest pitch.

(3,e,5,2,8)

Conclusion

With a large set such as this pentachord, it is difficult to quickly parse the normal form.

  1. Put the pcs in an ascending form. It does not matter on which pitch you start, only that the pitches are ascending. You can imagine this using a clock face–each pitch class must go in the order it would if you move clockwise around the clock without skipping any pitch classes. For this pentachord, let’s start with:
    1. (2,3,5,8,e)
  2. List every possible ascending arrangements of the pc set. There will always be the same number of arrangements as there are members of the pc set, because each integer will have one ascending arrangement with it as a starting pitch class. For a pentachord, there will always be five possibilities, and for this particular pc set, they are:
    1. (2,3,5,8,e)
    2. (3,5,8,e,2)
    3. (5,8,e,2,3)
    4. (8,e,2,3,5)
    5. (e,2,3,5,8)
  3. Subtract the first number from the last number for each ascending arrangement. This will give you the interval between the outer pitch classes expressed as a number of half-steps. If the last number is smaller than the first number, you must use mod12 to convert it. For most students, it is easiest to add 12 to the smaller number before subtracting, but after you have practiced, you can apply this to the result if you’d like. For example above:
    1. (2,3,5,8,e) e-2 = 11-2 = 9
    2. (3,5,8,e,2) 2-3 (requires mod 12) = 14-3 = 11
    3. (5,8,e,2,3) 3-5 (requires mod 12) = 15-5 = 10
    4. (8,e,2,3,5) 5-8 (requires mod 12) = 17-8 = 9
    5. (e,2,3,5,8) 8-e (requires mod 12) = 20-11 = 9
  4. The normal form is the ascending arrangement with the smallest outer interval. If you have multiple arrangements with the same outer interval, you must proceed to the first tie-breaker. In our above example, we have three options with an outer interval of 9 (M6). (At this point, we have eliminated the two arrangements that had the larger outer intervals of 10 and 11.) We must compare each of the tied arrangements against each other to determine which example has the next smallest outer interval. You can do this easily by subtracting the second-to-last number from the first number, so for our remaining three with the outer interval of 9, we get:
    1. (2,3,5,8,e) 8-2 = 6
    2. (8,e,2,3,5) 3-8 (requires mod12) = 15-8 = 7
    3. (e,2,3,5,8) 5-e (requires mod12) = 17-11 = 6
  5. If there is still a tie, repeat the process for the remaining ascending orders, but using the third-to-last pitch class for your subtraction. Sometimes, this can move through multiple tie-breakers, so just continue moving inward through the arrangements. The previous example still had a tie between the first and last arrangement, so we have to go one last round:
    1. (2,3,5,8,e) 5-2 = 3
    2. (e,2,3,5,8) 3-e (requires mod12) = 15-11 = 4
  6. Once you have only one smallest interval, you have found the normal form for that pc set. Remember that you must change your notation to include square brackets once you have determined normal form. For our example above, our normal form is:
    1. [2,3,5,8,e]

Note that there cannot be a definitive normal form for symmetrical sets such as diminished seventh chords, augmented triads, and whole-tone scales. For these, it is simplest to list the lowest integer first.