An Introductionto Pitch-Class Set Theory

The following slides give a brief introduction to some of the concepts and terminology necessary in discussing post-

tonal music.

Pitch-Class Notation

1. Pitch class (abbreviated, “pc”): The twelve different pitches (independent of octave displacement) of the equal tempered system.

Pitch-Class Notation

1. Pitch class (abbreviated, “pc”): The twelve different pitches (independent of octave displacement) of the equal tempered system.

2. The twelve pitch classes are: C (C=B=D=etc.), C, D, D, E, F, F, G, G, A, A, B)

Pitch-Class Notation1. Pitch class (abbreviated, “pc”): The twelve different pitches (independent

of octave displacement) of the equal tempered system.

2. The twelve pitch classes are: C (C=B=D=etc.), C, D, D, E, F, F, G, G, A, A, B)

3. Pitch classes may be expressed as integers: 0, 1 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. (0=C, 1=C, 2=D, . . . 9=A, 10=B, 11=B.)

Pitch-Class Notation

1. Pitch class (abbreviated, “pc”): The twelve different pitches (independent of octave displacement) of the equal tempered system.

2. The twelve pitch classes are: C (C=B=D=etc.), C, D, D, E, F, F, G, G, A, A, B)

3. Pitch classes may be expressed as integers: 0, 1 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. (0=C, 1=C, 2=D, . . . 9=A, 10=B, 11=B.)

4. Each integer representing the distance from 0 (or “C”) in half steps.

Pitch-Class Notation

1. Pitch class (abbreviated, “pc”): The twelve different pitches (independent of octave displacement) of the equal tempered system.

2. The twelve pitch classes are: C (C=B=D=etc.), C, D, D, E, F, F, G, G, A, A, B)

3. Pitch classes may be expressed as integers: 0, 1 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. (0=C, 1=C, 2=D, . . . 9=A, 10=B, 11=B.)

4. Each integer representing the distance from 0 (or “C”) in half steps.

5. Sometimes “A” is substituted for 10 and “B” substituted for 11. So the list of pc’s: 0123456789AB. (Less frequently “T” is substituted for 10 and “E” for 11: 0123456789TE)

Pitch-Class Notation

1. Pitch class (abbreviated, “pc”): The twelve different pitches (independent of octave displacement) of the equal tempered system.

2. The twelve pitch classes are: C (C=B=D=etc.), C, D, D, E, F, F, G, G, A, A, B)

3. Pitch classes may be expressed as integers: 0, 1 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. (0=C, 1=C, 2=D, . . . 9=A, 10=B, 11=B.)

4. Each integer representing the distance from 0 (or “C”) in half steps.

5. Sometimes “A” is substituted for 10 and “B” substituted for 11. So the list of pc’s: 0123456789AB. (Less frequently “T” is substituted for 10 and “E” for 11: 0123456789TE)

Modulo Math • The “modulo” operator takes the remainder of

an integer divided by some other integer, the modulo.

Modulo Math • The “modulo” operator takes the remainder of

an integer divided by some other integer, the modulo.

• For example: 14 modulo 12 = 2. (That is, 12 goes into 14 once with a remainder of 2.

Modulo Math • The “modulo” operator takes the remainder of

an integer divided by some other integer, the modulo.

• For example: 14 modulo 12 = 2. (That is, 12 goes into 14 once with a remainder of 2.

• Hence, we say 14 is “2 mod 12”. (That is, 14 is equivalent to 2 in a modulo 12 system.)

Modulo Math • The “modulo” operator takes the remainder of

an integer divided by some other integer, the modulo.

• For example: 14 modulo 12 = 2. (That is, 12 goes into 14 once with a remainder of 2.

• Hence, we say 14 is “2 mod 12”. (That is, 14 is equivalent to 2 in a modulo 12 system.)

“Clock” Math

When doing mod 12 arithmetic we can visualize the process easily by using the face of a clock.

Pitch Class Sets (PC Sets)

• A pc set is simply a list of pc’s between brackets.• C major triad: [0,4,7]• F minor triad: [5,8,0]

Pitch Class Sets (PC Sets)

• A pc set is simply a list of pc’s between brackets.• C major triad: [0,4,7]• F minor triad: [5,8,0]

• Note that only one representation of a pc is necessary in describing a pc set; octave doublings and displacements are ignored:– [0,4,7,12,19] → [0,4,7] (Since 12=“0 mod 12” and

19=“7 mod 12,” they are unnecessary in describing the pc set; [0,4,7] is sufficient.)

Pitch Class Sets (PC Sets)

• A pc set is simply a list of pc’s between brackets.• C major triad: [0,4,7]• F minor triad: [5,8,0]

• Note that only one representation of a pc is necessary in describing a pc set; octave doublings and displacements are ignored:– [0,4,7,12,19] → [0,4,7] (Since 12=“0 mod 12”

and 19=“7 mod 12,” they are unnecessary in describing the pc set; [0,4,7] is sufficient.)

– Also note that we always use the mod 12 designation of the pc: [5,8,12] → [5,8,0]

Pc sets, cont.

• Each one of the follow is merely one expression of a [0,5,6] pc set:

[0,5,6] [0,5,6] [0,5,6] [0,5,6]

• Sometimes the commas are left out. This allows for a very compact notation using A=10 and B =11, or T=10 and E=11. Thus [9, 10, 11] can be expressed [9AB] or [9TE]. (We’ll use the former in this class.)

Transposing PC Sets

• To transpose a pc set simply add or subtract the same integer (number of half steps) from each member of the set.

Example: Transpose [158] by a minor third up (+3): [1+3, 5+3, 8+3] = [4,8,11] = [48B]

Transposing PC Sets

• To transpose a pc set simply add or subtract the same integer (number of half steps) from each member of the set.Example: Transpose [158] by a minor third up (+3): [1+3, 5+3, 8+3] = [4,8,11] = [48B]

• Addition/subtraction must be by modulo 12 (mod 12)Example: Transpose [158] down by a perfect fifth (-7): [1-7, 5-7, 8-7] mod 12 = [6, 10, 11] = [6AB]

Transposing PC Sets

• To transpose a pc set simply add or subtract the same integer (number of half steps) from each member of the set.

Example: Transpose [158] by a minor third up (+3): [1+3, 5+3, 8+3] = [4,8,11] = [48B]

• Addition/subtraction must be by modulo 12 (mod 12)

Example: Transpose [158] down by a perfect fifth (-7): [1-7, 5-7, 8-7] mod 12 = [6, 10, 11] = [6AB]

• Note that in mod 12 terms, transposition down by 7 (-7) is the same as transposing up by 5 (+5). Both yield the same pc set.

Transposing PC Sets

• To transpose a pc set simply add or subtract the same integer (number of half steps) from each member of the set.

Example: Transpose [158] by a minor third up (+3): [1+3, 5+3, 8+3] = [4,8,11] = [48B]

• Addition/subtraction must be by modulo 12 (mod 12)

Example: Transpose [158] down by a perfect fifth (-7): [1-7, 5-7, 8-7] mod 12 = [6, 10, 11] = [6AB]

• Note that in mod 12 terms, transposition down by 7 (-7) is the same as transposing up by 5 (+5). Both yield the same pc set.

“Normal” Form

1. Arrange the pcs in ascending numerical order

“Normal” Form

1. Arrange the pcs in ascending numerical order

2. List all of the set’s rotations.

“Normal” Form

1. Arrange the pcs in ascending numerical order

2. List all of the set’s rotations.

3. Compute the ordered pc interval between the first and last pcs of each rotation. Compare these intervals. If one rotation has the smallest outside interval, it is the normal form.

“Normal” Form

1. Arrange the pcs in ascending numerical order2. List all of the set’s rotations.3. Compute the ordered pc interval between the first and last pcs of

each rotation. Compare these intervals. If one rotation has the smallest outside interval, it is the normal form.

4. If the smallest outside interval is shared by more than one rotation, compute and compare the interval between the first and next-to-last pcs for only those rotations.

“Normal” Form

1. Arrange the pcs in ascending numerical order2. List all of the set’s rotations.3. Compute the ordered pc interval between the first and last pcs of

each rotation. Compare these intervals. If one rotation has the smallest outside interval, it is the normal form.

4. If the smallest outside interval is shared by more than one rotation, compute and compare the interval between the first and next-to-last pcs for only those rotations.

5. If a tie still exists, compute and compare the interval between the first and next-to-the-next-to-the-last pcs of only those rotations that are still in contention.

“Normal” Form1. Arrange the pcs in ascending numerical order2. List all of the set’s rotations.3. Compute the ordered pc interval between the first and last pcs of

each rotation. Compare these intervals. If one rotation has the smallest outside interval, it is the normal form.

4. If the smallest outside interval is shared by more than one rotation, compute and compare the interval between the first and next-to-last pcs for only those rotations.

5. If a tie still exists, compute and compare the interval between the first and next-to-the-next-to-the-last pcs of only those rotations that are still in contention.

6. Continue this process as long as necessary to break a tie. If you have computed all possible intervals and a tie still exists, select the rotation that begins with the lowest pc number.

“Normal” Form1. Arrange the pcs in ascending numerical order2. List all of the set’s rotations.3. Compute the ordered pc interval between the first and last pcs of

each rotation. Compare these intervals. If one rotation has the smallest outside interval, it is the normal form.

4. If the smallest outside interval is shared by more than one rotation, compute and compare the interval between the first and next-to-last pcs for only those rotations.

5. If a tie still exists, compute and compare the interval between the first and next-to-the-next-to-the-last pcs of only those rotations that are still in contention.

6. Continue this process as long as necessary to break a tie. If you have computed all possible intervals and a tie still exists, select the rotation that begins with the lowest pc number.

7. Transpose the winner so that the first pc becomes 0.

“Normal” Form1. Arrange the pcs in ascending numerical order2. List all of the set’s rotations.3. Compute the ordered pc interval between the first and last pcs of

each rotation. Compare these intervals. If one rotation has the smallest outside interval, it is the normal form.

4. If the smallest outside interval is shared by more than one rotation, compute and compare the interval between the first and next-to-last pcs for only those rotations.

5. If a tie still exists, compute and compare the interval between the first and next-to-the-next-to-the-last pcs of only those rotations that are still in contention.

6. Continue this process as long as necessary to break a tie. If you have computed all possible intervals and a tie still exists, select the rotation that begins with the lowest pc number.

7. Transpose the winner so that the first pc becomes 0.

Inverting PC Sets

1. To invert a pc set, simply subtract each element from 0 mod 12. (Subtracting from 0 mod 12 is the same as subtracting from 12.)

Example: Invert [89B] = [0-8, 0-9, 0-11] = [4,3,1]

Inverting PC Sets1. To invert a pc set, simply subtract each element from 0

mod 12. (Subtracting from 0 mod 12 is the same as subtracting from 12.) Example: Invert [89B] = [0-8, 0-9, 0-11] = [4,3,1]

2. This creates a mirror image (that is, bi-laterally symmetry) around 0. (See clock face to the right.)

Inverting PC Sets1. To invert a pc set, simply subtract each element from 0

mod 12. (Subtracting from 0 mod 12 is the same as subtracting from 12.) Example: Invert [89B] = [0-8, 0-9, 0-11] = [4,3,1]

2. This creates a mirror image (that is, bi-laterally symmetry) around 0. (See clock face to the right.)

3. A real inversion, however, should be based upon the first pc of the original set. Thus we need to transpose the result we obtained in 1., above, so that the inverted set begins on pc 8: [4+4, 3+4,1+4]=[8,7,5]=[875].

Inverting PC Sets1. To invert a pc set, simply subtract each element from 0

mod 12. (Subtracting from 0 mod 12 is the same as subtracting from 12.) Example: Invert [89B] = [0-8, 0-9, 0-11] = [4,3,1]

2. This creates a mirror image (that is, bi-laterally symmetry) around 0. (See clock face to the right.)

3. A real inversion, however, should be based upon the first pc of the original set. Thus we need to transpose the result we obtained in 1., above, so that the inverted set begins on pc 8: [4+4, 3+4,1+4]=[8,7,5]=[875].

4. So, the inversion of [89B] is [875], a mirror image around the first pc of the original set--here pc 8. (See clock face to the right

Inverting PC Sets1. To invert a pc set, simply subtract each element from 0

mod 12. (Subtracting from 0 mod 12 is the same as subtracting from 12.) Example: Invert [89B] = [0-8, 0-9, 0-11] = [4,3,1]

2. This creates a mirror image (that is, bi-laterally symmetry) around 0. (See clock face to the right.)

3. A real inversion, however, should be based upon the first pc of the original set. Thus we need to transpose the result we obtained in 1., above, so that the inverted set begins on pc 8: [4+4, 3+4,1+4]=[8,7,5]=[875].

4. So, the inversion of [89B] is [875], a mirror image around the first pc of the original set--here pc 8. (See clock face to the right

“Prime” Form

• Determine the normal form of the pc set.

“Prime” Form

• Determine the normal form of the pc set.• Determine its inversion.

“Prime” Form

• Determine the normal form of the pc set.• Determine its inversion.• Arrange the inversion in ascending numerical order.

“Prime” Form

• Determine the normal form of the pc set.• Determine its inversion.• Arrange the inversion in ascending numerical order.• Transpose the inversion so that the first pc of the

inversion is 0

“Prime” Form

• Determine the normal form of the pc set.• Determine its inversion.• Arrange the inversion in ascending numerical order.• Transpose the inversion so that the first pc of the

inversion is 0• Compare the original and the inversion beginning with

the last pcs and working backwards toward the first pcs. When you find that one set has a lower number in a give place than the other set, that set is the “prime” form of the set.

“Prime” Form

• Determine the normal form of the pc set.• Determine its inversion.• Arrange the inversion in ascending numerical order.• Transpose the inversion so that the first pc of the

inversion is 0• Compare the original and the inversion beginning with

the last pcs and working backwards toward the first pcs. When you find that one set has a lower number in a give place than the other set, that set is the “prime” form of the set.

• Obviously, if the two forms are identical no comparison is necessary.

“Prime” Form

• Determine the normal form of the pc set.• Determine its inversion.• Arrange the inversion in ascending numerical order.• Transpose the inversion so that the first pc of the

inversion is 0• Compare the original and the inversion beginning with

the last pcs and working backwards toward the first pcs. When you find that one set has a lower number in a give place than the other set, that set is the “prime” form of the set.

• Obviously, if the two forms are identical no comparison is necessary.

PITCH INTERVALS

A “Pitch Interval” is the standard definition of an interval between any two pitches

• Ordered Intervals:– A3 to D5 is a Perfect 11th (ascending), also +17 half

steps– D5 to A3 is a Perfect 11th (descending; also -17 half

steps

• Unordered Intervals– Is strictly a measure of distance.– The distance between A3 and D5 is a perfect 11th (17

half steps)

Pitch-Class Interval Classes

• Ordered Pitch Class Intervals– Can only be from 0 to 11. Since we are dealing with

pcs, octave placement is irrelevant, so there is no need for negative numbers

• Unordered Pitch Class Intervals is the smallest distance between two pcs no matter what direction you are measuring. Informally referred to as “Interval Classes” or “ic”’s.

• There are 7: 0, 1, 2, 3, 4, 5, and 6.

Interval Vector. . . is the measure of the interval content of a pc set.

•To create an interval vector for a set we list the number of instances of each ic (pitch-class interval class) in order of size.

Interval Vector

Interval Classes: 1 2 3 4 5 6

1 0 0 0 1 1

. . . is the measure of the interval content of a pc set.

•To create an interval vector for a set we list the number of instances of each ic (pitch-class interval class) in order of size.

•For Instance, given the pc set (0,1,6), what is its interval vector?•It contains a single instance of the ic 1: (01)•It contains no instances of the ic 2.•It contains no instances of the ic 3.•It contains no instances of the ic 4•It contains one instance of the ic 5: (16).•It contains one instance of the ic 6: (06)•The interval vector for (016), then, is <100011>. The following table illustrates how to read an interval vector:

Interval Vectors, cont.

A set that contains N pcs, will contain (N*(N-1))/2 ics.

No. PCs No. ICs

2 1

3 3

4 6

5 10

6 15

7 21

8 28

9 36

10 45

11 55