Automatic Exercise Generation for Exploring Connections between Mathematics and Music

2.1 Arithmetic of Musical Intervals

Pythagoras was one of the first who discovered and formalized one of the connections between Mathematics and Music.

Pythagoras studied the arithmetic of musical intervals. He experimented with the tones produced by strings of different lengths and expressed musical intervals by numerical ratios. Pythagoras discovered a relationship between musical harmony and the mathematical harmony by relating ratios to consonance/dissonance notions. The Pythagorean tuning and the associated Pythagorean scale were derived based on arithmetic principles. The Pythagoreans discovered relations between musical intervals and numerical ratios on a monochord (see Figure 1).

Figure 1

Monochord.

Reducing a chord to the half of its total length, then the sound produced by the half chord will form an interval of octave with respect to the sound produced by the whole chord. The sound that we will hear is one octave higher than the sound obtained with the total chord. Defining the length ratio by dividing the length of the reduced chord, in this case half of the length, by the total length, one can determine the length ratio associated to the octave interval which is 1/2. Now, dividing the chord into 12 units of equal length and considering the chord reduced to 8 units (see Figure 1), then the sound produced with the reduced chord will form an interval of fifth with respect to the sound produced with the chord having total length. The length ratio 8/12 = 2/3 can therefore be associated with the interval of fifth. The numerical ratios defined by the length ratios are associated with the frequencies of tones. The frequency ratios are inversely related with the length ratios. For example, reducing the chord to one half of its total length implies that the frequency will go up by the factor 2, or, reducing the chord to 2/3 of its total length implies that the frequency will go up by the factor 3/2. Table 1 contains the length and frequency ratios associated with the intervals of octave, fifth and fourth, where the length ratios are illustrated on the monochord in Figure 1.

The musical intervals can be constructed using the arithmetic of proportions. The product of two proportions corresponds to the sum of two intervals. The division of two proportions corresponds to the difference of two intervals. Considering the frequency ratios, since the sum of an interval of fifth and an interval of fourth gives an octave, where we will write fifth + fourth = octave, in terms of ratios this equation corresponds to

Considering the difference of intervals, since octave-fourth = fifth, then dividing the frequency ratio of the octave by the frequency ratio of the fourth interval, one obtains the frequency ratio of the fifth:

In the Pythagorean tuning system, all intervals of the musical scale are constructed using the interval of fifth with ratio 3/2. For example, going up two fifths from the note C, one obtains the note D (which is separated by an interval of ninth from C), and going down one octave one obtains the note D that forms and interval of major second with respect to the initial note C. The frequency ratio associated with the interval of major second can therefore be obtained as follows:

Proceeding in the same way by going up intervals of fifth and going down octave intervals, one can determine the frequency ratios of the other intervals listed in Table 2. One can observe that the ratios of octave, fifth, and fourth are simple depending on low integers. In contrast, the other intervals, for example, the interval of major seventh, have ratios depending on higher integers. These differences were related by the Pythagoreans to the consonance and dissonance of the sound produced by the interval. Intervals with simple ratios, formed with the lowest integers, were perceived to be more consonant than intervals with frequency ratios formed with higher integers.

2.2 Geometric Approach to Musical Composition

Geometric and musical transformations are related as follows. The geometric transformations: translation, reflection at an horizontal axis and reflection at a vertical axis correspond to the musical transformations: transposition, inversion, and retrograde, respectively. Let T _n , I, and R denote the transposition of n halftones, the inversion, and the retrograde, respectively. These musical transformations are applied, for example, in the 12-tone compositional technique. The 12-tone technique uses themes that are sequences of 12 notes of the chromatic scale that must come from different pitch classes. Each pitch class occurs once and only once in the theme. The sequences are called tone rows. The composition consists of an original tone row and it can have additional rows that are created by applying T n , I, and R transformations, or combinations of those transformations ( T n I , T n R , T n IR ), to the original tone row. Modular arithmetic can be used to obtain the tones resulting from applying the transformations to a given pitch taking into account the equivalent pitch classes. If the difference of two integers a and b is a multiple of 12, then a and b are congruent modulo 12 (12 is a divisor of their difference). In arithmetic mod 12, one has, for example, 10 + 7 = 5 (mod 12).

The set Z ₁₂ = {0,1,…,11} contains the integers mod 12. These integers are assigned to the 12 pitch classes as indicated in the modular clock in Figure 2.

Figure 2

Modular clock with pitch class integers.

A transposition moves a sequence of pitches up (or down) by a given number of halftones, where T n , n ∈ N 0 , denotes the transposition moving n halftones up (and T − n , n halftones down). Consider a sequence of pitches X = ( x 1 , … , x K ) , x ₁,…, x _K ∈ Z ₁₂. The transposition of the sequence of pitches X is defined by T n ( X ) = ( x 1 + n , … , x K + n ) .

Example:

Consider the sequence of pitches given by X = (4, 7, 0, 9), in the first bar of Figure 3, then T 3 ( X ) = (7, 10, 3, 0) corresponds to the motif presented in the second bar.

Figure 3

Sequence of pitches X and transposition T 3 ( X ) .

A retrograde reverses the order of the pitches of a given sequence. Consider a sequence of pitches X = ( x 1 , … , x K ) , x ₁,…,x _K ∈ Z ₁₂. The retrograde of the sequence of pitches X is defined by R ( X ) = ( x K , x K − 1 , … , x 1 ) .

Example:

Given the sequence of pitches X = (4, 7, 0, 9), in the first bar of Figure 4, then, applying the retrograde to X, one obtains R ( X ) = (9, 0, 7, 4), corresponding to the motif presented in the second bar.

Figure 4

Sequence of pitches X and retrograde R ( X ) .

The inversion reflects the pitches of a given sequence about 0. Consider a sequence of pitches X = ( x 1 , … , x K ) , x ₁,…,x _K ∈ Z ₁₂. The inversion of X is defined by I ( X ) = ( 12 − x 1 , … , 12 − x K ) . On the modular clock, the inversion corresponds to a reflection about an axis passing through 0 and 6 (see Figure 5).

Figure 5

Inversion as a reflection on the modular clock.

Example:

Consider the sequence of pitches given by X = (4, 7, 0, 9), in the first bar of Figure 6. Applying the inversion to that sequence, one obtains I ( X ) = (12 – 4, 12 – 7, 12 – 0, 12 – 9) = (8, 5, 0, 3), corresponding to the motif in the second bar.

Figure 6

Sequence of pitches X and inversion I ( X ) .

The inversion can also be obtained by reflecting the original motif on the modular clock as illustrated in Figure 7, where the red lines connect the notes of the original motif and the reflected blue lines connect the notes of the inversion motif.

Figure 7

Inversion on the modular clock.

3.1 MVGEN

MVGEN is a multi-version question and test generation system, a Domain Specific Language, developed in PERL, that uses LaTeX as typographical language and allows the random generation of different question types and the random creation of a large number of different question and test versions. The system has the advantage that it is simple and self-contained. With MVGEN, one can generate, for example, multiple-choice, multiple-selection, numeric response, matching, ordering, and free response questions for different education domains. Applications to the generation of multiple-choice questions for higher education Mathematics have been presented in Brito et al. (2022). The reader is also referred to Brito et al. (2019), where a more detailed description about MVGEN and a presentation of various kinds of exercises that can be generated with this system can be found. Multiple-choice questions in MVGEN are written in LaTeX and stored in a database. These questions can easily be built by providing a stem together with a list containing one or more keys and a list with three or more distractors (considering questions with four options), where an additional table allows introducing associated random parameters in the stem and in the choices to generate different question versions. The questions in the database are structured as follows:

#qn

stem

#v

list with 1 or more keys

#f

list with 3 or more distractors

#tab

table with instances (values or text)

a :: b :: …

MVGEN extracts then randomly one correct option from #v and three distractors from #f and shuffles the options for each question version. The stem, key, and distractors can depend on random parameters defined in the table #tab. Together with the test file containing the questions, MVGEN also generates a verification file for each question. This file allows the instructor to verify if the exercise is correctly built up. Since the system uses LaTeX language, questions for Mathematics and Music can be generated with the additional package MusiXTeX that allows musical typesetting in LaTeX. In the following two subsections, examples of multiple-choice questions generated in MVGEN for Mathematics and Music about the topics introduced in the previous section are presented.

3.2 Arithmetic and Music

To study the relation between Arithmetic and Music, exercises about musical intervals and their associated frequency ratios can be constructed in the following examples, where the Pythagorean tuning system was considered.

Example 1

Consider the question in Figure 8.

Figure 8

Example 1 – multiple-choice question about intervals and frequency ratios.

In this exercise, the aim is to determine the interval formed by two notes, where the corresponding frequencies are known. Therefore, the frequencies of both notes should be divided and the simplified ratio can then be associated with the correct interval.

This multiple-choice question was generated with the following code:

#q2

Consider the Pythagorean tuning system. Determine the interval defined by the note of frequency $f=100$Hz and the note of frequency $f=$#freq Hz.

#v

#int

#f

major third

fifth

major seventh

octave

#tab

freq :: int

$\frac{900}{8}$ :: major second

$\frac{400}{3}$ :: fourth

$\frac{2700}{16}$:: major sixth

The three rows in the table contain three frequency values to be instanced in the stem (in #freq), defining the frequency of the second note. These frequency values are associated with the intervals listed in the second column int of the table, which define the key #int of the multiple-choice question. Each time the question is generated, a different row is randomly selected, and three distractors from the list of four options are randomly chosen. The key and the three distractors appear then in a shuffled form in the question. Figure 9 contains the output of the questions in the verification file, corresponding to the three rows in the table. The correct and false options are indicated by “v” and “f”, respectively.

Figure 9

Verification file for Example 1.

Example 2

Consider the question in Figure 10.

Figure 10

Example 2 – multiple-choice question about intervals and frequency ratios.

In this exercise, the objective is to establish an equation by identifying the given product with the frequency ratio of the interval of major seventh, to solve the equation for x and then to determine the interval associated with the obtained frequency ratio.

This multiple-choice question was generated with the following code:

#q2

Let $x$ be the frequency ratio of a musical interval in Pythagorean tuning. Knowing that multiplying $x$ by $#ratio$, one obtains an interval of #int, then $x$ corresponds to an interval of:

#v

#intx.

#f

#false.

fifth.

major sixth.

major seventh.

#tab

ratio :: int :: intx :: false

\frac{3}{2} :: major seventh :: major third :: major second

\frac{4}{3} :: fifth :: major second :: major third

\frac{9}{8} :: fifth :: fourth :: major second

In this exercise, the interval of x is identified by intx in the table #tab. One of the three intervals listed in the column will appear in the key through the random row selection when generating the exercise. Multiplying the corresponding interval ratio by the ratio provided in the table leads to the interval int, where both ratio and int are associated with the selected interval intx in the same row. Concerning the distractors, the intervals fifth, major sixth, and major seventh are common distractors for the three available configurations (the three rows in the table) and another specific distractor represented by false is added to the list of false options, which varies with the selection of the row. Figure 11 presents the three possible question configurations, where one can identify the associated row instances in each exercise.

Figure 11

Verification file for Example 2.

Example 3

In the following exercise, see Figure 12, given a note and its corresponding frequency, the frequency of a new note has to be determined considering the Pythagorean scale. Therefore, the interval has to be determined between the two notes and the frequency of the given note has to be multiplied by the interval’s corresponding frequency ratio. In the example in Figure 12, since F forms and interval of fourth with respect to C, then, according to the Pythagorean tuning, the frequency of C must be multiplied by the ratio of 4/3 to obtain the frequency of F.

Figure 12

Example 3 – multiple-choice question about intervals and frequency ratios.

This multiple-choice question was generated with the following code:

#q3

Consider the Pythagorean tuning and the musical note $C$ with frequency $f=261$Hz. Then, the musical note #N has frequency

#v

$#freq$Hz.

#f

$#false$Hz.

$\frac{522}{3}$Hz.

$\frac{783}{4}$Hz.

#tab

N :: freq :: false

G :: \frac{783}{2} :: \frac{1044}{3}

F :: \frac{1044}{3} :: \frac{783}{2}

In the stem of this multiple-choice question, note C and its frequency are given. The new note, represented by the parameter #N, is randomly selected from the table and can be F or G. The correct associated frequency is thus obtained by multiplying 261 Hz by 3/2 or 4/3, respectively, leading to the results in the second column of the table. The correct option will appear in #freq, as instance of the key. Two distractors in the list #f are common to both keys, and another distractor #false is obtained from the third column of the table. In this case, the key of the second row will be the distractor corresponding to the first key and conversely, so that all question versions will have displayed the same options (see Figure 13).

Figure 13

Verification file for Example 3.

3.3 Geometry and Music

The construction of exercises for studying the relation between Geometry and Music, where it is useful and more interesting to illustrate the geometric transformations in score notation, can be achieved with the LaTeX package MusiXTeX.

Example 4

Consider the question in Figure 14.

Figure 14

Example 4 – multiple-choice question about musical transformations.

This multiple-choice question was generated with the following code:

#q3

Determine the operator that transforms the musical motif presented in the first bar into the motif in the second bar of the following measure:

\seqnbar{\Notes \qa{egjh}\en\bar\Notes \qa{j_l_ig}\en}

#v

$RT_3$

$T_3 R$

#f

$RT_2$

$T_8R$

$T_8I$

In this exercise, the sequence of notes of the motif in the first bar and the transformed sequence in the second bar are indicated directly in the stem, where the command \seqnbar{…} was defined in the LaTeX layout file using the MusiXTeX instructions as follows:

\def\seqnbar#1{

\begin{music}

\setstaffs1{1} \setclef1{\treble} \generalmeter{\meterC}

\startextract #1 \endextract

\end{music}}

The second motif was obtained from the first one by applying a retrograde and a transposition of three halftones. Since the combination of retrograde and transposition is commutative, two correct transformations are indicated in the list of keys #v and then three distractors are listed in #f. The aim is then to determine the correct operator that transforms the motif of the first bar into the motif of the second bar. The output in the verification file (see Figure 15) permits the instructor to verify if the exercise is correctly build up.

Figure 15

Verification file for Example 4.

Example 5

In the following exercise, see Figure 16, a musical motif is presented in the stem and the aim is to determine its inversion by identifying the correct inversion motif in the given options.

Figure 16

Example 5 – multiple-choice question about musical transformations.

This multiple-choice question was generated with the following code:

#q4

Determine the Inversion of the motif in the following measure:

\seqnbar{\Notes \qa{#notes} \en}

#v

\seqnbar{\Notes \qa{#inversion} \en}

#f

\seqnbar{\Notes \qa{#false1} \en}

\seqnbar{\Notes \qa{#false2} \en}

\seqnbar{\Notes \qa{#false3} \en}

#tab

notes :: inversion :: false1 :: false2 :: false3

cfgi :: cgf^j :: jgfc :: dfhc :: ^f^cig

dfhc :: ^hg^dc :: chfd :: efgi :: ^h^c^d^f

The motif that will appear in the stem is represented by #notes. When the exercise is generated, a row is randomly selected from the table, where the notes in the first column determine the motif in the stem. The associated notes in the second column form the corresponding inversion motif that will appear in the key given at #v, and the incorrect options in the false columns will appear as distractors to be instanced in the list #f.

Figure 17 contains the output of the verification file corresponding to the instances in the first row of the table #tab and Figure 18, to the instances in the second row. In Figure 17, the original musical motif is given by the sequence X = (0, 5, 7, 11). One can confirm that the correct option is A, since I(X) = (12 – 0, 12 – 5, 12 – 7, 12 – 11) = (0, 7, 5, 1). Considering the second version in Figure 18, the original musical motif is given by the sequence X = (2, 5, 9, 0) and the correct option appears in A, since I(X) = (12 – 2, 12 – 5, 12 – 9, 12 – 0) = (10, 7, 3, 0).

Figure 17

Verification file for Example 5 – first exercise version.

Figure 18

Verification file for Example 5 – second exercise version.

Example 6

Consider the exercise in Figure 19, where the aim is to determine the new musical motif obtained by applying a combination of two transformations to a given motif.

Figure 19

Example 6 – multiple-choice question about musical transformations.

In this exercise, the musical motif in the stem is fixed and what will vary in the stem, when generating the exercise, is the transformation that should be applied to the motif. For that purpose, this multiple-choice question was generated with the following code:

#q6

Determine #tr of the following musical motif:

\seqnbar{\Notes \qa{fjhj}\en}

#v

\seqnbar{\Notes \qa{#true} \en}

#f

\seqnbar{\Notes \qa{#false1} \en}

\seqnbar{\Notes \qa{#false2} \en}

\seqnbar{\Notes \qa{#false3} \en}

#tab

tr :: true :: false1 :: false2 :: false3

$T_2R$ :: kikg :: gkik :: hjhf :: jhjf

$RT_4$ :: l_klh :: jhjf :: hl_kl :: _h_k_hh

In this example, one from two transformations provided in the first column of the table #tab will be randomly selected and substituted in #tr in the stem. Each row in the table contains, associated with the transformation, three distractors (false1, false2, false3) and the key (true).

Figure 20 contains the output of the verification file corresponding to the instances in the first row of the table. Since the original musical motif is given by the sequence X = (5, 0, 9, 0), then R(X) = (0, 9, 0, 5) and thus T 2 R ( X ) = (0 + 2, 9 + 2, 0 + 2, 5 + 2) = (2, 11, 2, 7). (Note that 12 = 0 (mod 12).) In the second version, see Figure 21, the correct option can be obtained as follows: T 4 ( X ) = (5 + 4, 0 + 4, 9 + 4, 0 + 4) = (9, 4, 1, 4), where 13 = 1 (mod 12), and, consequently, R T 4 ( X ) = (4, 1, 4, 9).

Figure 20

Verification file for Example 6 – first exercise version.

Figure 21

Verification file for Example 6 – second exercise version.