Mazzola, Guerino / Noll, Thomas / Lluis-Puebla, Emilio: Perspectives in Mathematical and Computational Music Theory

9.2 Example: Inversion with Retrograde

In music, retrogade means playing backwards, beginning with the last note and ending with the first note of the original composition. However, the time allotted by parameters such as » $Onset$ «, » $Duration$ «, etc. can’t be inverted so intuitively; the example of a recording played backwards is not musical retrograde.

We will work with the form $Onset-- >$ Simple $(R)$ . Here we can see a similarity with the inversion of $Pitch$ and the symmetry of retrograde: $s ks(x) = e .(-1)x = s- x$ , where $s$ is the total »size« of the piece, measured in $R$ . If, for example, we have a measure in common time (divided in four), $s = 4$ , and if $x = 1.5$ (a note that begins on $Onset = 1.5)$ in its retrograde the same note will begin on $Onset = 4- 1.5 = 2.5$ .

But what will happen when we have 2 notes with $Onset$ and $Offset$ , separated by $Onsetdistance = Offset -Onset$ ? That is, $note$ 1 has, for example, $Onset = 1.5$ and $note$ $2$ begins at $Onset+ Onsetdistance = 1.5+ .75 = 2.25$ . If we apply the symmetry $ks$ , we will have

{ks(Onset),ks(Offset)}= {s -Offset,s - Onset}

that, in our example, is ${4 - 2.25,4- 1.5}= {1.75,2.5}$ . Then the retrograde of $note$ $2$ begins exactly Onsetdistance $(.75)$ before the retrograde of $note$ $1$ . However, if the duration of $note$ 2 is longer than Onsetdistance (for example, if it is 1.5) then $ks($ Offset $)$ will not be finished when $ks(Onset)$ begins.

Therefore, we include the parameter $Duration$ when we define a retrograde symmetry, and we work in the two dimensional space $o+ Onset Duration$ . If we take two events $OD(Onset,Duration) and OD(Of fset,Duration)$ , first we must apply $s e .(- 1)$ to the parameter $Onset$ of both notes, i.e., $OD(ks(Onset),Duration)$ and $OD(ds(Offset),Duration)$ . Consider the duration of $note$ $1 = .5,note$ $2 = 1.0 > .75$ . Then we have $(2.5,.5),(1.75,1.0)$ . However, this transformation is intermediate, because we want to avoid overlaps. That’s why we take into account the parameter $Duration$ to obtain $OD(ks(Onset)- Duration1,Duration1)$ and $OD(ks(Offset)- Duration2,Duration2)$ , that is $(2.5- .5,.5) = (2.0,.5)$ and $(1.75 - 1.0,1.0) = (.75,1.0)$ . This way, the overlap from $2.5$ to $2.75$ is completely avoided, because the retrograde of $note$ $2$ ends at $1.75$ and $note$ 1 begins at $2.0$ . We define retrograde as $2 2 Ks : R -- > R ,(o,d) '---> (s- o - d,d)$ , that is,

[ ] (s,0) - 1 -1 Ks = e . 0 1 .

It is clear that in this example we are not considering the parameter $Loudness$ which is also subject to retrograde. Finally, the inversion with retrograde, as used in serial dodecafonic music, involves three parameters, that is, $Onset$ , $Pitch$ , and $Duration$ . Then we have:

|_ - 1 0 -1 _| KUs,t = e(s,t,0) . |_ 0 -1 0 _| 0 0 1

and if the original $Onset$ , $Pitch$ , and $Duration$ values are $o,p,d (- R$ , then: