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

Figure 1:

For such parameters, there is no exact solution. This shows an approximate solution with four errors, indicated by the arrows.

The first constraint states that inside a voice, a pattern a the correct number of onsets, that is, there are not two onsets at the same time. It can be written $alldiff(Vi,1...Vi,ni)$ , so that the pattern is not degenerated. We could eliminate this constraint by taking the durations as variables and removing $0$ from the domains, but it would make the second constraint more complex (as a capacity constraint).

Second constraint, we must never hear two onsets at the same time, during a fixed duration $D$ . It can be written : for two different voices, $i1 /= i2$ , two onsets $k1$ and $k2$ and for all $p1$ such that $Vi1,k1 + p1 * l1 \< D$ (the onset is repeated while it doesn’t exceed the duration),

Vi1,k1 + p1 * l1[l2] /= Vi2,k2

An important parameter of this CSP is the average density of onsets. A density of $2$ means that we hear in average one onset per two time units. Of course, it fixes the number of onsets by pattern, that is the number of variables for our CSP. There is obviously no solution when the density of onsets becomes too big.

Independently of the CP resolution, we can study this problem thanks to the Chinese Remainder Theorem. In a particular case with two voices, and only one onset by voice, the goal is to find $V1$ et $V2$ such that

V1 + p1 *l1 /= V2 +p2 *l2

This exactly the negation of the chinese theorem, ie solve $x = V1[l1]$ et $x = V2[l2]$ with $x \< l1$ et $l2$ . So there exists a unique solution $x0$ in $[0,ppcm(l1,l2)]$ , given by the formula $x0 = g1l1V2+g2l2V1 pgcd(l1l2)$ , where $g1$ and $g2$ are the Bezout coefficient of $l1$ and $l 2$ , ie $g l+ g l = pgcd(l,l) 11 2 2 1 2$ . These two values are given, so we simply have to find $V 1$ and $V 2$ such that $x < D 0$ , ie solve

g1l1V2 + g2l2V1 > pgcd(l1l2)* D

It is possible to generalize to $n$ voices, although it is unfortunately not the same generalization as the chinese theorem with $n$ equations. Our goal is now to solve the system $x = Vi1[li1]$ $x = Vi2[li2]$ for $i1 /= i2 \< n$ .