Finite fields and musical phrasings

posted by on 2013.12.20, under Supercollider
20:

In this post I want to talk about musical phrasings in algorithmic composition and algebra over finite fields. Yep, algebra, so brace yourself  :-).
I am pretty sure you have been exposed to clock arithmetic: basically, you consider integer numbers modulo a natural number p. Now, in the case p is a prime, the commutative ring \mathbb{Z}_{p} is actually a field with finite elements. There is a name for that, and it is finite field, or Galois* field.
Now consider a pxp-matrix A with entries in \mathbb{Z}_{p}\:: A will give us a map \varphi_{A} from \mathbb{Z}_{p} to itself via

\varphi_{A}(i):=\sum_{j=0\ldots p-1}(A_{ij}\:j)\:{\rm mod}\:p

where A_{ij} are the entries of A. Now, consider a function f from \mathbb{Z}_{p} to a finite set S, and consider the subgroup P of the group of permutations of p objects given by elements \pi satifying

f(\pi(i)) = f(i)

If the function f is not injective, this subgroup will be not empty. Notice that for any function f of the type above, the matrix A gives us another function A^{*}f defined as**

A^{*}f(i):=f(\varphi_{A}(i))

Now we can ask ourselves: given a function f, can we find a matrix A such that

(A^{*})^{k}f = f

for some natural number k? What this means is that we want to find a matrix such that applied k-times to the function f returns the function itself. This problem is equivalent to finding a matrix A such that

A^{k} = \pi,\quad\pi\in{P}

In the case in which the function f is injective, this coincides with the problem of finding idempotents matrices in a finite field: the fantastic thing is that they are known to exist***, and even better their number is known in many situations!
“Ok, now, what all of this has to do with musical phrasing?! No, seriously, I’m getting annoyed, what really? ”
I hear your concern, but as with almost everything in life, it’s just a matter of perspective. So, consider the set S as a set of pairs (pitch, duration) for a note. A function f above will tell us in which order we play a note and what is the duration of each note: in other words, it is a musical phrasing. By picking a matrix A, we can generate from f a new phrasing, given by the function A^{*}f, and we can reiterate the process. If A is such that it satisfies the condition above, after a finite number of steps, the phrasing will repeat. Hence, the following Supercollider code

s.boot;

SynthDef(\mall,{arg out=0,note, amp = 1;
    var sig=Array.fill(3,{|n| SinOsc.ar(note.midicps*(n+1),0,0.3)}).sum;
    var env=EnvGen.kr(Env.perc(0.01,1.2), doneAction:2);
    Out.ar(out, sig*env*amp!2);
}).add;

(
var matrix, index, ind, notes, times, n, a;

notes = [48, 53, 52, 57, 53, 59, 60] + 12;
times = [1/2, 1/2, 1, 1/2, 1/2, 1, 1]*0.5;
n = 7;
matrix = Array.fill(n,{Array.fill(n, {rrand(0, n-1);})});
index = (0..(n-1));

a = Prout({
     inf.do({
            ind = [];
           
            matrix.collect({|row|
               ind = ind ++ [(row * index).sum % n];
               ((row * index).sum % n).yield;
            });
           index = ind;
         
     });
   }); 
       
   

Pbind(*[\instrument: \mall, \index: a, \note: Pfunc({|ev| notes[ev[\index]];}), \dur:Pfunc({|ev|
 times[ev[\index]];})]).trace.play;
)

s.quit;

represents a “sonification” of the probability distribution of finding a matrix A with the properties above, for p=7. We are assuming that the various matrices have equal probability to be generated in the code above.
So, how does this sound?
Like this

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*By the way, you should check out the life of Galois, just to crush all your stereotypes about mathematicians. 😉
**Mathematicians love these “dual” definitions.
***Apart from the identity matrix and elements of P themselves, clearly.

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