quotile sequencer/rhythmik nach euklid

[jane doe]

eine neue datei euclid.bsh im verzeichnis commands anlegen und folgenden inhalt einfügen.

das ergebnis ist im sequencer eine belegung der steps nach dem euklidischen algorithmus.

quellen:

• http://www.quotile-sequencer.com/
• http://cgm.cs.mcgill.ca/~godfried/publications/banff.pdf
• http://ruinwesen.com/blog?id=216

/*
 * euclid.bsh
 *
 * copyright by nina scholz, 2009
 * feel free to e-mail me at nina [at] the-abc.de
 *
 * for quotile, found at http://www.quotile-sequencer.com/
 *
 * sources:
 *    http://cgm.cs.mcgill.ca/~godfried/publications/banff.pdf
 *    http://ruinwesen.com/blog?id=216
 *
 * comment: i dislike the string solutions (as in the sources described), so i changed it to a more algorithmic way.
 *
 */

signature = "pattern,row,int,int";

documentation = "Generates the euclidean rhythm for a row based on 16 steps. (Nina Scholz)";
syntax = "euclid [pattern-name] [row: 1 - 16] [amount: 1 - 127] [pulses: 1 - 16]";
example = "euclid p6 2 127 13";

run() {

   euclid(int pulses, int steps) {
      int [] sequence = new int[pulses];
      int seqCount = 0;
      int repeats = ggt(pulses, steps);
      pulses /= repeats;
      steps /= repeats;
      int remainder = steps % pulses;
      int interval = (steps - remainder) / pulses;

      if (remainder <= 1) {
         while (seqCount < pulses - 1) {
            sequence[seqCount] = interval;
            seqCount++;
         }
         sequence[seqCount] = interval + remainder;
         seqCount++;

      } else {
         int aMax = remainder;
         int bMax = pulses - remainder;
         int maxGGT = ggt(aMax, bMax);
         aMax /= maxGGT;
         bMax /= maxGGT;
         int bFill = 0;
         int addLast = 0;

         if (aMax > bMax && aMax % bMax == 1) {
            addLast = 1;
            aMax--;
            maxGGT = ggt(aMax, bMax);
            aMax /= maxGGT;
            bMax /= maxGGT;
         }

         if (bMax > aMax) {
            bFill = bMax % aMax;
            bMax = (bMax - bFill) / aMax;
            if (bFill > 1) bMax++;
            aMax = 1;
         }

         boolean isACount = true;
         int count = aMax - 1;

         while (seqCount + addLast < pulses - bFill) {
            if (isACount) {
               if (count < aMax) {
                  sequence[seqCount] = interval + 1;
                  count++;
               } else {
                  sequence[seqCount] = interval;
                  count = 1;
                  isACount = false;
               }
            } else {
               if (count < bMax) {
                  sequence[seqCount] = interval;
                  count++;
               } else {
                  sequence[seqCount] = interval + 1;
                  count = 1;
                  isACount = true;
               }
            }
            seqCount++;
         }

         while (seqCount + addLast < pulses) {
            sequence[seqCount] = interval;
            seqCount++;
         }
      }
      if (addLast == 1) {
         sequence[seqCount] = interval + 1;
         seqCount++;
      }
      if (repeats > 1) {
         for (int i = 1; i < repeats; i++) {
            for (int j = 0; j < seqCount; j++) {
               sequence[i * seqCount + j] = sequence[j];
            }
         }
      }
      return sequence;
   }

   int ggt(int a1, int a2) {
      if (a1 < a2)
         return ggt(a2, a1);
      else if (a2 != 0 && a1 % a2 != 0)
         return ggt(a1 % a2, a2);
      else
         return a2;
   }


   pattern = arg1;
   row = arg2 - 1;
   amount = arg3;
   pulses = arg4;

   int [] sequence = euclid(pulses, 16);
   int seqOffset = 0;
   pattern.clear();

   for (int i = 0; i < pulses; i++) {
      pattern.set(seqOffset, row, amount);
      pattern.set(seqOffset + 16, row, amount);
      seqOffset += sequence[i];
   }
   pattern.draw();
}

 

[Orphu]

Hallo Jane Doe,

ist das Script von dir ? Weil Nina Scholz -> Jane Doe 8)

Finde den Seq ebenfalls sehr interessant und lese mich gerade ein in die Möglichkeiten.

Hast du ne Idee wie man mittels Scripte nen Arpeggio realisieren könnte?

Gruß,
Christian

 

[jane doe]

ja, es ist von mir.

alles was programmierbar ist, müßte gehen. ansonsten bleibt nur löten (kleiner scherz).

 

[Gomonoa]

Eine Frau die Musik produziert und auch noch programmiert. Gibts nicht allzu oft sowas, nicht schlecht