I believe Processing's core strength is its pixel [] array. As a result I am always trying to find cool ways to use noise functions to manipulate images. Here is Ken Perlins algorithm formatted to use floats instead of doubles

1.  static int p[] = new int[512], permutation[] = { 151,160,137,91,90,15,
     131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23,
      190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33,
      88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166,
      77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244,
      102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196,
      135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123,
      5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42,
      223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9,
      129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228,
      251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107,
      49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254,
      138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180
      };
   static public float impnoise(float x, float y, float z) {
         int X = (int)Math.floor(x) & 255,                  // FIND UNIT CUBE THAT
             Y = (int)Math.floor(y) & 255,                  // CONTAINS POINT.
             Z = (int)Math.floor(z) & 255;
         x -= Math.floor(x);                                // FIND RELATIVE X,Y,Z
         y -= Math.floor(y);                                // OF POINT IN CUBE.
         z -= Math.floor(z);
         float u = fade(x),                                // COMPUTE FADE CURVES
                v = fade(y),                                // FOR EACH OF X,Y,Z.
                w = fade(z);
         int A = p[X  ]+Y;
         int AA = p[A]+Z;
      int   AB = p[A+1]+Z;
         // HASH COORDINATES OF
           int  B = p[X+1]+Y;
           int  BA = p[B]+Z;
           int  BB = p[B+1]+Z;      // THE 8 CUBE CORNERS,
   
         return lerpu(w, lerpu(v, lerpu(u, grad(p[AA  ], x  , y  , z   ),  // AND ADD
                                        grad(p[BA  ], x-1, y  , z   )), // BLENDED
                                lerpu(u, grad(p[AB  ], x  , y-1, z   ),  // RESULTS
                                        grad(p[BB  ], x-1, y-1, z   ))),// FROM  8
                        lerpu(v, lerpu(u, grad(p[AA+1], x  , y  , z-1 ),  // CORNERS
                                        grad(p[BA+1], x-1, y  , z-1 )), // OF CUBE
                                lerpu(u, grad(p[AB+1], x  , y-1, z-1 ),
                                        grad(p[BB+1], x-1, y-1, z-1 ))));
      }
      static float fade(float t) { return t * t * t * (t * (t * 6 - 15) + 10); }
      static float lerpu(float t, float a, float b) { return a + t * (b - a); }
      static float grad(int hash, float x, float y, float z) {
         int h = hash & 15;                      // CONVERT LO 4 BITS OF HASH CODE
         float u = h<8 ? x : y,                 // INTO 12 GRADIENT DIRECTIONS.
                v = h<4 ? y : h==12||h==14 ? x : z;
         return ((h&1) == 0 ? u : -u) + ((h&2) == 0 ? v : -v);
      }