Java: Mixing together two double bitstrings for Genetic Algorithm crossover -


i implementing evolutionary neural network. ran problem when comes crossover of 2 double values. evolving weights of links in neural network.

    //get weights want crossover     double weighta = a.getweight();     double weightb = b.getweight();     //round 6 decimal numbers.     weighta = (double)math.round(weighta * 1000000) / 1000000;     weightb = (double)math.round(weightb * 1000000) / 1000000;     //convert doubles binary strings     string binarya = long.tobinarystring(double.doubletorawlongbits(weighta));     string binaryb = long.tobinarystring(double.doubletorawlongbits(weightb));     //define random crossover point.     int crossoverpoint = randint(0, binarya.length());     //put strings based on crossover point.     string newbinary = binarya.substring(0,crossoverpoint) + binaryb.substring(crossoverpoint+1,binaryb.length());     double newweight = double.longbitstodouble(new biginteger(newbinary, 2).longvalue());

the problem encountering getting large or small weights after crossover result of how many bits used in each string decimal places. how should values after crossover similar 2 parents?

i had workaround problem gave me decent results sure not correct approach, finds average between 2 values , adds gaussian noise standard deviation based on interval of original 2 values.

    double interval = math.abs(weighta-weightb);     double newweight = (weighta+weightb)*0.5 + r.nextgaussian()*interval*2;

i'm not familiar genetic algorithms, know treatment of doubles doesn't seem way of approaching it:

i assume here want use first crossoverpoint bits of binary representation of first double , last (64-crossoverpoint) bits of second double (correct me if i'm wrong). if use strings you'll have make sure include leading 0s. simpler approach combine binary representations of longs using bit operations:

long weightalong = double.doubletorawlongbits(weighta); long weightblong = double.doubletorawlongbits(weightb); long mask = -1l; // bits set 1 int crossoverpoint = randint(0, long.size); long combined; // treat special cases because of modulo long.size of second parameter of shifting operations if (crossoverpoint == 0) {     combined = weightblong; } else if (combined == long.size) {     combined = weightalong; } else {     combined = (weightalong & (mask << (long.size - crossoverpoint))) |                (weightblong & (mask >>> crossoverpoint)); } double newweight = double.longbitstodouble(combined);

however binary representation of doubles guess combining binary representations way may not best way combine doubles:

  • if first bits different, right choice of crossoverpoint (1) can change sign.
  • the exponent comes weighta in (52 / 64) of cases.
  • nan, positive_infinity, , negative_infinity can produced values different of these 3 if unlucky combination in mantissa.

i guess workaround seems better choice. (maybe should ask question on https://cs.stackexchange.com/)


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