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Jul 30th, 2012
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  1. import std.stdio;
  2. import std.algorithm;
  3. import std.random;
  4. import std.array;
  5. import std.math;
  6.  
  7. struct city{
  8.     int x;
  9.     int y;
  10. }
  11.  
  12. struct chromosome{
  13.     city[] dna;
  14. }  
  15.  
  16. chromosome[2] crossOver(chromosome mother, chromosome father){
  17.     assert(mother.dna.length==father.dna.length);
  18.    
  19.     foreach(a;mother.dna)
  20.         assert(mother.dna.count(a)==father.dna.count(a));
  21.        
  22.     //we map 1 dna unit from the mother to a corresponding unit of the father (and vice versa)
  23.     // 4 1 2 3 5
  24.     // 5 3 1 2 4
  25.     //
  26.     // if we map 1 => 3
  27.     // 4 3 2 1 5
  28.     // 5 1 3 2 4
  29.        
  30.     int mapIndex=uniform(0,mother.dna.length);
  31.    
  32.     auto point1=mother.dna.dup[mapIndex];
  33.     auto point2=father.dna.dup[mapIndex];
  34.    
  35.     //we're mapping point1 to point1
  36.     swap(mother.dna[mother.dna.countUntil(point1)],mother.dna[mother.dna.countUntil(point2)]);
  37.     swap(father.dna[father.dna.countUntil(point1)],father.dna[father.dna.countUntil(point2)]);
  38.    
  39.     return [mother,father];
  40. }
  41.  
  42. chromosome mutate(chromosome victim){
  43.     int from=uniform(0,victim.dna.length);
  44.     int to=uniform(0,victim.dna.length);
  45.     swap(victim.dna[from],victim.dna[to]);
  46.     return victim;
  47. }
  48.  
  49. chromosome[] reproduce(chromosome mother,chromosome father){
  50.     //we reproduce
  51.     //mother's and father's dna are preserved because they contain a good solution
  52.     //we also add a mutation of mother/father
  53.     //a crossing over of mother/father
  54.     //and a crossing over + mutation
  55.     chromosome[] returnvalue;
  56.     returnvalue~=mother;
  57.     returnvalue~=father;
  58.    
  59.     returnvalue~=mutate(mother);
  60.     returnvalue~=mutate(father);
  61.    
  62.     chromosome[] crossedOver=crossOver(mother,father);
  63.     chromosome[] crossedOverAndMutated=array(map!(x => mutate(x))(crossedOver));
  64.    
  65.     returnvalue~=crossedOver;
  66.     returnvalue~=crossedOverAndMutated;
  67.    
  68.     return returnvalue;
  69. }
  70. void main(){
  71.     city[] cities=[city(20,0),city(50,50),city(75,30),city(0,10),city(25,25),city(10,65)];
  72.     chromosome[] workers;
  73.     while(workers.length<60){
  74.         randomShuffle(cities);
  75.         if(!canFind(workers,chromosome(cities)))
  76.             workers~=chromosome(cities.dup);
  77.     }
  78.    
  79.     bool fitnessCompare(chromosome first,chromosome second){
  80.         return fitness(first)>fitness(second);
  81.     }
  82.    
  83.     while(true){
  84.         workers=array(sort!(fitnessCompare)(workers));
  85.        
  86.         writeln(workers[0]);
  87.         stdout.flush();
  88.        
  89.         chromosome[] newWorkers;
  90.         for(int x=0;x<10;x+=2)
  91.             newWorkers~=reproduce(workers[x],workers[x+1]);
  92.        
  93.         workers=newWorkers;
  94.     }
  95.    
  96. }
  97.  
  98. float fitness(chromosome victim){
  99.     city[] cities=victim.dna;
  100.    
  101.     //we need to start from home and return to home
  102.     cities=city(0,0) ~ cities;
  103.     cities~=city(0,0);
  104.    
  105.     float travelled=0f;
  106.     for(int x=0;x<cities.length-1;x++)
  107.         travelled+=distance(cities[x],cities[x+1]);
  108.     writeln(100/travelled);
  109.     return 100/travelled;
  110. }
  111.  
  112. float distance(city from,city to){
  113.     return sqrt(cast(float)(pow(to.x-from.x,2) + pow(to.y-from.y,2)));
  114. }
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