Jenetics: how to interpret and get the result of a chromosome?

Question

I practice using this library, Jenetics.io. In their manual they have an example, Traveling salesman. The example only shows the minimum path found, but does not show the point set or point order of the solution. How do I get the set of points of the solution?

the manual.. page 129 enter link description here

to run the code, the result:

[[D@79fc0f2f|[D@37a71e93|[D@4783da3f|[D@6e2c634b|[D@49097b5d|[D@378fd1ac|[D@5e9f23b4|[D@2dda6444|[D@50040f0c] -> 16.46284073991415

the code that I use:

public class EncotrarRuta implements Problem<ISeq<double[]>,EnumGene<double[]>,Double>{

private final ISeq<double[]> _points;
private ArrayList<Double> puntosX = new ArrayList<>();
private ArrayList<Double> puntosY = new ArrayList<>();


public EncotrarRuta(ArrayList<Double> puntosX, ArrayList<Double> puntosY) {
    this._points = null;
    this.puntosX = puntosX;
    this.puntosY = puntosY;
}


public EncotrarRuta(ISeq<double[]> points) {
    this._points = requireNonNull(points);
}

@Override
public Function<ISeq<double[]>, Double> fitness() {
    return p -> IntStream.range(0, p.length())
        .mapToDouble(i -> {
            final double[] p1 = p.get(i);
            final double[] p2 = p.get((i+1)%p.size());
            return hypot(p1[0]-p2[0],p1[1]-p2[1]);
        }).sum();
}

@Override
public Codec<ISeq<double[]>, EnumGene<double[]>> codec() {
    return Codecs.ofPermutation(_points);
}

public static EncotrarRuta of(ArrayList<Double> puntosX, ArrayList<Double> puntosY){
    final MSeq<double[]> points = MSeq.ofLength(puntosX.size());
    
    for (int i = 0; i < puntosX.size(); i++) {
        points.set(i, new double[]{puntosX.get(i),puntosY.get(i)});
    }

    return new EncotrarRuta(points.asISeq());
}

public void run() {

    EncotrarRuta encotrarRuta = EncotrarRuta.of(puntosX, puntosY);

    Engine<EnumGene<double[]>,Double> engine = Engine
        .builder(encotrarRuta)
        .optimize(Optimize.MINIMUM)
        .maximalPhenotypeAge(11)
        .populationSize(500)
        .alterers(
            new SwapMutator<>(0.2),
            new PartiallyMatchedCrossover<>(0.35))
        .build();

    EvolutionStatistics<Double, ?> statistics = EvolutionStatistics.ofNumber();

    Phenotype<EnumGene<double[]>,Double> best = engine.stream()
        .limit(bySteadyFitness(25))
        .limit(250)
        .peek(statistics)
        .collect(toBestPhenotype());


    System.out.println(best);
   

}

}

Answer 1

I found the solution, or what I wanted to obtain,

best.genotype().chromosome().iterator().forEachRemaining(i -> {
        for (double ds : i.allele()) {
            System.out.println(ds);
        }
    });

with that I get the list of points of the solution

[[D@7cd84586|[D@30dae81|[D@1b2c6ec2|[D@4769b07b|[D@4edde6e5|[D@66a29884|[D@1e80bfe8|[D@70177ecd] -> 12.65685424949238
1.0
1.0
2.0
2.0
3.0
3.0
3.0
1.0
4.0
1.0
6.0
1.0
6.0
3.0
5.0
4.0