我有一个实现模拟退火的程序。我对接受概率有疑问,可能是因为我不明白为什么将欧拉数提高到(能量 - 能量)的幂是有用的。
即使温度非常低,概率也始终超过 1.0 (100%),这使得这有效地成为随机搜索。如何将我的接受概率固定为 sA 的正常率(开始时接受较差解决方案的可能性很高,接近尾声的可能性很低)?
下面是方法代码:
if (mutatedSolutionFitness > originalSolutionFitness) {
return 1.0;
} else {
System.out.println("Original solution fitness: "+originalSolutionFitness);
System.out.println("Mutated solution fitness: "+mutatedSolutionFitness);
System.out.println("Temperature: "+this.temperature);
final double chance = Math.exp((originalSolutionFitness - mutatedSolutionFitness) / this.temperature);
System.out.println("Math.exp((originalSolutionFitness - mutatedSolutionFitness) / this.temperature): "+chance);
System.out.println();
return chance;
}
这是几次输出:
Original solution fitness: 0.6666666666666666
Mutated solution fitness: 0.5555555555555556
Temperature: 999998.000001
Math.exp((originalSolutionFitness - mutatedSolutionFitness) / this.temperature): 1.0000001111113395
Original solution fitness: 0.6666666666666666
Mutated solution fitness: 0.6666666666666666
Temperature: 999997.000003
Math.exp((originalSolutionFitness - mutatedSolutionFitness) / this.temperature): 1.0
Original solution fitness: 0.6666666666666666
Mutated solution fitness: 0.6666666666666666
Temperature: 999996.000006
Math.exp((originalSolutionFitness - mutatedSolutionFitness) / this.temperature): 1.0
Original solution fitness: 0.6666666666666666
Mutated solution fitness: 0.5555555555555556
Temperature: 999995.00001
Math.exp((originalSolutionFitness - mutatedSolutionFitness) / this.temperature): 1.0000001111116728
Original solution fitness: 0.6666666666666666
Mutated solution fitness: 0.4444444444444444
Temperature: 999994.0000149999
Math.exp((originalSolutionFitness - mutatedSolutionFitness) / this.temperature): 1.0000002222235802
Original solution fitness: 0.6666666666666666
Mutated solution fitness: 0.5555555555555556
Temperature: 999993.0000209998
Math.exp((originalSolutionFitness - mutatedSolutionFitness) / this.temperature): 1.000000111111895