我应该实现一个具有2个输入、2个隐藏和1个输出神经元的人工神经网络(ANN),它可以解决XOR问题。应该使用进化算法来优化网络的权重。给出了每个神经元的激活函数和每个神经网络的适应度函数。下图总结了这个问题,并介绍了我使用的变量名:
现在我尽了最大的努力来解决这个问题,但即使使用1000个ANN和2000代的进化算法,我的最佳适应度也永远不会超过0.75。我的代码包括一个带有神经元、激活和适应度函数的ANN类,以及一个包含进化算法并优化ANN权重的主类。代码如下:
每个ANN用-1和1之间的随机权重进行初始化,并且能够变异,即返回一个随机选择的权重不同的变异。
public class ANN implements Comparable<ANN> {
private Random rand = new Random();
public double[] w = new double[6]; //weights: in1->h1, in1->h2, in2->h1, in2->h2, h1->out, h2->out
public ANN() {
for (int i=0; i<6; i++) //randomly initialize weights in [-1,1)
w[i] = rand.nextDouble() * 2 - 1;
}
//calculates the output for input a & b
public double ann(double a, double b) {
double h1 = activationFunc(a*w[0] + b*w[2]);
double h2 = activationFunc(a*w[1] + b*w[3]);
double out = activationFunc(h1*w[4] + h2*w[5]);
return out;
}
private double activationFunc(double x) {
return 2.0 / (1 + Math.exp(-2*x)) - 1;
}
//calculates the fitness (divergence to the right output)
public double fitness() {
double sum = 0;
//test all possible inputs (0,0; 0,1; 1,0; 1,1)
sum += 1 - Math.abs(0 - ann(0, 0));
sum += 1 - Math.abs(1 - ann(0, 1));
sum += 1 - Math.abs(1 - ann(1, 0));
sum += 1 - Math.abs(0 - ann(1, 1));
return sum / 4.0;
}
//randomly change random weight and return the mutated ANN
public ANN mutate() {
//copy weights
ANN mutation = new ANN();
for (int i=0; i<6; i++)
mutation.w[i] = w[i];
//randomly change one
int weight = rand.nextInt(6);
mutation.w[weight] = rand.nextDouble() * 2 - 1;
return mutation;
}
@Override
public int compareTo(ANN arg) {
if (this.fitness() < arg.fitness())
return -1;
if (this.fitness() == arg.fitness())
return 0;
return 1; //this.fitness > arg.fitness
}
@Override
public boolean equals(Object obj) {
if (obj == null)
return false;
ANN ann = (ANN)obj;
for (int i=0; i<w.length; i++) { //not equal if any weight is different
if (w[i] != ann.w[i])
return false;
}
return true;
}
}
主类具有进化算法,并使用精英主义和基于排名的选择来创建每个种群的下一代,即复制100个最好的人工神经网络,其余900个是先前成功的人工神经网络的突变。
//rank-based selection + elitism
public class Main {
static Random rand = new Random();
static int size = 1000; //population size
static int elitists = 100; //number of elitists
public static void main(String[] args) {
int generation = 0;
ArrayList<ANN> population = initPopulation();
print(population, generation);
//stop after good fitness is reached or after 2000 generations
while(bestFitness(population) < 0.8 && generation < 2000) {
generation++;
population = nextGeneration(population);
print(population, generation);
}
}
public static ArrayList<ANN> initPopulation() {
ArrayList<ANN> population = new ArrayList<ANN>();
for (int i=0; i<size; i++) {
ANN ann = new ANN();
if (!population.contains(ann)) //no duplicates
population.add(ann);
}
return population;
}
public static ArrayList<ANN> nextGeneration(ArrayList<ANN> current) {
ArrayList<ANN> next = new ArrayList<ANN>();
Collections.sort(current, Collections.reverseOrder()); //sort according to fitness (0=best, 999=worst)
//copy elitists
for (int i=0; i<elitists; i++) {
next.add(current.get(i));
}
//rank-based roulette wheel
while (next.size() < size) { //keep same population size
double total = 0;
for (int i=0; i<size; i++)
total += 1.0 / (i + 1.0); //fitness = 1/(rank+1)
double r = rand.nextDouble() * total;
double cap = 0;
for (int i=0; i<size; i++) {
cap += 1.0 / (i + 1.0); //higher rank => higher probability
if (r < cap) { //select for mutation
ANN mutation = current.get(i).mutate(); //no duplicates
if (!next.contains(mutation))
next.add(mutation);
break;
}
}
}
return next;
}
//returns best ANN in the specified population
public static ANN best(ArrayList<ANN> population) {
Collections.sort(population, Collections.reverseOrder());
return population.get(0);
}
//returns the best fitness of the specified population
public static double bestFitness(ArrayList<ANN> population) {
return best(population).fitness();
}
//returns the average fitness of the specified population
public static double averageFitness(ArrayList<ANN> population) {
double totalFitness = 0;
for (int i=0; i<size; i++)
totalFitness += population.get(i).fitness();
double average = totalFitness / size;
return average;
}
//print population best and average fitness
public static void print(ArrayList<ANN> population, int generation) {
System.out.println("Generation: " + generation + "\nBest: " + bestFitness(population) + ", average: " + averageFitness(population));
System.out.print("Best weights: ");
ANN best = best(population);
for (int i=0; i<best.w.length; i++)
System.out.print(best.w[i] + " ");
System.out.println();
System.out.println();
}
}
尽管我在这方面花了相当多的心思,并使用了我学到的技术,但结果并不令人满意。由于某些原因,对于每个权重,最优权重似乎会漂移到-1。这怎么说得通呢?权重的范围从-1到1是否是一个好的选择?除了突变之外,我还应该引入交叉吗?我知道这是一个非常具体的问题,但我会非常感谢一些帮助!
发布于 2015-06-04 16:19:02
网络结构不正确。如果没有每个节点的偏差或阈值,这个网络就不能解决XOR问题。
一个隐藏节点应该对OR进行编码,另一个隐藏节点应该对AND进行编码。然后,对于XOR问题,输出节点可以编码OR隐藏节点为正,而and隐藏节点为负。只有当OR隐藏节点被激活而and隐藏节点未被激活时,这才会产生积极的结果。
我还会增加权重的边界,让EA自己找出来。但这取决于网络结构,如果有必要的话。
如果要将此网络与隐藏节点和阈值一起使用,请参阅:http://www.heatonresearch.com/online/introduction-neural-networks-java-edition-2/chapter-1/page4.html
如果您想使用另一个带有偏差的网络,请参阅:http://www.mind.ilstu.edu/curriculum/artificial_neural_net/xor_problem_and_solution.php
https://stackoverflow.com/questions/27379864
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