# waysun一路阳光

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http://hi.baidu.com/litertiger/blog/item/3798546625f86224aa184c30.html

-3.0<x1<12.1

4.1<x2<5.8

1%的变异

25%交叉

/**
* 实现Michalewicz
*
* @author not attributable
* @version 1.0
*/

public class JGA {

bestindival bd = null;

String[] ipop = new String[10];

int gernation = 0;

public JGA() {
this.ipop = inialPops();
}

double calculatefitnessvalue(String str) { // str为染色体，前面18个为x1表示部分，后面15个为x2表示部分
String str1 = str.substring(0, 18);
// System.out.println(str1);
String str2 = str.substring(18);
// System.out.println(str2);
int b1 = Integer.parseInt(str1, 2);
// System.out.println(b1);
int b2 = Integer.parseInt(str2, 2);
// System.out.println(b2);
double x1 = -3.0 + b1 * (12.1 - (-3.0)) / (Math.pow(2, 18) - 1);
// System.out.println(x1);
double x2 = 4.1 + b2 * (5.8 - 4.1) / (Math.pow(2, 15) - 1);
// System.out.println(x2);
double fitness = 21.5 + x1 * Math.sin(4 * 3.1415926 * x1) + x2
* Math.sin(20 * 3.1415926 * x2);
//System.out.println("eval=f(" + x1 + "," + x2 + ")=" + fitness);
return fitness;
}

String inialPop() { // 初始化10个字符串
String res = "";
for (int i = 0; i < 33; i++) {
if (Math.random() > 0.5) {
res += "0";
} else {
res += "1";
}

}
return res;
}

String[] inialPops() {
String[] ipop = new String[10];
for (int i = 0; i < 10; i++) {
ipop[i] = inialPop();
}
return ipop;

}

void select() {
double evals[] = new double[10];// 所有染色体适应值
double p[] = new double[10];// 各染色体选择概率
double q[] = new double[10];// 累计概率
double F = 0;
for (int i = 0; i < 10; i++) {
evals[i] = calculatefitnessvalue(ipop[i]);
if (bd == null) {
bd = new bestindival();
bd.fitness = evals[i];
bd.generations = 0;
bd.str = ipop[i];
} else {
if (evals[i] > bd.fitness)// 最好的记录下来
{
bd.fitness = evals[i];
bd.generations = gernation;
bd.str = ipop[i];
}

}
F = F + evals[i];// 所有染色体适应值总和

}
for (int i = 0; i < 10; i++) {
p[i] = evals[i] / F;
if (i == 0)
q[i] = p[i];
else {
q[i] = q[i - 1] + p[i];
}
}
for (int i = 0; i < 10; i++) {

double r = Math.random();
if (r <= q[0]) {
ipop[i] = ipop[0];

} else {
for (int j = 1; j < 10; j++) {
if (r < q[j]) {
ipop[i] = ipop[j];
break;
}
}
}
}

}

void cross() { // 交叉率为25%，平均为25%的染色体进行交叉
String temp1, temp2;
for (int i = 0; i < 10; i++) {
if (Math.random() < 0.25) {
double r = Math.random();
int pos = (int) (Math.round(r * 1000)) % 33;
if (pos == 0) {
pos = 1;
}
temp1 = ipop[i].substring(0, pos)
+ ipop[(i + 1) % 10].substring(pos);
temp2 = ipop[(i + 1) % 10].substring(0, pos)
+ ipop[i].substring(pos);
ipop[i] = temp1;
ipop[(i + 1) / 10] = temp2;

}

}
}

void mutation() {
// 1%基因变异m*pop_size 共330个基因，为了使每个基因都相投机会发生变异，需要产生[1--330]上均匀分布的
for (int i = 0; i < 4; i++) {
int num = (int) (Math.random() * 330 + 1);
int chromosomeNum = (int) (num / 33) + 1; // 染色体号
int mutationNum = num - (chromosomeNum - 1) * 33; // 基因号
if (mutationNum == 0)
mutationNum = 1;
//System.out.println(num + "," + chromosomeNum + "," + mutationNum);
String temp;
if (ipop[chromosomeNum].charAt(mutationNum - 1) == '0') {
if (mutationNum == 1) {
} else {
if (mutationNum != 33) {
.substring(0, mutationNum - 1)
+ "1"
} else {
.substring(0, mutationNum - 1)
+ "1";
}
}
} else {
if (mutationNum == 1) {
} else {
if (mutationNum != 33) {
.substring(0, mutationNum - 1)
+ "0"
} else {
.substring(0, mutationNum - 1)
+ "1";
}
}

}

}

}

void process()
{
for(int i=0;i<1000000;i++)
{
select();
cross();
mutation();
gernation=i;

}
System.out.println("最优值"+bd.fitness+",代数"+bd.generations);
}

public static void main(String args[]) {
JGA j = new JGA();
// System.out.println(j.calculatefitnessvalue("000001010100101001101111011111110"));
j.process();

}
}

class bestindival { // 存储最佳的
public int generations;

public String str;

public double fitness;

}

litertiger
2006-12-09
posted on 2009-04-15 23:19 weesun一米阳光 阅读(239) 评论(0)  编辑  收藏 所属分类: JAVA源码总结备用

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