标签：问题 random nec 小数点 遗传算法 dispose rtti oid 多少
1.问题描述
针对如下问题，设计遗传算法进行求解。
2.Java源代码
GA.java
package GA;
import java.util.Random;
class GA {
public static final int varnum = 4;//变量的个数
public static final double[] lower = new double[varnum];
public static final double[] uper = new double[varnum];
public static final int POP_SIZE = 200; //种群数目
public static final double[][] initpop = new double[varnum][POP_SIZE];
public static final int M = 22; //每一个变量编码位数
public static String[] pop = new String[POP_SIZE];//种群编码
public static double[][] result = new double[varnum][POP_SIZE];//种群代表的结果
public static final int LENGTH=M * varnum;//编码长度，因为要精确到小数点后六位，所以编为22位长，22*i,i为变量个数
public static final int MJ2 = 4194304;//2^22
public static double[] fitness = new double[POP_SIZE];//存放种群适应度
public static final double PC = 0.99;//交叉率
public static final double PM = 0.2;//变异率
public static double[] p = new double[POP_SIZE];//轮盘赌方法个体适应度概率(按比例的适应度分配)
public static double[] q=new double[POP_SIZE];//q[i]是前n项p之和(累积概率)
public static Random random=new Random();//用于产生随机数的工具
public static Best best=new Best();//记录最佳答案的对象
public void encoding() //编码
{
for (int i = 0; i < POP_SIZE; i++) {
pop[i]="";
for(int j=0;j
double d1=((initpop[j][i]-lower[j])/(uper[j]-lower[j]))*(MJ2-1);
String GeneCode=Integer.toBinaryString((int)d1);
if(GeneCode.length()
int k=M-GeneCode.length();
for(int l=0;l
GeneCode="0"+GeneCode;
}
}
pop[i] += GeneCode; //将最终的编码存入pop[i]
}
}
}
public void decoding()//解码，将2进制编码转换为10进制
{
for (int i = 0; i < pop.length; i++) {
for(int j=0;j
int k = Integer.parseInt((pop[i].substring(j*22, (j+1)*22)), 2); //注意括号中的值!!!
if(j==1 || j==3){
result[j][i] = lower[j]+k*(uper[j]-lower[j])/(MJ2-1);
result[j][i] = (int)result[j][i];
//System.out.print("打印变量");
//System.out.print(result[j][i]);
}else{
result[j][i]=lower[j]+k*(uper[j]-lower[j])/(MJ2-1);
//System.out.print("打印变量");
//System.out.print(result[j][i]);
}
}
}
}
public void fitness()
{
for (int i = 0; i < POP_SIZE; i++) {
fitness[i] = 1000;
double a = 127 - 2*result[0][i]*result[0][i] - 3*result[1][i]*result[1][i]*result[1][i]*result[1][i] - result[2][i] - 4*result[3][i]*result[3][i];
if(a>=0){
fitness[i]= 100000 - ((result[0][i]-10)*(result[0][i]-10) + 5*(result[1][i]-12)*(result[1][i]-12) + result[2][i]*result[2][i]*result[2][i]*result[2][i] + 3*(result[3][i]-11)*(result[3][i]-11));
}
//System.out.print("打印适值" + i + " ");
//System.out.print(fitness[i]);
}
}
public void crossover(){//单点交叉
for(int i=0;i
double d=random.nextDouble();
if(d
int k1=random.nextInt(POP_SIZE);
int k2=random.nextInt(POP_SIZE);
int position=random.nextInt(LENGTH);
String s11="",s12="",s21="",s22="";
s11=pop[k1].substring(0,position);
s12=pop[k1].substring(position,LENGTH);
s21=pop[k2].substring(0,position);
s22=pop[k2].substring(position, LENGTH);
pop[k1]=s11+s22;
pop[k2]=s21+s12;
}
}
}
public void mutation() //变异
{
for (int i = 0; i < pop.length; i++) {
for (int j = 0; j < LENGTH; j++) {
double k=random.nextDouble();
if(PM>k)
{
mutation(i,j);
}
}
}
}
public void mutation(int i,int j) //变异
{
String s=pop[i];
StringBuffer sb=new StringBuffer(s);
if(sb.charAt(j)==‘0‘)
sb.setCharAt(j, ‘1‘);
else
sb.setCharAt(j, ‘0‘);
pop[i]=sb.toString();
}
public void roulettewheel()
{
decoding();
fitness();
double sum=0;
for (int i = 0; i
sum=fitness[i]+sum;
}
for (int i = 0; i < POP_SIZE; i++) {
p[i]=fitness[i]/sum;
q[i]=0;
}
for (int i = 0; i < POP_SIZE; i++) {
for (int j = 0; j <= i; j++) {
q[i]+=p[j];
}
}
double[] ran = new double[POP_SIZE];
String[] tempPop=new String[POP_SIZE];
for (int i = 0; i < ran.length; i++) {
ran[i]=random.nextDouble();
}
for (int i = 0; i < ran.length; i++) {
int k = 0;
for (int j = 0; j < q.length; j++) {
if(ran[i]
k=j;
break;
}
else continue;
}
tempPop[i]=pop[k];
}郑州人流手术多少钱 http://mobile.chnk120.com/
for (int i = 0; i < tempPop.length; i++) {
pop[i]=tempPop[i];
//System.out.print("输出种群!");
//System.out.print(pop[i] + " ");
}
}
public void evolution() //进化
{
encoding();
crossover();
mutation();
decoding();
fitness();
roulettewheel();
findResult();
}
public void dispose(int n) //对进化进行迭代
{
for (int i = 0; i < n; i++) {
evolution();
System.out.println("第" + i + "次迭代!");
}
}
public double findResult()
{
if(best == null) best=new Best();
double max = best.fitness;
for (int i = 0; i < fitness.length; i++) {
if(fitness[i] >= max)
{
best.fitness = fitness[i];
for(int m=0;m
best.x[m]=result[m][i];
}
best.str = pop[i];
}
}
return max;
}
public static void main(String[] args) {
//d为初试种群
lower[0] = 0;
uper[0] = 8.28;
lower[1] = -10;
uper[1] = 10;
lower[2] = -10;
uper[2] = 10;
lower[3] = -5;
uper[3] = 5;
//初始化种群
for(int i=0;i
for(int j=0;j
result[i][j]=lower[i]+random.nextDouble()*(uper[i]-lower[i]);
}
}
//初始化其它参数
GA ga = new GA();
//进化，这里进化10000次
long starttime=System.currentTimeMillis();
ga.dispose(10000);
long endtime=System.currentTimeMillis();
System.out.println("进化耗时："+(endtime-starttime)+"ms");
System.out.println("结果为：");
for(int i=0;i
System.out.print("x["+(i+1)+"]="+best.x[i]+"t");
}
System.out.println();
System.out.println("约束条件1的值："+(127 - 2*best.x[0]*best.x[0] - 3*best.x[1]*best.x[1]*best.x[1]*best.x[1] - best.x[2] - 4*best.x[3]*best.x[3]));
System.out.println("目标函数值：" + ((best.x[0]-10)*(best.x[0]-10) + 5*(best.x[1]-12)*(best.x[1]-12) + best.x[2]*best.x[2]*best.x[2]*best.x[2] + 3*(best.x[3]-11)*(best.x[3]-11)));
System.out.println("Function="+(100000 - best.fitness));
}
}
Best.java
package GA;
class Best { // 存储最佳的
public int generations;
public String str;
public double fitness;
public int varnum=5;
public double []x=new double[varnum];
}
3.运行结果
进化耗时：12938ms
结果为：
x[1]=2.7331778176254793t
x[2]=2.0t
x[3]=-6.508828761297991E-4t
x[4]=4.0t
约束条件1的值：0.06012891735616677
目标函数值：699.8067046302506
Function=699.8067046302458
标签：问题 random nec 小数点 遗传算法 dispose rtti oid 多少
原文地址：https://www.cnblogs.com/gnz49/p/12120131.html
