本文目录一览:
- 1、《Java遗传算法编程》pdf下载在线阅读全文,求百度网盘云资源
- 2、如何用Java实现遗传算法?
- 3、用java编程遗传算法怎样记录每一代的值呢?
- 4、使用java来实现在智能组卷中的遗传算法(急急急)
《Java遗传算法编程》pdf下载在线阅读全文,求百度网盘云资源
《Java遗传算法编程》百度网盘pdf最新全集下载:
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简介:本书简单、直接地介绍了遗传算法,并且针对所讨论的示例问题,给出了Java代码的算法实现。全书分为6章。第1章简单介绍了人工智能和生物进化的知识背景,这也是遗传算法的历史知识背景。第2章给出了一个基本遗传算法的实现;第4章和第5章,分别针对机器人控制器、旅行商问题、排课问题展开分析和讨论,并给出了算法实现。在这些章的末尾,还给出了一些练习供读者深入学习和实践。第6章专门讨论了各种算法的优化问题。
如何用Java实现遗传算法?
通过遗传算法走迷宫。虽然图1和图2均成功走出迷宫,但是图1比图2的路径长的多,且复杂,遗传算法可以计算出有多少种可能性,并选择其中最简洁的作为运算结果。
示例图1:
示例图2:
实现代码:
import java.util.ArrayList;
import java.util.Collections;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Random;
/**
* 用遗传算法走迷宫
*
* @author Orisun
*
*/
public class GA {
int gene_len; // 基因长度
int chrom_len; // 染色体长度
int population; // 种群大小
double cross_ratio; // 交叉率
double muta_ratio; // 变异率
int iter_limit; // 最多进化的代数
Listboolean[] individuals; // 存储当代种群的染色体
Labyrinth labyrinth;
int width; //迷宫一行有多少个格子
int height; //迷宫有多少行
public class BI {
double fitness;
boolean[] indv;
public BI(double f, boolean[] ind) {
fitness = f;
indv = ind;
}
public double getFitness() {
return fitness;
}
public boolean[] getIndv() {
return indv;
}
}
ListBI best_individual; // 存储每一代中最优秀的个体
public GA(Labyrinth labyrinth) {
this.labyrinth=labyrinth;
this.width = labyrinth.map[0].length;
this.height = labyrinth.map.length;
chrom_len = 4 * (width+height);
gene_len = 2;
population = 20;
cross_ratio = 0.83;
muta_ratio = 0.002;
iter_limit = 300;
individuals = new ArrayListboolean[](population);
best_individual = new ArrayListBI(iter_limit);
}
public int getWidth() {
return width;
}
public void setWidth(int width) {
this.width = width;
}
public double getCross_ratio() {
return cross_ratio;
}
public ListBI getBest_individual() {
return best_individual;
}
public Labyrinth getLabyrinth() {
return labyrinth;
}
public void setLabyrinth(Labyrinth labyrinth) {
this.labyrinth = labyrinth;
}
public void setChrom_len(int chrom_len) {
this.chrom_len = chrom_len;
}
public void setPopulation(int population) {
this.population = population;
}
public void setCross_ratio(double cross_ratio) {
this.cross_ratio = cross_ratio;
}
public void setMuta_ratio(double muta_ratio) {
this.muta_ratio = muta_ratio;
}
public void setIter_limit(int iter_limit) {
this.iter_limit = iter_limit;
}
// 初始化种群
public void initPopulation() {
Random r = new Random(System.currentTimeMillis());
for (int i = 0; i population; i++) {
int len = gene_len * chrom_len;
boolean[] ind = new boolean[len];
for (int j = 0; j len; j++)
ind[j] = r.nextBoolean();
individuals.add(ind);
}
}
// 交叉
public void cross(boolean[] arr1, boolean[] arr2) {
Random r = new Random(System.currentTimeMillis());
int length = arr1.length;
int slice = 0;
do {
slice = r.nextInt(length);
} while (slice == 0);
if (slice length / 2) {
for (int i = 0; i slice; i++) {
boolean tmp = arr1[i];
arr1[i] = arr2[i];
arr2[i] = tmp;
}
} else {
for (int i = slice; i length; i++) {
boolean tmp = arr1[i];
arr1[i] = arr2[i];
arr2[i] = tmp;
}
}
}
// 变异
public void mutation(boolean[] individual) {
int length = individual.length;
Random r = new Random(System.currentTimeMillis());
individual[r.nextInt(length)] ^= false;
}
// 轮盘法选择下一代,并返回当代最高的适应度值
public double selection() {
boolean[][] next_generation = new boolean[population][]; // 下一代
int length = gene_len * chrom_len;
for (int i = 0; i population; i++)
next_generation[i] = new boolean[length];
double[] cumulation = new double[population];
int best_index = 0;
double max_fitness = getFitness(individuals.get(best_index));
cumulation[0] = max_fitness;
for (int i = 1; i population; i++) {
double fit = getFitness(individuals.get(i));
cumulation[i] = cumulation[i - 1] + fit;
// 寻找当代的最优个体
if (fit max_fitness) {
best_index = i;
max_fitness = fit;
}
}
Random rand = new Random(System.currentTimeMillis());
for (int i = 0; i population; i++)
next_generation[i] = individuals.get(findByHalf(cumulation,
rand.nextDouble() * cumulation[population - 1]));
// 把当代的最优个体及其适应度放到best_individual中
BI bi = new BI(max_fitness, individuals.get(best_index));
// printPath(individuals.get(best_index));
//System.out.println(max_fitness);
best_individual.add(bi);
// 新一代作为当前代
for (int i = 0; i population; i++)
individuals.set(i, next_generation[i]);
return max_fitness;
}
// 折半查找
public int findByHalf(double[] arr, double find) {
if (find 0 || find == 0 || find arr[arr.length - 1])
return -1;
int min = 0;
int max = arr.length - 1;
int medium = min;
do {
if (medium == (min + max) / 2)
break;
medium = (min + max) / 2;
if (arr[medium] find)
min = medium;
else if (arr[medium] find)
max = medium;
else
return medium;
} while (min max);
return max;
}
// 计算适应度
public double getFitness(boolean[] individual) {
int length = individual.length;
// 记录当前的位置,入口点是(1,0)
int x = 1;
int y = 0;
// 根据染色体中基因的指导向前走
for (int i = 0; i length; i++) {
boolean b1 = individual[i];
boolean b2 = individual[++i];
// 00向左走
if (b1 == false b2 == false) {
if (x 0 labyrinth.map[y][x - 1] == true) {
x--;
}
}
// 01向右走
else if (b1 == false b2 == true) {
if (x + 1 width labyrinth.map[y][x + 1] == true) {
x++;
}
}
// 10向上走
else if (b1 == true b2 == false) {
if (y 0 labyrinth.map[y - 1][x] == true) {
y--;
}
}
// 11向下走
else if (b1 == true b2 == true) {
if (y + 1 height labyrinth.map[y + 1][x] == true) {
y++;
}
}
}
int n = Math.abs(x - labyrinth.x_end) + Math.abs(y -labyrinth.y_end) + 1;
// if(n==1)
// printPath(individual);
return 1.0 / n;
}
// 运行遗传算法
public boolean run() {
// 初始化种群
initPopulation();
Random rand = new Random(System.currentTimeMillis());
boolean success = false;
while (iter_limit-- 0) {
// 打乱种群的顺序
Collections.shuffle(individuals);
for (int i = 0; i population - 1; i += 2) {
// 交叉
if (rand.nextDouble() cross_ratio) {
cross(individuals.get(i), individuals.get(i + 1));
}
// 变异
if (rand.nextDouble() muta_ratio) {
mutation(individuals.get(i));
}
}
// 种群更替
if (selection() == 1) {
success = true;
break;
}
}
return success;
}
// public static void main(String[] args) {
// GA ga = new GA(8, 8);
// if (!ga.run()) {
// System.out.println("没有找到走出迷宫的路径.");
// } else {
// int gen = ga.best_individual.size();
// boolean[] individual = ga.best_individual.get(gen - 1).indv;
// System.out.println(ga.getPath(individual));
// }
// }
// 根据染色体打印走法
public String getPath(boolean[] individual) {
int length = individual.length;
int x = 1;
int y = 0;
LinkedListString stack=new LinkedListString();
for (int i = 0; i length; i++) {
boolean b1 = individual[i];
boolean b2 = individual[++i];
if (b1 == false b2 == false) {
if (x 0 labyrinth.map[y][x - 1] == true) {
x--;
if(!stack.isEmpty() stack.peek()=="右")
stack.poll();
else
stack.push("左");
}
} else if (b1 == false b2 == true) {
if (x + 1 width labyrinth.map[y][x + 1] == true) {
x++;
if(!stack.isEmpty() stack.peek()=="左")
stack.poll();
else
stack.push("右");
}
} else if (b1 == true b2 == false) {
if (y 0 labyrinth.map[y - 1][x] == true) {
y--;
if(!stack.isEmpty() stack.peek()=="下")
stack.poll();
else
stack.push("上");
}
} else if (b1 == true b2 == true) {
if (y + 1 height labyrinth.map[y + 1][x] == true) {
y++;
if(!stack.isEmpty() stack.peek()=="上")
stack.poll();
else
stack.push("下");
}
}
}
StringBuilder sb=new StringBuilder(length/4);
IteratorString iter=stack.descendingIterator();
while(iter.hasNext())
sb.append(iter.next());
return sb.toString();
}
}
用java编程遗传算法怎样记录每一代的值呢?
在实例化一个数组
没循环一次往数组里添加一个值
这样就可以了
使用java来实现在智能组卷中的遗传算法(急急急)
题目好像是让你做个增强版的List ,简单的都实现了 程序架子大概是这样,排序查找什么的百度搜下 算法很多,套着每样写个方法就行了,测试就在main‘方法里写
public class MyList {
private String[] arr;
private int count ;
public MyList (int count){
arr = new String[count];
this.count = count;
}
public MyList (int[] intArr){
arr = new String[intArr.length];
this.count = intArr.length;
for(int i=0;iintArr.length;i++){
arr[i] = intArr[i]+"";
}
}
public MyList (String[] stringArr){
arr = stringArr;
this.count = stringArr.length;
}
public int getLength(){
return count;
}
//清空容器内的数组。
public void clearAll(){
arr = new String[count];
}
//通过给定元素下标来删除某一元素
public void removeBySeqn(int seqn){
if(seqn = 0 seqncount){
arr[seqn] = null;
}
}
public static void main(String[] args){
MyList list = new MyList (40);
MyList list1 = new MyList ({3,2,125,56,123});
MyList list2 = new MyList ({"123",""ad});
list2.removeBySeqn(0);
list1.clearAll();
}
}