各位大神大哥小弟求帮助关于遗传算法代码

朱威斯 · 发表于 2023-3-26 19:12:05

如题在做毕设目前遇到困难

深度大脑 · 发表于 2023-3-26 19:12:49

function main()
clear;
clc;
%种群大小
popsize=100;
%二进制编码长度
chromlength=10;
%交叉概率
pc = 0.6;
%变异概率
pm = 0.001;
%初始种群
pop = initpop(popsize,chromlength);

for i = 1:100
%计算适应度值（函数值）
objvalue = cal_objvalue(pop);
fitvalue = objvalue;
%选择操作
newpop = selection(pop,fitvalue);
%交叉操作
newpop = crossover(newpop,pc);
%变异操作
newpop = mutation(newpop,pm);
%更新种群
pop = newpop;
%寻找最优解
[bestindividual,bestfit] = best(pop,fitvalue);
x2 = binary2decimal(bestindividual);
x1 = binary2decimal(newpop);
y1 = cal_objvalue(newpop);
if mod(i,10) == 0
      figure;
      fplot('10*sin(5*x)+7*abs(x-5)+10',[0 10]);
      hold on;
      plot(x1,y1,'*');
      title(['迭代次数为n=' num2str(i)]);
      %plot(x1,y1,'*');
end
end
fprintf('The best X is --->>%5.2f\n',x2);
fprintf('The best Y is --->>%5.2f\n',bestfit);
%初始化种群大小
%输入变量：
%popsize：种群大小
%chromlength：染色体长度-->>转化的二进制长度
%输出变量：
%pop：种群
function pop=initpop(popsize,chromlength)
pop = round(rand(popsize,chromlength));
%rand(3,4)生成3行4列的0-1之间的随机数
% rand(3,4)
%
% ans =
%
%    0.8147 0.9134 0.2785 0.9649
%    0.9058 0.6324 0.5469 0.1576
%    0.1270 0.0975 0.9575 0.9706
%round就是四舍五入
% round(rand(3,4))=
% 1 1 0 1
% 1 1 1 0
% 0 0 1 1
%所以返回的种群就是每行是一个个体，列数是染色体长度
%二进制转化成十进制函数
%输入变量：
%二进制种群
%输出变量
%十进制数值
function pop2 = binary2decimal(pop)
[px,py]=size(pop);
for i = 1:py
pop1(:,i) = 2.^(py-i).*pop(:,i);
end
%sum(.,2)对行求和，得到列向量
temp = sum(pop1,2);
pop2 = temp*10/1023;
%计算函数目标值
%输入变量：二进制数值
%输出变量：目标函数值
function [objvalue] = cal_objvalue(pop)
x = binary2decimal(pop);
%转化二进制数为x变量的变化域范围的数值
objvalue=10*sin(5*x)+7*abs(x-5)+10;
%如何选择新的个体
%输入变量：pop二进制种群，fitvalue：适应度值
%输出变量：newpop选择以后的二进制种群
function [newpop] = selection(pop,fitvalue)
%构造轮盘
[px,py] = size(pop);
totalfit = sum(fitvalue);
p_fitvalue = fitvalue/totalfit;
p_fitvalue = cumsum(p_fitvalue);%概率求和排序
ms = sort(rand(px,1));%从小到大排列
fitin = 1;
newin = 1;
while newin<=px
if(ms(newin))<p_fitvalue(fitin)
      newpop(newin,:)=pop(fitin,:);
      newin = newin+1;
else
      fitin=fitin+1;
end
end
%交叉变换
%输入变量：pop：二进制的父代种群数，pc：交叉的概率
%输出变量：newpop：交叉后的种群数
function [newpop] = crossover(pop,pc)
[px,py] = size(pop);
newpop = ones(size(pop));
for i = 1:2:px-1
if(rand<pc)
      cpoint = round(rand*py);
      newpop(i,:) = [pop(i,1:cpoint),pop(i+1,cpoint+1:py)];
      newpop(i+1,:) = [pop(i+1,1:cpoint),pop(i,cpoint+1:py)];
else
      newpop(i,:) = pop(i,:);
      newpop(i+1,:) = pop(i+1,:);
end
end
%关于编译
%函数说明
%输入变量：pop：二进制种群，pm：变异概率
%输出变量：newpop变异以后的种群
function [newpop] = mutation(pop,pm)
[px,py] = size(pop);
newpop = ones(size(pop));
for i = 1:px
if(rand<pm)
      mpoint = round(rand*py);
      if mpoint <= 0;
         mpoint = 1;
      end
      newpop(i,:) = pop(i,:);
      if newpop(i,mpoint) == 0
         newpop(i,mpoint) = 1;
      else newpop(i,mpoint) == 1
         newpop(i,mpoint) = 0;
      end
else newpop(i,:) = pop(i,:);
end
end
%求最优适应度函数
%输入变量：pop:种群，fitvalue:种群适应度
%输出变量：bestindividual:最佳个体，bestfit:最佳适应度值
function [bestindividual bestfit] = best(pop,fitvalue)
[px,py] = size(pop);
bestindividual = pop(1,:);
bestfit = fitvalue(1);
for i = 2:px
if fitvalue(i)>bestfit
      bestindividual = pop(i,:);
      bestfit = fitvalue(i);
end
end