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1、 背包问题-遗传算法解决 function Population1=GA_copy(Population,p,w0,w)%复制算子%Population为种群n=length(Population(:,1);fvalue=zeros(1,n);for i=1:n fvalue(i)=GA_beibao_fitnessvalue(Population(i,:),p,w0,w);endfval=fvalue/sum(fvalue);F(1)=0;for j=1:n F(j+1)=0; for k=1:j F(j+1)=F(j+1)+fval(k); endendfor i=1:n test=ran
2、d; for j=1:n if(test>=F(j)&&(test<F(j+1) Population1(i,:)=Population(j,:); end endend function Population1=GA_exchange(Population,pc)%遗传算法交换算子%pc为交换概率Population1=Population;POP=;n=length(Population(:,1);%k=floor(n*pc); %用于交换的染色体数目%采用单点交换算子j=1;l=length(Population(1,:);for i=1:n test(i)=
3、rand; if test(i)<pc for z=1:l POP(j,z)=Population(i,z); end POP(j,l+1)=i; p(j)=randint(1,1,1 l-1); j=j+1; endendk0=j-1;k=floor(k0/2);if k>=1 for m=1:k for t=p(2*m-1)+1:l s=POP(2*m-1,t); POP(2*m-1,t)=POP(2*m,t); POP(2*m,t)=s; end end for m=1:k0 for i=1:l Population1(POP(m,l+1),i)=POP(m,i); end
4、endend function fitnessvalue=GA_fitnessvalue(x,p,w0,w)%使用惩罚法计算适应度值 %x为染色体%p为背包问题中每个被选物体的价值%w0为背包问题中背包总容积%w为背包问题中每个被选物品的容积l=length(x);for i=1:l a(i)=p(i).*x(i);endf=sum(a);b=min(w0,abs(sum(w)-w0);for i=1:l wx(i)=w(i).*x(i);endif abs(sum(wx)-w0)>b*0.99 p=0.99;else p=abs(sum(wx)-w0)/b;endfitnessvalu
5、e=f*(1-p)*(1-p)*(1-p);function Population=GA_Initial(n,P)%n为染色体长度，即所供选择的物品总数%P为初始种群大小for i=1:P for j=1:n temp=rands(1,1); if temp>0 Population(i,j)=1; else Population(i,j)=0; end endend function Population1=GA_tubian(Population,pe_tubian)%遗传算法突变算子%pe为突变概率Population1=Population;n=length(Population(:,1);m=length(Population(1,:);for i=1:n for j=1:m test=rand; if test<pe_tubian Population1(i,j)=1-Population1(i,j); end endend
matlab、lingo程序代码3背包问题遗传算法
