?? mutation.m
字號(hào):
function [newpop]=mutation(oldpop,stringlength,a,b,pm)
pz=size(oldpop,1);
for i=1:pz
if(rand<pm)
mpoint=round(rand*(stringlength-1))+1;
newpop=oldpop;
newpop(i,mpoint)=abs(oldpop(i,mpoint)-1);
sum=0;
for j=1:stringlength
sum=sum+newpop(j)*2^(j-1);
end
newpop(i,stringlength+1)=sum;%解碼
x(i)=newpop(i,stringlength+1);
newpop(i,stringlength+2)=fitness(x,pz,stringlength);
else
newpop=oldpop;
end
end
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