jcross417.m

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% 交叉操作-见陈国良的书
function [A,B] = jcross417(oldpop,mate1,mate2,loccross,d)

popsize = length(oldpop);
lchrom = length(oldpop(1).chrom);

newpop1(1:lchrom-loccross+1) = oldpop(mate2).chrom(loccross:lchrom);
newpop2(1:lchrom-loccross+1) = oldpop(mate1).chrom(loccross:lchrom);
for i = 1:lchrom
    if oldpop(mate1).chrom(i) == oldpop(mate2).chrom(loccross)
        Tempoldpop1 = arrayshift(oldpop(mate1).chrom,i);
        break
    end
end
for i = 1:lchrom
    if oldpop(mate2).chrom(i) == oldpop(mate1).chrom(loccross)
        Tempoldpop2 = arrayshift(oldpop(mate2).chrom,i);
        break
    end
end
for i = 1:lchrom
    already1 = 0;
    already2 = 0;
    % 查找存在的重复元素
    for j = 1:lchrom-loccross+1
        % 若找到 already1 置1
        if Tempoldpop1(i) == newpop1(j)
            already1 = 1;
            break
        end
    end
    for j = 1:lchrom-loccross+1
        if Tempoldpop2(i) == newpop2(j)
            already2 = 1;
            break
        end
    end    
    % 若没有找到，置入个体
    if already1 == 0;
        newpop1(length(newpop1)+1) = Tempoldpop1(i);
    end
    if already2 == 0;
        newpop2(length(newpop2)+1) = Tempoldpop2(i);
    end
end

Afit = objfunc(newpop1,d);
Bfit = objfunc(newpop2,d);

% 在子代中取出适应度大的一个
if Afit > Bfit
    A = newpop1;
else
    A = newpop2;
end   

% 在父代中取出适应度大的一个
if oldpop(mate1).fitness > oldpop(mate2).fitness
    B = oldpop(mate1).chrom;
else
    B = oldpop(mate2).chrom;
end

if sum(A) ~= sum(oldpop(mate1).chrom) | sum(B) ~= sum(oldpop(mate1).chrom)
    A = oldpop(mate1).chrom;
    B = oldpop(mate2).chrom;
end