智能算法之遗传算法
1.背景
2.算法
3.案例
3.1 案例求解二元函数的最大值
例1:计算二元函数 f ( x , y ) = 20 + x 2 + y 2 − 10 ∗ ( c o s ( 2 π x ) + c o s ( 2 π y ) ) f(x,y)=20+x^2+y^2-10*(cos(2\pi x)+cos(2\pi y)) f(x,y)=20+x2+y2−10∗(cos(2πx)+cos(2πy)) 的最大值,其中 x ∈ [ 0 , 10 ] , y ∈ [ 0 , 5 ] x\in[0,10],y\in [0,5] x∈[0,10],y∈[0,5]。
目标函数:
% -------------------适应度函数--------------------目标函数
function y = fitnessfcn(x)
y = 20+x(1)^2+x(2)^2-10*(cos(2*pi*x(1))+cos(2*pi*x(2)));
end
选择/复制:
% ----------------------------选择子函数selection--------------------------------
function population = selection(population, best, fitness, N)
newpopulation = zeros(N, size(population, 2)); % 先预先分配内存 可以不要
p = fitness./sum(fitness);
q = cumsum(p);
for i = 1:(N-1)
r = rand;
tmp = find(r <= q);
newpopulation(i, :) = population(tmp(1), :);
end
newpopulation(N, :) = best; % 保留最优
population = newpopulation;
end
交叉:
% -------------------交叉子函数crossover-------------------
function population = crossover(population, best, pc, L, N)
for i = 1:2:(N-1)
cc = rand;
if cc < pc
point = 1 + ceil(rand*(L-2)); % 取得一个2到L-1的整数
ch = population(i, :);
population(i, point+1:L) = population(i+1, point+1:L);
population(i+1, point+1:L) = ch(1, point+1:L);
end
end
population(N, :) = best; % 保留最优
end
变异:
% -------------------------变异子函数mutation----------------------
function population = mutation(population, pm, L, N)
mm = rand(N, L)<pm; % 小于变异概率的赋值为1,其他为0;也可以取反变异,这里的mm是逻辑值
mm(N, :) = zeros(1, L); % 最优保留,不变异
population(mm) = 1-population(mm); % 变异发生
end
二进制转换为十进制:
% ----------------解码子函数decode------------------
function [xx, fitness] = decoding(population, lb, ub, Li, N)
for i = 1:N
for k = 1:length(Li)
s(k) = 0;
for j = 1:Li(k)
s(k) = s(k) + population(i, sum(Li(1:k))-j+1)*2^(j-1); % 二进制转十进制
end
x(k) = (ub(k) - lb(k))*s(k)/(2^Li(k)-1)+lb(k); % 映射到取值范围内
end
fitness(i) = fitnessfcn(x);
xx(i, :) = x;
end
end
主函数:
clc;clear;close all;
% ----------------初始化参数--------------------
e = 0.01; % 计算精度
lb = [0; 0]; % 自变量下界
ub = [10; 5]; % 自变量上界
Li = ceil(log2((ub-lb)./e));
L = sum(Li); % 变量字串长度
N = 30; % 群体规模
T = 30; % 最大遗传代数
pc = 0.9; % 交叉概率
pm = 0.05; % 变异概率
% -----------------初始种群----------------------
population = round(rand(N, L)); % 初始种群,二进制编码(0,1)之间随机数,然后四舍五入为0/1
[xx, fitness] = decoding(population, lb, ub, Li, N); % 解码,计算适应度
[optfit, indmax] = max(fitness); % 初始种群最佳个体的适应度和索引
best = population(indmax, :); % 初始种群最优染色体 二进制的
optx = xx(indmax, :); % 初始种群最优变量值 十进制的
fitcurve = zeros(1, T+1); % 有多少代最优适应度
fitcurve(1) = optfit; % 第一代的最优适应度
% -------------------迭代求解----------------------
for ii = 1:T
population = selection(population, best, fitness, N); % 选择(复制)
population = crossover(population, best, pc, L, N); % 交叉
population = mutation(population, pm, L, N); % 变异
[xx, fitness] = decoding(population, lb, ub, Li, N); % 解码,计算适应度
[fmax, indmax] = max(fitness); % 当代最佳个体
if fmax >= optfit
best = population(indmax, :); % 到目前为止最优染色体
optx = xx(indmax, :); % 到目前为止最优变量值
optfit = fmax; % 到目前为止最优适应度
end
fitcurve(ii+1) = optfit; % 存储每代的最优适应度
end
optx % 最优变量
optfit % 最优适应度
% ------------------------画图---------------------------
subplot(1,2,1);
plot(0:T, fitcurve);
title('最优适应度曲线');
xlabel('遗传代数'); ylabel('最优适应度');
subplot(1,2,2);
x = linspace(0, 10);
y = linspace(0, 5);
[x, y] = meshgrid(x, y);
f = 20+x.^2+y.^2-10*(cos(2*pi*x)+cos(2*pi*y));
mesh(x, y, f);
hold on;
scatter3(optx(1), optx(2), optfit, 'red', 'filled');
xlabel('x'); ylabel('y'); zlabel('f');title('f(x,y)=20+x^2+y^2-10*(cos(2\pi x)+cos(2\pi y))')
得到最优解为: ( 9.5406 , 4.5303 ) (9.5406,4.5303) (9.5406,4.5303),最大值为 f ( 9.5406 , 4.5303 ) = 151.0422 f(9.5406,4.5303)=151.0422 f(9.5406,4.5303)=151.0422
转载:https://blog.csdn.net/ymengm/article/details/128140071
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