我在前面已經(jīng)給出了模擬退火法的完整知識點(diǎn)和源碼實(shí)現(xiàn)：智能優(yōu)化算法—蟻群算法（Python實(shí)現(xiàn)）

模擬退火和蒙特卡洛實(shí)驗(yàn)一樣，全局隨機(jī)，由于沒有自適應(yīng)的過程(例如向最優(yōu)靠近、權(quán)重梯度下降等)，對于復(fù)雜函數(shù)尋優(yōu)，很難會找到最優(yōu)解，都是近似最優(yōu)解；然而像蝙蝠算法、粒子群算法等有向最優(yōu)逼近且通過最優(yōu)最差調(diào)整參數(shù)的步驟，雖然對于下圖函數(shù)易陷入局部最優(yōu)，但是尋優(yōu)精度相對較高。如果理解這段話應(yīng)該就明白了為什么神經(jīng)網(wǎng)絡(luò)訓(xùn)練前如果初步尋優(yōu)一組較好的網(wǎng)絡(luò)參數(shù)，會使訓(xùn)練效果提高很多，也會更快達(dá)到誤差精度。

1.2 sko.SA 實(shí)現(xiàn)

#===========1導(dǎo)包================
import matplotlib.pyplot as plt
import pandas as pd
from sko.SA import SA
 
#============2定義問題===============
fun = lambda x: x[0] ** 2 + (x[1] - 0.05) ** 2 + x[2] ** 2
 
#=========3運(yùn)行模擬退火算法===========
sa = SA(func=fun, x0=[1, 1, 1], T_max=1, T_min=1e-9, L=300, max_stay_counter=150)
best_x, best_y = sa.run()
print('best_x:', best_x, 'best_y', best_y)
 
#=======4畫出結(jié)果=======
plt.plot(pd.DataFrame(sa.best_y_history).cummin(axis=0))
plt.show()
 
 
 
 
#scikit-opt 還提供了三種模擬退火流派: Fast, Boltzmann, Cauchy.
 
#===========1.1 Fast Simulated Annealing=====================
from sko.SA import SAFast
 
sa_fast = SAFast(func=demo_func, x0=[1, 1, 1], T_max=1, T_min=1e-9, q=0.99, L=300, max_stay_counter=150)
sa_fast.run()
print('Fast Simulated Annealing: best_x is ', sa_fast.best_x, 'best_y is ', sa_fast.best_y)
 
#===========1.2 Fast Simulated Annealing with bounds=====================
from sko.SA import SAFast
 
sa_fast = SAFast(func=demo_func, x0=[1, 1, 1], T_max=1, T_min=1e-9, q=0.99, L=300, max_stay_counter=150,
                 lb=[-1, 1, -1], ub=[2, 3, 4])
sa_fast.run()
print('Fast Simulated Annealing with bounds: best_x is ', sa_fast.best_x, 'best_y is ', sa_fast.best_y)
 
#===========2.1 Boltzmann Simulated Annealing====================
from sko.SA import SABoltzmann
 
sa_boltzmann = SABoltzmann(func=demo_func, x0=[1, 1, 1], T_max=1, T_min=1e-9, q=0.99, L=300, max_stay_counter=150)
sa_boltzmann.run()
print('Boltzmann Simulated Annealing: best_x is ', sa_boltzmann.best_x, 'best_y is ', sa_fast.best_y)
 
#===========2.2 Boltzmann Simulated Annealing with bounds====================
from sko.SA import SABoltzmann
 
sa_boltzmann = SABoltzmann(func=demo_func, x0=[1, 1, 1], T_max=1, T_min=1e-9, q=0.99, L=300, max_stay_counter=150,
                           lb=-1, ub=[2, 3, 4])
sa_boltzmann.run()
print('Boltzmann Simulated Annealing with bounds: best_x is ', sa_boltzmann.best_x, 'best_y is ', sa_fast.best_y)
 
#==================3.1 Cauchy Simulated Annealing==================
from sko.SA import SACauchy
 
sa_cauchy = SACauchy(func=demo_func, x0=[1, 1, 1], T_max=1, T_min=1e-9, q=0.99, L=300, max_stay_counter=150)
sa_cauchy.run()
print('Cauchy Simulated Annealing: best_x is ', sa_cauchy.best_x, 'best_y is ', sa_cauchy.best_y)
 
#==================3.2 Cauchy Simulated Annealing with bounds==================
from sko.SA import SACauchy
 
sa_cauchy = SACauchy(func=demo_func, x0=[1, 1, 1], T_max=1, T_min=1e-9, q=0.99, L=300, max_stay_counter=150,
                     lb=[-1, 1, -1], ub=[2, 3, 4])
sa_cauchy.run()
print('Cauchy Simulated Annealing with bounds: best_x is ', sa_cauchy.best_x, 'best_y is ', sa_cauchy.best_y)

2 Matlab實(shí)現(xiàn)

2.1 模擬退火法

clear
clc
T=1000; %初始化溫度值
T_min=1; %設(shè)置溫度下界
alpha=0.99; %溫度的下降率
num=1000; %顆粒總數(shù)
n=2; %自變量個(gè)數(shù)
sub=[-5,-5]; %自變量下限
up=[5,5]; %自變量上限
tu
for i=1:num
for j=1:n
x(i,j)=(up(j)-sub(j))*rand+sub(j);
    end
    fx(i,1)=fun(x(i,1),x(i,2));
end
 
%以最小化為例
[bestf,a]=min(fx);
bestx=x(a,:);
trace(1)=bestf;
while(T>T_min)
for i=1:num
for j=1:n
            xx(i,j)=(up(j)-sub(j))*rand+sub(j);
        end
        ff(i,1)=fun(xx(i,1),xx(i,2));
        delta=ff(i,1)-fx(i,1);
if delta<0
            fx(i,1)=ff(i,1);
x(i,:)=xx(i,:);
else
            P=exp(-delta/T);
if P>rand
                fx(i,1)=ff(i,1);
x(i,:)=xx(i,:);
            end
        end  
    end
if min(fx)<bestf
        [bestf,a]=min(fx);
        bestx=x(a,:);
    end
    trace=[trace;bestf];
    T=T*alpha;
end
disp('最優(yōu)解為：')
disp(bestx)
disp('最優(yōu)值為：')
disp(bestf)
hold on
plot3(bestx(1),bestx(2),bestf,'ro','LineWidth',5)
figure
plot(trace)
xlabel('迭代次數(shù)')
ylabel('函數(shù)值')
title('模擬退火算法')
legend('最優(yōu)值')

function z=fun(x,y)
z = x.^2 + y.^2 - 10*cos(2*pi*x) - 10*cos(2*pi*y) + 20;

function tu
[x,y] = meshgrid(-5:0.1:5,-5:0.1:5);
z = x.^2 + y.^2 - 10*cos(2*pi*x) - 10*cos(2*pi*y) + 20;
figure
mesh(x,y,z)%建一個(gè)網(wǎng)格圖，該網(wǎng)格圖為三維曲面，有實(shí)色邊顏色，無面顏色
hold on
xlabel('x')
ylabel('y')
zlabel('z')
title('z =  x^2 + y^2 - 10*cos(2*pi*x) - 10*cos(2*pi*y) + 20')

這里有一個(gè)待嘗試的想法，先用蒙特卡洛/模擬退火迭代幾次全局去找最優(yōu)的區(qū)域，再通過其他有向最優(yōu)逼近過程的算法再進(jìn)一步尋優(yōu)，或許會很大程度降低產(chǎn)生局部最優(yōu)解的概率。

下面是模擬退火和蒙特卡洛對上述函數(shù)尋優(yōu)的程序，迭代次數(shù)已設(shè)為一致，可以思考下兩種程序?qū)懛ǖ男省⒐餐c(diǎn)、缺點(diǎn)。理論研究講究結(jié)果好，實(shí)際應(yīng)用既要保證結(jié)果好也要保證程序運(yùn)算效率。

2.2 蒙特卡諾法

clear
clc
num=689000; %顆?？倲?shù)
n=2; %自變量個(gè)數(shù)
sub=[-5,-5]; %自變量下限
up=[5,5]; %自變量上限
tu
x=zeros(num,n);
fx=zeros(num,1);
for i=1:num
for j=1:n
x(i,j)=(up(j)-sub(j))*rand+sub(j);
    end
    fx(i,1)=fun(x(i,1),x(i,2));
end
 
[bestf,a]=min(fx);
bestx=x(a,:);
 
disp('最優(yōu)解為：')
disp(bestx)
disp('最優(yōu)值為：')
disp(bestf)
hold on
plot3(bestx(1),bestx(2),bestf,'ro','LineWidth',5)