模拟退火算法-旅行商问题-matlab实现

整理一下数学建模会用到的算法，供比赛时候参考食用。

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旅行商问题（TSP）：

给定一系列城市和每对城市之间的距离，求解访问每一座城市一次并回到起始城市的最短回路。

它是组合优化中的一个NP困难问题，在运筹学和理论计算机科学中非常重要。

以下两个程序，在不同数据集合情况下表现有所差别，理论上第一个程序的解更为优化。

clear
clc
a = 0.99;       %温度衰减函数的参数
t0 = ;        %初始温度
tf = ;         %终止温度
t = t0;
Markov_length = ;  %Markov链长度
 
% load data.txt
% x = data(:, ::); x = x(:);
% y = data(:, ::); y = y(:);
% data = [,;x, y];
% coordinates = data;
coordinates = [
        565.0   575.0;      25.0   185.0;     345.0   750.0;
        945.0   685.0;     845.0   655.0;     880.0   660.0;
         25.0   230.0;     525.0  1000.0;     580.0  1175.0;
       650.0  1130.0;   1605.0   620.0;   1220.0   580.0;
      1465.0   200.0;   1530.0     5.0;    845.0   680.0;
       725.0   370.0;    145.0   665.0;    415.0   635.0;
       510.0   875.0;    560.0   365.0;    300.0   465.0;
       520.0   585.0;    480.0   415.0;    835.0   625.0;
       975.0   580.0;   1215.0   245.0;   1320.0   315.0;
      1250.0   400.0;    660.0   180.0;    410.0   250.0;
       420.0   555.0;    575.0   665.0;   1150.0  1160.0;
       700.0   580.0;    685.0   595.0;    685.0   610.0;
       770.0   610.0;    795.0   645.0;    720.0   635.0;
       760.0   650.0;    475.0   960.0;     95.0   260.0;
       875.0   920.0;    700.0   500.0;    555.0   815.0;
       830.0   485.0;   1170.0    65.0;    830.0   610.0;
       605.0   625.0;    595.0   360.0;   1340.0   725.0;
      1740.0   245.0;
    ];
coordinates(:,) = [];
amount = size(coordinates,);        %城市的数目
%通过向量化的方法计算距离矩阵
dist_matrix = zeros(amount,amount);
coor_x_tmp1 = coordinates(:,) * ones(,amount);
coor_x_tmp2 = coor_x_tmp1';
coor_y_tmp1 = coordinates(:,) * ones(,amount);
coor_y_tmp2 = coor_y_tmp1';
dist_matrix = sqrt((coor_x_tmp1 - coor_x_tmp2).^ + (coor_y_tmp1 - coor_y_tmp2).^);
 
sol_new = :amount;         %产生初始解，sol_new是每次产生的新解
sol_current = sol_new;      %sol_current是当前解
sol_best = sol_new;         %sol_best是冷却中的最好解
E_current = inf;            %E_current是当前解对应的回路距离
E_best = inf;               %E_best是最优解
p = ;
 
rand('state', sum(clock));
 
for j = :
    sol_current = [randperm(amount)];
    E_current = ;
    for i=:(amount-)
        E_current = E_current+dist_matrix(sol_current(i), sol_current(i+));
    end
    if E_current<E_best
        sol_best = sol_current;
        E_best = E_current;
    end
end
 
while t >= tf
   for r = :Markov_length      %Markov链长度
    %产生随机扰动
    if(rand < 0.5)
        %两交换
        ind1 = ;
        ind2 = ;
        while(ind1 == ind2)
           ind1 = ceil(rand * amount);
           ind2 = ceil(rand * amount);
        end
        tmp1 = sol_new(ind1);
        sol_new(ind1) = sol_new(ind2);
        sol_new(ind2) = tmp1;
    else
        %三交换
        ind=ceil(amount*rand(,));
        ind = sort(ind);
        sol_new = sol_new(, [:ind()-, ind()+:ind(),ind():ind(),ind()+:end]);
 
    end
 
    %检查是否满足约束
 
    %计算目标函数值（即内能）
    E_new = ;
    for i = :(amount - )
        E_new = E_new + dist_matrix(sol_new(i),sol_new(i + ));
    end
    %再算上从最后一个城市到第一个城市的距离
    E_new = E_new + dist_matrix(sol_new(amount),sol_new());
 
    if E_new < E_current
        E_current = E_new;
        sol_current = sol_new;
        if E_new < E_best
            E_best = E_new;
            sol_best = sol_new;
        end
    else
        %若新解的目标函数值大于当前解，
        %则仅以一定概率接受新解
        if rand < exp(-(E_new - E_current) / t)
            E_current = E_new;
            sol_current = sol_new;
        else
            sol_new = sol_current;
        end
 
    end
   end
 
   t = t * a;      %控制参数t（温度）减少为原来的a倍
end
 
E_best = E_best+dist_matrix(sol_best(end), sol_best());
 
disp('最优解为:');
disp(sol_best);
disp('最短距离:');
disp(E_best);
 
data1 = zeros(length(sol_best), );
for i = :length(sol_best)
    data1(i, :) = coordinates(sol_best(,i), :);
end
 
data1 = [data1; coordinates(sol_best(,),:)];
 
figure
plot(coordinates(:,)', coordinates(:,2)', '*k', data1(:,)', data1(:, 2)', 'r');
title( [ '近似最短路径如下，路程为' , num2str( E_best ) ] ) ;

另一种根据《数学建模算法与应用—司守奎》中算法改编：

clc;
clear;
close all;
 
coordinates = [
         25.0   185.0;     345.0   750.0;
        945.0   685.0;     845.0   655.0;     880.0   660.0;
         25.0   230.0;     525.0  1000.0;     580.0  1175.0;
       650.0  1130.0;   1605.0   620.0;   1220.0   580.0;
      1465.0   200.0;   1530.0     5.0;    845.0   680.0;
       725.0   370.0;    145.0   665.0;    415.0   635.0;
       510.0   875.0;    560.0   365.0;    300.0   465.0;
       520.0   585.0;    480.0   415.0;    835.0   625.0;
       975.0   580.0;   1215.0   245.0;   1320.0   315.0;
      1250.0   400.0;    660.0   180.0;    410.0   250.0;
       420.0   555.0;    575.0   665.0;   1150.0  1160.0;
       700.0   580.0;    685.0   595.0;    685.0   610.0;
       770.0   610.0;    795.0   645.0;    720.0   635.0;
       760.0   650.0;    475.0   960.0;     95.0   260.0;
       875.0   920.0;    700.0   500.0;    555.0   815.0;
       830.0   485.0;   1170.0    65.0;    830.0   610.0;
       605.0   625.0;    595.0   360.0;   1340.0   725.0;
      1740.0   245.0;
    ];
coordinates(:,) = [];
data = coordinates;
 
% 读取数据
% load data.txt;
 
% x = data(:, ::); x = x(:);
% y = data(:, ::); y = y(:);
x = data(:, );
y = data(:, );
start = [565.0   575.0];
data = [start; data;start];
 
% data = [start; x, y;start];
% data = data*pi/;
 
% 计算距离的邻接表
count = length(data(:, ));
d = zeros(count);
for i = :count-
    for j = i+:count
%         temp = cos(data(i, )-data(j,))*cos(data(i,))*cos(data(j,))...
%             +sin(data(i,))*sin(data(j,));
         d(i,j)=( sum( ( data( i , : ) - data( j , : ) ) .^  ) ) ^ 0.5 ;
%         d(i,j) = *acos(temp);
    end
end
d =d + d';  % 对称  i到j==j到i
 
S0=[];          % 存储初值
Sum=inf;     % 存储总距离
 
rand('state', sum(clock));
 
% 求一个较为优化的解，作为初值
for j = :
    S = [ +randperm(count-), count];
    temp = ;
    for i=:count-
        temp = temp+d(S(i), S(i+));
    end
    if temp<Sum
        S0 = S;
        Sum = temp;
    end
end
 
e = 0.1^;   % 终止温度
L = ;     % 最大迭代次数
at = 0.999999;     % 降温系数
T = ;             % 初温
 
% 退火过程
for k = :L
    % 产生新解
    c =+floor((count-)*rand(,));
 
    c = sort(c);
    c1 = c();  c2 = c();
    if c1==
        c1 = c1+;
    end
    if c2==
        c2 = c2+;
    end
    % 计算代价函数值
    df = d(S0(c1-), S0(c2))+d(S0(c1), S0(c2+))-...
        (d(S0(c1-), S0(c1))+d(S0(c2), S0(c2+)));
    % 接受准则
    if df<
        S0 = [S0(: c1-), S0(c2:-:c1), S0(c2+:count)];
        Sum = Sum+df;
    elseif exp(-df/T) > rand()
        S0 = [S0(: c1-), S0(c2:-:c1), S0(c2+:count)];
        Sum = Sum+df;
    end
    T = T*at;
    if T<e
        break;
    end
end
 
data1 = zeros(, count);
% y1 = [start; x, y; start];
for i =:count
   data1(:, i) = data(S0(,i), :)';
end
 
figure
plot(x, y, 'o', data1(, :), data1(, :), 'r');
title( [ '近似最短路径如下，路程为' , num2str( Sum ) ] ) ;
disp(Sum);
S0