模拟退火算法解决旅行商问题(matlab)

模拟退火算法解决旅行商问题。

根据概率产生新解主要包含两个途径：二交换和三交换

二交换是在TSP回路中选择两个城市直接交换

三交换是在TSP回路中选择三个点，p1,p2,p3，然后将p1,p2之间的城市直接与p3之前对应长度的城市交换

这里产生新解的方法不唯一，只要能够保证产生的新解可以包含最优解所在的解空间即可

是否接受新解主要包含两种情况：

新解比历史最优解好，则百分百接受新解

新解比当前解好，没历史最优解好，则以一定概率接受新解，并且随着温度的降低。接受的概率也会降低。

如下是TSP代码。

clear
clc
a = 0.99;   % 温度衰减函数的参数
t0 = 97; tf = 3; t = t0;
Markov_length = 10000;  % Markov链长度
coordinates = [
    1    565.0   575.0; 2     25.0   185.0; 3    345.0   750.0;
    4    945.0   685.0; 5    845.0   655.0; 6    880.0   660.0;
    7     25.0   230.0; 8    525.0  1000.0; 9    580.0  1175.0;
    10   650.0  1130.0; 11  1605.0   620.0; 12  1220.0   580.0;
    13  1465.0   200.0; 14  1530.0     5.0; 15   845.0   680.0;
    16   725.0   370.0; 17   145.0   665.0; 18   415.0   635.0;
    19   510.0   875.0; 20   560.0   365.0; 21   300.0   465.0;
    22   520.0   585.0; 23   480.0   415.0; 24   835.0   625.0;
    25   975.0   580.0; 26  1215.0   245.0; 27  1320.0   315.0;
    28  1250.0   400.0; 29   660.0   180.0; 30   410.0   250.0;
    31   420.0   555.0; 32   575.0   665.0; 33  1150.0  1160.0;
    34   700.0   580.0; 35   685.0   595.0; 36   685.0   610.0;
    37   770.0   610.0; 38   795.0   645.0; 39   720.0   635.0;
    40   760.0   650.0; 41   475.0   960.0; 42    95.0   260.0;
    43   875.0   920.0; 44   700.0   500.0; 45   555.0   815.0;
    46   830.0   485.0; 47  1170.0    65.0; 48   830.0   610.0;
    49   605.0   625.0; 50   595.0   360.0; 51  1340.0   725.0;
    52  1740.0   245.0;
    ];
coordinates(:,1) = [];
amount = size(coordinates,1);   % 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(:,1) * ones(1,amount);
coor_x_tmp2 = coor_x_tmp1‘;
coor_y_tmp1 = coordinates(:,2) * ones(1,amount);
coor_y_tmp2 = coor_y_tmp1‘;
dist_matrix = sqrt((coor_x_tmp1-coor_x_tmp2).^2 + ...
    (coor_y_tmp1-coor_y_tmp2).^2);

sol_new = 1:amount;         % 产生初始解
% sol_new是每次产生的新解；sol_current是当前解；sol_best是冷却中的最好解；
E_current = inf;E_best = inf;       % E_current是当前解对应的回路距离；
% E_new是新解的回路距离；
% E_best是最优解的
sol_current = sol_new; sol_best = sol_new;
p = 1;

while t>=tf
    for r=1:Markov_length       % Markov链长度
        % 产生随机扰动
        if (rand < 0.5) % 随机决定是进行两交换还是三交换
            % 两交换
            ind1 = 0; ind2 = 0;
            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
            ind1 = 0; ind2 = 0; ind3 = 0;
            while (ind1 == ind2) || (ind1 == ind3) ...
                    || (ind2 == ind3) || (abs(ind1-ind2) == 1)
                ind1 = ceil(rand.*amount);
                ind2 = ceil(rand.*amount);
                ind3 = ceil(rand.*amount);
            end
            tmp1 = ind1;tmp2 = ind2;tmp3 = ind3;
            % 确保ind1 < ind2 < ind3
            if (ind1 < ind2) && (ind2 < ind3)
                ;
            elseif (ind1 < ind3) && (ind3 < ind2)
                ind2 = tmp3;ind3 = tmp2;
            elseif (ind2 < ind1) && (ind1 < ind3)
                ind1 = tmp2;ind2 = tmp1;
            elseif (ind2 < ind3) && (ind3 < ind1)
                ind1 = tmp2;ind2 = tmp3; ind3 = tmp1;
            elseif (ind3 < ind1) && (ind1 < ind2)
                ind1 = tmp3;ind2 = tmp1; ind3 = tmp2;
            elseif (ind3 < ind2) && (ind2 < ind1)
                ind1 = tmp3;ind2 = tmp2; ind3 = tmp1;
            end

            tmplist1 = sol_new((ind1+1):(ind2-1));
            sol_new((ind1+1):(ind1+ind3-ind2+1)) = ...
                sol_new((ind2):(ind3));
            sol_new((ind1+ind3-ind2+2):ind3) = ...
                tmplist1;
        end

        %检查是否满足约束

        % 计算目标函数值（即内能）
        E_new = 0;
        for i = 1 : (amount-1)
            E_new = E_new + ...
                dist_matrix(sol_new(i),sol_new(i+1));
        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(1));

        if E_new < E_current
            E_current = E_new;
            sol_current = sol_new;
            if E_new < E_best
                % 把冷却过程中最好的解保存下来
                E_best = E_new;
                sol_best = sol_new;
            end
        else
            % 若新解的目标函数值小于当前解的，
            % 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
for i=1:length(coordinates)
    plot(coordinates(i,1),coordinates(i,2),‘r*‘);
    hold on;
end;
x=coordinates([sol_best sol_best(1)],1);
y=coordinates([sol_best sol_best(1)],2);
plot(x,y);
disp(‘最优解为：‘)
disp(sol_best)
disp(‘最短距离：‘)
disp(E_best)

最后得到的图像如下，在目测条件下，应该是满足条件(不一定是最优的，但也差不多了)