文件列表:

ANN_Num (0, 2017-05-20)
ANN_Num\ann_methodA.m (579, 2017-05-20)
ANN_Num\ann_methodB.m (2004, 2017-05-20)
ANN_Num\ann_methodC.m (211, 2017-05-20)
ANN_Num\ann_train.m (392, 2017-05-20)
ANN_Num\getImages.m (401, 2017-05-20)
ANN_Num\getTestData.m (741, 2017-05-20)
ANN_Num\main.m (1192, 2017-05-20)
ANN_Num\numS (0, 2017-05-20)
ANN_Num\numS\0numS.bmp (694, 2017-05-20)
ANN_Num\numS\1numS.bmp (694, 2017-05-20)
ANN_Num\numS\2numS.bmp (694, 2017-05-20)
ANN_Num\numS\3numS.bmp (694, 2017-05-20)
ANN_Num\numS\4numS.bmp (694, 2017-05-20)
ANN_Num\numS\5numS.bmp (694, 2017-05-20)
ANN_Num\numS\6numS.bmp (694, 2017-05-20)
ANN_Num\numS\7numS.bmp (694, 2017-05-20)
ANN_Num\numS\8numS.bmp (694, 2017-05-20)
ANN_Num\numS\9numS.bmp (694, 2017-05-20)
ANN_Num\numSnoise (0, 2017-05-20)
ANN_Num\numSnoise\0numSn.bmp (694, 2017-05-20)
ANN_Num\numSnoise\1numSn.bmp (694, 2017-05-20)
ANN_Num\numSnoise\2numSn.bmp (694, 2017-05-20)
ANN_Num\numSnoise\3numSn.bmp (694, 2017-05-20)
ANN_Num\numSnoise\4numSn.bmp (694, 2017-05-20)
ANN_Num\numSnoise\5numSn.bmp (694, 2017-05-20)
ANN_Num\numSnoise\6numSn.bmp (694, 2017-05-20)
ANN_Num\numSnoise\7numSn.bmp (694, 2017-05-20)
ANN_Num\numSnoise\8numSn.bmp (694, 2017-05-20)
ANN_Num\numSnoise\9numSn.bmp (694, 2017-05-20)
ANN_Num\numV (0, 2017-05-20)
ANN_Num\numV\0numV.bmp (694, 2017-05-20)
ANN_Num\numV\1numV.bmp (694, 2017-05-20)
ANN_Num\numV\2numV.bmp (694, 2017-05-20)
ANN_Num\numV\3numV.bmp (694, 2017-05-20)
ANN_Num\numV\4numV.bmp (694, 2017-05-20)
ANN_Num\numV\5numV.bmp (694, 2017-05-20)
ANN_Num\numV\6numV.bmp (694, 2017-05-20)
ANN_Num\numV\7numV.bmp (694, 2017-05-20)
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# GA4TSPProblem 在matlab上使用遗传算法解决TSP旅行者问题 > TSP问题指的是从一个节点开始遍历其他所有节点并回到初始节点,构成一个哈密顿回路,节点与节点之间距离不同,目标是找到一条回路使得总路程最短,也即就是走最短的路遍历所有节点回到起点。 > 遗传算法模仿达尔文进化论中优胜劣汰的思想,从随机初始总群开始,不断进化最终选出接近最优解的一代,从而求解出近似最优解 ### 问题描述 下图矩阵展示了不同城市之间的距离,城市到自身的距离为0,现要求从Hong Kong出发,找一条最短的旅游顺序,使得游览所有城市后回到Hong Kong。  ### 基本思路 主要是问题的编码阶段,对于TSP问题在遗传算法中编码使用整数编码,使用整数来代表每一个城市,比如这里可以依次使用1,2,3,...,13表示这13个城市,9则代表Hong Kong。 |城市|编码基因| |---|---| |Amsterdam|1| |Athens|2| |Auckland|3| |Bahrain|4 |Bangkok|5 |Colombo|6 |Dubai|7 |Frankflurt|8 |HK|9 |Jakarta|10 |Kuala Lumpur|11 |London|12 |Manila|13 这13个数字的一个排列即是一种路径方案,但注意这条路径是一个环,收尾相接,因此起点是哪个城市是无所谓的,只要数字的相对位置确定,那13种(谁是起点)归并为同一种方案。因此所有可能的方案数为:13!/13 (全排列除以13) 示例染色体:[1 2 3 4 5 6 7 8 9 10 11 12 13],同[2 3 4 5 6 7 8 9 10 11 12 13 1]等属于同一种方案。 每种方案的路程代价即个体的适应度,路程越短代价越小适应度越高。 ### 函数解释  - cost.m:用来计算一种方案的路程代价,参数为方案数列染色体和代价矩阵。算法中优选代价小的,淘汰代价大的; - crossover.m:染色体交叉,参数为两个父代染色体,交叉失败返回0,否则返回交叉后的两个子代染色体; - crosscheck.m:检查两个父代是否可交叉得到有效的子代染色体,因为交叉后如果染色体中出现重复的数字是无效的方案,等于某个城市走了两次同时有的城市没遍历到; - generate.m:产生一个随机的染色体,实际是将实力染色体中的基因顺序随机打乱得到; - mutation.m:按照一定基因突变概率使参数染色体发生基因突变得到突变后的子代染色体; - TSP.m:遗传算法主体,其中采用了精英选择法,将父代的优秀个体加入到下一代的竞争中。 ### matlab代码清单 cost.m ``` function [value]=cost(parent,costM) [NumP,tmp]=size(parent); for i=1:NumP parent_i = parent(i,:); value_i = 0; for j=1:12 value_i = value_i + costM(parent_i(j),parent_i(j+1)); end value_i = value_i + costM(parent_i(13),parent_i(1)); value(i,1) = value_i; end end ``` crosscheck.m ``` %check if the two parent can generate valid children after crossover function res = crosscheck(parent1,parent2) res = 0; for i = 1:12 p1 = [parent1(:,1:i) parent2(:,i+1:13)]; p2 = [parent2(:,1:i) parent1(:,i+1:13)]; if length(p1)==length(unique(p1)) && length(p2)==length(unique(p2)) res = 1; end end end ``` crossover.m ``` function [child1, child2]=crossover(parent1, parent2) child1 = 0; child2 = 0; if crosscheck(parent1,parent2) == 1 p1 = [1 1 1 1 1 1 1 1 1 1 1 1 1]; p2 = [2 2 2 2 2 2 2 2 2 2 2 2 2]; %make sure the children are valid while length(p1)>length(unique(p1)) || length(p2)>length(unique(p2)) %randomly select a cross point crossPoint = randi([1 12]); p1 = [parent1(:,1:crossPoint) parent2(:,crossPoint+1:13)]; p2 = [parent2(:,1:crossPoint) parent1(:,crossPoint+1:13)]; end child1 = p1; child2 = p2; end end ``` generate.m ``` function pop = generate(num) res = []; %generate num random populations for i = 1:num tmp = [0 0 0 0 0 0 0 0 0 0 0 0 0]; tmp0 = [1 2 3 4 5 6 7 8 9 10 11 12 13]; for j = 1:13 %random index index = randi([1 14-j]); tmp(j) = tmp0(index); %delete randomly selected element tmp0(:,index) = []; end res = [res;tmp]; end pop = res; end ``` mutation.m ``` function [child]=mutation(parent,probability) if rand() <= probability P = parent; %select two random cities to interchange random1 = 0; random2 = 0; while random1 == random2 random1 = randi([1 13]); random2 = randi([1 13]); end %mutation P(:,[random1,random2])=P(:,[random2,random1]); child=P; else child = 0; end end ``` TSP.m ``` function tmp = TSP() close all; clc; %----------------------------data area------------------------------------- % distance from a city to others Amsterdam = [0 2.2 18.1 4.8 9.2 8.4 5.2 0.4 9.3 11.3 10.2 0.4 10.4]; %1 Athens = [2.2 0 17.5 2.8 7.9 6.6 3.3 1.8 8.5 9.8 8.7 2.4 9.6]; %2 Auckland = [18.1 17.5 0 14.7 9.6 10.9 14.2 18.2 9.1 7.6 8.7 18.3 8.0]; %3 Bahrain = [4.8 2.8 14.7 0 5.4 3.8 0.5 4.4 *** 7.0 6.0 5.1 7.4]; %4 Bangkok = [9.2 7.9 9.6 5.4 0 2.4 4.9 9.0 1.7 2.3 1.2 9.5 2.2]; %5 Colombo = [8.4 6.6 10.9 3.8 2.4 0 3.3 8.1 4.1 3.3 2.5 8.7 4.6]; %6 Dubai = [5.2 3.3 14.2 0.5 4.9 3.3 0 4.8 6.0 6.6 5.5 5.5 6.9]; %7 Frankflurt = [0.4 1.8 18.2 4.4 9.0 8.1 4.8 0 9.2 11.1 10.0 0.6 10.3]; %8 HK = [9.3 8.5 9.1 *** 1.7 4.1 6.0 9.2 0 3.3 2.5 9.6 1.1]; %9 Jakarta = [11.3 9.8 7.6 7.0 2.3 3.3 6.6 11.1 3.3 0 1.2 11.7 2.8]; %10 KualaLumpur = [10.2 8.7 8.7 6.0 1.2 2.5 5.5 10.0 2.5 1.2 0 10.5 2.5]; %11 London = [0.4 2.4 18.3 5.1 9.5 8.7 5.5 0.6 9.6 11.7 10.5 0 10.7]; %12 Manila = [10.4 9.6 8.0 7.4 2.2 4.6 6.9 10.3 1.1 2.8 2.5 10.7 0]; %13 costM = [Amsterdam;Athens;Auckland;Bahrain;Bangkok;Colombo;Dubai;Frankflurt;HK;Jakarta;KualaLumpur;London;Manila]; %mutation probability pmutation = 1.0; %max generation MaxGeneration = 200; %poputation size popsize = 20; %select popsize parents from randomsize generated possible parents randsize = 200; %parent generations parentpop = []; %best parent of every generation best_cost = []; %-------------------generate parent generation----------------------------- preparentpop = generate(randsize); [A,index] = sort(cost(preparentpop,costM),1,'ascend'); %orderd preparentpop orderedpreparentpop = preparentpop(index,:); %selected top popsize parentpop parentpop = orderedpreparentpop([1:popsize],:); %---------------------main revolution loop--------------------------------- for igen = 1:MaxGeneration childpop = []; childpopsize = [0 0]; %generate enough children while childpopsize(1) < popsize % To generate the random index for crossover and mutation ind=randi(popsize,[1 2]) ; parent1 = parentpop(ind(1),:); parent2 = parentpop(ind(2),:); [child1,child2] = crossover(parent1,parent2); [child3] = mutation(parent1,pmutation); if child1~=0 childpop = [childpop;child1]; end if child2~=0 childpop = [childpop;child2]; end if child3~=0 childpop = [childpop;child3]; end childpopsize = size(childpop); end % Elite: parentpop and childpop are added together before sorting for the best popsize to continue allpop = [parentpop;childpop]; [A,index] = sort(cost(allpop,costM),1,'ascend'); orderdallpop = allpop(index,:); %parentpop of current generation parentpop = orderdallpop([1:popsize],:); best_cost(igen)=A(1); end %display display('the best parentpop:') parentpop display('the lowest cost of every generation:') best_cost' figure,plot(1:igen,best_cost,'b') title('GA algorithm for TSP problem') end ``` ### 实例结果 进行了200代进化,大约在50代以后收敛,得到近似最优方案的最小路程代价为**40.2**,最优染色体为:**[2 8 12 1 9 13 3 10 11 5 6 7 4]**, 并不是每次运行都会得到绝对最优解,遗传算法容易早熟陷入局部最优解,需要调参优化:  > Github源码下载:[https://github.com/jiangxh1992/GA4TSPProblem](https://github.com/jiangxh1992/GA4TSPProblem)

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