%該文件為gacm算法的主程序clear;clear all;clc;  %初始化population=100; %種群數(shù)量  可用于外設(shè)I=imread('圖像地址');%讀圖像hist=imhist(I);  %獲取圖像的灰度圖a=size(hist,1);  %計(jì)算hist的第一維長(zhǎng)度c=2;  %可以做成界面的傳輸量 ，聚類數(shù)V00=zeros(population,c); %對(duì)聚類中心進(jìn)行的編碼for i=1:population   V00(i,:)=round(rand(1,c)*(a-1));end adapt0=zeros(population,1);adapt1=zeros(population,1);adapt2=zeros(population,1);for i=1:population    %調(diào)用函數(shù),輸出適應(yīng)度 [V00(i,:),adapt0(i)]=fcm(hist,c,V00(i,:));  adapt_averge=mean(adapt0); %計(jì)算平均適應(yīng)度V1=V00;   %保留初始化聚類中心adapt1=adapt0;  adapt_average1=adapt_average0; generation=100;    for k=1:generation        s1=select(V1,adapt_value1,population);           %制作屬于自己的選擇算子     s_code1=dec2bin(s1,8);  s_code=mat2str(s_code1);for i=1:population                     %轉(zhuǎn)化為二進(jìn)制編碼      s_code1(i)=strcat(s_code(2*i-1,:),s_code(2*i,:)); 這里可能有問題，還要繼續(xù)觀察end  c1=cross(s_code1,population);                  %制作屬于自己的交叉算子      v1=mutation(c1,population);             %制作屬于自己的變異算子        V2=(bin2dec(v1))';                %轉(zhuǎn)化為十進(jìn)制編碼            for i=1:population           [V2(i,:),adapt2(i)]=fcm(hist,c,V2(i,:));    %計(jì)算經(jīng)選擇、交叉、變異后的適應(yīng)度        end            adapt_average2=mean(adapt2);     %計(jì)算平均適應(yīng)度        if abs(adapt_average2-adapt_average1)<=0.0001    %算法停止條件            break;        else            V1=V2;                adapt1=adapt2;             adapt_average1=adapt_average2;         endend    [max_adapt,i]=max(adapt2);   %選取最大適應(yīng)度值    number=find(adapt2==max_adapt);     %具有最大適應(yīng)度的種群所占比例                                            %           ?如何進(jìn)行劃分     V=zeros(1,c);   V=V2(i,:);     dist = zeros(a,c);  %產(chǎn)生一個(gè)a*c矩陣   for k=1:a       for i=1:c                                                       dist(k,i)=sqrt(k-V(i)); %計(jì)算歐式距離 先列后行                              (8)式       end   end    %求隸屬度    m=2;%m為加權(quán)指數(shù)    有可能的話可用于外設(shè)                                               u = zeros(a,c);   b=0;    ht=0;  %ht=zeros(a,1);   for k=1:a       for i=1:c           a_a=dist(k,i);           if a_a==0               temp=0;           else               temp=(1/a_a)^(2/(m-1));           end           ht=ht+temp;   %將分母的和算出 ht(k)=ht(k)+temp;         end   end         for k=1:a       for i=1:c           if dist(k,i)==0               u(k,i)=0;           else               u(k,i)=abs(ht/(dist(k,i)^(2/(m-1))));% u(k,i)=abs(ht(k)/(dist(k,i)^(2/(m-1))));           end       end   end