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function [c10,c11]=cross_2d(s_code10,s_code11) 
    
   %交叉算子 
   
   pc=0.8;       %交叉概率取0.6 
   population=20; 
    
   %(1,2)/(3,4)/(5,6)进行交叉运算，(7,8)/(9,10)复制 
    
   ww0=s_code10; 
   ww1=s_code11; 
    
   for i=1:(pc*population/2) 
       r0=abs(round(rand(1)*10)-3); 
       r1=abs(round(rand(1)*10)-3); 
       for j=(r+1):8 
           temp0=ww0(2*i-1,j); 
           ww0(2*i-1,j)=ww0(2*i,j); 
           ww0(2*i,j)=temp; 
       end 
   end 
    
   c10=ww0; 
   c11=ww1;