clear,close all

A11=[1 0 0 0 0
    0 1 0 0 0
    0 0 1 0 0
    0 0 0 1 0
    0 0 0 0 1]
A12=[0 1 0 0 0
    0 0 1 0 0
    0 0 0 1 0
    0 0 0 0 1
    1 0 0 0 0]
A13=[1 0 0 0 0
    0 1 0 0 0
    0 0 1 0 0
    0 0 0 1 0
    0 0 0 0 1]
A21=[0 0 0 0 0
    0 0 0 0 0
    0 0 0 0 0
    0 0 0 0 0
    0 0 0 0 0]
A22=[0 0 0 1 0
    0 0 0 0 1
    1 0 0 0 0
    0 1 0 0 0
    0 0 1 0 0]
A23=[1 0 0 0 0
    0 1 0 0 0
    0 0 1 0 0
    0 0 0 1 0
    0 0 0 0 1]


Hqc=[A11,A12,A13;A21,A22,A23]
[br,bc]=size(A11)
[cb,tb]=size(Hqc)
D=[A12,A13;A22,A23]
Hqc1=em(Hqc)
D1=em(D)
rd1=rank(D1)
r=rank(Hqc1)
delta_r=rank(Hqc1)-rank(D1)

switch delta_r
    case 0
        D=[D,eye(cb)]
        D=em(D)
        D=rrem(D) 
        invD=D(1:cb,cb+1:2*cb)%求得D的逆陣
        M=Hqc(1:cb,1:tb-cb)
        TEM=mod(invD*M,2)
        I=eye(tb-cb)
        Z=mod(TEM*I,2)
        P=conj(Z)'
        Gqc=[I,P]
        
        
    otherwise %如果D非滿秩，則將Hqc序列重排
        tmp=Hqc(1:cb,tb-delta_r+1:tb)
        Hqc(1:cb,1+delta_r:tb)=Hqc(1:cb,1:tb-delta_r)
        Hqc(1:cb,1:delta_r)=tmp
        per_D=Hqc(1:cb,tb-cb+1:tb)
        per_D1=em(per_D)
        delta_l=0
        %如果D仍然非滿秩，則增加列數
        
       while (r-rank(per_D1))>0
            delta_l=delta_l+1
            per_D=Hqc(1:cb,tb-cb+1-delta_l*br:tb)
            per_D1=em(per_D),end
            lb=delta_l*br+cb, %得到新的列數
    end
    
    
  %per_D=[A11,A12,A13;A21,A22,A23]
  %per_D1=em(per_D)
  %per_D1=rrem(per_D1)