6
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enter image description here Let given
$M =$

 1     0     1
 0     1     1
 1     1     1

and $b =$

 1
 0
 1

How to find the solution $x_3$ where $x=${$x_1,x_2,x_3$}?

Solution: Based on Wiedemann algorithm we has $u_i=[0; 0; 1]$. I am getting confused in the Step2. How to find $M_s$? Is it just the $3^th$ columns of $M_t$ as $M_s=[1 1 1]$ or it is created from Krylov sequence? Please help me find M_s and minimal polynomial?

Matlab code:

%% Input Generation Matrix M, b and index ith
%% Output single symbol x
M=[1 0 1;0 1 1;1 1 1];
b=[1; 0; 1];
index=2;
[K N]=size(M);
Mt=M';
Imatrix=eye(K,K);
x=[];
for index=1:K
u_i=Imatrix(:,index);
se_Krylov=[];
se_Krylov(:,1)=u_i;
    for i=2:(2*K) % due to i=1 is u_i
       se_Krylov(:,i)=mod((Mt^i)*u_i,2);
    end
%% M_s is the operator Mt retricted to S    
M_s=se_Krylov(index,:);

%% Compute the polynominal using Berlekmap Massey
[f, LCP] = Berlekamp_Massey2(M_s);
d=size(f,2)-1; %deg of f: x^d+x^(d-1)+....

x_comma=zeros(K,1);
    for i=d:-1:1
        x_comma=mod(x_comma+f(d).*(Mt)^(i-1)*u_i,2);
    end
x_single=mod(x_comma'*b,2) ;%Inner product 
x(index)=x_single;
end
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migrated from stackoverflow.com Dec 30 '14 at 13:28

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  • $\begingroup$ -1 for the text-as-image. $\endgroup$ – Federico Poloni Dec 31 '14 at 9:44

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