Solving XOR Equation Systems

Question

I have to solve a system consisting of 32 xor equations, each of which includes 15 of 32 variables. It would be like this:

i[0] = p[0] ^ p[4] ^ p[5] ^ p[10] ^ p[11] ^ p[20] ^ p[21] ^ p[22] ^ p[23] ^ p[25] ^ p[26] ^ p[27] ^ p[28] ^ p[30] ^ p[31]

i [n] and p [n] are 16-bit integers.

So, I understand that I will get a 32x32 matrix (containing only 1s and 0s) and a 32-vector result.

Apparently eliminating Gauss is what I need, but I can't think about the problem, could someone give me some idea of ​​how to solve such a problem?

EDIT: here is the solution in C ++

void SolveLinearSystem(u8 A[32][32], u8 B[32])
{
int N = 32;
for (int K = 0; K < N; ++K) 
{
    if( A[K][K] == 0 ) 
    {
        for(int i = K + 1; i<N ; ++i ) 
        { 
            if(A[i][K] != 0) 
            {
                for( int L = 0; L < N; ++L )
                {
                    int s = A[K][L];
                    A[K][L] = A[i][L];
                    A[i][L] = s;
                }
                int s = B[i];
                B[i] = B[K];
                B[K] = s;
                    break;
            }
        }
    }

    for( int I = 0; I<N; ++I)
    {
        if( I!=K )
        { 
            if( A[I][K] ) 
            { 
                int M = 0;
                for( M=K; M<N; ++M ) 
                { 
                    A[I][M] = A[I][M] ^ A[K][M]; 
                }
                B[I] = B[I] ^ B[K];
            }
        }
    }
}
}

Answer 1

Yes, you can use a Gaussian exception to solve this problem. The key is to recognize that the XOR operation is equivalent to adding modulo 2. Thus, the equation you wrote is equivalent to

i[0] = (p[0] + p[4] + ... ) mod 2

M*p=i mod 2

, , , 2. 0, , , .

Answer 2

, . :

[a b; c d] -> [a b; 0 d-(b*c/a)] -> [a 0; 0 d-(b*c/a)] -> [1 0; 0 1]

: MATLAB .

, (.. , , ) , . , , , , , ..

xor. 16- xor'd . , xor'd , , :

p[0] = a[1] ^ a[2]

:

p[0][0] = a[1][0] ^ a[2][0]
p[0][1] = a[1][1] ^ a[2][1]
…

bit 16- . , 16 .

xor . "" xor. :

1 + 0 = 0 + 1 = 1; 1 + 1 = 0 + 0 = 0
1 * 0 = 0 * 1 = 0 * 0 = 0; 1 * 1 = 1

1 ( ), .

xor. 1 0. gauss-jordan ( ), . .

, ++-, , . , ! Github: XorSystem. invert() ANMatrix.