Question

我正在尝试生成随机二进制矩阵及其逆模q，其中q是2的幂。有时当我的矩阵的行列式是可逆的模q（因此Z_q上的矩阵是可逆的）时，我得到了错误＆＃34; InvMod：inverse undefined Aborted（core dumped）＆＃34;和其他时间计算逆。我做错了什么？

#include <iostream>
//NTL files
#include <NTL/ZZ_p.h>
#include <NTL/vec_vec_ZZ_p.h>
#include <NTL/LLL.h>
#include <NTL/matrix.h>
#include <NTL/vector.h>
#include <NTL/tools.h>
#include <NTL/ZZ.h>
#include <NTL/vec_vec_ZZ.h>
using namespace std;
using namespace NTL;

int main(){//task generate a random matrix S with 0/1 entries stored as a ZZ_p matrix, then generate a random, invertible S 
int nn = 8;
ZZ n = ZZ(nn);
ZZ N = ZZ(0);
ZZ q; power2(q, 4);
ZZ_p::init(q);

mat_ZZ S; S.SetDims(nn,nn); 
for(int i = 0; i<nn; i++){
    for(int j = 0; j<nn; j++){
        S[i][j] = RandomBits_ZZ(1);     
    }
}
mat_ZZ_p S1; S1.SetDims(nn,nn);//copy to ZZ_P
mat_ZZ_p R; R.SetDims(nn,nn);//will set to inverse if 

cout<<"The random matrix is S = "<<endl; //print S
for(int i = 0; i<nn; i++){
    for(int j=0; j<n;j++){
        cout<<S[i][j]<<", ";
    } cout<<endl;
}

ZZ d; determinant(d,S); ZZ_p d1; conv(d1, d % q);
if(GCD(q,d) == 1){//convert to mod q datatype
    for(int i = 0; i<nn; i++){
        for(int j = 0; j<nn; j++){
            conv(S1[i][j], S[i][j]);        
        }
    }
    //let's invert the matrix and print it!
    cout<<"The random matrix is R = "<<endl; //print R
    R = inv(S1); //mul(R,R,S1);
    for(int i = 0; i<nn; i++){
        for(int j=0; j<n;j++){
            cout<<R[i][j]<<", ";
        } cout<<endl;
    }
}

cout<<endl<<"det of S is "<<d<<" and this mod q is "<<d1<<endl;
cout<<"Our modulus is "<< q <<endl;

return 0;
}

Answer 1

如果行列式是可逆的mod q，这仅表示存在逆矩阵。但是计算这个矩阵的算法仍然可以达到需要计算没有一个元素的逆的算法。

如果#include <iostream> //NTL files #include <NTL/mat_ZZ_p.h> using namespace std; using namespace NTL; int main() {//task generate a random matrix S with 0/1 entries stored as a ZZ_p matrix, then generate a random, invertible S int nn = 8; ZZ q; power2(q, 4); ZZ_p::init(q); mat_ZZ_p S; S.SetDims(nn, nn); for(int i = 0; i < nn; i++) { for(int j = 0; j < nn; j++) { S[i][j] = conv<ZZ_p>(RandomBits_ZZ(1)); } } mat_ZZ_p R; R.SetDims(nn, nn);//will set to inverse if cout << "The random matrix is S = " << endl << S; ZZ_p d; determinant(d, S); cout << endl << "det(S) = " << d << endl; cout << "q = " << q << endl; if(GCD(conv<ZZ>(d), q) == 1) { // let's invert the matrix and print it! R = inv(S); cout << "The random matrix is R = " << R << endl; } return 0; }为素数，则没有此问题。

顺便说一句，这是代码的简化版本。

InputStream is = getClass().getResourceAsStream("/resources/file.txt");

在NTL中反转ZZ_p矩阵

1 个答案: