Exact Large Finite Field Linear Algebra Library (e.g. GF(2^128) / GF(2^256) )


The NTL library seems to work, using this (sorry I'm quite unable to program in C++) code

#include <NTL/GF2E.h>
#include <NTL/GF2EX.h>
#include <NTL/GF2X.h>
#include <NTL/GF2XFactoring.h>

NTL_CLIENT

int main()
{
    GF2X P = BuildIrred_GF2X(256);
    GF2E::init(P);

    GF2E zero = GF2E::zero();
    GF2E one;
    GF2E r = random_GF2E();
    GF2E r2 = random_GF2E();
    conv(one, 1L);
    cout << "Cardinality: " << GF2E::cardinality() << endl;
    cout << "ZERO: " << zero << " --> " << IsZero(zero) << endl;
    cout << "ONE:  " << one  << " --> " << IsOne(one)   << endl;
    cout << "1/r:  " << 1/r  << ", r * (1/r): " << (r * (1/r)) << endl;
    cout << "1/r2:  " << 1/r2  << ", r2 * (1/r2): " << (r2 * (1/r2)) << endl;
}

it seems to work, proof (output of this program):

Cardinality: 115792089237316195423570985008687907853269984665640564039457584007913129639936
ZERO: [] --> 1
ONE:  [1] --> 1
1/r:  [0 1 0 1 1 0 1 1 1 0 1 1 1 0 0 1 1 0 1 1 1 0 1 1 0 0 0 0 0 1 0 1 0 1 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 0 1 0 1 0 0 0 1 1 1 0 1 1 1 1 0 1 0 1 0 1 1 0 1 1 1 0 0 0 1 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 0 1 1 0 1 0 1 0 0 0 0 0 1 0 0 1 1 0 0 1 0 0 1 0 1 1 1 1 0 0 1 1 0 1 0 1 1 1 1 1 1 0 1 1 0 0 0 0 0 1 1 0 1 0 0 1 1 1 0 1 1 1 1 1 0 1 0 1 0 0 0 1 1 0 0 1 1 0 0 1 0 1 1 1 0 1 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 0 1 0 1 0 0 1 1 0 1 1 0 1 1 1 1 1 0 0 1 1 0 1 0 1 0 0 0 0 1 1 0 0 1 1 1 0 1], r * (1/r): [1]
1/r2:  [1 0 1 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 0 0 1 1 1 0 0 0 1 1 1 1 1 0 1 0 1 1 0 0 1 1 1 0 1 0 1 0 0 1 0 0 0 0 1 1 1 0 0 0 1 1 1 1 1 0 0 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 0 1 0 1 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 1 0 1 0 0 0 1 0 1 0 1 1 0 0 0 1 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 1 1 0 1 1 0 0 0 0 1 1 0 1 1 1 0 1 0 0 0 0 0 1 1 0 1 1 1 0 0 0 0 1 1 0 1 0 0 0 0 1 0 0 0 0 1 1 1 1 1 0 1 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 1 0 1 1 1 0 1 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 1 1 0 1 0 0 1 0 1 0 0 1 1], r2 * (1/r2): [1]

Even inverting seems to work (scroll as right as possible in the output sample above) :-)

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