Computing echelon form of smaller matrix takes longer than larger matrix?

[user202729]

Use flint's algorithm through SageMath's interface (which uses fmpq_mat_rref under the hood).

sage: entry_size = 10000
....: num_col = 20
....: num_row = 20
....: A = matrix(QQ, [[randint(1, 2^entry_size) for col in range(num_col)] for row in range(num_row)
....: ])
....: t = walltime()
....: A.echelonize(algorithm="flint")
....: t = walltime(t)
....: t
1.3925843238830566
sage: entry_size = 10000
....: num_col = 40
....: num_row = 40
....: A = matrix(QQ, [[randint(1, 2^entry_size) for col in range(num_col)] for row in range(num_row)
....: ])
....: t = walltime()
....: A.echelonize(algorithm="flint")
....: t = walltime(t)
....: t
0.022356271743774414

Why is it that in the first case with a 20 × 20 matrix it takes >1 second while in the second case it is instant?

In both cases the matrix is invertible, thus the echelon form is identity.

(larger context: sagemath/sage#39197 , if flint is the fastest in all cases it would be easiest to just switch to flint, but this is not the case at the moment)

[albinahlback]

Can you replicate this in C, that is, make a MWE? I am not familiar with how Python works under the hood.

[albinahlback]

Perhaps the tuning parameters could be changed for fmpz_mat_rref. Nonetheless, they should probably be tuned on a hardware-level.

[user202729]

Sure:

#include <flint/flint.h>
#include <flint/fmpz.h>
#include <flint/fmpz_mat.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

void random_fmpz_mat(fmpz_mat_t mat, slong entry_size) {
    slong i, j;
    flint_rand_t state;
    flint_randinit(state);
    for (i = 0; i < fmpz_mat_nrows(mat); i++) {
        for (j = 0; j < fmpz_mat_ncols(mat); j++) {
            fmpz* tmp = fmpz_mat_entry(mat, i, j);
            fmpz_randbits(tmp, state, entry_size);
            fmpz_add_ui(tmp, tmp, 1);
        }
    }
}

int main() {
    slong entry_size = 10000;
    slong num_col = 20, num_row = 20;
    //slong num_col = 40, num_row = 40;

    fmpz_mat_t A;
    fmpz_mat_init(A, num_row, num_col);

    random_fmpz_mat(A, entry_size);

    clock_t start = clock();
    fmpz den;
    fmpz_init(&den);
    fmpz_mat_rref(A, &den, A);
    clock_t end = clock();

    double time_taken = (double)(end - start) / CLOCKS_PER_SEC;
    printf("Time taken: %f seconds\n", time_taken);

    fmpz_mat_clear(A);

    return 0;
}

Change the parameter between 20 and 40 at the commented out line.

[albinahlback]

I would guess fmpz_mat_rref is suboptimal in the sense that it does not take the entry sizes into consideration. I don't know about the complexity of the operation, but it looks like the the entry sizes where not taken into consideration.

[albinahlback]

I suppose @fredrik-johansson is the expert here. Perhaps he can answer when he is back from vacation and has time.