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the gnfs is the most efficient algorithm for factoring numbers formed of equal composites.

But it’s sequential/Linea Algebra parts mean (If I’m not wrong), that it requires at least 10 minutes on current hardware to factor semi‑primes formed of composites of equal length (in the case of 382 bits semi‑primes).

Are there less efficient algorithms but by being more parallelizable, that would allow to solve batch of such semi‑primes faster using more resources ? (the batch is sequential : a number needs to be factored by a participant to know the next number to solve)…

For more information, here’s in pseudocode how the semiprime needs to be chosen by the person who submits both the semiprime/factors (the real program contains code so that the semiprime is chosen efficiently and that factoring is the issue) :

uint1024 w = hash(common_challenge_to_participants);

//Check that |common_challenge_to_participants| <= \tilde{n} = 16 * |n|_2.
uint64_t abs_offset = (chosen_wOffset > 0) ? common_challenge_to_participants.wOffset : -common_challenge_to_participants.wOffset;


if (abs_offset > 16 * common_challenge_to_participants.common_target_nBits) { //where target_nbits is between 350 to 400
    LogPrintf("invalid wOffset\n");
    return false;
}


//Get the semiprime from random seed
mpz_t n, W;
mpz_init(n);
mpz_init(W);
mpz_import(W, 16, -1, 8, 0, 0, w.u64_begin()); //cast w to W


//Add the offset to w to find the semiprime submitted: n = w + offset
if (common_challenge_to_participants.wOffset >= 0) {
    mpz_add_ui(n, W, abs_offset);
} else {
    mpz_sub_ui(n, W, abs_offset);
}


//Clear memory for W.
mpz_clear(W);


//Check the number n has nBits
if (mpz_sizeinbase(n, 2) != common_challenge_to_participants.common_target_nBits) {
    LogPrintf("invalid nBits");
    mpz_clear(n);
    return false;
}


//Divide the factor submitted by N
mpz_t nP1, nP2;
mpz_init(nP1);
mpz_init(nP2);
mpz_import(nP1, 16, -1, 8, 0, 0, common_challenge_to_participants.nP1.u64_begin());
mpz_tdiv_q(nP2, n, nP1);
//Check the bitsizes are as expected
const uint16_t nP1_bitsize = mpz_sizeinbase(nP1, 2);
const uint16_t expected_bitsize = (common_challenge_to_participants.common_target_nBits >> 1) + (common_challenge_to_participants.common_target_nBits & 1);


if (nP1_bitsize != expected_bitsize) {
    LogPrintf("nP1 expected bitsize=%s, actual size=%s\n", nP1_bitsize, expected_bitsize);
    mpz_clear(n);
    mpz_clear(nP1);
    mpz_clear(nP2);
    return false;
}


//Check nP1 is a factor
mpz_t n_check;
mpz_init(n_check);
mpz_mul(n_check, nP1, nP2);


//Check that nP1*nP2 == n.
if (mpz_cmp(n_check, n) != 0) {
    LogPrintf("nP1 does not divide N.  N=%s nP1=%s\n", mpz_get_str(NULL, 10, n), mpz_get_str(NULL, 10, nP1));
    mpz_clear(n);
    mpz_clear(nP1);
    mpz_clear(nP2);
    mpz_clear(n_check);
    return false;
}


//Check that nP1 <= nP2.
if (mpz_cmp(nP1, nP2) > 0) {
    LogPrintf("error: nP1 must be the smallest factor. N=%s nP1=%s\n", mpz_get_str(NULL, 10, n), mpz_get_str(NULL, 10, nP1));
    mpz_clear(n);
    mpz_clear(nP1);
    mpz_clear(nP2);
    mpz_clear(n_check);
    return false;
}


//Clear memory
mpz_clear(n);
mpz_clear(n_check);


//Test nP1 and nP2 for primality.
int is_nP1_prime = mpz_probab_prime_p(nP1, params.MillerRabinRounds);
int is_nP2_prime = mpz_probab_prime_p(nP2, params.MillerRabinRounds);


//Clear memory
mpz_clear(nP1);
mpz_clear(nP2);


//Check they are both prime
if (is_nP1_prime == 0 || is_nP2_prime == 0) {
    LogPrintf("At least 1 composite factor found, rejected.\n");
    return false;
}


return true;
  • Update : I know about the other question. But My problem is doing it in a specific very short timeframe.
user2284570
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2 Answers2

2

One thought is the highly parallelisable elliptic curve method. There is a page of the largest prime factors found using ECM and the corresponding composite. By both metrics, the problem size seems to fit in range.

ETA 20240313: Given the data provided by Samuel Neves below, it seems technically possible to factor such numbers using ECM in 40,000 4 GHz core minutes with a minimum run time of one minute. After that the question becomes one of economics. You might investigate how much 3000 or so Raspberry Pi 5 cards ($4\times$ 2.4Ghz) cost when bought in bulk and amortise that upfront cost over the number of instances that you win less the running cost of the rig. I'd suggest not considering FPGAs as the cosenter link description heret of the elliptic curve method is dominated by high-precision multiplies which benefit from specialised circuitry on CPUs. Your number of wins may be bounded by how long it take someone to buy a bigger setup.

Note that the expected time between solutions will be exponentially distributed, but this is not affected by the introduction of a new target number if you fail to win a given instance.

Daniel S
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-2

I don’t fully understand how to do it but mseive can perform the linear Algebra/krylov part on several ɢᴘᴜ which solve the sequential problem : the remaining is in theory to just add more cores.

Just the matter to modify ᴄᴀᴅᴏ‑ɴꜰꜱ default script.

However, turns out you have to find the correct number as a result of a hash : so the real effort is to factor something like 20,000 such numbers on that period and not just 1 : won’t happen for less $100…

user2284570
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