Alright, it's time to play the guessing game.

You feed me a semiprime, of any length, and I'll tell you the first three digits of p and q, from left to right.

I get no hints besides the semiprime itself.

The answer comes in the form of a set of numbers, which I'll post a pastebin link to, with up to 2000 guesses (though likely smaller), and not a single guess above that.

If your pair of 3 digit numbers is present in that set, I win.

If not, you win.

Any takers?

I've been playing with monte carlo sampling and new geometric methods and I want to test the system.

1500379577

@antigermgerm thats a pretty small number bruh, but I'll give it a go.

@Wisecrack my wife told me size don't matter

@antigermgerm Your wife is lying.

Also your pair, with the first 3 digits of p and q (working from left to right) is in this list:

https://rentry.co/9dbznxdf

Generally I need semiprimes with at least 6 digits per factor, can be more, as long as both p and q are 6 digits, but this is a good sign.

I hadn't tested on smaller numbers yet.

Anyway, its 1436 pairs, out of 1 million possible, or 810k possible depending on how you calculate it.

Also, thank you for testing. And post your P and Q factors after I've posted my answer if you could so anyone watching or participating can check it for themselves.

Bonus points for people posting the hash of p or q, along with the value of N, so trolling (by posting non-semiprimes, or posting public keys from known websites) can be mostly avoided.

I'm aiming for verifiable results.

@Wisecrack True! It was 26479*56663

@antigermgerm It's a tree-generating algorithm with some semi-novel geometric interpretations on the opening of coppersmith attacks.

Generally for the first digit of p and q, it pairs it down to about 20% the length (some of the filters pair it down further).

For the second digit, growth is sub x10,

For the third digit, as you see, we hit a lower limit of 1-2% of the total possible pairs, though I've seen lower limits with various independant filters.

I'm currently working on the false negative and false positive rates and have a solid way of dealing with those that I just need to test on a large scale.

@Wisecrack would u be kind enough to vulgarize that a little? So you're discovering the factors out of mathematical interpolation? How?

@antigermgerm its complicated but it amounts to term rewriting.

I'll post more when I've confirmed more. Also need more public tests. Not just for eyeballs, but to isolate the results of the code from the answers so I know theres no cross-contamination where the code is cheating somehow.

@antigermgerm

> my wife told me size don't matter

If your wife also told you that grammar doesn’t matter, she lied.

@Wisecrack you are devRant's math wizard! :)

@Lensflare you're so rule-based mate. No offense but you might be retarded in a piagetian sense. Do you even know about Jean Piaget?

https://simplypsychology.org/piaget...

@Lensflare I know she's lying. She said that but then she got scared of my giant dong.

@antigermgerm "I know she's lying. She said that but then she got scared of my giant dong."

I've heard of ending a war with guns. I've heard of ending a war with tanks.

But never a giant dong.

@Lensflare post a large semiprime, lets test if I can retrieve the first three digits of its factors in sub 2k samples.

@antigermgerm normally I‘m not such a pedantic grammar nazi but your inability to learn despite many corrections from my side leaves me with no other choice but to let you know that you are the retarded one.
I‘m sorry.

It's too hot for me to understand what you're talking about, but I'm leaving a comment so that hopefully when it's less hot I can re-read the rant

@BordedDev boils down to this:

for any number n, which is the product of two prime numbers p and q, I can approximate p and q out to some k number of digits.

For example if we're working with

p = 80721930301

q = 316975007579

n = pq = 25586834468950984751279

Ck=3 = [807, 316]

Ck=4 = [8072, 3169], etc

It turns out with a little bit of work, I can get a subset of pairs that are likely approximations of p and q.

For ck=3, the total number of possible pairs of 3 digit numbers that approximate p and q is about 1 million.

Of that 1 million I can always get a subset of that, about 22% at the limit, or <2000 possible pairs, that is always guaranteed to contain Ck=3= ~p and ~q.

I can likewise do this for Ck=4, Ck=5, etc, for any arbitrary length of n.

Theres some clever exploits of geometric interpretations of multiplication, and underlying novel properties, but thats the gist.

Still a lot of work to do but I'd be happy to field more tests for those interested.

3374984681642906029895764478867930155957

You'll fail this one.

@whimsical your pair will be in this data:

https://rentry.co/zokbyx9q

what were your factors?

I'd like to add it reduces the search space to about 0.1% to 0.22% not 1-2% or 22% as I mistakenly posted.

@Wisecrack Sounds very interesting, but what about 6? Or is that cheating? Then 10941738641570527421809707322040357612003732945449205990913842131476349984288934784717997257891267332497625752899781833797076537244027146743531593354333897

@BordedDev

No, 6 digits isn't cheating, I'm just not there in terms of performance, though I know exactly how I'm getting to it. Probably be another week.

Anyway,

your pair is in this dataset:

https://rentry.co/xpuudpfq

Thank you for participating.

Fun fact: So far all of the numbers picked by other users ended in a 7.

@Wisecrack will you crack the twin prime conjecture at some time?

@Lensflare Not any times soon. I did take a crack at goldbach, and ended up with new ways of stating the problem as-we-know-it, which is new avenues of attack, but it'd take 20 fucking pages to explain it.

Basically for goldbach to be proven wrong, there would have to exist a very strangely constructed number that, through a conspiracy of all the smaller primes, just manages to evade being constructed by a sum of primes.

As the length of the number of sums constructed becomes a covered set of all permutations of the sum of any two primes under any given n, each pair becomes a witness against the probability of finding one of these 'strange' integers, for some value of 'strange'.

You can represent the set of all even numbers under n as G(n), and all primes as p, and show through construction that for all G(n)-p, there exists no difference G(n)-p, that is not prime, but thats as far as I got.

It actually has a deep relationship to the TWC thats expressible in this way.

@Wisecrack too bad.

With the semiprimes I guess it goes more into the direction of proving N = PN

@Wisecrack Ah no I meant the semiprime 6 (2x3)

BTW the other one was RSA-155

102639592829741105772054196573991675900716567808038066803341933521790711307779

× 106603488380168454820927220360012878679207958575989291522270608237193062808643

Though now that lensflare mentioned the 7 thing, I guess RSA-160 would be interesting as well

@BordedDev [102, 106] is in the set.

https://rentry.co/xpuudpfq

I'm aiming for RSA-896, and if the full solution works on RSA-896 then it is known that by default it will work on RSA-2048.

@Lensflare good catch on the 7.

I hadn't considered that might be a determining factor, though by the amount of numbers I've already tested on, if it were an issue it would have already come up.

@BordedDev "6 (2x3) "

Ah, I see.

I dunno. At smaller bitlengths geometric attacks break down. Sampler doesn't like them.

I'll have to go into my magic black box and sprinkle more of my fairy god mother's pixie dust/cremated ashes on it and see what results pop out!

Consult a doctor if your RSA lasts more than 4 hours.

Mega disappointed though because it's not enough to destroy modern civilization with yet.

@Wisecrack Yeah, was just sharing for the others ;P

For me it was mostly a tease since you can't get a 3 digit number for something that just uses single digit numbers (since leading 0s don't count ;P)

@BordedDev "(since leading 0s don't count ;P)"

You've probably already figured out the baseline mechanism, but actually, with term rewriting, and chunking, they do.

Otherwise it'd still be exponential complexity rather than subexponential.

It's in GNFS territory, and I'm looking at it as an accelerator for coppersmith for reference.

@Wisecrack You're giving me a little too much credit, I've always enjoyed math and have decent intuition. Unfortunately at the end of my schooling the advanced math we had a terrible teacher, and she scrapped the whole advanced math class because nobody showed up on the first day after holiday (even though no other after school stuff started that first week). Though my last normal math teacher was nice, enjoyed "arguing" with him how 0/0 should equal 1.

All in all what I'm trying to say, is what you're doing is way above my level XD

@BordedDev

0/0 =1?

Ha, thats a fun one. Probably drove them to pull their hair out.

Math is mostly about developing a deep intuition and familiarity with the properties of numbers, rather than the rules.

Do cryptography. You'll encounter infinite dead ends but learn a lot about the numbers along the way.

@Wisecrack It did indeed, he has also did a class where he explained how 1 + 1 = 3

Haven't considered doing anything in cryptography, but I'll have a look at it