So I cracked prime factorization. For real.

I can factor a 1024 bit product in 11hours on an i3.
No GPU acceleration, no massive memory overhead. Probably a lot faster with parallel computation on a better cpu, or even on a gpu.

4096 bits in 97-98 hours.

Verifiable. Not shitting you. My hearts beating out of my fucking chest. Maybe it was an act of god, I don't know, but it works.

What should I do with it?

I messaged a professor at MIT and surprisingly got a response back.

He told me that "generating primes deterministically is a solved problem" and he would be very surprised if what I wrote beat wheel factorization, but that he would be interested if it did.

It didnt when he messaged me.

It does now.

Tested on primes up to 26 digits.
Current time tends to be 1-100th to 2-100th of a second.

Seems to be steady.

First n=1million digits *always* returns false for composites, while for primes the rate is 56% true vs false, and now that I've made it faster, I'm fairly certain I can get it to 100% accuracy.

In fact what I'm thinking I'll do is generate a random semiprime using the suspected prime, map it over to some other factor tree using the variation on modular expotentiation several of us on devrant stumbled on, and then see if it still factors. If it does then we know the number in question is prime. And because we know the factor in question, the semiprime mapping function doesnt require any additional searching or iterations.

The false negative rate, I think goes to zero the larger the prime from what I can see. But it wont be an issue if I'm right about the accuracy being correctable.

I'd like to thank the professor for the challenge. He also shared a bunch of useful links.

That ones a rare bird.

POSTMORTEM

"4096 bit ~ 96 hours is what he said.

IDK why, but when he took the challenge, he posted that it'd take 36 hours"

As @cbsa wrote, and nitwhiz wrote "but the statement was that op's i3 did it in 11 hours. So there must be a result already, which can be verified?"

I added time because I was in the middle of a port involving ArbFloat so I could get arbitrary precision. I had a crude desmos graph doing projections on what I'd already factored in order to get an idea of how long it'd take to do larger
bit lengths

@p100sch speculated on the walked back time, and overstating the rig capabilities. Instead I spent a lot of time trying to get it 'just-so'.
Worse, because I had to resort to "Decimal" in python (and am currently experimenting with the same in Julia), both of which are immutable types, the GC was taking > 25% of the cpu time.

Performancewise, the numbers I cited in the actual thread, as of this time:
largest product factored was 32bit, 1855526741 * 2163967087, took 1116.111s in python.
Julia build used a slightly different method, & managed to factor a 27 bit number, 103147223 * 88789957 in 20.9s,
but this wasn't typical.

What surprised me was the variability. One bit length could take 100s or a couple thousand seconds even, and a product that was 1-2 bits longer could return a result in under a minute, sometimes in seconds.

This started cropping up, ironically, right after I posted the thread, whats a man to do?

So I started trying a bunch of things, some of which worked. Shameless as I am, I accepted the challenge. Things weren't perfect but it was going well enough. At that point I hadn't slept in 30~ hours so when I thought I had it I let it run and went to bed. 5 AM comes, I check the program. Still calculating, and way overshot. Fuuuuuuccc...
So here we are now and it's say to safe the worlds not gonna burn if I explain it seeing as it doesn't work, or at least only some of the time.

Others people, much smarter than me, mentioned it may be a means of finding more secure pairs, and maybe so, I'm not familiar enough to know.

For everyone that followed, commented, those who contributed, even the doubters who kept a sanity check on this without whom this would have been an even bigger embarassement, and the people with their pins and tactical dots, thanks.

So here it is.
A few assumptions first.
Assuming p = the product,
a = some prime,
b = another prime,
and r = a/b (where a is smaller than b)
w = 1/sqrt(p)
(also experimented with w = 1/sqrt(p)*2 but I kept overshooting my a very small margin)

x = a/p
y = b/p

1. for every two numbers, there is a ratio (r) that you can search for among the decimals, starting at 1.0, counting down. You can use this to find the original factors e.x. p*r=n, p/n=m (assuming the product has only two factors), instead of having to do a sieve.

2. You don't need the first number you find to be the precise value of a factor (we're doing floating point math), a large subset of decimal values for the value of a or b will naturally 'fall' into the value of a (or b) + some fractional number, which is lost. Some of you will object, "But if thats wrong, your result will be wrong!" but hear me out.

3. You round for the first factor 'found', and from there, you take the result and do p/a to get b. If 'a' is actually a factor of p, then mod(b, 1) == 0, and then naturally, a*b SHOULD equal p.
If not, you throw out both numbers, rinse and repeat.

Now I knew this this could be faster. Realized the finer the representation, the less important the fractional digits further right in the number were, it was just a matter of how much precision I could AFFORD to lose and still get an accurate result for r*p=a.

Fast forward, lot of experimentation, was hitting a lot of worst case time complexities, where the most significant digits had a bunch of zeroes in front of them so starting at 1.0 was a no go in many situations. Started looking and realized
I didn't NEED the ratio of a/b, I just needed the ratio of a to p.

Intuitively it made sense, but starting at 1.0 was blowing up the calculation time, and this made it so much worse.

I realized if I could start at r=1/sqrt(p) instead, and that because of certain properties, the fractional result of this, r, would ALWAYS be 1. close to one of the factors fractional value of n/p, and 2. it looked like it was guaranteed that r=1/sqrt(p) would ALWAYS be less than at least one of the primes, putting a bound on worst case.

The final result in executable pseudo code (python lol) looks something like the above variables plus

while w >= 0.0:
if (p / round(w*p)) % 1 == 0:
x = round(w*p)
y = p / round(w*p)
if x*y == p:
print("factors found!")
print(x)
print(y)
break
w = w + i

Still working but if anyone sees obvious problems I'd LOVE to hear about it.

As you can see from the screenshot, its working.
The system is actually learning the associations between the digit sequence of semiprime hidden variables and known variables.

Training loss and value loss are super high at the moment and I'm using an absurdly small training set (10k sequence pairs). I'm running on the assumption that there is a very strong correlation between the structures (and that it isn't just all ephemeral).

This initial run is just to see if training an machine learning model is a viable approach.

Won't know for a while. Training loss could get very low (thats a good thing, indicating actual learning), only for it to spike later on, and if it does, I won't know if the sample size is too small, or if I need to do more training, or if the problem is actually intractable.

If or when that happens I'll experiment with different configurations like batch sizes, and more epochs, as well as upping the training set incrementally.

Either case, once the initial model is trained, I need to test it on samples never seen before (products I want to factor) and see if it generates some or all of the digits needed for rapid factorization.

Even partial digits would be a success here.
And I expect to create multiple training sets for each semiprime product and its unknown internal variables versus deriable known variables. The intersections of the sets, and what digits they have in common might be the best shot available for factorizing very large numbers in this approach.

Regardless, once I see that the model works at the small scale, the next step will be to increase the scope of the training data, and begin building out the distributed training platform so I can cut down the training time on a larger model.

I also want to train on random products of very large primes, just for variety and see what happens with that. But everything appears to be working. Working way better than I expected.

The model is running and learning to factorize primes from the set of identities I've been exploring for the last three fucking years.

Feels like things are paying off finally.

Will post updates specifically to this rant as they come. Probably once a day.

Still on the primenumbers bender.
Had this idea that if there were subtle correlations between a sufficiently large set of identities and the digits of a prime number, the best way to find it would be to automate the search.

And thats just what I did.

I started with trace matrices.
I actually didn't expect much of it. I was hoping I'd at least get lucky with a few chance coincidences.

My first tests failed miserably. Eight percent here, 10% there. "I might as well just pick a number out of a hat!" I thought.

I scaled it way back and asked if it was possible to predict *just* the first digit of either of the prime factors.

That also failed. Prediction rates were low still. Like 0.08-0.15.

So I automated *that*.

After a couple days of on-and-off again semi-automated searching I stumbled on it.

[1144, 827, 326, 1184, -1, -1, -1, -1]

That little sequence is a series of identities representing different values derived from a randomly generated product.

Each slots into a trace matrice. The results of which predict the first digit of one of our factors, with a 83.2% accuracy even after 10k runs, and rising higher with the number of trials.

It's not much, but I was kind of proud of it.

I'm pushing for finding 90%+ now.

Some improvements include using a different sort of operation to generate results. Or logging all results and finding the digit within each result thats *most* likely to predict our targets, across all results. (right now I just take the digit in the ones column, which works but is an arbitrary decision on my part).

Theres also the fact that it's trivial to correctly guess the digit 25% of the time, simply by guessing 1, 3, 7, or 9, because all primes, except for 2, end in one of these four.

I have also yet to find a trace with a specific bias for predicting either the smaller of two unique factors *or* the larger. But I haven't really looked for one either.

I still need to write a generate that takes specific traces, and lets me mutate some of the values, to push them towards certain 'fitness' levels.
This would be useful not just for very high predictions, but to find traces with very *low* predictions.

Why? Because it would actually allow for the *elimination* of possible digits, much like sudoku, from a given place value in a predicted factor.

I don't know if any of this will even end up working past the first digit. But splitting the odds, between the two unique factors of a prime product, and getting 40+% chance of guessing correctly, isn't too bad I think for a total amateur.

Far cry from a couple years ago claiming I broke prime factorization. People still haven't forgiven me for that, lol.

I've been away a while, mostly working 60-70 hour weeks.

Found a managers job and the illusion of low-level stability.

Also been exploring elliptic curve cryptography and other fun stuff, like this fun equation...

i = log(n, 2**0.5)
base = (((int((n/(n*(1-(n/((((abs(int(n+(n/(1/((n/(n-i))+(i+1)))))+i)-(i*2))/1))/1/i)))))*i)-i)+i))

...as it relates to A143975 a(n) = floor(n*(n+3)/3)

Most semiprimes n=pq, where p<q, appear to have values k in the sequence, where k is such that n+m mod k equals either p||q or a multiple thereof.

Tested successfully up to 49 bits and counting. Mostly haven't gone further because of work.

Theres a little more math involved, and I've (probably incorrectly) explained the last bit but the gist is the factorization doesn't turn up anything, *however* trial lookups on the sequence and then finding a related mod yields k instead, which can be used to trivially find p and q.

It has some relations to calculating on an elliptic curve but thats mostly over my head, and would probably bore people to sleep.

So I promised a post after work last night, discussing the new factorization technique.

As before, I use a method called decon() that takes any number, like 697 for example, and first breaks it down into the respective digits and magnitudes.

697 becomes -> 600, 90, and 7.

It then factors *those* to give a decomposition matrix that looks something like the following when printed out:

offset: 3, exp: [[Decimal('2'), Decimal('3')], [Decimal('3'), Decimal('1')], [Decimal('5'), Decimal('2')]]

offset: 2, exp: [[Decimal('2'), Decimal('1')], [Decimal('3'), Decimal('2')], [Decimal('5'), Decimal('1')]]

offset: 1, exp: [[Decimal('7'), Decimal('1')]]

Each entry is a pair of numbers representing a prime base and an exponent.

Now the idea was that, in theory, at each magnitude of a product, we could actually search through the *range* of the product of these exponents.

So for offset three (600) here, we're looking at

2^3 * 3 ^ 1 * 5 ^ 2.

But actually we're searching

2^3 * 3 ^ 1 * 5 ^ 2.
2^3 * 3 ^ 1 * 5 ^ 1
2^3 * 3 ^ 1 * 5 ^ 0
2^3 * 3 ^ 0 * 5 ^ 2.
2^3 * 3 ^ 1 * 5 ^ 1

etc..

On the basis that whatever it generates may be the digits of another magnitude in one of our target product's factors.

And the first optimization or filter we can apply is to notice that assuming our factors pq=n,
and where p <= q, it will always be more efficient to search for the digits of p (because its under n^0.5 or the square root), than the larger factor q.

So by implication we can filter out any product of this exponent search that is greater than the square root of n.

Writing this code was a bit of a headache because I had to deal with potentially very large lists of bases and exponents, so I couldn't just use loops within loops.

Instead I resorted to writing a three state state machine that 'counted down' across these exponents, and it just works.

And now, in practice this doesn't immediately give us anything useful. And I had hoped this would at least give us *upperbounds* to start our search from, for any particular digit of a product's factors at a given magnitude. So the 12 digit (or pick a magnitude out of a hat) of an example product might give us an upperbound on the 2's exponent for that same digit in our lowest factor q of n.

It didn't work out that way. Sometimes there would be 'inversions', where the exponent of a factor on a magnitude of n, would be *lower* than the exponent of that factor on the same digit of q.

But when I started tearing into examples and generating test data I started to see certain patterns emerge, and immediately I found a way to not just pin down these inversions, but get *tight* bounds on the 2's exponents in the corresponding digit for our product's factor itself. It was like the complications I initially saw actually became a means to *tighten* the bounds.

For example, for one particular semiprime n=pq, this was some of the data:

n - offset: 6, exp: [[Decimal('2'), Decimal('5')], [Decimal('5'), Decimal('5')]]

q - offset: 6, exp: [[Decimal('2'), Decimal('6')], [Decimal('3'), Decimal('1')], [Decimal('5'), Decimal('5')]]

It's almost like the base 3 exponent in [n:7] gives away the presence of 3^1 in [q:6], even
though theres no subsequent presence of 3^n in [n:6] itself.

And I found this rule held each time I tested it.

Other rules, not so much, and other rules still would fail in the presence of yet other rules, almost like a giant switchboard.

I immediately realized the implications: rules had precedence, acted predictable when in isolated instances, and changed in specific instances in combination with other rules.

This was ripe for a decision tree generated through random search.

Another product n=pq, with mroe data
q(4)
offset: 4, exp: [[Decimal('2'), Decimal('4')], [Decimal('5'), Decimal('3')]]

n(4)
offset: 4, exp: [[Decimal('2'), Decimal('3')], [Decimal('3'), Decimal('2')], [Decimal('5'), Decimal('3')]]

Suggesting that a nontrivial base 3 exponent (**2 rather than **1) suggests the exponent on the 2 in the relevant
digit of [n], is one less than the same base 2 digital exponent at the same digit on [q]

And so it was clear from the get go that this approach held promise.

From there I discovered a bunch more rules and made some observations.
The bulk of the patterns, regardless of how large the product grows, should be present in the smaller bases (some bound of primes, say the first dozen), because the bulk of exponents for the factorization of any magnitude of a number, overwhelming lean heavily in the lower prime bases.

It was if the entire vulnerability was hiding in plain sight for four+ years, and we'd been approaching factorization all wrong from the beginning, by trying to factor a number, and all its digits at all its magnitudes, all at once, when like addition or multiplication, factorization could be done piecemeal if we knew the patterns to look for.

I may have accidentally found a legit factorization method that converts factoring to a combinatorics problem over a graph, with a time complexity that is the factorial of the logarithm of the semi prime being factored.

I don't know if this is supposed to be good or not, and I don't want to post it prematurely like I almost always do. Not at least until I study its properties better, but it's still a pretty interesting find I think.

Math question time!

Okay so I had this idea and I'm looking for anyone who has a better grasp of math than me.

What if instead of searching for prime factors we searched for a number above p?
One with a certain special property. BEAR WITH ME. I know I make these posts a lot and I'm a bit of a shitposter, but I'm being genuine here.

Take this cherry picked number, 697 for example.

It's factors are 17, and 41. It's trivial but just for demonstration.

If we represented it's factors as a bit string, where each bit represents the index that factor occurs at in a list of primes, it looks like this

1000001000000

When converted back to an integer that number becomes 4160, which we will call f.

And if we do 4160/(2**n) until the result returns
a fractional component, then N in this case will be 7.

And 7 is the index of our lowest factor 17 (lets call it A, and our highest factor we'll call B) in our primes list.

So the problem is changed from finding a factorization of p, to finding an algorithm that allows you to convert p into f. Once you have f it's a matter of converting it to binary, looking up the indexes of all bits set to 1, and finding the values of those indexes in the list of primes.

I'm working on doing that and if anyone has any insights I'm all ears.

How resource calculations for software services like code analysis, monitoring, etc are done:

Opening fridge, putting all the beer one can find in it.

Opening the necessary tools, e.g Excel, Accounting software, ....

Drinking the first beer.

Starting to aggregate the monthly costs - cause you can never trust the reports written by someone else...

First beer poof.

Looking at the monthly cost, adding columns "Intended use", "Actual usage pattern", "Usage factor"...

Opening next beer...

Usage factor is btw a factor of 0.1 ... 1.0 - to give an estimate how much the products feature are actually used, for further analysis if the invest is justified or not...

Oh. Another half bottle gone...

Filling in the columns...

Oh. Bottle empty and the next one toooooooooooooooo...

*burping*
*cracking finger joints*

Now let's get to the sad part...

Next worksheet, adding infrastructure costs...

Cost and description as columns.

Hehe. Column sounds like gollum.

Another beer...

Ugh. Need the paper reports, manually typing off things for stuff that was e.g. tax deductible.

Many beers die during this task. Poor little beers, dying for such an boring and mundane task...

SUM is a real useful function. I don't think I can add numbers anymore.

Now we can add another sheet.

Hehe. Sheet sounds like shit. And yes, everything in this file is shit.

Summing up costs from both sheets and including the cost factor from 1
... Beeeeeeeer Beeeeer beer we need more beer here... Beer beer beer...

Where was I. Oh yeah. Cost factorization total vs effective.

Why do I want to get even more drunk.

Oh yeah. Most software is completely underused and the costs aren't justified.

Let's add some colored highlighting ...

Uuuuh. ,Too much red. Better change the highlights.

Too much red.

More beer.

Don't give a fuck.

Hm.

Time for some whiskey.

What else is there to do....

Oh yeah.

Diagrams.

The bloody wankers from accounting need diagrams as numbers are too boring.

Not that everything in accounting is boring, no matter how much you paint colors on it... *sigh*

Hm. More whiskey...

Hehe. Whiskey rhymes with frisky.

Uff. Now just need to write mail. Mail mail mail....

"Copy paste the last mail from last month"

Hm.

Ah.

*sipping whiskey*

Spell check extension - to the rescue.

Thesaurus *burps*.

Let's change a few words here and there... Maybe another paragraph there.

Uh....

Trying to attach file...

*fucking mouse is pretty constantly crashing into empty beer bottles*

Done.

Damn.

Need to press send button.

*Creating mess on the desk by just randomly crashing the beer bottles*

Done.

*Pressing computers power button*

Mwahahahaha. No mouse needed.

*regretting to stand up too quickly, nearly barfing on the floor*

Couch ... Where Couch...

After hitting several doors, frames and other stuff, the glorious mission ended successfully with a most graciously executed gut buster on the couch.

(Regretting next morning to have emptied two 6 packs and a few glasses of whiskey)

Let some number P be the product of two factors, A and B

Let the iterations of A*1..2..3..N up to B+1 be a directed graph.

Would this graph be eularian?

If so then it should be possible to use the BEST algorithm to count the number of eularian circuits, yielding B, no?

Edit: this is supported by the following text:

An arborescence can equivalently be defined as a rooted digraph in which the path from the root to any other vertex is unique.

* * *

Where the product of any two primes is unique, the path to it across our graph here, is also unique.

I'm teaching a couple of classes where students (~18 years old) work on their own projects. I just deleted two of those from my machine: one Angular and one Spring Boot, but just boilerplate. Together, they were about 500 MB. I spent 2-3 hours working on a little Go tool to make concurrent HTTP requests and to report statistics on the response time. The entire repository is roughly 500 kB in size, but solves a genuine problem. My students have a bloat ratio of 1000 compared to me as a baseline, but my stuff actually does things. Today, I programmed prime factorization in PHP for some load tests (mod_php vs. PHP-FPM). The PHP script is 1148 bytes long (but the file system reports 4 kB). My students could learn more from such a script than from their overblown "projects", but "PHP sucks" I hearsay, so let's bloat on.

I am trying to decompose a 3D matrix using python library scikit-tensor. I managed to decompose my Tensor (with dimensions 100x50x5) into three matrices. My question is how can I compose the initial matrix again using the decomposed matrix produced with Tensor factorization? I want to check if the decomposition has any meaning. My code is the following:

import logging

from scipy.io.matlab import loadmat

from sktensor import dtensor, cp_als

import numpy as np

//Set logging to DEBUG to see CP-ALS information

logging.basicConfig(level=logging.DEBUG)

T = np.ones((400, 50))

T = dtensor(T)

P, fit, itr, exectimes = cp_als(T, 10, init='random')

// how can I re-compose the Matrix T? TA = np.dot(P.U[0], P.U[1].T)

I am using the canonical decomposition as provided from the scikit-tensor library function cp_als. Also what is the expected dimensionality of the decomposed matrices.