Do all the things like ++ or -- rants, post your own rants, comment on others' rants and build your customized dev avatarSign Up
From the creators of devRant, Pipeless lets you power real-time personalized recommendations and activity feeds using a simple APILearn More
Search - "factorization"
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?250
"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
@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:
w = w + i
Still working but if anyone sees obvious problems I'd LOVE to hear about it.39
So I accidentally reinvented Pollard's algorithm.
CONTEST - Win big $$$ straight from Wisecrack!
For all those who participated in my original "cracking prime factorization" thread (and several I decided to add just because), I'm offering a whopping $5 to anyone who posts a 64 bit *product* of two primes, which I cant factor. Partly this is a thank you for putting up with me.
FIVE WHOLE DOLLARS! In 1909 money thats $124 dollars! Imagine how many horse and buggy rides you could buy with that back then! Or blowjobs!
Probably not a lot!
So the contest rules are simple:
Enter 32 for the number of bits per prime, and generate a 64 bit product.
Post it here to enter the contest.
Products must be 64 bits, and the result of just *two* prime numbers. Smaller or larger bit lengths for products won't be accepted at this time.
I'm expecting a few entries on this. Entries will generally be processed in the order of submission, but I reserve the right to wave this rule.
After an entry is accepted, I'll post "challenge accepted. Factoring now."
And from that point on I have no more than 5 hours to factor the number, (but results usually arrive in 30-60 minutes).
If I fail to factor your product in the specified time (from the moment I indicate I've begun factoring), congratulations, you just won $5.
Payment will be made via venmo or other method at my discretion.
One entry per user. Participants from the original thread only, as well as those explicitly mentioned.
Limitations: Factoring shall be be done
1. without *any* table lookup of primes or equivalent measures, 2. using anything greater than an i3, 3. without the aid of a gpu, 4. without multithreading. 5. without the use of more than one machine.
To claim your prize, post the original factors of your product here, after the deadline has passed.
And then I'll arrange payment of the prize.
You MUST post the factors of your product after the deadline, to confirm your product and claim your prize.99
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
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.9
Is there any exact way to get the product of all primes under n multiplied together, without explicitly knowing what those primes are?
Lets call this number V.
Because hypothethetically, if we calculate from the *base* of V, then we can derive easy divisibility rules for V-1 and V+1, as laid out
And then, unless I've misunderstood something, the problem of factorization has been changed from division into an addition and subtraction problem.12
Taught myself how to program a Texas Instruments 84 graphical calculator to solve the common mathematical equations in high school. Shamelessly copied a lot from the internet but learned quite quickly. Things like solving 2 and 3 degree polynomials, prime factorization or calculating integrals and stuff.
I quite liked how I could make life easier for myself and eventually class mates. Just rolled into software engineering afterwards I guess. No regrets thus far...1
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.
Hang with me! This is *not* a math shitpost, I repeat, it is NOT a math shitpost, not entirely anyway.
It appears there is for products of two non-trivial factors, a real number n (well a rational number anyway) such that p/n = i (some number in the set of integers), whos factor chain is apparently no greater than floor(log(log(p))**2)-2, and whos largest factor is never greater than p^(1/4).
And that this number is at least derivable, laboriously with the following:
And assuming you have the factors of p/z = jkl..
then instead of doing
p/(jkl..) = z
you can do
p-(jkl) to get the value of [result] whos index is a-1
Getting the actual factor tree of p/z is another matter, but its a start.
Edit: you have to provide your own product.
Preferably import Decimal first.3
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:
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
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, P.U.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.1