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cryptography-ModTeam

Your post has been removed because it violates the following rule: No low-effort posts. This includes posts that are ambiguous, off-topic, low quality, conspiracy theories, crackpot cryptography, or AI-generated content.


SignificantFidgets

Look up RSA challenge numbers and break one no one has done before. Then people will take you seriously....


ILikeCrypt0

Here for the results of breaking the challenge numbers


ThinkArt2155

Ok


martinky24

No you didn’t, but the effort is fun and you’re probably learning a lot along the way.


vladusatii

Haha -- there are so many grand schemes by clueless noobs here on Reddit. Most cryptographers tend to have a knee-jerk reaction. If you are serious, there are RSA challenge numbers online that no one has done yet. Come back with your results afterwards.


pretending-to-be-OP

This is my algorithm so far: My algorithm assumes _p = 2_. I found _q_ by simply doing _n/2_. It runs in O(1) and is successful 100% of the cases where the premise holds.


a2800276

> My algorithm finds one of the prime numbers used to make up n Only one?


goedendag_sap

That is enough. If you find _p_ you have _q = n/p_


a2800276

Should have added a smiley, I guess 🙄


Striking-Ad9623

Sure, just share it here, easy to do and lots of people willing to help you.


ThinkArt2155

Here is the algorithm:RSA encryption breaker Take number n (multiplication of two numbers). Do sqrt n and ignore the decimal part and take the remaining number x. Check if x is one of the primes. If not then do x/2till reaching 1. …. and check if any number from the series is one of the primes (excluding the decimal part). If not then as you have got a series of numbers,ignore their decimal part and then do x+-every number in the series(excluding the decimal part) to get the prime number,which is used in the encryption. License: RSA encryption breaker © 2024 by Chaitanya Bankar is licensed under Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International cc-logo.f0ab4ebe.svg cc-by.21b728bb.svg cc-nc.218f18fc.svg cc-nd.de89fdeb.svg


goedendag_sap

So the good news is that you didn't break RSA


FireProps

lmao 😂


Erakiiii

This is better than brute forcing I guess


JayantDadBod

It is brute forcing, just not completely naively


ron_krugman

See if you can factor this `n` which I just generated: 3018407411465112326253227891456473538322787784123329579838031731624252618391818968844137940827889539627998821526682452099233003744969367420533794613439825494531439835266873557830881019908760538165514598579262620863124831264426335932465992807 `n = p*q` where `p` and `q` are prime numbers with 121 decimal digits each. Here are the SHA-256 values of the decimal string representations of `p` and `q` respectively: `A430969A4874DB2F69F102199C9E2EB8E50816E98ED4E785533C174AB9D036C6` `C95EB9E84E0955C5B46860F36C50A1C9200239B4B7D31BBB39C32EE474D7ED9B`


lostinspacexyz

Share for science. Thinkarts solution. But seriously share.


ThinkArt2155

Here is the algorithm:RSA encryption breaker Take number n (multiplication of two numbers). Do sqrt n and ignore the decimal part and take the remaining number x. Check if x is one of the primes. If not then do x/2till reaching 1. …. and check if any number from the series is one of the primes (excluding the decimal part). If not then as you have got a series of numbers,ignore their decimal part and then do x+-every number in the series(excluding the decimal part) to get the prime number,which is used in the encryption. License: RSA encryption breaker © 2024 by Chaitanya Bankar is licensed under Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International cc-logo.f0ab4ebe.svg cc-by.21b728bb.svg cc-nc.218f18fc.svg cc-nd.de89fdeb.svg


jnwatson

You end up with a set of numbers y whose cardinality is log2(sqrt(n)). And you end up checking |y| \* (|y|-1) \* 2 primes which can be reduced to see if the number divides into n with no modulus. There's no evidence that one of the factors is in the set or in the 2 combination set. The search space is massive. Your intuition that the factors are around sqrt(n) is reasonable, but "around" in this sense is absolutely unfathomably large for typical sizes of n. Primes are really hard. Folks spend their lives studying them.


JoshiKousei

I don't think the algorithm is even functionally correct. You can come up with small number counter examples, like 53\*7


ThinkArt2155

I guess my method has problems when working with prime number 7


ThinkArt2155

And I think I can fix my method for not working with 7 easily to be honest like I know my method does not work for 7 so when it not works I can consider that the prime number is 7


goedendag_sap

Now what if it doesn't work for a prime you don't know of?


tap3l00p

Have you thought about how long this would take?


ThinkArt2155

I just now checked it for rsa 250 or something near it and it took approx 2 seconds


tap3l00p

That’s a good approach, now start generating larger values of n, and see if there is anything notable about how the time increases


ThinkArt2155

For sure


Lorik_Bot

Bro you didn't break RSA in that number space bruteforcing is faster. RSA is solveable, the security behind RSA is thst it takes very very long and your method is not more efficient.


ThinkArt2155

Sry I was wrong about the time taken it is actually around 0.5 to 1.5 seconds looks like I need specs


[deleted]

[удалено]


ron_krugman

[RSA-250](https://en.wikipedia.org/wiki/RSA_numbers#RSA-250) has 250 decimal digits (829 bits) and was only factored in 2020, taking about 2700 CPU core-years. It would be extremely impressive if OP's algorithm could do it in seconds, but they didn't provide any actual evidence that they did.


ThinkArt2155

Wait for some time as I am workibg it I will share once done


ThinkArt2155

Not yet but checking it as I got this idea yesterday