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+--- Forum: Coding (https://sinister.li/Forum-Coding--71)
+--- Thread: Tutorial RSA (/Thread-Tutorial-RSA--93520)
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RE: RSA - phyrrus9 - 12-25-2017
(12-25-2017, 09:30 PM)CC.py Wrote: Wouldn't CS logic fall under mathematical logic?
Technically it's still math. It's boolean algebra
RE: RSA - Blink - 12-25-2017
(12-25-2017, 10:02 PM)phyrrus9 Wrote:(12-25-2017, 09:30 PM)CC.py Wrote: Wouldn't CS logic fall under mathematical logic?
Technically it's still math. It's boolean algebra
It's all math, everything's math, so much math... You are math, I'm math, there's nothing else...
https://www.xkcd.com/435
RE: RSA - Sun0228 - 01-03-2018
Thank you, I think that I've overcome my fear from crypto and math! Actually, it doesn't even look too hard.
RE: RSA - Blink - 01-04-2018
(01-03-2018, 04:32 PM)Sun0228 Wrote: Thank you, I think that I've overcome my fear from crypto and math! Actually, it doesn't even look too hard.
Wait till you get to ECC, it's a bit hard to grasp.
If you wanna see another simple cryptographically function, there's a neat hash function called CubeHash that you can take a look at.
RE: RSA - Sun0228 - 01-04-2018
(01-04-2018, 04:08 PM)Ender Wrote:(01-03-2018, 04:32 PM)Sun0228 Wrote: Thank you, I think that I've overcome my fear from crypto and math! Actually, it doesn't even look too hard.
Wait till you get to ECC, it's a bit hard to grasp.
If you wanna see another simple cryptographically function, there's a neat hash function called CubeHash that you can take a look at.
It didn't take long and my crypto-math fear level is back at normal.
RE: RSA - Dr. Keter - 01-10-2018
Neat! I'll probably be looking more into cryptography. The math is pretty fascinating. I'm still not quite sure why this is secure though. You make it sound like you can compute the private key from the public key. Am I missing something?
(01-04-2018, 04:25 PM)Sun0228 Wrote:(01-04-2018, 04:08 PM)Ender Wrote:(01-03-2018, 04:32 PM)Sun0228 Wrote: Thank you, I think that I've overcome my fear from crypto and math! Actually, it doesn't even look too hard.
Wait till you get to ECC, it's a bit hard to grasp.
If you wanna see another simple cryptographically function, there's a neat hash function called CubeHash that you can take a look at.
It didn't take long and my crypto-math fear level is back at normal.
In my experience, new things tend to look way intimidating before you actually learn them.
RE: RSA - Shinoa - 01-10-2018
(01-10-2018, 04:59 AM)Dr. Keter Wrote: I'm still not quite sure why this is secure though. You make it sound like you can compute the private key from the public key. Am I missing something?That's only assuming you know p and q, and in turn t, without that it's practically impossible. In general you only release your public key and nothing else.
Also glad you enjoyed the tutoral
RE: RSA - Dr. Keter - 01-16-2018
(01-10-2018, 05:19 AM)CC.py Wrote:(01-10-2018, 04:59 AM)Dr. Keter Wrote: I'm still not quite sure why this is secure though. You make it sound like you can compute the private key from the public key. Am I missing something?That's only assuming you know p and q, and in turn t, without that it's practically impossible. In general you only release your public key and nothing else.
Also glad you enjoyed the tutoral
If all you have is the public key, how can you encrypt? You said that the formula to encrypt is (m^e) % n = c where n is p*q. How do you get n?
RE: RSA - Shinoa - 01-16-2018
(01-16-2018, 05:19 AM)Dr. Keter Wrote:(01-10-2018, 05:19 AM)CC.py Wrote:(01-10-2018, 04:59 AM)Dr. Keter Wrote: I'm still not quite sure why this is secure though. You make it sound like you can compute the private key from the public key. Am I missing something?That's only assuming you know p and q, and in turn t, without that it's practically impossible. In general you only release your public key and nothing else.
Also glad you enjoyed the tutoral
If all you have is the public key, how can you encrypt? You said that the formula to encrypt is (m^e) % n = c where n is p*q. How do you get n?
Wait my bad. During the key exchange you give e and n, and factoring n for p and q is super unrealistic because it's such a big number, meaning you can't really compute the private key
RE: RSA - Dr. Keter - 01-16-2018
(01-16-2018, 08:03 AM)CC.py Wrote:(01-16-2018, 05:19 AM)Dr. Keter Wrote:(01-10-2018, 05:19 AM)CC.py Wrote: That's only assuming you know p and q, and in turn t, without that it's practically impossible. In general you only release your public key and nothing else.
Also glad you enjoyed the tutoral
If all you have is the public key, how can you encrypt? You said that the formula to encrypt is (m^e) % n = c where n is p*q. How do you get n?
Wait my bad. During the key exchange you give e and n, and factoring n for p and q is super unrealistic because it's such a big number, meaning you can't really compute the private key
Okay okay that makes quite a bit more sense then.