+Billy Rebecchi its realy good to watch some one explaining things in this manner ...

Billy Rebecchi agreed

me too preparing for final exam...

Not for a final, but I used this so i can test out my program to see if it was written correctly.

I enjoy the background ambience of the audience, it's normal for them to make some noise, it also makes it feel less lonely, unlike many other such videos. Eddie Woo's enthusiasm is the absolute best though!

7:50 "Is it really that hard? :(" This dude just casually destroying kids while teaching RSA, amazing.

He might be saying "Is it really that hard to STFU?"

such a smooth way to burn

@@Subtitles00 That was encrypted message for "Stfu bit*h!". Only people with some processing power are able to decipher such messages.

Obviously this class has mentally challenged people who should not be there.

same, I had such an amazing teacher, but he only teaches secondary and i miss his classes now that i m doing my degree, gems like these teachers are needed at each uni or school

wait, its a secondary school?

@@tanveerhasan2382 yh

what? secondary school, learning rsa wtf? Some people are so lucky. Damn i wish i had that in my school

I fell u bro

That's how it is when a teacher is actually passionate about what they're teaching. :P

Feels like he is teaching to me. The whiteboard is a giveaway

same

That is the case very often. Quite a lot of teachers/professors want to seem smarter by making their topic appear more complex than it actually is. If you understood everything and can simplify what they are saying and they react insulted, you know they're just trying to show off instead of teaching.

I am in a cryptography class right now, and this is literally the easiest explanation I have found... I agree with you 100%

Academic institutions make a profit off making things seem more difficult than they are

he kinda skipped over the most important steps like: using euclid's algorithm to solve de = 1 (mod n), fermat's little theorem and euler's theorem . There is at least half of a semester of learning about discrete math before you could truly learn RSA in a rigorous manner. Understandable for HS teaching but in uni u need to understand the proof behind things, not just the equations.

Yeah hes great

lol

Right? I think most teachers care more about being in power than their students learning. One time all the students in our class collected our notes and shared them on Google Drive so that we could learn Chemistry better and the teacher found out and threatened that she would fail us if we did it again. She cared more about us struggling to write down notes + pay attention rather than us learning the material and getting better grades. Compare that to Eddie who happily uploads his lectures on youtube! Sometimes the current school system / current school teachers are so backwards.

if Eddie Woo explains it better than your teacher, than Eddie is the leader in the Theoretical Computer Science.

Well, he only explained a very simplified RSA. Though understanding simplified way first probably would help to understand underlying formulas and stuff better. Btw maybe he does in the next video I did not check.

​@@gytisx13 No I think he leaves it at this level without going deeper that because he's teaching high schoolers. In all of Eddie's videos it seems like he struggles to dumb down his lessons as to not intimidate his young students and keep them engaged. It's sad that he's such a caring teacher that even when he tries to do that students talk during class (4:10) or give up saying it's so hard (7:54). Like he says its to complicated explain the mod calculator "trick" that takes about 2 seconds to understand. Or why the the number of coprimes of the product of two primes is just the product of 1 minus each prime (which is explained in a simple understanding of prime numbers). I feel like his highschool students not really caring is what has made him into such a great teacher. Compare that to college where the students are so compelled to learn for the high tuition or job prospects but it causes the opposite effect: professors can bullshit their lecture slides and put little to no effort into class.

​@@wonderfulworldofmarkets9033 you said it's very easy to understand why the amount of coprimes with pq between 1 and pq is (p-1)(q-1) but i have tried so hard and still can't figure it out, can you explain for me why that is the case? Edit: nvm i figured it out after asking

Thank you very much, I need it for a Covid App and your explanation was very useful!! 😁

what if we choose 17 of 11? using 17 for the secret key decrypts the message. so can there be multiple "d" for the same "e"? wouldnt that make different private keys?

You do realise that he's teaching a class of high school kids yeah? If you throw your equations at them they will be bored the hell out. Keep your explanation to the simplest form like Eddie was doing here

how can we apply this for huge numbers, we can't cycle through millions of possibilities when e is too long, it'll take way too much time

thanks, also does using extended euclidean algorithm have similar pattern?

exactly!!!!

I agree

well i dont completely agree his explanation is good no doubt about it but do u consider how bored the students might be ? other teachers may not be good at teaching ? they were exhausted and wanted a break but didnt get 1. So its only natural to talk with others or do something that may disturb teacher

@@silverzero9524 Thats all assumption. Why are you speaking for them 😂

So trueee!

In uni and have a professor who sits and reads verbatim from slides for an hour. I would love to have someone as enthusiastic as Eddie teach me.

Nothing sadder than a sad Eddie ☹️

I thought a thug life meme would start at that point

aw :c

did someone leave? or was a student talking again?

I thought the student was talking again

Students are teenagers. This is high school

When you're a teenager, you got more important things to worry about than coprime numbers and modular arithmetics :)

dawg you assume everyone on earth has the same mindset/priorities as you do lmao. you have the benefit of hindsight

@@ArtyomPalvelev Oh yeah I am sure they are discussing vital issues much more important than their education. Give me a break.

@@ArtyomPalvelev poor mentality

I was thinking the exactly same.

He sort of said that when he showed that d was added to the encryption formula e(mod phi(N)). What he did not explain is why d*e*(mod phi(N)) has to be 1. Also, he didn't tell them the maximum length of the message is limited to N. So, for instance, with his numbers it would not have been possible to encrypt the letter 'Z' in one step.

@@iycgtptyarvg why not? If my N is 247. What is the highest number I can encrypt/decrypt?

@@MrNoxiium [0,246]. But, with his numbers you cannot do more than [0,13].

Nerd

QuantuMuffin yes

If you're doing it on paper, you could expect it would take a long time. Luckily, computers are REALLY good (and fast) at doing multiplication and modulo arithmetic. Using massive numbers, with the aid of a computer that has already been given the process, it shouldn't take more than a few seconds. Even with extremely massive numbers. The power of RSA comes from one thing that this brilliant teacher didn't cover in this video, and that is the fact that factoring larger and larger numbers take exponentially more time. One easy way to get the value of D is if you have de(mod phi(N)), d = e*phi(N)-1. The number 29 would work equally well as 11. We can check 5*29=145. 145/6~24.167... or 14 45 mod 6 = 1. With small numbers, there seem to be a lot of options for D. But as we'll see, the bigger we make P and Q (and by extension, phi(PQ)), the more disparate and hard-to-guess D becomes. Once you decide on prime values for P and Q, the rest of the math is trivial for a computer. But, you never tell anyone those values. They only get to know that the product of P*Q=14. Let's say you chose larger numbers, and their product is 143. It's not super hard, but it's not trivial (11 and 13). Now try 44,747,819... it's getting harder (6599 and 6781). Now try that with two prime numbers that are 400+ digits long. It will take all the computers in the world longer than the age of the Universe to figure that out... UNLESS you already know one or both of the factors! The largest known prime number has over 23 million digits. Good luck, Eve!

The point of RSA is an asymmetric encryption. Where sender and receiver of the message didn't have same key. If he use (5,14) for both sender and receiver will have same key. This coincident can happen because he use 2, and 7 as 'p' and 'q' for the sake of simplicity in calculation. If we use bigger prime number for 'p' and 'q', most of the time sender and receiver will have different key (asymmetrical). for example if we use (7,33) to encrypt , we can't use (7,33) to decrypt the message. let says we send B or 2. 2^7 mod 33 = 29 if we tried to decrypt using (7,33) we got : 29^7 mod 33 = 17 (it's wrong message, it should be 2). but if we use (3,33) to decrypt the message : 29^3 mod 33 = 2.

@@illiacvie I think that is not what he meant. He's not trying to use the same key, he's asking why at 11:12 he didn't chose 5 , as he went for 11. I've tried to look for an answer, but I've only found that the number "e" can be calculated with another method, but nothing on the "reasons it's a little harder to explain". Maybe it's in the extended euclidean algorithm, but that i haven't checked myself

I know this comment comes from 3 years ago... But oh well I can't help to reply.. On 09:00 there's step 5, de(mod Φ(N))=1, where the value of e is already known as 5 5.d(mod 6)=1, in mod 6, every 6th step will always gives remainder of 0 So to complete 5.d(mod 6)=1 equation, we'll pick 6th step minus 1 The d value can be (6,12,18,24,30) minus 1 = 5,11,17,23,29 So the decryption key can be (5,14) or (11,14) or (17,14) and so on. To proof if above decryption keys are correct, let's calculate back: 1) key (5,14) -> 4^5 (mod 14) = 1,024-(74x14) = 2 -> (5,14) is correct 2) key (11,14) -> 4^11(mod14) = 4,194,304-(299,593x14) = 2 -> correct 3) key (17,14) -> 17,179,869,184-(1,227,133,513x14) = 2 -> correct So the conclusion is: He choose 11 because the decryption keys can be 5,11,17,23,29 Hope this helps someone in the future.. :D

@@auroraaa._. You didn't understand the question. If the keys can be any of the set {5, 11, 17...} why choose 11 versus 17, etc.? The question is: What actually defines a public versus private key? Is this arbitrary? Computational speed limit? Crypto-strength? How do you weigh this?

@@AverageJoe8686 He chose 11 over 5 (or 17) because he wants to shows that on simpler math, the key can be multiple. This happens because he used smaller number on p & q. What defines a public key? The logic is like this, when you lock your door on your house, you will always choose to use complicated one. So thieves will have longer time to break through your door, even though the thief can just use hacksaw, or TNT as 'bruteforce' way to break your lock. In real usage, the decryption key usually only has 1 or several possible key due to usage of high modulus number, on my previous comment, there's mod 6, so in every 6th thep, there will be a valid key. So if we use (for example) mod 1,000,000,000 the key is only valid for every 1,000,000,000th step. I hope you get what I mean

@@auroraaa._. You still didn't get it. 11,000th prime versus 17,000th prime. Is this arbitrary? Computational speed limit? Crypto-strength? How do you weigh this?

@@AverageJoe8686 Nowadays, the minimum acceptable crypto strength for RSA is 1024-bit, or equivalent to 80-bit symmetric key. RSA is an asymmetric system, while an example of a symmetric system is AES (Advanced Encryption Standard). The strength of RSA 1024-bit key is more or less equivalent to AES 80-bit key, it is also approximately equivalent to another encryption algorithm known as Triple DES. The 1024-bit here is not the string length, where as 'abcdef' is 6 string length. The algorithm use Base64 encoding to cope with the string length. Below is the strength equivalency between symmetric key and asymmetric key: 80 = 1024, 112 = 2048, 128 = 3072, 192 = 7680, 256 = 15360 The difficulty of breaking 112-bit compared to 80-bit strength is 2^48 times more difficult. A 80-bit key strength is recommended to be obsolete at 2010 by the National Institute of Standard and Technology (further read: csrc.nist.gov/CSRC/media/Projects/Key-Management/documents/transitions/Transitioning_CryptoAlgos_070209.pdf) RSA 512-bit can be broken by a normal 2GHz PC in about 73 days, requiring 5 GB of diskspace and 2 GB of RAM. NIST determine the obsoletion of a key strength by calculating the speed and computational power needed for a key to be broken. And also, English is not my first language, so if your question is not answered yet, perhaps you can rephrase it.

He means to share a common factor of 14 from my understanding. Because 2 is a factor of 14 and it also is a factor of 8, 10 and 12. Therefore they share a common factor. :-)

Thanks a lot mate!

I also did not understand what he meant to say. Perhaps, +Eddie Woo can make it clear.

He mainly meant common factor with 2; all even numbers (here, because we are dealing with 14 = even) will share at least common factor, i.e., the number 2. Therefore, they're all eliminated.

Aparna Mahalingam where does q=7 came from couldnt follow it.

Hi Mr. Thadeus, i'm Enoch. I'm not good at English but i really like Cryptography, i found these videos are highly helpful so i watched these by reading the English subtitles and translate to my language where necessary. Unfortunately there're not subtitles, so could you please help me to add subtitles on this video? Thanks