Finally. A good explanation on this topic with no skipped steps nor assumption about the viewers' previous knowledge. Amazing professor!

I've watched and looked at so much for someone to finally break it down for a smooth brain like myself. Thanks for actually breaking things down into simple steps, with a simple example.

This guy is super easy to understand and teaches clearly with no skipped steps. Thanks Shad. If there was no covid and you lived close by I'd take you to lunch

this is the way to teach !!!!!!!!!! lovely system of visuals ,clear speech ,clear reppresentation etc etc this is paradise after watching loads of other videos that are messy, jumbled up and only fit as a reminder for people that already know the stuff!

This one so far is the best explanation on RSA I have come across. Thank you, Prof!

This video is so underrated, but the best on this topic I've seen. And I've seen A LOT. Thank you so much for this.

The best tutorial on this encryption method. Thank you

Brilliant, so easy to understand. Amazing. Thank you so much!

Sir, my hat off to you. Nor my book or my professor was able to explain this with such simplicity!

Hands down the best video and explanation I’ve seen thus far! Thank you!!!! 🥲

THANK YOU!!....went to 5 other sources and they were crap! You are a life saver!

Now that's a good explanation. Unless one is able to encrypt and decrypt (with pen and paper) they will not really understand it. And this example doesn't skip any steps. Best explanation so far!

It's very easy to understand, thank you very much for your work!

Awesone explanation, Thanks so much! I'm not sure what I would've done without this tutorial

Great job explaining this!!! I would take a class under this professor anytime.

What a clear presentation of RSA. THANKS so much.

I simply loved the explanation, this was great. Your way of explaining the algo was quite smooth and smile. Thank you.

thank you! helping me study for my cryptography final

Professional as usual! Thanks!

Very simple to understand! Keep up the great job!

Very informative and useful. Also explained in simple terms. Thank you very much Sir.

omg so simple and comprehensive explanation great job!

great teacher, now I can solve my assignment after watching your video

Thank you for the thorough explanation of RSA. Very helpful.

Great explanation, thanks a lot!

Illuminating! Thank you

Your walkthrough in excell was very helpful. Thank you for the effort!

Brilliant explanation !!

Amazing! Thank you very much

Thanks for your video Professor Sluiter! I'am a brazilian student, and yours videos helpe me study!!

Step by Step makes sense!!

Great Work Shad!!

Great lecture, thanks!

Great stuff Professor!

Fantastic!

Very good explanation, thanks!

Nice video. FYI Rivest, Sharmir & Adelman only patented RSA, they were not the first to discover it

thanks for teaching, very clear now.

Thank you sir! Comprehensible lecture!

Best explanation on the internet 🎉

really helpful man thank you

You are such a genius creature on the planet earth Prof Sluiter. Hats off!

teacher thank's very much for having explained them carefully and well-deatiled, everything was clear ! , break a leg with your videos !

best video till now

THANK YOU SO MUCH YOU ARE A LIFE SAVER

A great video.

Thank you so much, now everythig become clear)

Best teacher 🙌

How come these smart people not have millions of views. Thank you so much , very clear an precise explaination

Great video

Thank you very much. You help me solve the issue bothering me for a long time

Thanks for thr nice explanation. One of the best and easy to understand

Great explanation

Amazing... keep it up

I love you Professor Sluiter

Thank you so much. I'm taking a class for this same topic and you break it down so easily.

Nice video dude

confused during the explanation of the decryption where does the 32 mod 14 come from?

Amazing

So I get that we are calling message "B" as 2, but what's the typical way of converting a longer message to a numeric value ? If I wanted the send the message "Hi" how would I get an "m" to feed into the algorithm

Really good.

Thanks!

So as far as I understand there can be more than 1 private keys d that work the exact same way to decrypt a message.Isn't that a vulnerability?

Thanks very much for posting very helpful , I have a few questions please as there are a few things about the video I do not quite grasp. Can you please explain the follows: In Column 'A' of the spreadsheet the rule is 'must be a coprime of 6 and 14' when I look up the definition of coprime it says "Co-prime numbers are the numbers whose common factor is only 1" You rules out the number 2 and 4, I can see why you rule out 2 because 6 / 2 = 3 and 14 / 2 = 7 (so their only factor is not 1). However taking the 4, 6 /4 = 1.5 and 14 /4 = 3.5 (in other words not pure integers, and I though the result had to be a pure integer i.e. not remainder)? or do you always remove even numbers before starting the maths? Can you please explain/clarify the above for me. One a related note, I did not realise the private / public keys are actually 2 numbers (e.g. two numbers each key) working together. Given a public key do not know of a script (preferably using PoweShell) where I can split the representation of a public key (e.g. byte array) into the two distant values to see that they are ? Thank you very much again for posting :)

thanks mn

I don't get one thing, the decrypt number "d" can be 5, 11 and 17. And 5 is the same as the encryption key. What stops an attacker from using 5 instead of 11 to decrypt the message, since all of these 3 values can decrypt it?

Can anyone explain why the decrypt step works or where to find that information? I understand the steps but what rules in modular arithmetic lead to the original data being the result of the equation.

Hi, question. How would you encrypted a message that is a word or even a sentence e.g. "hello"?

i am still lost on how to find the e and d. at the part where you choose a co-prime, and the entire process for d.

Does anyone know how to get the d variable as fast as possible in python?

Nice video. Must e and d be integers? Are there special rules that these two numbers need to follow? In Excel you said that e < T and e must be coprime with T and N. But you didn't say that in the beginning.

for finding e, whyd you choose 14?? and how is 4 coprime for 6 and 14???

Hi, do the primes you encrypt with have to be much larger than the values you are encrypting? I am trying to encrypt the number 77 with your cipher: m = 77 77^5 mod14 = 7 (encrypting) C = 7 7^11 mod14 = 7 (decrypting) M = 7

why does "de (modN) =1" - what's the basis of this?

yoo does someone know how would u do it with actual text? I've read that you must convert it into numbers and in the video he says that u must use ASCII but how? I would appreciate if u had some material about how to do it without the B=2 thing ( min 11:30 ) .

This was so easy to understand. Can you do one for elliptical curves primarily for Bitcoin?

encypting o with (5,14) give a, and decrypting a with (11,14) gives a. how to get o instead of a after decryption?

Is it a problem the decrypted “2” is not exactly integer?

Im lost how on how 5 is a number co prime with 6 and 14?

I tried some different messages using the keys you used here and found any message that exceeds n(14 in this case) will not decrypt correctly. Is this an error on my part or is this a normal limitation of rsa?

good explanation, however from your example you could have chosen multiple keys. This means that for each public key there are multiple private keys?

I understand what's happening here, however nobody encrypts anything just to decrypt it again afterwards. You want to send the encrypted message to someone who can decrypt it. And you certainly don't send them your private key. And the receiver has their own private key, which is most probably based on different prime numbers. So, how does that work together with the public key?

Can any one please reply me..? I've got d value as 0 , what to do if we get as such... Please reply faster no time... 😭

very interesting. I wonder how many 'secret' documents from the 1990's are now unsecure?

Great video - Could some one explain - at 8.08 he mentions you can't use 4 because it is even, leaving us with 5. In my head though the factors of 6 and 14 are 1,2,7,14 and 1,2,3,6. Why not even?

buy how does an computer decides which two huge prime numbers to chose? How does he know they are prime?

For I = 2 to (amount - 1): if (( amount mod I) = 0) then print I :next i

but when the message is encrypted using the public key and once we have the encrypted message, one can always reverse engineer to breakdown the message. This method of encryption is only useful when the public key generates different encryption text even after giving it the same input message.

In the real world, don't you need to make sure that N is large enough to cover the symbol range you wish to encrypt? E.g. for alpha text, 1.. 26 --> T >= 27. And then you need two primes whose multiple is >= 27, say (3,11), (5,7). Of course it'll be (Yuge prime, Bigly prime), but I'm just asking about the "min" value for p, q. Would 0.. 25 be allowed? It would handle the number of symbols, but I'm not sure if 0 would break any of the math.

Very well explained sir, but if you use the encoding as A=1 , it will not get encrypted since 1^e mod N = 1 ... so i guess just using normal ascii code is a better way of doing things?

tankes pro

bilale anlatır gibi anlatmış. helal olsun!

1. for the part of choosing e and d, we got the 6 is T and the 14 is N. Is this correct? 2. how was 5 coprime with 6 and 14? Can you elaborate this 2, 3, 4 removal process and reasons?

Can you explain why decrypting RSA without knowing the privately held prime factor is difficult?

where does 14 came from ??????

Great content sir, but let me try coding the letter m by its code ASCII=109 Coding RSA gives 109^5(mod 14) = 9 Decoding RSA gives 9^11(mod 14) = 11 We know that 109 (mod 14) = 11 Decoding gives multiple solutions, so we cannot retreive only original number that is 109 Could you clarify ? Regards

why dislike?

I presume this video explains RSA simple way but this method crypts letters always into same number, which makes it valuable to letter frequency decryption (extremely easy to solve). How is it sorted?

where the 14 came from ? minute 7.58

I think there is a mistake in your example. (kzfaq.info/get/bejne/oJh-ddySm6eog40.html) When I send an encrypted message (encrypted with his public key) to a friend, he doesn't need my public key to decrypt it. Because he can only decrypt it with his private key. So there is not any relation between my public key and the message which I send to him. What do you think about it?