RSA algorithm step by step example

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  • čas přidán 2. 06. 2024
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    RSA example step by step shows how RSA encryption works mathematically as well as with data encryption tools such as GPG. RSA encryption is used in many public and private key encryption systems. This video will show public and private key examples and then the mechanics of using the RSA algorithm.
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    Index
    0:00 What is RSA Encryption?
    0:17 Who are Ronald Rivest Adi Shamir and Adelman?
    0:35 Public and Private Keys
    0:50 GPG encryption example
    2:10 Example of factors of numbers
    2:45 William Stanley Jevons prime number challenge
    3:15 Examples of finding prime factors
    4:25 How to make a public RSA key
    5:50 Simple example of a public and private key
    11:20 How to encrypt a letter with the public key
    13:45 Decrypt a letter using the private key
    15:50 What are RSA number challenges?
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Komentáře • 151

  • @jessycagonzalez2666
    @jessycagonzalez2666 Před rokem +36

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

    • @ANGOLARICK
      @ANGOLARICK Před rokem

      Minus the use of excel to complete mod and the explanation of figuring out mod.

  • @brandonm5295
    @brandonm5295 Před 3 lety +26

    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.

  • @eliaswachira5381
    @eliaswachira5381 Před 3 lety +56

    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

    • @gagadaddy8713
      @gagadaddy8713 Před rokem

      Euler Totient ...... why every thing name after Euler in Math arena ..... 😝

    • @gagadaddy8713
      @gagadaddy8713 Před rokem +1

      @Elias , By the way, if you really have lunch with Professor, make sure you pay cash, no credit card. You know that he keep both the public and private key 😂😂😂

    • @vengeance3102
      @vengeance3102 Před rokem

      @@gagadaddy8713 stop talking please

    • @gagadaddy8713
      @gagadaddy8713 Před rokem

      @@vengeance3102 WTF you are doing here ....

  • @vimcfy5216
    @vimcfy5216 Před 2 lety

    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!

  • @NGUYENTHIKIMNGAN_
    @NGUYENTHIKIMNGAN_ Před rokem

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

  • @zgb3l
    @zgb3l Před rokem

    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.

  • @chakalonzote
    @chakalonzote Před 2 lety

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

  • @paulotcj
    @paulotcj Před rokem +1

    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!

  • @elenesokhadze255
    @elenesokhadze255 Před 3 lety +2

    The best tutorial on this encryption method. Thank you

  • @danielrozanski9594
    @danielrozanski9594 Před 2 lety

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

  • @marianaaa24
    @marianaaa24 Před 2 lety +1

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

  • @shashikantdivekar7839
    @shashikantdivekar7839 Před 2 lety +2

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

  • @becool1270
    @becool1270 Před rokem

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

  • @An.Individual
    @An.Individual Před 3 lety +6

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

  • @alexhope77
    @alexhope77 Před 2 lety +1

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

  • @user-ogdbalsuzf6
    @user-ogdbalsuzf6 Před rokem +1

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

  • @jeffreycanipe1936
    @jeffreycanipe1936 Před rokem

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

  • @terryloo
    @terryloo Před 2 lety

    What a clear presentation of RSA. THANKS so much.

  • @quochuybui4693
    @quochuybui4693 Před 3 lety +2

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

  • @lmaoty
    @lmaoty Před 3 lety +5

    Very simple to understand! Keep up the great job!

  • @ahmedaj2000
    @ahmedaj2000 Před 3 lety +3

    thank you! helping me study for my cryptography final

  • @rosannaisaksson3579
    @rosannaisaksson3579 Před 8 měsíci

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

  • @McBhayan
    @McBhayan Před rokem

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

  • @kennguyen6802
    @kennguyen6802 Před 2 lety

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

  • @FagneTReges
    @FagneTReges Před rokem

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

  • @egorutkin6347
    @egorutkin6347 Před rokem

    omg so simple and comprehensive explanation
    great job!

  • @laxmirajani1876
    @laxmirajani1876 Před 3 lety

    Great Work Shad!!

  • @attilaguba856
    @attilaguba856 Před rokem

    Professional as usual! Thanks!

  • @dimitarkostov8301
    @dimitarkostov8301 Před rokem

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

  • @spongebobsquarepants4576

    Best explanation on the internet 🎉

  • @ricp
    @ricp Před rokem

    Great explanation, thanks a lot!

  • @pewpewwepwep9296
    @pewpewwepwep9296 Před rokem

    Brilliant explanation !!

  • @jeffvolini2186
    @jeffvolini2186 Před 2 lety

    Very good explanation, thanks!

  • @hotbird3
    @hotbird3 Před rokem

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

  • @YeloPartyHat
    @YeloPartyHat Před 2 lety

    Great stuff Professor!

  • @netlogis
    @netlogis Před 3 měsíci

    Step by Step makes sense!!

  • @dickybear
    @dickybear Před 2 lety

    Thank you sir! Comprehensible lecture!

  • @rob651
    @rob651 Před rokem

    Amazing! Thank you very much

  • @rajabhattacharjee6982
    @rajabhattacharjee6982 Před 2 lety

    Illuminating! Thank you

  • @mirsyedabdulhadi6912
    @mirsyedabdulhadi6912 Před 2 lety

    best video till now

  • @fe4000
    @fe4000 Před 2 lety

    Great lecture, thanks!

  • @Jarjarbean75
    @Jarjarbean75 Před 3 lety

    thanks for teaching, very clear now.

  • @the_wizard_exe
    @the_wizard_exe Před 4 měsíci

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

  • @aimatters5600
    @aimatters5600 Před rokem

    really helpful man thank you

  • @user-pt9yu8ll8m
    @user-pt9yu8ll8m Před 7 měsíci

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

  • @Reroutmd5
    @Reroutmd5 Před 3 lety

    Amazing... keep it up

  • @kartz2010
    @kartz2010 Před 3 lety +2

    Fantastic!

  • @jason54953
    @jason54953 Před 2 lety +1

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

  • @khanthein1740
    @khanthein1740 Před 2 lety

    THANK YOU SO MUCH YOU ARE A LIFE SAVER

  • @Velvet_hippo
    @Velvet_hippo Před 3 lety

    Great explanation

  • @CO-wd3pc
    @CO-wd3pc Před 2 lety

    Best teacher 🙌

  • @charliemelga7445
    @charliemelga7445 Před 2 lety +5

    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 :)

    • @lezmkasd
      @lezmkasd Před 2 lety +1

      I was stuck on this too and had to do some googling. When he mentioned the rule that the numbers must be a coprime of 6 AND 14, he meant that 2,3,4,5 has to be coprime with BOTH 6 and 14, or they must share a greatest common factor of only 1.
      In the case of 4, it will never have a greatest common factor of only 1 because 2 is a factor of 4. It is not asking if 4 is a factor of 6 and 14, which is where I think you were confused about.
      In the case of 3, its only factor is 1 and itself, but 3 is a factor of 6, which is ruled out.
      5 is the only coprime of both 6 and 14, where they all share a greatest common factor of only 1.
      For your second question, it is extremely hard to find the private key based on the public key because you would need to find two large prime numbers just like he mentioned in the beginning of the video. Modern computers today are not capable of finding those primes but maybe someday when we reach quantum computing.
      Hope this helped.

  • @albanec4702
    @albanec4702 Před 2 lety

    Thank you so much, now everythig become clear)

  • @KnowledgeThroughScholars

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

  • @marianopatino-paul5062

    Great video

  • @VehlaEngineer_
    @VehlaEngineer_ Před rokem

    A great video.

  • @natisnotanerd
    @natisnotanerd Před 3 dny

    I love you Professor Sluiter

  • @updatedotexe
    @updatedotexe Před 3 lety +2

    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.

    • @eggtimer2
      @eggtimer2 Před 2 lety

      This is for entertainment. Not a msth course.

  • @dubiousinsights4008
    @dubiousinsights4008 Před 2 lety +1

    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

  • @eclypze_
    @eclypze_ Před 2 lety

    Nice video dude

  • @crowforcast3583
    @crowforcast3583 Před 3 lety +2

    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

    • @robbywelch4716
      @robbywelch4716 Před 2 lety +1

      hello felix I feel your pain!! I don't know for sure, however this should always work. My theory is that the larger the number you used the more accurate that your calculator or whatever you are using has to be before rounding.

    • @gd44481
      @gd44481 Před 2 lety

      Hello, m can not be bigger than n

  • @XXBrunX
    @XXBrunX Před 2 lety +3

    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?

    • @captainnemo6873
      @captainnemo6873 Před rokem

      I have the same question, using the public key(5,14) can also decrypt the letter "D" out to "B", looking forward to an answer too.

  • @Woktavius
    @Woktavius Před rokem +1

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

  • @authenticknowledge9349
    @authenticknowledge9349 Před rokem +1

    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?

  • @daboyz6106
    @daboyz6106 Před 7 měsíci

    Really good.

  • @samlitwin6117
    @samlitwin6117 Před 2 lety

    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.

  • @east4205
    @east4205 Před 2 lety

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

  • @oliviervanlier4947
    @oliviervanlier4947 Před 3 lety

    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?

    • @albertroswell
      @albertroswell Před 3 lety +1

      it IS encrypted, it just happens to be encrypted to the same value but that is unknown to the public, next letters will be encoded to something different

  • @aurelio4601
    @aurelio4601 Před rokem

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

  • @MobilTemp
    @MobilTemp Před 2 lety

    Amazing

  • @kartibok001
    @kartibok001 Před 3 lety +1

    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?

    • @Jamie-ef9dl
      @Jamie-ef9dl Před 2 lety +6

      Bit late but the requirement is they need to be Coprime, so they cannot share factors, as 4 has factors 1,2,4 they all share a factor of 2 which is why

    • @kartibok001
      @kartibok001 Před 2 lety

      @@Jamie-ef9dl Much appreciated.

  • @najeebbenawa
    @najeebbenawa Před 3 lety

    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?

    • @elisavetkonstantopoulou5386
      @elisavetkonstantopoulou5386 Před 3 lety +3

      1. N=14 and T=6, that is correct, yes
      2. Two numbers are co prime if they have no positive integer that can divide both, except for 1.
      - we remove 2 because the factors of 2 are: 1, 2 , the factors of 6 are: 1, 2, 3 and the factors of 14 are: 1, 2, 7, 14. We can see that 2 and 6 are not co prime, since the have two common factors (1 and 2). For the same reason 2 and 14 are not co prime.
      - we remove 3 because the factors of 3 are: 1, 3. We can see that 3 and 6 are not co prime, since the have two common factors (1 and 3). Even though 3 is co prime with 14, we need a number that is co prime with both 6 and 14.
      - we remove 4 because the factors of 4 are: 1, 2, 4. We can see that 4 and 6 are co prime, since the only common factor is 1 and the same thing goes for 14. But 4 is an even number and if we choose even numbers there is a possibility that (e x d) mod T will not be 1. (the same logic could be applied to 2)
      - we are left with 5 and we keep it, because 5 has no common factors with either 6 or 14, except 1, so 5 and 6 are co prime and 5 and 14 are co prime.
      I hope this cleared things out for you!

    • @KnowledgeThroughScholars
      @KnowledgeThroughScholars Před 3 lety

      Thanks, i was also this question in my mind. Now it clear to me

  • @natuu4648
    @natuu4648 Před 3 lety

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

    • @shadsluiter
      @shadsluiter  Před 3 lety

      Thank you. Not planning on the elliptical curves. However, can you suggest a resource?

  • @enoname3091
    @enoname3091 Před 2 lety +1

    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 ) .

    • @gd44481
      @gd44481 Před 2 lety

      There are countless text to ascii converters on the internet. But rsa can only encrypt up to the size of the key, so you would have to generate huge primes to encrypt any meaningful message.
      (What you can do is encrypt the text with AES and then encrypt the AES key with rsa)

    • @gd44481
      @gd44481 Před 2 lety

      Also, you dont have to use ASCII.
      If you really want to encrypt with only rsa then convert every letter of the text to ascii and encrypt each one and maybe store them in an array or something.

  • @Kevin-kk2rx
    @Kevin-kk2rx Před 2 lety +1

    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?

    • @gd44481
      @gd44481 Před 2 lety +1

      Thats a limitation

  • @rohanhasani807
    @rohanhasani807 Před 2 lety

    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.

  • @KDmadness18
    @KDmadness18 Před rokem

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

  • @triquangnguyen7579
    @triquangnguyen7579 Před rokem

    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.

  • @zeropain9319
    @zeropain9319 Před 3 lety +1

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

  • @KrassIankov
    @KrassIankov Před 3 lety

    Thanks!

    • @shadsluiter
      @shadsluiter  Před 3 lety

      You bet! Are you studying computer science?

  • @ayeshaali6462
    @ayeshaali6462 Před rokem

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

  • @ishimweeric7399
    @ishimweeric7399 Před 3 lety

    thanks mn

  • @davep8221
    @davep8221 Před rokem

    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.

  • @Astro-Markus
    @Astro-Markus Před rokem

    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?

  • @stupossibleify
    @stupossibleify Před 3 lety

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

    • @wadallat2
      @wadallat2 Před 3 lety

      Without that prime factor you can not decrypt. Find that factor is not at all simple, there isn't a simple and efficient way of doing that. While it may not be apparent when working with small numbers it becomes very difficult with numbers that are hundreds digits long. There is a section in the video starting from 15:37 which is related to that. For example a 232 decimal digits number was factored in 2009 and an "only" 8 digits longer (240 digits) were factored in 2019. Adding 8 digits took another 10 years to solve. Now, the most recent RSA numbers have hundreds more digits. Unless something revolutionary happens (e.g. success of quantum computers) I think we are safe.

  • @lionel4450
    @lionel4450 Před 3 lety

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

  • @dominiquerossignol2212
    @dominiquerossignol2212 Před 2 lety +1

    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

    • @gd44481
      @gd44481 Před 2 lety

      Hey i might be late but you cant encrypt where n < message

    • @ryklin1
      @ryklin1 Před rokem

      @@gd44481 in my experience, when implementing any hash function, the mod value must be > than the length of the array holding the data.

    • @ayeshaali6462
      @ayeshaali6462 Před rokem

      @@ryklin1 encypting o with (5,14) give a, and decrypting a with (11,14) gives a.
      how to get o instead of a after decryption?

  • @otooleger
    @otooleger Před 3 lety +1

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

  • @MrHardrocker98
    @MrHardrocker98 Před 3 lety

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

    • @wadallat2
      @wadallat2 Před 3 lety +1

      32 mod 4 is exactly 2. In the video what you see is a an approximation since the calculator can not be that precise.
      Another way of calculating that is that 299,593 x 14 = 4,194,302 which is exactly 2 less than the number in question (4,194,304).

  • @fablefallen
    @fablefallen Před rokem

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

  • @stevenshrii
    @stevenshrii Před rokem

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

  • @romansiry637
    @romansiry637 Před 3 lety +1

    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?

  • @shashikiranburugala9141
    @shashikiranburugala9141 Před 2 lety +1

    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... 😭

  • @orhanaygun7298
    @orhanaygun7298 Před rokem

    bilale anlatır gibi anlatmış. helal olsun!

  • @_i_m_
    @_i_m_ Před rokem

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

  • @ardaakyigit7939
    @ardaakyigit7939 Před 3 lety +1

    I think there is a mistake in your example. (czcams.com/video/j2NBya6ADSY/video.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?

    • @wadallat2
      @wadallat2 Před 3 lety +1

      I think you are right. If I send a message to you I would need your public key so I can encrypt the message and you will need your own private key to decrypt that.

    • @MichaelApproved
      @MichaelApproved Před 3 lety +1

      You’re right, he mixed up examples in his explanation. He was referring to a cryptographic signature. The signature process is a way of proving the message came from you.
      To sign a message, you encrypt it with your private key. That encrypted message can be decrypted by your public key, which everyone should have. The point of that encryption isn't to hide the contents, it's to prove that it was written by you. If the message gets decrypted properly with your public key, then it must've come from you because only you (should) have the private key it was encrypted with.
      Getting back to this mixup. When you send someone a message, you want to encrypt it with their public key but you ALSO want to sign it with your private key, so your friend knows it really came from you and not an imposter.
      That's why your friend would need their private key and your public key. Their private key would be used to decrypt and your public key would be used to authenticate the encrypted signature you included.

    • @adrianvmois9426
      @adrianvmois9426 Před rokem

      @@MichaelApproved for digital signature is not about the encryption of the whole document, is about obtaining a hash (a short string of the document) and that is signature and is obtained with the use of the private key (the sender uses his private key to create the digital signature hash). Then the receiver use sender's public key to verify that the hash is autentic, to confirm the signature was created with the real private key.

  • @RajasthaniLadka
    @RajasthaniLadka Před 3 lety

    How it's not breakable... We can calculate 'd'(private key) with the help of e(public key) as we did in first time to genrate 'd' and totient of n also we can calculate cause we know n is the multiplication of two prime number so first we will break n into those two prime number and will calculate toient of n easily... So now we have everything to calculate d then why it's not breakable?

    • @entityone0x193
      @entityone0x193 Před 2 lety

      In order to guess d you would need e and t. T consist of (p-1)*(q-1) and although we now n which is the result of p*q we can not derive the original prime numbers p and q from just knowing n. The RSA Algorithm is easily breakable when you use fairly small numbers p and q, but when you use numbers with over 100 digits, brute forcing would take years to guess the correct factors. The security of RSA lies in the fact, that although you can easily multiple two very large prime numbers, you can't guess those originally used numbers by just looking at the result.

    • @entityone0x193
      @entityone0x193 Před 2 lety

      "...

    • @RajasthaniLadka
      @RajasthaniLadka Před 2 lety

      @@entityone0x193 This point raise another question if its so difficult to calculate pq from N then how we generate keys for first time, as to generate public and private at some point we have to do all these calculation at once to generate these keys. How we did at first time??

  • @samueldomb110
    @samueldomb110 Před 8 měsíci

    where the 14 came from ? minute 7.58

  • @abeladerese8593
    @abeladerese8593 Před 2 lety

    tankes pro

  • @learnwithcode4211
    @learnwithcode4211 Před 2 lety +1

    why dislike?