Cryptography | The Mathematics of RSA and the Diffie-Hellman Protocol

Поділитися
Вставка
  • Опубліковано 27 сер 2024
  • Click here to enroll in Coursera's "Cryptography I" course (no pre-req's required): click.linksyne...
    STEMerch Store: stemerch.com/
    Support the Channel: / zachstar
    PayPal(one time donation): www.paypal.me/...
    If you missed part 1: • The Mathematics of Cry...
    Instagram: / zachstar
    Twitter: / imzachstar
    Join Facebook Group: / majorprep
    ►My Setup:
    Space Pictures: amzn.to/2CC4Kqj
    Magnetic Floating Globe: amzn.to/2VgPdn0
    Camera: amzn.to/2RivYu5
    Mic: amzn.to/2BLBkEj
    Tripod: amzn.to/2RgMTNL
    Equilibrium Tube: amzn.to/2SowDrh
    ►Check out the MajorPrep Amazon Store: www.amazon.com...

КОМЕНТАРІ • 108

  • @zachstar
    @zachstar  5 років тому +29

    In case you missed part 1: ua-cam.com/video/uNzaMrcuTM0/v-deo.html

    • @zeuscesar
      @zeuscesar 5 років тому +1

      Good 2 part video dude! thanks for that!

    • @cardcode8345
      @cardcode8345 3 роки тому

      Which college major should I go for
      Computer science or Electrical engineering?

    • @ajaydabral5672
      @ajaydabral5672 3 роки тому

      Wanna discuss with you something ,if possible to contact you

    • @Michael-vf2mw
      @Michael-vf2mw 6 місяців тому

      @@cardcode8345Mechanical

  • @raadwan
    @raadwan 5 років тому +201

    Now I'm smarter than I ever intended to be. Thank You.

  • @zajec11
    @zajec11 5 років тому +71

    After trying to understand this algorithm for about 4 months, this single video has effectively set in motion the change of my life

  • @digitalconsciousness
    @digitalconsciousness 3 роки тому +12

    The new thing in cryptography now is lattice cryptography. The idea is that you form a multidimensional lattice with vectors that are mostly orthogonal, read in your byte from your plaintext (ex: 'g'), find a lattice coordinate that is also 'g' (you associate L points with bytes), but instead of choosing the cardinal coordinate of that lattice point, you choose a point near it; that point is then written to your ciphertext. It is impossible for an attacker to read the point from the ciphertext and match it to the lattice to discover the byte it is because they do not have the lattice basis (vectors) to reconstruct the lattice. All they have is a random point in space. They cannot even reconstruct the lattice from many sample points because the points were not chosen at the exact lattice coordinate: they were only chosen near them. So the sample of all points is completely random.
    The math part of this comes in with cryptographers wanting to represent a string of vector coordinates as a polynomial. They are, as I understand it, able to write coordinates (4,7,3,5,etc) as a polynomial, then they appear to write the exponents of the polynomial + coefficient backwards and this is the final thing that is written to the ciphertext. Anyway, if you ever feel like exploring the polynomial aspect of it and doing a video about it, I would love that. It's cutting edge stuff, mainly because lattice encryption is resistant to quantum computing attacks so far.

  • @JamesPerez328
    @JamesPerez328 5 років тому +35

    I learned the Caesar Cipher and RSA Encryption in my Hardware Security class. It's honestly so cool.

  • @mhh5002
    @mhh5002 5 років тому +59

    Second here. U r such an underrated UA-camr. Another great video again

  • @fiNitEarth
    @fiNitEarth 5 років тому +57

    Wtf i just randomly got the first part recommended, watched it and landed here 20 minutes after the upload xD So thats kinda creepy :D

  • @vovan101
    @vovan101 3 роки тому +3

    The only explanation of correctness I found after 2 days internet search. Thank you very much.

  • @baraaali4147
    @baraaali4147 4 роки тому +6

    Really good videos! I'm watching this to educate myself on what I want to major in and this by far has made computer science a really interesting and fun field.

  • @thedoublehelix5661
    @thedoublehelix5661 4 роки тому +6

    The proof for Euler's Theorem is so nice. You should do a video on that!

  • @qigongandthemartialarts3273
    @qigongandthemartialarts3273 5 років тому +6

    you do an excellent job explaining things in your videos keep up the good work

  • @arwaaldurehim2727
    @arwaaldurehim2727 Рік тому +1

    After studying number theory in my math major classes I’m glad to know all its applications and how cool it actually is

  • @sky-sight
    @sky-sight 5 років тому +6

    You know a video is good when there are more then 9k views but 0 dislikes there is usually that 1 hater who dislikes and leave so good job.

  • @mbjelica947
    @mbjelica947 5 років тому +19

    Great videos on cryptography! How about maybe sometime doing a video on Nikola Tesla and his inventions, including some of the technical aspects? So fascinating and revolutionary.

    • @zachstar
      @zachstar  5 років тому +10

      Thanks! And could definitely be a good video idea. I’ll see what I can do with that.

    • @shivashankar28
      @shivashankar28 5 років тому +3

      Please we need more Electrical engineering videos not CS stuff

  • @AjayKumar-fd9mv
    @AjayKumar-fd9mv 4 роки тому +1

    Omg, this is great. Keep posting such great videos.

  • @samgallon1273
    @samgallon1273 5 років тому +18

    First video I have ever seen with 0 dislikes despite having more than 10 k views

    • @chinkeehaw9527
      @chinkeehaw9527 5 років тому

      Sam Gallon
      Now it has 4 dislikes

    • @aashitashyam6060
      @aashitashyam6060 4 роки тому +2

      People probably saw your comment and disliked the video just to prove you wrong.

    • @hassanm.1887
      @hassanm.1887 4 роки тому

      @@aashitashyam6060 true

  • @matt-in8td
    @matt-in8td 4 роки тому +1

    You actually saved my exam, I did not understand DH protocol at all. Thanks a lot!!

  • @zdravkotraykov3752
    @zdravkotraykov3752 5 років тому +4

    Thank you so much! You explain it so simple and easy!

  • @stevenshrii
    @stevenshrii 5 місяців тому

    If you know e, but to find d.. d=1:while(d< some-number){if ((e*d) mod n)= 1){print d}:d++} it will show all the possible of d

  • @dreaminfinity7716
    @dreaminfinity7716 5 років тому +12

    Mindboggling 🙄😮😮

  • @SzechSauce
    @SzechSauce 4 роки тому +2

    Awesome thank you so much for the super clear explanation!

  • @user-um7tw6kx4r6
    @user-um7tw6kx4r6 2 роки тому +1

    That Stanford course is NOT for beginners, unless "beginners" means "advanced math students, who never used their advanced math for Cryptography specifically"!

  • @OneCatholicSpeaks
    @OneCatholicSpeaks 5 років тому +1

    Interesting video. I was just thinking that if the uncracked code was a string of letters, then I could enter that string into a computer programmed with the formula. Have the computer cycle through possible cypher numbers until a string of three letters to a viable word such as or and see if the rest of the cypher follows and translates.

  • @kid_kulafu_1727
    @kid_kulafu_1727 5 років тому +2

    We want more!!!!!

  • @jayh5992
    @jayh5992 5 років тому +1

    You should maybe mention what is considered a "safe prime number" on the Diffie-Hellman.

  • @rocksonrong8401
    @rocksonrong8401 4 роки тому +2

    Hey I have a question. I noticed that there can be multiple values of d(private key)
    (e.g.- d=63 also works for the above example). But isn't the private key supposed to be unique?? How do you explain that??

  • @SamvitAgarwal
    @SamvitAgarwal 5 років тому +2

    Doesn't the entire security of RSA rely on the value m then? Since an eavesdropper knows e,n and the value of m^e mod n, couldn't they just brute-force to find the value of m?

    • @hughjazz4936
      @hughjazz4936 5 років тому +2

      Yes, they can. That's why those values are chosen such that bruteforcing it would take 10 to 15 years with the best computers available.
      Every code can be cracked given enough time, but the point is that this time is long enough for the message to not be relevant anymore.
      Once computers get fast enough to crack codes in a reasonable time, you simply choose bigger values.

    • @MSneberger
      @MSneberger 4 роки тому +1

      It is thought that the number of atoms in the universe is around 10^80, which is roughly equivalent to 2^265. This means that brute forcing a 256-bit encryption key would be equivalent to counting every atom in the universe, which while theoretically possible, is not "really" possible.

  • @AakashKumar-gl2fk
    @AakashKumar-gl2fk 3 роки тому

    Today I felt: Prime numbers r prime for many reasons. Salute to all prime numbers serving for humanity and cryptography

  • @rickyleung5890
    @rickyleung5890 5 років тому +2

    why can't the eavesdropper solve the equation?
    e.g. 5^a mod 23 = 8
    i.e. 5^a = 23n + 8 where n is an integer
    i.e. a = log(23n + 8, 5)

    • @zachstar
      @zachstar  5 років тому +9

      How are you gonna find n? n is gonna be so big that even a super computer wouldn’t be able to do it fast enough.

    • @mathematicalninja2756
      @mathematicalninja2756 5 років тому

      I would use quantum computer to find the discrete logarithms.

    • @Darticus42
      @Darticus42 5 років тому

      Mathematical Ninja good luck finding a quantum computer with more than 5 qubits and can run Shor's Algorithm

    • @Darticus42
      @Darticus42 5 років тому +1

      Ricky Leung also keep in mind that the primes everyone knows (23 and 5 in the example, and thus 8 as well) are also extremely large. Taking a log of a number like that is also going to be very difficult with a classical computer when you have to compute it so many times to find a and n

  • @jamesedward9306
    @jamesedward9306 2 роки тому +1

    I realize this is an old video, with the most recent comment being several months old but I have a specific question on some confusion I have. First let me say I viewed part one and this video and I'm with it all except for a small part at the 1:30 -1:50 mark. As you summarize how it all works you say that what you and the friend did was: (g^b)^a (mod p) (g^a)^b (mod p) and that = g^(a*b) (mod p).
    But that's not what you actually did a little earlier. You did g^a (mod p) and sent THAT result to your friend, who then applied his secret key, b to it. And you did vice versa. Thus yielding your common secret key, or 2. With the notation (g^b)^a (mod p) are you saying 5^4^6 (mod p) ??
    5^4^6 is an enormous number. Shouldn't that notation read read: ((g^b) mod p)^a (mod p) = 2 the shared secret key.
    I only raise the issue because you talk about g^(ab) and I don't see you doing that exact calculation anywhere earlier. It could be I just don't understand mod notation.
    Zach or anyone else, help here would be appreciated.
    Big fan Zach, love your stuff.

  • @andreavecchio4674
    @andreavecchio4674 4 роки тому +1

    at 7:23, shouldn’t x and n(so m and n in the demonstration) be coprime to apply euler’s formula? Great stuff btw

  • @tomtian895
    @tomtian895 4 роки тому +3

    turns out your professor may not care more about your grade than an youtuber.

  • @F_F_C-16
    @F_F_C-16 Рік тому

    when I divided the value of G to the power of A by the Value of P i got a remainder of 0. is that a problem?

  • @rizolli-bx9iv
    @rizolli-bx9iv 3 роки тому

    Generally euler theorem is the fundamental of cryptography

  • @gamesniper98
    @gamesniper98 5 років тому +1

    You’re awesome

  • @imagineaworldwhereallyourw7859
    @imagineaworldwhereallyourw7859 2 роки тому

    5:00 Not explained how one chooses 7 and 23; is 23 easy to calculate knowing 7 and Fi(N)?

  • @carlfels2571
    @carlfels2571 3 роки тому

    I take Cryptography I at my university this semester :)

  • @gazfilm693
    @gazfilm693 3 роки тому

    2:27 ".. and this means they either need the value of a or b. Either one works because once they have it they can figure out what this entire value is." Um, what? Huh? Why? How? Thank you

  • @user-dg8rr8pl6f
    @user-dg8rr8pl6f 6 місяців тому

    I have a question, what do e and d mean?

  • @aizhigitmamatov5214
    @aizhigitmamatov5214 2 роки тому

    why do the values you and your friend select for diffie hellman protocol have to be less than the mod divisor (in this case 23)?

  • @benoit__
    @benoit__ 5 років тому

    So, the key is chosen by the Prime number and number on the Diffie-Hellman Protocol or is it random?

  • @alleygh0st
    @alleygh0st 3 роки тому

    So what you are saying is p and g are the public key, a is my own private key and g^ab (mod p) is our secret key? Or am I missing something?

  • @BederikStorm
    @BederikStorm 10 місяців тому

    Is it Dan Boneh's course?

  • @inshafahmed8656
    @inshafahmed8656 4 роки тому

    Ur awesome man!!!

  • @1Backi
    @1Backi 2 роки тому +1

    Thank you for the Video - im trying to get warm for potential studies next year and even for someone that didnt study much of anything durring the last decade it really gives some understanding in how these encryptions can work - so big credits for that. I actually have a hard time understanding why the mod(N) is not really "reflected" in the term m^ed at 6:39 if some smart people want to explain it to me :-)

  • @gamereditor59ner22
    @gamereditor59ner22 5 років тому +1

    Interesting notion. Can one time pad work with end to end encryption to pass the secret key successfully without "Eve" detecting it?

    • @zachstar
      @zachstar  5 років тому +2

      Technically yeah it could but you need a key for the one time pad anyway so if you were able to establish that key somehow beforehand then you wouldn't need to send one.

    • @gamereditor59ner22
      @gamereditor59ner22 5 років тому +1

      @@zachstar Thanks for the information and keep up the good work!

    • @woobilicious.
      @woobilicious. 4 роки тому

      Addition & subtraction under modulo 128/256 in binary is just xor, and most symmetric cyphers just generate a fake one time pad that is xor'd with the plaintext. And again xor'd to decrypt.

  • @skyline7532
    @skyline7532 5 років тому

    This is awesome

  • @BrunoValleBR
    @BrunoValleBR 4 роки тому

    How did you get to congruent to 8 in the first calculation?

  • @UMAIRKHAN-cz3pn
    @UMAIRKHAN-cz3pn 5 років тому +3

    Make a video on blockchain also.

  • @SeeingGreenDevils
    @SeeingGreenDevils 4 роки тому

    ok so do we use RSA OR DH? or are they used together? i know DH exchanges secret keys so why need RSA? I'm missing something there obviously.

    • @woobilicious.
      @woobilicious. 4 роки тому +1

      Both are used, DH is used to exchange a key for symmetric algorithm (because RSA is slow), but RSA is used for identity authentication, so you're not doing DH with Eve pretending to be Bob (man in the middle attack)

  • @steventran739
    @steventran739 5 років тому

    Make video about architecture

  • @erdemyilmaz6172
    @erdemyilmaz6172 2 роки тому

    Can you make a video on sha256?

  • @Gotta270101
    @Gotta270101 3 роки тому

    Great video! But please somebody help me out:
    at 6:32 you raise (m^e) to the power of d.
    But how?
    You don't know m^e, you just know (m^e MOD N).
    Shouldn't (m^e MOD N) be raised to d?

    • @1Backi
      @1Backi 2 роки тому

      same question, could you find an answer?

    • @Tejas-mm6tu
      @Tejas-mm6tu Рік тому

      Hey did you get ans?

    • @Tejas-mm6tu
      @Tejas-mm6tu Рік тому

      @@1Backi hey did you get ans?

    • @1Backi
      @1Backi Рік тому

      @@Tejas-mm6tu ua-cam.com/video/wXB-V_Keiu8/v-deo.html&ab_channel=ArtoftheProblem i guess this helped me understand better

  • @PandaBros863
    @PandaBros863 5 років тому

    Can someone explain how he got 5^6(mod23)=8, I saw his first video but I don’t understand the math behind this specific problem I’m lost

    • @moveonvillain1080
      @moveonvillain1080 5 років тому

      5^6mod(23)=remainder of ((5^6)÷23) which is 8

    • @moveonvillain1080
      @moveonvillain1080 5 років тому

      This mod is what u use in programming language , its the algebraic operator denoted by % in programming language

    • @PandaBros863
      @PandaBros863 5 років тому

      Tanay Verma thank you

  • @zajec11
    @zajec11 4 роки тому +1

    How do you select e and d? It seems that you've picked them somewhat at random, but what are the rules for picking those numbers computationally?

    • @jameswang7362
      @jameswang7362 4 роки тому

      You can select e relatively prime to phi(n) and then compute d (the multiplicative inverse of e) with the Extended Euclidean Algorithm.

  • @tsunami5884
    @tsunami5884 5 років тому

    why is the 0 with a line in it 5?

  • @maddyIncubus
    @maddyIncubus 4 роки тому

    where is the link to the course ?!

    • @zachstar
      @zachstar  4 роки тому +1

      top of description

  • @PhysicsBro-xb8qx
    @PhysicsBro-xb8qx 5 років тому +1

    WHAT IS YOUR DEGREE OR PROFESSION IN MY OPINION YOUR A ENGINEER

    • @Thmyris
      @Thmyris 5 років тому +1

      He is an engineer

    • @adrianokano8949
      @adrianokano8949 5 років тому +6

      He is an Electrical Engineer

    • @Roosyer
      @Roosyer 5 років тому +2

      He is a Math wizard

  • @unitedstatesofindia1460
    @unitedstatesofindia1460 3 роки тому

    Only 44k made it .... Not-at-all-strange

  • @rickh3714
    @rickh3714 5 років тому

    I have just posted my new Zodiac Killer Z340 decryption. It is in a Billowy wave format not in the dreaded cryptographical grid beloved of some mathematical types! Please click on the round abstract humanoid profile icon to see. Thanks.

    • @Darticus42
      @Darticus42 5 років тому

      Rick H rule 1 of crypto: never roll your own crypto

    • @harjitsingh7308
      @harjitsingh7308 5 років тому

      Darticus the Great that doesn't mean you can't make your own crypto and play with it ;)

    • @rickh3714
      @rickh3714 5 років тому

      Darticus the not so great? What rot! This case has not been 'solved' for 50 years because people have not thought outside the box. There are no explicit rules for the cryptographical methods of the esoteric and unhinged. They subvert the existing orthodoxies don't they?
      I have solved- to a coherent English sentence and pertinent appended German word , by an 'Archimedean' spiral algorithm, the Ebeorietemethhpiti letter remainder of the Z408.
      Also the Riverside Confession letter's 50 year hidden Morse transcript.
      'Intestis as I hone Z... Mete Stine I ensure it fund... etc etc.
      My Z34O solution may not satisfy an exclusively (myopic?) mathematical mind that is blind to the visual and geometric clues of an esoteric thinker.
      Godel himself showed that solving something and being able to provide a mathematical proof are not necessarily co-evident.
      What is wrong with finding a ' liquid' state solution to the Z340?
      'Zodiac' hints at this a multiplicity of times with his wordplay (some in Latin). There are more clues to the initial 340 keyword in another communique that I have yet to present.
      Throw away Occam's rusty old razor and conventionality when it comes to esoteric coding and highly complex personalities.

  • @rakra4551
    @rakra4551 4 роки тому

    what's with the annoying background music ?

  • @alwaysincentivestrumpethic6689
    @alwaysincentivestrumpethic6689 5 років тому +1

    Difficult

  • @TheRealInky
    @TheRealInky 4 роки тому

    Mind your ps and qs :D

  • @matthewto7406
    @matthewto7406 5 років тому +2

    First again?

  • @xamael1989
    @xamael1989 5 років тому

    Your looks more like 5

  • @ShopperPlug
    @ShopperPlug 3 роки тому

    Coursera is garbage, Udemy.com is way better.

  • @shivashankar28
    @shivashankar28 5 років тому

    Please we need more Electrical engineering videos, not CS stuff

    • @Darticus42
      @Darticus42 5 років тому +7

      Shiva Shankar Nah this is fine too