It's good in that it's simple, however unlike the discreet logarithm problem, you would be able to make a good estimate of the secret colour based on the starting colour and one of the mixed colours. For example if Eve received the starting colour yellow, and a mixed colour green, she can infer that the secret colour mixed in must be some shade of blue, which makes her search much easier. Recognizing this threw me off a bit at first.
unfortunately, public key is completely different than key exchange. public key requires different keys to encrypt and decrypt, so there's no need for diffie hellman to agree on a secret key.
Martin Hellman said: The system...has since become known as Diffie-Hellman key exchange. While that system was first described in a paper by Diffie and me, it is a public key distribution system, a concept developed by Merkle, and hence should be called 'Diffie-Hellman-Merkle key exchange' if names are to be associated with it. I hope this small pulpit might help in that endeavor to recognize Merkle's equal contribution to the invention of public key cryptography.
A great illustration. Diffie-Hellman has a well-known, fun vulnerability. Spoilers: Eve, knowledgeable herself on color theory, intercepts messages between Alice and Bob not letting their messages go directly to them. Instead she creates a color of her own. Mixing it twice with each of Alice and Bob's colors she creates two keys. She can now read Bob's message, re-encrypt, and send to Alice and pose as Bob. Same goes in the other direction. If only Alice could trust Bob's color comes from him.
I've watched a few videos on public key cryptography, but never really understood how it worked until I heard this colour analogy. Absolutely phenomenal video!
My background in advanced math concepts is somewhat limited, and so it's always been difficult for me to intuitively grasp how DH worked. After years of struggling, this is the one video that really drove the point home for me. Thank you!
Very nice! Hat off! One of the best explanations I have seen, and nice put into the story. however, when you swap those powers, you should use parenthesis, that is because generally, powering is not commutative. That is, a^b^c is not equal to a^c^b, modular or non modular powering. Powering is right-associative. But (a^b)^c=a^b*a^b*...a^b (c times) which is a^(b*c)=a^(c*b)=a*a*a*a.... (b*c times), which is (a^c)^b always, modular or not. This is due to the commutativity of the _multiplication_ operation. Not the powers.
Funny you say that, i'm working on developing a podcast right now. I was town between just using the audio from these or doing a new conversational approach. can you listen to the demo I posted last week and give feedback? ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-1w4Y_sCDeCE.html
Time hardened Encryption just like safe hardening how much time is needed to open it. I love this, this is the best way to explain encryption ever. I love how they have IBM sage running for this video also. Amazing
Brilliant explanation about key exchange for those of you interested in how your data is encrypted over the web. Ok, when the maths comes you need to pay attention but all in all the best explanation I've found.
I'm reading wiki trying to understand how public-key encryption works (I'm told its better than symmetrical encryption). I remember someone tried to explain this before using colors, so a quick search--and I find your video. This is a great video.
I really really like the music in this video. It mysterious. I like the fact that you take your time to explain and it is also visual. Nice creative video.
Can't thank you enough. Awesome video. I wish you also explained how the digital signature works in order to avoid Eve pretending to be either Bob or Alice.
Fantastic. I've watched many videos on this same topic; nevertheless, this is The Best one. A million thanks for breaking down difficult concepts in an easy, understandable way. Kudos!
wow finaly the video i was looking for with the best explanation and number proving examples thank you very much I also checked your chanel realy awesome
Great video, you described it in a perfect way to understand. Though I'm not sure if it was clear for everyone that this was merely for calculating a mutual key to use with a cipher, and not really for actually communicating information itself.
Excellent video! My only complaint is the explanation of "how Alice did the same calculation as Bob" from 7:27 to about 7:40. Starting at 7:27, we see that "12 = 3^13mod17". Then conveniently, right at 7:34, when that figure is substituted into Alice's original expression, the "mod 17" DISAPPEARS and the 12 is simply replaced by "3^13". Although this IS mathematically correct, it REQUIRES a rather advanced principle of modular arithmetic: namely, that [(a*mod c)^ b]*mod c = (a^b)mod c. (In the example from the video, a = 3^13, b is 15, and c is 17). So, you effectively CAN simply remove the extra "mod c" term, but the video glosses over this difficult but crucial step. My sister and I just spent 2 hours figuring out the proof for this principle. If anyone's interested I can share a photo of the completed proof. (It can be found online also).
For a few months, my teacher didn't manage to explain this to a class. In 8 minutes, this video can explain it to every dummy. If it's simple, keep it simple.
Good explanation, better than those explanations given by the professors in lectures... My tutors can explain this to me for 1 day and I still don't get it. Now I find this concept extremely simple.
Really helpful and very interesting..thank you very much.😁😄 it helped me in my exam...it is tomorrow by the way😋 please can you do more videos..on like Rc6,Fiestel structure,Aes,Des,ceasar cipher,etc. but please don't stop making videos..keep doing it. We really appreciate it. thank you soo much again.
The trick in a nutshell: ( G^*a* mod P )^*b* mod P = G^*a*^*b* mod P = ( G^*b* mod P)^*a* mod P = *key* *a* and *b* - private numbers *key* - private key (same for both) G - public generator P - public prime module ( G^*a* mod P ) = *A* ( G^*b* mod P) = *B* *A* and *B* - public numbers both sites do: *A*^*b* mod P = *B*^*a* mod P = *key*
I try to calculate in Javascript but found it not the same, is there any wrong? According to the fomula "( G^a mod P )^b mod P = G^a^b mod P", Assume G = 3, a = 13, P = 17, b = 15 Math.pow(Math.pow(3, 13) % 17, 15) % 17 = 10 Math.pow(Math.pow(3, 13), 15) % 17 = 2 Math.pow(Math.pow(3, 15) % 17, 13) % 17 = 10 But 10 is not equal to 2