could you please link an explanation to this? i've been trying to understand how g^a and g^b were made public when in actuality it was those numbers _mod n_ that were made public. also don't understand how you can calculate (g^a)b mod n without first having g^a, which again, wasn't made public because of the modulo term
@@znefas I never saw an explanation myself, but consider the following: imagine we are working in base N, and not binary or decimal. Then, modN just keeps the last digit. So, since we know that it's possible to start by calculating the last digit when doing long addition/multiplication, as the carry only travels left, we can discard the left digits at any intermediate steps and still have the same result as if we only did it at the end. Signed addition and negative powers (division) break this, but wrapping subtraction of unsigned integers actually survives afaik. This is the same reason why we can know the last digit of 7^44444 will be 1, without having to calculate the whole number.
@@qm3ster the only thing i understood is how the carry only traveling left may be useful when trying to calculate a number smaller than the entire expression; i don't understand how the whole mod N thing works i also hate modular arithmetic notation because "13 = 5 mod 8" is so much more annoying than "13 % 8 = 5" for me as a programmer
@@znefas for all unsigned integers a, b, c, n: (a + b)%n == (a%n + b%n)%n (a * b)%n == ((a%n)*(b%n))%n ((a ^ b) ^c)%n == (((a%n)^b)%n)^c)%n As long as at the end you were dropping the stuff left of n, the result won't be changed by dropping it earlier, as it cannot affect what is on the right. The intermediate values will in fact be different, which is what makes this a computational saving. But outside cryptography and adjacent subjects like checksumming you usually want the most significant bits, not the least significant, they're called that for a reason, so this doesn't get used *too* often.
@@znefas g^a and g^b wasn't made public, only g^a%n and g^b%n - Mike got a bit sloppy when presenting (just as we do when we write down the maths). If we knew g^a, and since we know g, it would be trivial to work out a (and similarly for b).
I just took a crypto class in university and my professor got so wrapped up in the specifics of group theory that I didn't even understand how Diffie-Hellman worked, but this made everything so clear without really being any less mathematical.
There is quite some math behind it about rings, fields and other group theory. But the plain formulas in crypto are pretty nice and easy. The computation is quite a mess as numbers get pretty big, often even bigger than your memory and power and multiply are not the chips best friend.
I find there are a lot of these types of algorithms; Radix Sort comes to mind... It seems like voodoo, then you look at the algorithm, and probably at first think 'that can't possibly work...?' but viola...
@@MrHaggyythat's still somewhat elegant, square-multiply-modulo is a genius and simple solution to easily calculate such an otherwise gargantuant number. I find the truly hard (and amazing) part is the theory behind which numbers are safe to use, even more fascinatingly and terrifyingly genius are the ways people find to crack it
@@justalonelypoteto square-multiply is a genius algorithm as it has view operations,. Lattice and Russian Peasant are awesome too. A lot more operations but most of them are cheap additions that can be done in parallel. My interest is hard-realtime systems. We want algorithms with fixed calculation times. Square-multiply can vary depending on the numbers. And on cheap and small hardware you might be able to eavedrop with an oscilloscope.
@@MrHaggyy Montgomery's ladder might be up your alley, it performs both operations each time with the only distinguishable difference being a possible timing variation because the needed operation affects which memory address is read, apparently this is fixed by scattering the data around to ensure cache misses
Came in expecting some incomprehensible esoteric math but this was very easy to understand. Reminds me of that Doctor Strange quote "It's a simple spell but quite unbreakable"
awesome video, I see some comments point out its actually (((g^a) mod n) ^ g)) mod n, so here is why its the same as (g^a)^b mod n: lets say: g^a = k*n+x where x is an unknown number because: (u + v) mod n =(u mod n + v mod n) mod n and because: k*n mod n = 0 we get: (g^a) mod n = (k*n+x) mod n = x ((g^a) mod n)^b mod n = x^b mod n i used the binomial theorem nCr to get: (g^a)^b = (k*n+x)^b = (k*n)^b + (nC1)*(k*n)^(b-1)*x + ... + (nC(b-1))*( (k*n)^(1)*x^(b-1) + x^b since n appears in every part except in x^b every other part is set to 0 when we take the modulo: (g^a)^b mod n = x^b mod n so therefore (g^a)^b mod n = ((g^a) mod n)^b mod n
Interesting proof. But a few corrections: First what you are trying to prove is that (g^a)^b mod n == (g^a mod n)^b You made a mistake, in (((g^a) mod n) ^ g)) mod n . You put an extra "g", it's a "b" Also to simplify...or clarify your proof, and make it more understandable, organized: 1) supposing: g^a = n*k+x ( *WHERE "x" IS NOT A MULTIPLE OF "n"* , in other words "x" is the "remainder" of the division by "n" ) This is an important purposeful "setting up" so that: n*k mod n == 0 and x mod n == x 2) mod n of any number which is a multiple of n will equal zero --> n*k mod n == 0 3) Proof that: (g^a mod n)^b == x^b a) Which would mean --> (g^a mod n) == x a) substitute "n*k+x" for "g^a", from step "1" (n*k + x) mod n == x --> given that: n*k mod n == 0 (any multiple of "n" will be zero) b) If (n*k + x) mod n == x, then (g^a mod n) == x, c) meaning --> (g^a mod n)^b == x^b 4) (g^a)^b mod n a) substitute (n*k+x) for g^a, given in step "1" --> (n*k+x)^b mod n
b) Expanding (n*k+x)^b, using binomial theorem, *IMPORTANT MAGIC* : *all terms will be multiples of "n"* except the last term which is x^b c) Mod n of all terms that are multiples of "n" will be zero, leaving only the last term: x^b d) Thus (g^a)^b mod n = x^b 5) (g^a)^b mod n == (g^a mod n)^b , as both equal to x^b .
Technically you could have n on the clock but you'd have to omit 0. You can have n or you can have 0, but not both. Having n on the clock would actually be more like counting units of 1 up to n and then starting back at 1 again for the next unit until you run out of units. Same result (specifically in this case because the number itself doesn't matter), but a different perspective on it.
Having a 0 instead of the n is very useful when you work with such a construct though. This clock face is what mathematicians call a cyclic group, and its "neutral element" is 0 (or n, they are the same in the cyclical group). That means that (a+0)mod n=a mod n, so the 0 doesn't change an element by being added. That is very obvious if you actually write it as a 0 instead of an n.
The idea of diffie hellman is actually really clever. Its actually pretty hard to come up with a way to do that, and not have anyone be able to reverse how they did it or needing to share something that can be reversed. I can see why we still use this today it would be very difficult to replace it, and its very ingenious how it works and its not incredibly complicated either.
It looks really simple, but it unfortunately doesn't explain whether "(g^a mod n)^b" and "(g^a)^b mod n" are the same. I mean obviously it must be the same for it to work.
The simple answer as to why they're the same is that Modulo arithmetic=Finite fields, and Finite fields are consistent. Proving that finite fields are consistent would probably be another couple videos...
Short answer: It's because the product in modulo/clock arithmetic is defined that way. If you wanna know more I'll explain it in more detail: Maybe it's easier if you think of g^a mod n just as a number between 0 and n-1 (actually if we use this notation we should start from 1 and go up to n-1, but it's not extremely important so let's forget about it). Think of the multiplication between two such numbers as a standard multiplication, but if the result is bigger than n-1 we take the remainder. So basically the product is just: (a mod n)*(b mod n) = a*b mod n (THIS IS THE DEFINITION OF PRODUCT IN MODULO ARITHMETIC) Elevating g to the power a is just repeated multiplication, so what you do is apply the definition of product: (g mod n)^a = (g mod n)*(g mod n)*...*(g mod n) = g*g*g*...*g mod n You can easly see that we get a similar thing if we elevate g^a to the power b: (g^a mod n)*(g^a mod n)*...*(g^a mod n) = (g^a)*(g^a)*...*(g^a) mod n = (g^a)^b mod n So yeah, (g^a mod n)^b = (g^a)^b mod n Hope it's clear now ;)
The "mod n" isn't an operation in this case. Appending "mod n" to a term just signals that you're calculating in clock arithmetic, meaning you always stay between 0 and n-1. If you do 2+3 in "mod 4", the answer is 5, or 1, or even 9, because in "mod 4" 1 is equal to 5 by definition. Calculating a modulo is not an operation that needs to be performed if you define yourself to be working in clock arithmetic
Actually g^(ab) mod n = (g^a mod n)^ b mod n, but yes, it is the same. The proof is just long and kinda gross which is why he didn't get into it. Here it is if you want to see it though. Proof: Any number m mod n is defined as the remainder left after dividing m by n, so we have m=q*n+r where q is the whole number of times n goes into m, and m mod n=r, which is the remainder. So g^a=q_a*n+r_a, and g^a mod n=r_a. (I'm using the underscore to denote a subscript.) (g^a mod n)^b=r_a^b=q_1*n+r_1, and (g^a mod n)^b mod n=r_1. g^(a*b)=q_(ab)*n+r_(ab), and g^(a*b) mod n =r_(ab). But we also know g^(a*b)=(g^a)^b, and g^a=q_a*n+r_a, so g^(a*b)=(q_a*n+r_a)^b. Expanding that out into a sum we get (q_a*n+r_a)^b=sum((q_a*n)^(b-j)*r_a^j,j=0..b) The last term in this sum is r_a^b and every other term has at least one factor of n, so we can rewrite it as (q_a*n+r_a)^b=n*sum(q_a^(b-j)*n^(b-j-1)*r_a^j , j=0..b-1)+r_a^b. Now let q_2=sum(q_a^(b-j)*n^(b-j-1)*r_a^j , j=0..b-1), so we have (q_a*n+r_a)^b=q_2*n+r_a^b. Remember that this is equal to g^(a*b). So we have g^(a*b)=q_2*n+r_a^b. We already have r_a^b=q_1*n+r_1, so we have g^(a*b)=(q_1+q_2)*n+r_1 and g^(a*b) mod n = r_1. We already had that (g^a mod n)^b mod n = r_1 so we have now proved that g^(a*b) mod n = (g^a mod n)^b mod n.
This made so much sense and was super easy to grasp. Mike pound really is a great teacher. Also props to Diffie Hellman for creating such an easy but quite unbreakable algorithm.
It would be awesome if you guys did some podcasts. I learn so much from you guys and it would be great to listen while in the car or at school. Thanks guys.
Hi Mike, Alice or Bob never shared "g to the power of a or b" but they shared "g to the power of a (b) mod n". But as we continued the example we said they shared the power of a or b. Can you please help explain that?
For anyone who got confused with this, we start with the left-hand side of the equation: (g^a mod p)^b mod p = (g^ab mod p) mod p [by the property of modular exponentiation] Now, we move on to the right-hand side of the equation: (g^b mod p)^a mod p = (g^ba mod p) mod p [by the property of modular exponentiation] Since (g^ab mod p) mod p = (g^ba mod p) mod p, we can say that: (g^a mod p)^b mod p = (g^b mod p)^a mod p Hope this helps.
It's interesting that he doesn't mention that n should be prime in addition of being large. That is an aspect equally as important as its size which could easily break the encryption (if n isn't prime). There was even a defcon panel (2016 I think) that discussed this very problem and how it is often ignored.
I really was hoping he would explain why n has to be prime. I'm thinking the reason is that the search space of [numbers] mod n would be much smaller than n if it weren't a prime number, but that's just my intuition and I'm not sure if I'm right.
@@tissuepaper9962 The goal is to have the denominator and numerator being coprime i.e. their greatest common divisor is 1 as otherwise, you'd end up getting 0 as the result. The only way to guarantee that is with a prime number which is naturally coprime to any number.
The clock-face analogy is really powerful as it really helps explain why it is not very reversible. "Undoing" g^a%n is effectively going backwards in steps of n and checking to see if g factors into this number an integer number of times. Since a is generally (very) large, we'd have to go backwards an awful lot of steps. There can be more efficient ways of doing it, but conceptually it's the same. Note 2048 bits --> N < 2^2048 ~ 10^616. The world's best supercomputer has a power of ~1exaflop or 10^18 floating point calculations per second. Assuming every attempt could be achieved in 1 operation, that would take 10^598 sec = *10^591 years* I always find it difficult to understand how "difficult" something is when we say p is a really big number, until it is put into a time-to-calculate number.
At 4:00 that exponentiation is flawed, because you need to calculate (g^a mod n)^b, not (g^a)^b. Those are not the same, and the explanation he gave is not correct. He should have said that because of the modulus exponentiation property (g^a mod n)^b mod n is the same as (g^b mod n)^a mod n. He omitted a really important step when he jumped to (g^a)^b is equal to (g^b)^a. That is something that you arrive at when you apply the modulus exponentiaion property.
thank you for putting this video up. Such a good teacher and elegant explanation of public private key at its core. There are 100s of videos super complicated.. this is very nice, one can implement themselves right off the this video..
not only that, but the primality of g is completely irrelevant - the n needs to be prime if you want the n in the 2 to 4k range (and DH to remain secure), best if n is actually a safe prime (then g can be any number but 0, 1 and n-1) - a prime for which (n-1)/2 is also prime
The whole time I am thinking about the clock-drawing test, after seeing those clockface numbers beeing squished together. PS. I find this video even simpler to understand than the original "simple" version with the food colouring.
Yes, the thing with math is that it is a language which you need to learn. Once you do, the concepts behind the maths become not that complicated, when you know them. Implementation could still be complicated though. Devils in the details...
3:44 Bob is sending (g^b mod n) and not g^b without modulo n. Else it would be simple to extract b. But the result of "(g^b)^a mod n" is the same as "(g^b mod n)^a mod n". Was that him saying "I change notation to save space"?
Great video. One thing I’m not sure on is, how is g^b raised to the power of ^a, because it seems that the value shared is just “g^b mon n” as a discrete value ?? So how is the g^b disaggregated? 3:25
Finally a decent video on computerphile... all of them should be AT LEAST this caliber instead of resorting to oversimplified (and hence incorrect) analogies.
After watching all three videos, its much easier to understand and glad you used 'n' here that you didn't use in the last video with colors. This was an amazing series but if you label them as part of a series or make 1 full video combining all three or the first and last video, more viewers might decide to stick to the end. Anyway, Thanks a lot for this.
Yeah, I wanted to hear that too, but considering the technique wouldn't work otherwise, it probably is. It also kind of makes intuitive sense in the terms of significant bits. Modulo tosses away all most-significant bits, and only keeps the least significant bits. Multiplication will push the least significant bits into the most significant bits (two 32 bit numbers multiplied for example, gives a 64 bit number). So tossing away those most significant bits (which would've been moved to even higher significant bits) shouldn't matter.
The modulo just takes the remainder of the division: being the multiplication commutative, g^(ab) = g^(ba), you'll get the same number. You can then do modulo n, but the number will be the same anyways.
Let's say g^{a} = X + Y, where X is divisible by n and Y < n (therefore g^{a} mod n = Y). Then g^{ab} is (X^b + c1*X{b-1}Y + c2*X{b-2}Y^{2} + ... + Y^b). Since X is a multiple of n, all but the last term will be zero mod n, so we are left with Y^b mod n = ( g^{a} mod n)^b mod n after we take the modulo. Therefore g^{ab} mod n = g^{a} mod n)^b mod n. I hope that's clear enough.
This is why he put this in the Maths video and not the colour video. As you should know this basic thing to know the theory. And he mentioned it in the video, very short.
I was wondering why Diffie-Hellman hasn't been deprecated in favor of a better algorithm since it's almost 50 years old now. Now I know why. You could always increase the bit count of N to compensate for faster computers and still get a shared secret in as low as 2 messages. Neat!
The missing part is why (g^a mod n) is fast to compute even though the reverse is not. Tl;dr; g^a can be computed by starting with x_0=1 and and defining x_1=a_0^2 or x_1=x_0^2 * g and then doing the same for x_2 and so on. Making the right choices at each step (which is basically reading out the bits of a) you get g^a in at most log2(a)+1 steps.
thanks your video is very helpful, but i have a question let's say in signal or realtime chat webapp WhatsApp using Diffie Hellman, on server side only public key is saved. So I have a question where is the private key stored for each user when they log in so that they can be retrieved and every time you log in on another device, how do you get that user's private key? thanks and hope you can help me answer my question
Yeah! Best explanation on key exchange ever. Thanks to both videos I understood it instantaneously. A test on paper and mental arithmetic worked like a charm. Thanks :-)
At 4:47 it’s been said that we know g^a and g^b, but shouldn’t it say we know g^a mod n and g^b mod n? If g is public and g^a would be public too, we could calculate the value of a using the logarithm of g for g^a, couldn’t we?
3:47 "Alice is going to take the g^b that Bob sent" Doesn't Bob actually send b^g mod n ? If yes, how does Alice reverse that into b^g? And if not (Bob actually sends b^g), then it's very easy to compute b, Bob's private key.
Is it true about what he says around 4:42 that we know what g^a and g^b is? I can't find evidence of this elsewhere, and it would seem to be to be an incorrect statement, as if you knew, g^b, and g, then it would be trivial to compute the log of g^b, and find out what b was. I think he must mean: g^a Mod n and g^b Mod n.
@03:45 -> Does Bob send g^b or [(g^b) mod n]? Both are completely different numbers, so how does Alice extract g^b from [(g^b) mod n] if it is indeed the latter that got sent by Bob, since Alice has no knowledge of b?
Bob sends (g^b) mod n. however, Alice does not need g^b. when working with the mod operator, it does not matter if you apply it to the intermediary results as well or only the final result. he probably should have mentioned it.
@@danielf.7151 thanks yeah looked it up and it's the reduction property modular exponentiation : (a^b) mod n = ((a mod n)^b) mod n which can be fairly easily deduced (when a is represented as k*n + r) with binomial expansion of ((k*n + r)^b) mod n which leaves behind r^b in both cases.
Does Alice and Bob gonna share the exponent values with the modulo operation or without it at the first exchange ? Cuz he first said said that both of them gonna use the modulo operation but he completed the analogy with only the exponents values (g^a and g^b)
Making g really big does improve security, but also drastically increases the computing and memory resource requirements since g^a will be (even more) ridiculously large.
Just to caveat, in modulo we deal with numbers from 0 to n-1. We exclude n since the remainder can't be equal n - it must be less than n, otherwise when we're dividing by n and we see that there's a remainder n it means that n could fit one more time in that number leaving us with the remainder 0.
I like the Explanation it help me understand lot better but I have two Questions 1. You mention G^a would be huge number. Did you mention the full equation G^a Mod n would give you a huge number or you just meant G^a Only. 2. you mention N should be a huge number, 2000 to 4000, so that it's not easy to brute force attack. There are computers these days that can calculate number between 4000. in these case is algorithm is secrete or N is secrete?
1. Only g^a is a huge number. g^a Mod n is always a number between 0 and n-1 inclusive. 2. N is not secret, only a and b are. also, 2000 to 4000 is not the bumber itself, but the length of the number in bits. a 2000 bit number has around 600 digits in decimal.
X mod n will be in the range of 0 to n-1, not 0 to n. I'm also a little unclear as to how we get g^(a*b) mod n. Bob, for instance would have g, n, b, and g^a mod n. I think it might be the same as (g^a mod n)^b mod n, which Bob could compute, but I'm not sure if that's correct, or if so, why.
That "mod n" is important for keeping things secret. Remember that it is the *discrete* logarithm problem that is hard. If I give you `g` and `g^a` you can find `a` you could find `a` by log_g (g^a). That is what a logarithm is: the inverse of exponentiation. The 'mod n' piece is what make the problem hard because you don't know how many times you have looped back to 0 again. For any given g and n there actually might be a lot of values for a that would work, *but* if g and n are relatively prime (they have no common factors) then there is exactly one value a less than n that will result in g^ mod n. That is why the subtle mention that g and n are large *prime* numbers. The math still works if they are not, but it makes the problem of finding a much easier.
4:45 when he writes g^a and g^b as being made public, he should have said and written g^a mod N, and g^b mod N, were made public. thanks for explanation.
I don't understand the part in 3:45. So do they send the answer to the equations of (g^a mod n)? Or do they send over each part separately? Because if they sent it separately then that'd be useless because a and b would be public. And if they send (g^a mod n) then how does bob change that in (g^a^b mod n)? How is that possible? Because the answer is the mod, and that would be ((g^a mod n)^b) which is a different answer. And if they sent (g^a) then if bob multiplied (g^a)^b it'd give you the right answer but the answer is different from literally getting the answer of (g^a) then powering it to ^b. So how do they send the data over without making it public?
bob sends g^b mod n alice sends g^a mod n a and b are not revealed (g^a mod n)^b mod n = (g^b mod n)^a mod n = the shared secret Alice doesnt need to know b and Bob doesnt need to know a, those are private.
Another thing that I noticed (but doesn't matter), is that actually, Bob can't compute (g^a)^b mod n, because he has to calculate (g^a mod n)^b mod n. The beauty of modulo arithmetic is that the result is the same.
This is a really nice explanation and thank you for talking about it. I wish you talked a bit about man in the middle weakness and the ElGamal improvement.
It is quite misleading that after they exchange (g^a)mod n and (g^b)mod n Mike goes on as if alice knew the value of g^b and bob g^a and even saying that those values would now be public, which is simply wrong.
I'm glad I'm not the only one I thought this, I literally couldn't go on watching the video, this bothered me so much. At 3:47 He says Bob sends Alice g^b, well now everyone knows b. If he meant to say Bob sends Alice g^b mod n, then he doesn't explains how Alice is able to extract g^b from that (why can't people listening in do this then??) in order to raise it to a. Very frustrating.
When we start to go inside the cryptography's math we always face the formula : a^b mod n. But if a and b are huge number, how in the world can we compute a^b ? To work out a^b mod n ?
I don't really get it why it's difficult to crack the a or b. Assume calculating big 'A' takes 1ms, a=1000,000, a brute force to crack little 'a' will take 1000,000 rounds, which takes 1000,000ms=1000s. Is it the bigger the number, the slower the calculate process? Then it will be gradually slower and slower from 1 to 1000,000?
Yes, you can do that pretty efficiently. It's called modular exponentiation. Basically, ga^b is calculated (ga is (g^a)mod n) the way it normally would, but after every little step of that calculation, the mod n of the current result so far is calculated.
Everywhere are for example purpose this little numbers. But what number range do a b g n actual need to be in a real life use case? What would be a small prime number 'g' be? Are 2000-4000 some legit values in a real implementation or are there some more standardized values?
Can I get some clarification on the math? I understand how g^ab % n is the same as g^ba % n but after the private numbers are used (g^a%n and g^b%n) and shared, when the other person's private number is used, using Alice for this example, instead of g^ba % n wouldn't it be (g^b%n)^a? This order of operations is confusing me
@@evankivolowitz4788 This is not true. What you are thinking of is prime factorization (getting the product in primes of a number) but that does not come into play here. Also you never send g^a or g^b but the remainder of that number after mod n. The reason for prime numbers is, that you distribute possible secrets on the whole range 0, ..., n-1.
Q: If you had a 100 petaflop supercomputer and you discovered some workarounds in how diffie hellman key exchange can be sped up, say 5x, how long would it take on average to discover the private key given the size of 4096 bit RSA? You can make some assumptions about how many operations it would take to compute one key pair.
Is it important for the encryption key how many times we go around the clock to get to the answer? Otherwise surely you would only have to brute force numbers zero to n?¿
If alice can take g^b then it is public right and that would mean you could find out b right and then just put (g^a)^b) mod n and you got the key??? But also you can't get ((g^a)^b) if the only number you get from alice is (g^a) mod n...
don't we usually choose n to be prime as well in order to make Z_n a field? Otherwise the largest field we can construct using modulo n is Z_(phi(n)), which is not necessarily large even if n is large.
In previous video Mike said the math is very complex so separated out to another video. Now by the end of this video Mike says the math is not very complex it's simple. 🙃 Actually it was hard for me to understand how (g^a mod n)^b became (g^ab mod n) . It felt so counter intuitive to me that I had to test the math with several sample values before I could convince my brain it is true.
I think the fact that addition, multiplication, and exponentiation maintain their properties under modulo arithmetics was worth mentioning.
could you please link an explanation to this? i've been trying to understand how g^a and g^b were made public when in actuality it was those numbers _mod n_ that were made public.
also don't understand how you can calculate (g^a)b mod n without first having g^a, which again, wasn't made public because of the modulo term
@@znefas I never saw an explanation myself, but consider the following: imagine we are working in base N, and not binary or decimal. Then, modN just keeps the last digit.
So, since we know that it's possible to start by calculating the last digit when doing long addition/multiplication, as the carry only travels left, we can discard the left digits at any intermediate steps and still have the same result as if we only did it at the end.
Signed addition and negative powers (division) break this, but wrapping subtraction of unsigned integers actually survives afaik.
This is the same reason why we can know the last digit of 7^44444 will be 1, without having to calculate the whole number.
@@qm3ster the only thing i understood is how the carry only traveling left may be useful when trying to calculate a number smaller than the entire expression; i don't understand how the whole mod N thing works
i also hate modular arithmetic notation because "13 = 5 mod 8" is so much more annoying than "13 % 8 = 5" for me as a programmer
@@znefas for all unsigned integers a, b, c, n:
(a + b)%n == (a%n + b%n)%n
(a * b)%n == ((a%n)*(b%n))%n
((a ^ b) ^c)%n == (((a%n)^b)%n)^c)%n
As long as at the end you were dropping the stuff left of n, the result won't be changed by dropping it earlier, as it cannot affect what is on the right.
The intermediate values will in fact be different, which is what makes this a computational saving.
But outside cryptography and adjacent subjects like checksumming you usually want the most significant bits, not the least significant, they're called that for a reason, so this doesn't get used *too* often.
@@znefas g^a and g^b wasn't made public, only g^a%n and g^b%n - Mike got a bit sloppy when presenting (just as we do when we write down the maths). If we knew g^a, and since we know g, it would be trivial to work out a (and similarly for b).
I just took a crypto class in university and my professor got so wrapped up in the specifics of group theory that I didn't even understand how Diffie-Hellman worked, but this made everything so clear without really being any less mathematical.
love the way this guy explains.. never imagined i could binge watch cryptography videos
Right?!
All these years of hearing how "diffie helman is extremely complex" yet such a simple and easily understandable algorithm. Thank you so much for this!
As soon as someone says "It's extremely complex", usually means they don't understand it, and therefore don't want to be asked to explain it...
It’s amazing how beautifully elegant and simple cryptography really is.
5:02 THIS! IS! CRYPTOGRAPHY! *_*kicks you down a 4096-bit deep well*_*
Alice and Bob only communicate in Pub. Alice and Bob are probably undergrads.
This is the most interesting channel about computer science. I’ll never be as much grateful as I would like to be. Thanks a lot for sharing these.
Wow. I've always thought D-H was some esoteric math magic. It is extremely clever and elegant.
There is quite some math behind it about rings, fields and other group theory. But the plain formulas in crypto are pretty nice and easy. The computation is quite a mess as numbers get pretty big, often even bigger than your memory and power and multiply are not the chips best friend.
I find there are a lot of these types of algorithms; Radix Sort comes to mind... It seems like voodoo, then you look at the algorithm, and probably at first think 'that can't possibly work...?' but viola...
@@MrHaggyythat's still somewhat elegant, square-multiply-modulo is a genius and simple solution to easily calculate such an otherwise gargantuant number. I find the truly hard (and amazing) part is the theory behind which numbers are safe to use, even more fascinatingly and terrifyingly genius are the ways people find to crack it
@@justalonelypoteto square-multiply is a genius algorithm as it has view operations,. Lattice and Russian Peasant are awesome too. A lot more operations but most of them are cheap additions that can be done in parallel. My interest is hard-realtime systems. We want algorithms with fixed calculation times. Square-multiply can vary depending on the numbers. And on cheap and small hardware you might be able to eavedrop with an oscilloscope.
@@MrHaggyy Montgomery's ladder might be up your alley, it performs both operations each time with the only distinguishable difference being a possible timing variation because the needed operation affects which memory address is read, apparently this is fixed by scattering the data around to ensure cache misses
I love these math versions. They help me understand it so much better than just the computer science bits. Please do more of these!
Great clock analogy. This should be the main video
A Parker clock, one might say!
It should not, as they used the colour analogy there.
The clock is not strictly an analogy, it's a common learning technique for modular arithmetic
It explains why n needs to be a big value.
The clock analogy I think is one of the best ways to explain modulo arithmetic...
Came in expecting some incomprehensible esoteric math but this was very easy to understand. Reminds me of that Doctor Strange quote "It's a simple spell but quite unbreakable"
If you watched Attack on Titan ,one of the character used. Only his words and hacked the entire cosmos to make it seem real at that moment
awesome video, I see some comments point out its actually (((g^a) mod n) ^ g)) mod n, so here is why its the same as (g^a)^b mod n:
lets say:
g^a = k*n+x
where x is an unknown number
because: (u + v) mod n =(u mod n + v mod n) mod n
and because: k*n mod n = 0
we get:
(g^a) mod n = (k*n+x) mod n = x
((g^a) mod n)^b mod n = x^b mod n
i used the binomial theorem nCr to get:
(g^a)^b = (k*n+x)^b = (k*n)^b + (nC1)*(k*n)^(b-1)*x + ... + (nC(b-1))*( (k*n)^(1)*x^(b-1) + x^b
since n appears in every part except in x^b every other part is set to 0 when we take the modulo:
(g^a)^b mod n = x^b mod n
so therefore (g^a)^b mod n = ((g^a) mod n)^b mod n
Yup.
Thanks for this, felt like a logical leap without this proof.
At the beginning there, did you put a g, where you meant a b?
@@nelsblair2667
He did, yes.
Interesting proof. But a few corrections:
First what you are trying to prove is that (g^a)^b mod n == (g^a mod n)^b
You made a mistake, in (((g^a) mod n) ^ g)) mod n . You put an extra "g", it's a "b"
Also to simplify...or clarify your proof, and make it more understandable, organized:
1) supposing: g^a = n*k+x
( *WHERE "x" IS NOT A MULTIPLE OF "n"* , in other words "x" is the "remainder" of the division by "n" )
This is an important purposeful "setting up" so that: n*k mod n == 0 and x mod n == x
2) mod n of any number which is a multiple of n will equal zero --> n*k mod n == 0
3) Proof that: (g^a mod n)^b == x^b
a) Which would mean --> (g^a mod n) == x
a) substitute "n*k+x" for "g^a", from step "1"
(n*k + x) mod n == x --> given that: n*k mod n == 0 (any multiple of "n" will be zero)
b) If (n*k + x) mod n == x, then (g^a mod n) == x,
c) meaning --> (g^a mod n)^b == x^b
4) (g^a)^b mod n
a) substitute (n*k+x) for g^a, given in step "1" --> (n*k+x)^b mod n
b) Expanding (n*k+x)^b, using binomial theorem,
*IMPORTANT MAGIC* : *all terms will be multiples of "n"* except the last term which is x^b
c) Mod n of all terms that are multiples of "n" will be zero, leaving only the last term: x^b
d) Thus (g^a)^b mod n = x^b
5) (g^a)^b mod n == (g^a mod n)^b , as both equal to x^b
.
Wow, that is way more simple than I was expecting.
Wouldn't a mod n only return values from 0 to _n_-1? You wrote _n_ on the clock.
Correct that was an error on their part since of course if a divides by n exactly the remainder is 0 not n.
Technically you could have n on the clock but you'd have to omit 0. You can have n or you can have 0, but not both. Having n on the clock would actually be more like counting units of 1 up to n and then starting back at 1 again for the next unit until you run out of units. Same result (specifically in this case because the number itself doesn't matter), but a different perspective on it.
Having a 0 instead of the n is very useful when you work with such a construct though. This clock face is what mathematicians call a cyclic group, and its "neutral element" is 0 (or n, they are the same in the cyclical group). That means that (a+0)mod n=a mod n, so the 0 doesn't change an element by being added.
That is very obvious if you actually write it as a 0 instead of an n.
And multiply by zero also give nice properties. :-)
But clocks are 1-based.
For some reason I understand all his videos but not many of the other tutors on Comphile.. Quite a teacher
The idea of diffie hellman is actually really clever. Its actually pretty hard to come up with a way to do that, and not have anyone be able to reverse how they did it or needing to share something that can be reversed. I can see why we still use this today it would be very difficult to replace it, and its very ingenious how it works and its not incredibly complicated either.
The simplicity of this algorithm is so amazing!
It looks really simple, but it unfortunately doesn't explain whether "(g^a mod n)^b" and "(g^a)^b mod n" are the same. I mean obviously it must be the same for it to work.
Fabian Neundorf (g^a mod n)^b mod n is the same as (g^a)^b mod n, I believe he also mentioned it briefly in the video.
The simple answer as to why they're the same is that Modulo arithmetic=Finite fields, and Finite fields are consistent. Proving that finite fields are consistent would probably be another couple videos...
Short answer: It's because the product in modulo/clock arithmetic is defined that way. If you wanna know more I'll explain it in more detail:
Maybe it's easier if you think of g^a mod n just as a number between 0 and n-1 (actually if we use this notation we should start from 1 and go up to n-1, but it's not extremely important so let's forget about it). Think of the multiplication between two such numbers as a standard multiplication, but if the result is bigger than n-1 we take the remainder.
So basically the product is just:
(a mod n)*(b mod n) = a*b mod n (THIS IS THE DEFINITION OF PRODUCT IN MODULO ARITHMETIC)
Elevating g to the power a is just repeated multiplication, so what you do is apply the definition of product:
(g mod n)^a = (g mod n)*(g mod n)*...*(g mod n) = g*g*g*...*g mod n
You can easly see that we get a similar thing if we elevate g^a to the power b:
(g^a mod n)*(g^a mod n)*...*(g^a mod n) = (g^a)*(g^a)*...*(g^a) mod n = (g^a)^b mod n
So yeah, (g^a mod n)^b = (g^a)^b mod n
Hope it's clear now ;)
The "mod n" isn't an operation in this case. Appending "mod n" to a term just signals that you're calculating in clock arithmetic, meaning you always stay between 0 and n-1. If you do 2+3 in "mod 4", the answer is 5, or 1, or even 9, because in "mod 4" 1 is equal to 5 by definition. Calculating a modulo is not an operation that needs to be performed if you define yourself to be working in clock arithmetic
Actually g^(ab) mod n = (g^a mod n)^ b mod n, but yes, it is the same. The proof is just long and kinda gross which is why he didn't get into it. Here it is if you want to see it though.
Proof:
Any number m mod n is defined as the remainder left after dividing m by n, so we have m=q*n+r where q is the whole number of times n goes into m, and m mod n=r, which is the remainder.
So g^a=q_a*n+r_a, and g^a mod n=r_a. (I'm using the underscore to denote a subscript.)
(g^a mod n)^b=r_a^b=q_1*n+r_1, and (g^a mod n)^b mod n=r_1.
g^(a*b)=q_(ab)*n+r_(ab), and g^(a*b) mod n =r_(ab).
But we also know g^(a*b)=(g^a)^b, and g^a=q_a*n+r_a, so g^(a*b)=(q_a*n+r_a)^b.
Expanding that out into a sum we get
(q_a*n+r_a)^b=sum((q_a*n)^(b-j)*r_a^j,j=0..b)
The last term in this sum is r_a^b and every other term has at least one factor of n, so we can rewrite it as
(q_a*n+r_a)^b=n*sum(q_a^(b-j)*n^(b-j-1)*r_a^j , j=0..b-1)+r_a^b.
Now let q_2=sum(q_a^(b-j)*n^(b-j-1)*r_a^j , j=0..b-1), so we have
(q_a*n+r_a)^b=q_2*n+r_a^b.
Remember that this is equal to g^(a*b). So we have
g^(a*b)=q_2*n+r_a^b.
We already have r_a^b=q_1*n+r_1, so we have
g^(a*b)=(q_1+q_2)*n+r_1
and g^(a*b) mod n = r_1.
We already had that (g^a mod n)^b mod n = r_1 so we have now proved that
g^(a*b) mod n = (g^a mod n)^b mod n.
I actually understood the math for once!
yeah, modular arithmetics is easy. :D ...
And theeeeres the party pooper.
Didn't work for me a=11 b=13 g=7 n=199
The math is easier to understand, because it's more technical.
Jyoti Das lol.
You are why I am confident I won't fail my Safety and Security module. Mike is exceptional at communicating these concepts!
I've finished this video after the colour example version of the diffie hellman and once again Dr.Mike on top This concept is really understandable.
This made so much sense and was super easy to grasp. Mike pound really is a great teacher. Also props to Diffie Hellman for creating such an easy but quite unbreakable algorithm.
I finally understand how this public-private key system works 🎉
Thanks!
It would be awesome if you guys did some podcasts. I learn so much from you guys and it would be great to listen while in the car or at school. Thanks guys.
Agree, but then we would miss the paper!
Hi Mike, Alice or Bob never shared "g to the power of a or b" but they shared "g to the power of a (b) mod n". But as we continued the example we said they shared the power of a or b. Can you please help explain that?
For anyone who got confused with this, we start with the left-hand side of the equation:
(g^a mod p)^b mod p
= (g^ab mod p) mod p [by the property of modular exponentiation]
Now, we move on to the right-hand side of the equation:
(g^b mod p)^a mod p
= (g^ba mod p) mod p [by the property of modular exponentiation]
Since (g^ab mod p) mod p = (g^ba mod p) mod p, we can say that:
(g^a mod p)^b mod p = (g^b mod p)^a mod p
Hope this helps.
Lmao everyone calling him out for not stopping at n-1. Give the dude a break!
☞ nope.avi
they need to feel smart
If you're not familiar with what he's explaining, this is enough to confuse you and this is not the only thing he says that can create confusion.
he kinda of implicitly stopped at n-1 when he mentioned there is the element 0 in mod operations
😂
great video, just to clarify modulo math, ((g^a)^b) mod n = ( ( (g^a) mod n )^b )mod n,
by using this we can generate same value from both side
This is the most important fact that makes this explanation work.
3:32 How can Alice perform (g^b)^a when she receives ((g^b) mod n) from Bob? Alice does not know the value of (g^b).
it's not mentioned, but (((g^b) mod n)^a) mod n is the same as ((g^b)^a) mod n. this applies zo adfition and multiplication as well
It's interesting that he doesn't mention that n should be prime in addition of being large. That is an aspect equally as important as its size which could easily break the encryption (if n isn't prime). There was even a defcon panel (2016 I think) that discussed this very problem and how it is often ignored.
In the previous video regarding this topic, with the color mixing, he mentions that n is a prime number.
I really was hoping he would explain why n has to be prime. I'm thinking the reason is that the search space of [numbers] mod n would be much smaller than n if it weren't a prime number, but that's just my intuition and I'm not sure if I'm right.
@@tissuepaper9962 The goal is to have the denominator and numerator being coprime i.e. their greatest common divisor is 1 as otherwise, you'd end up getting 0 as the result. The only way to guarantee that is with a prime number which is naturally coprime to any number.
The clock-face analogy is really powerful as it really helps explain why it is not very reversible. "Undoing" g^a%n is effectively going backwards in steps of n and checking to see if g factors into this number an integer number of times. Since a is generally (very) large, we'd have to go backwards an awful lot of steps. There can be more efficient ways of doing it, but conceptually it's the same.
Note 2048 bits --> N < 2^2048 ~ 10^616. The world's best supercomputer has a power of ~1exaflop or 10^18 floating point calculations per second. Assuming every attempt could be achieved in 1 operation, that would take 10^598 sec = *10^591 years*
I always find it difficult to understand how "difficult" something is when we say p is a really big number, until it is put into a time-to-calculate number.
At 4:00 that exponentiation is flawed, because you need to calculate (g^a mod n)^b, not (g^a)^b. Those are not the same, and the explanation he gave is not correct. He should have said that because of the modulus exponentiation property (g^a mod n)^b mod n is the same as (g^b mod n)^a mod n. He omitted a really important step when he jumped to (g^a)^b is equal to (g^b)^a. That is something that you arrive at when you apply the modulus exponentiaion property.
Fantastic explanation of the mathematics behind this encryption!
This is the best explanation I've ever found about it. Incredible concise and very illustrative. Thank you guys.
Thanks for explaining, you do such a great job. Always my favourite computerphile presenter.
4:05 - To be more accurate, it's the same as long as g, a, and b are all positive. Throw in negative numbers and it doesn't always hold.
thank you for putting this video up. Such a good teacher and elegant explanation of public private key at its core. There are 100s of videos super complicated.. this is very nice, one can implement themselves right off the this video..
It shouldn't be 'n' on the clock but 'n-1' because 'n'==0 in modolo
I'm not the only one who is botherd by this ^^
This is very true. Easy mistake to make, though.
same thought here - was about to write a comment.
There are 3 hard things in programming:
1 off by one errors and
2 naming things
I was about to write the same xd
At 1:51 you don't need the n as you have 0. n (mod n) = 0, so the clock would only go from 0, to n-1
The clockface only goes up to n-1, (n mod n = 0). He forgot to change that too when he added the zero afterwards
Thomas Wigele Yeah I came down to point it out as well :)
not only that, but the primality of g is completely irrelevant - the n needs to be prime if you want the n in the 2 to 4k range (and DH to remain secure), best if n is actually a safe prime (then g can be any number but 0, 1 and n-1) - a prime for which (n-1)/2 is also prime
666Tomato666 I don't get why n would have to be prime at all?
+Π. Καράπας This way you are 100% sure that the remainder is in [0, n-1].
Александър Савев but doesn't modulo always return something in the range of [0,n-1]?
I thought "solving the discrete log problem" was what happens when you clog the toilet at your in-laws' house?
...Sorry, I'll show myself out.
...yeah you should really log off
You didn’t get the video.... tbh he actually did explain it well
@@ivanarabome4172 and you didn't get the joke
@Jesse Duncan yup, been watching on instaflixxer for since november myself =)
Nice
The whole time I am thinking about the clock-drawing test, after seeing those clockface numbers beeing squished together.
PS. I find this video even simpler to understand than the original "simple" version with the food colouring.
Same. With the color one I was like "what do you mean I can't figure out what the other color was? We just subtract the public color from it!"
The actual process is more difficult as instead of adding, it's multiplication to the power of another colour. Imagine Yellow^red x blue.
Me too. It's so easy to understand math when it doesn't have weird symbols that I have no idea what the heck they mean.
Yes, the thing with math is that it is a language which you need to learn. Once you do, the concepts behind the maths become not that complicated, when you know them. Implementation could still be complicated though. Devils in the details...
3:44 Bob is sending (g^b mod n) and not g^b without modulo n. Else it would be simple to extract b. But the result of "(g^b)^a mod n" is the same as "(g^b mod n)^a mod n".
Was that him saying "I change notation to save space"?
Great video. One thing I’m not sure on is, how is g^b raised to the power of ^a, because it seems that the value shared is just “g^b mon n” as a discrete value ?? So how is the g^b disaggregated? 3:25
It actually doesn't matter whether you use g^b or g^b mod n, you get the same result. It's a property of the modulo operator.
Finally a decent video on computerphile... all of them should be AT LEAST this caliber instead of resorting to oversimplified (and hence incorrect) analogies.
Please do a video on Elliptic-curve cryptography and Elliptic-curve Diffie-Hellman , pretty please :D
Great explanations folks, keep pumping these videos out!
This is the video I've been waiting for!
Thanks Guys, awesome job!
After watching all three videos, its much easier to understand and glad you used 'n' here that you didn't use in the last video with colors. This was an amazing series but if you label them as part of a series or make 1 full video combining all three or the first and last video, more viewers might decide to stick to the end. Anyway, Thanks a lot for this.
Yeah but you never talked about if (g^{a} mod n)^{b} is the same as (g^{b} mod n)^{a}. As in, does g^{a^{b}} = g^{b^{a}} hold in the modulo field.
Yeah, I wanted to hear that too, but considering the technique wouldn't work otherwise, it probably is. It also kind of makes intuitive sense in the terms of significant bits. Modulo tosses away all most-significant bits, and only keeps the least significant bits. Multiplication will push the least significant bits into the most significant bits (two 32 bit numbers multiplied for example, gives a 64 bit number). So tossing away those most significant bits (which would've been moved to even higher significant bits) shouldn't matter.
The modulo just takes the remainder of the division: being the multiplication commutative, g^(ab) = g^(ba), you'll get the same number. You can then do modulo n, but the number will be the same anyways.
Let's say g^{a} = X + Y, where X is divisible by n and Y < n (therefore g^{a} mod n = Y). Then
g^{ab} is (X^b + c1*X{b-1}Y + c2*X{b-2}Y^{2} + ... + Y^b). Since X is a multiple of n, all but the last term will be zero mod n, so we are left with Y^b mod n = ( g^{a} mod n)^b mod n after we take the modulo. Therefore g^{ab} mod n = g^{a} mod n)^b mod n. I hope that's clear enough.
This is why he put this in the Maths video and not the colour video. As you should know this basic thing to know the theory. And he mentioned it in the video, very short.
I was wondering why Diffie-Hellman hasn't been deprecated in favor of a better algorithm since it's almost 50 years old now. Now I know why. You could always increase the bit count of N to compensate for faster computers and still get a shared secret in as low as 2 messages. Neat!
The missing part is why (g^a mod n) is fast to compute even though the reverse is not.
Tl;dr; g^a can be computed by starting with x_0=1 and and defining x_1=a_0^2 or x_1=x_0^2 * g and then doing the same for x_2 and so on. Making the right choices at each step (which is basically reading out the bits of a) you get g^a in at most log2(a)+1 steps.
thanks your video is very helpful, but i have a question let's say in signal or realtime chat webapp WhatsApp using Diffie Hellman, on server side only public key is saved. So I have a question where is the private key stored for each user when they log in so that they can be retrieved and every time you log in on another device, how do you get that user's private key? thanks and hope you can help me answer my question
Yeah! Best explanation on key exchange ever. Thanks to both videos I understood it instantaneously. A test on paper and mental arithmetic worked like a charm.
Thanks :-)
At 4:47 it’s been said that we know g^a and g^b, but shouldn’t it say we know g^a mod n and g^b mod n?
If g is public and g^a would be public too, we could calculate the value of a using the logarithm of g for g^a, couldn’t we?
3:47 "Alice is going to take the g^b that Bob sent"
Doesn't Bob actually send b^g mod n ? If yes, how does Alice reverse that into b^g? And if not (Bob actually sends b^g), then it's very easy to compute b, Bob's private key.
no, _a_ and _b_ are *the exponents* while _g_ is *the base*
bob actually sends g^b mod n
Thank you so very much for these videos. I find them invaluable in understanding the ins and outs of cryptography.
Is it true about what he says around 4:42 that we know what g^a and g^b is? I can't find evidence of this elsewhere, and it would seem to be to be an incorrect statement, as if you knew, g^b, and g, then it would be trivial to compute the log of g^b, and find out what b was. I think he must mean: g^a Mod n and g^b Mod n.
That really is a beautiful solution to exchanging keys. I really love that. Fantastic video as always!
g must be the generator of cyclic group of n (the set 0..n-1) by repeatedly applying the group operation (exponent, modulo)
@03:45 -> Does Bob send g^b or [(g^b) mod n]? Both are completely different numbers, so how does Alice extract g^b from [(g^b) mod n] if it is indeed the latter that got sent by Bob, since Alice has no knowledge of b?
Bob sends (g^b) mod n. however, Alice does not need g^b. when working with the mod operator, it does not matter if you apply it to the intermediary results as well or only the final result. he probably should have mentioned it.
@@danielf.7151 thanks yeah looked it up and it's the reduction property modular exponentiation : (a^b) mod n = ((a mod n)^b) mod n
which can be fairly easily deduced (when a is represented as k*n + r) with binomial expansion of ((k*n + r)^b) mod n which leaves behind r^b in both cases.
Does Alice and Bob gonna share the exponent values with the modulo operation or without it at the first exchange ? Cuz he first said said that both of them gonna use the modulo operation but he completed the analogy with only the exponents values (g^a and g^b)
0:42 Why is the g a prime number? And also why it is small? If g was big and could be a non prime, wouldn't it improve the security?
Making g really big does improve security, but also drastically increases the computing and memory resource requirements since g^a will be (even more) ridiculously large.
Just to caveat, in modulo we deal with numbers from 0 to n-1. We exclude n since the remainder can't be equal n - it must be less than n, otherwise when we're dividing by n and we see that there's a remainder n it means that n could fit one more time in that number leaving us with the remainder 0.
It's beautiful how simple and elegant it is.
I like the Explanation it help me understand lot better but I have two Questions
1. You mention G^a would be huge number. Did you mention the full equation G^a Mod n would give you a huge number or you just meant G^a Only.
2. you mention N should be a huge number, 2000 to 4000, so that it's not easy to brute force attack. There are computers these days that can calculate number between 4000. in these case is algorithm is secrete or N is secrete?
1. Only g^a is a huge number. g^a Mod n is always a number between 0 and n-1 inclusive.
2. N is not secret, only a and b are. also, 2000 to 4000 is not the bumber itself, but the length of the number in bits. a 2000 bit number has around 600 digits in decimal.
There is no better explanation of the "Diffie-Hellman key exchange".
X mod n will be in the range of 0 to n-1, not 0 to n.
I'm also a little unclear as to how we get g^(a*b) mod n. Bob, for instance would have g, n, b, and g^a mod n. I think it might be the same as (g^a mod n)^b mod n, which Bob could compute, but I'm not sure if that's correct, or if so, why.
According to Congruence modulo, (a mod n) * (b mod n) = a*b mod n, so (g^a mod n)^b mod n = (g^a)^b mod n
But isn't the public value "g^a mod n" instead of "g^a" or is that irrelevant, because "(g^a mod n)^b mod n" is the same as "(g^b mod n)^a mod n"?
was wondering the same thing, can someone answer
You can put "mod n" anywhere or nowhere. It doesn't make any difference as long as the end result is mod n.
That "mod n" is important for keeping things secret. Remember that it is the *discrete* logarithm problem that is hard. If I give you `g` and `g^a` you can find `a` you could find `a` by log_g (g^a). That is what a logarithm is: the inverse of exponentiation. The 'mod n' piece is what make the problem hard because you don't know how many times you have looped back to 0 again. For any given g and n there actually might be a lot of values for a that would work, *but* if g and n are relatively prime (they have no common factors) then there is exactly one value a less than n that will result in g^ mod n. That is why the subtle mention that g and n are large *prime* numbers. The math still works if they are not, but it makes the problem of finding a much easier.
Tested with a couple values, (g^a mod n)^b mod n seems to equal (g^b mod n)^a mod n .
Yes, it is.
And yes, "(g^a mod n)^b mod n" is the same as "(g^b mod n)^a mod n"
4:45 when he writes g^a and g^b as being made public, he should have said and written g^a mod N, and g^b mod N, were made public. thanks for explanation.
2:12 0 to n-1 not n
Technical question. Does the value of a, b, g, n are permanent, or they get changed periodically.
@3:55 this doesnt make sense, g^a isnt public, g^a mod n is public. for instance, bob would be doing this: g^(g^a mod n)^b mod n?
Thank you for the amazing explaination! One thing confused me about the circle... it should go from 0 to n-1 and not n right?
This video is a paradox, in which your head embodies the problem and the solution at the same time.
I don't understand the part in 3:45. So do they send the answer to the equations of (g^a mod n)? Or do they send over each part separately? Because if they sent it separately then that'd be useless because a and b would be public. And if they send (g^a mod n) then how does bob change that in (g^a^b mod n)? How is that possible? Because the answer is the mod, and that would be
((g^a mod n)^b) which is a different answer. And if they sent (g^a) then if bob multiplied (g^a)^b it'd give you the right answer but the answer is different from literally getting the answer of (g^a) then powering it to ^b. So how do they send the data over without making it public?
bob sends g^b mod n
alice sends g^a mod n
a and b are not revealed
(g^a mod n)^b mod n = (g^b mod n)^a mod n = the shared secret
Alice doesnt need to know b and Bob doesnt need to know a, those are private.
Another thing that I noticed (but doesn't matter), is that actually, Bob can't compute (g^a)^b mod n, because he has to calculate (g^a mod n)^b mod n. The beauty of modulo arithmetic is that the result is the same.
Watched both videos. Beautifully explained.
This is a really nice explanation and thank you for talking about it. I wish you talked a bit about man in the middle weakness and the ElGamal improvement.
Amazing video, but how do we know that (x^a mod n)^b is the same as (x^a)^b mod n? 🙂
It is quite misleading that after they exchange (g^a)mod n and (g^b)mod n Mike goes on as if alice knew the value of g^b and bob g^a and even saying that those values would now be public, which is simply wrong.
I'm glad I'm not the only one I thought this, I literally couldn't go on watching the video, this bothered me so much. At 3:47 He says Bob sends Alice g^b, well now everyone knows b. If he meant to say Bob sends Alice g^b mod n, then he doesn't explains how Alice is able to extract g^b from that (why can't people listening in do this then??) in order to raise it to a.
Very frustrating.
Is it weird that I found this easier to understand than the one with the colors?
When we start to go inside the cryptography's math we always face the formula : a^b mod n.
But if a and b are huge number, how in the world can we compute a^b ?
To work out a^b mod n ?
Thats where floating point and mantissa stuff comes in iirc
@@maybelbdidit You don't recall correctly. a^b for large a and b is arbitrary precision integer arithmetic, not floating point.
I don't really get it why it's difficult to crack the a or b. Assume calculating big 'A' takes 1ms, a=1000,000, a brute force to crack little 'a' will take 1000,000 rounds, which takes 1000,000ms=1000s.
Is it the bigger the number, the slower the calculate process? Then it will be gradually slower and slower from 1 to 1000,000?
I love how this all works on the fact that (((a^b)%x)^c)%x=(((a^c)%x)^b)%x
How is (g^a)^b mod n calculated efficiently? If 1
Yes, you can do that pretty efficiently. It's called modular exponentiation.
Basically, ga^b is calculated (ga is (g^a)mod n) the way it normally would, but after every little step of that calculation, the mod n of the current result so far is calculated.
Everywhere are for example purpose this little numbers. But what number range do
a
b
g
n
actual need to be in a real life use case? What would be a small prime number 'g' be?
Are 2000-4000 some legit values in a real implementation or are there some more standardized values?
g^a%n seems to form repeating patterns for the values of g,a,n I select.. is this not always the case and special values are unpredictable?
Can you (Dr. Mike Pound) please do a video on Linux security, and its flaws for being open-source?
Can I get some clarification on the math? I understand how g^ab % n is the same as g^ba % n but after the private numbers are used (g^a%n and g^b%n) and shared, when the other person's private number is used, using Alice for this example, instead of g^ba % n wouldn't it be (g^b%n)^a? This order of operations is confusing me
why do it have to be prime numbers?
Because if you can easily factor g^a or g^b, then you can find a or b in polynomial time (easy to do on technology we have today).
It also makes the number modulo n more uniformly distributed.
Because otherwise it wouldn't be a field.
@@evankivolowitz4788 This is not true. What you are thinking of is prime factorization (getting the product in primes of a number) but that does not come into play here. Also you never send g^a or g^b but the remainder of that number after mod n. The reason for prime numbers is, that you distribute possible secrets on the whole range 0, ..., n-1.
at 3:47 , why is (g^b mod n)^a mod n = g^ab mod n ? Is that so obvious?
(((((a^b mod x)^c mod x)^d mod x)^e mod x)^f mod x)^g mod x == a^bcdefg mod x
The first is just easier to compute since the numbers are smaller
Q:
If you had a 100 petaflop supercomputer and you discovered some workarounds in how diffie hellman key exchange can be sped up, say 5x, how long would it take on average to discover the private key given the size of 4096 bit RSA? You can make some assumptions about how many operations it would take to compute one key pair.
Is it important for the encryption key how many times we go around the clock to get to the answer? Otherwise surely you would only have to brute force numbers zero to n?¿
Hi, I tried to implement this in Javascript, then in Java but it doesn't seem to be working with large numbers generated by Alice and Bob.
How can Alice do [g^(ab) mod n] if she gets [n^b mod n] from Bob? Shouldn't she do [(g^b mod n)^a mod n]?
If alice can take g^b then it is public right and that would mean you could find out b right and then just put (g^a)^b) mod n and you got the key??? But also you can't get ((g^a)^b) if the only number you get from alice is (g^a) mod n...
I really loved this vid. Great work guys!
don't we usually choose n to be prime as well in order to make Z_n a field? Otherwise the largest field we can construct using modulo n is Z_(phi(n)), which is not necessarily large even if n is large.
In previous video Mike said the math is very complex so separated out to another video. Now by the end of this video Mike says the math is not very complex it's simple. 🙃
Actually it was hard for me to understand how (g^a mod n)^b became (g^ab mod n) . It felt so counter intuitive to me that I had to test the math with several sample values before I could convince my brain it is true.