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3 So far, pretty straightforward. Now, how can we get digital signatures? So in the early days, researchers-- and it's actually great computer scientists-- they proposed a digital signature can be implemented as the inverse of public key encryption. What does that mean? So I'll use RSA as the example. So RSA encryption is m to the e mod n. The encryption is c to the d mod n. So the first attempt is we will just use this as our sign function and use this as our verify function. So now the symbol is a little bit confusing. So now I'm signing a message [INAUDIBLE] c. Let me actually change it. This is RSA encryption. I'm going to transform it into signature scheme where sign signs a message, and verify, raise the signature sigma to the power of e, and checks whether or not I get back my message. So this actually makes a lot of sense. Why? Because think of m as a ciphertext. Then if I decrypt it, and then re-encrypt it, I should get back my ciphertext. So correctness-- we have correctness. And why is it unforgeable? Because an attacker does not have the secret key, so he should not be able to decrypt this m here. He cannot run this algorithm. That's the reasoning behind it. So far so good? But, unfortunately, it is broken. And so I'll give you, say, seven minutes to think about it. Can you come up with an attack, a forgery? You can see a bunch of messages and then output a forgery for a message you haven't seen before. So is the algorithm clear? [INAUDIBLE]. Can you speak louder? Is it just the product of any messages? Exactly. So if an adversary had seen a bunch of messages-- because RSA has this sometimes good, sometimes nice, sometimes bad property, that is multiplicative, homomorphic, or to use a less fancy word-- malleable. So if they, an adversary, sees this message, it can set m star to be m1 times m2 and sigma star to sigma 1 times sigma 2. You can check. This is a valid signature, message signature pair. You take this entire thing raised to d. They are raised to d individually and then multiply

4 You take this entire thing raised to d. They are raised to d individually and then multiply together, and that's exactly this message here. Attack one. OK. There's actually another attack. That's even simpler and tells you this scheme is even more broken. So all I want to do is to come up with a sigma when it's raised to e, that's equal to m. So I'm going to select the sigma, compute, m sigma raised to e, because e is my public key, mod m. I can do that. And then output sigma m-- oh sorry, m sigma. I select the signature first, and I raise it to the power of e. I get a very strange message, but it doesn't matter. That's my forgery. OK, so now you can see this scheme is basically totally broken. But they actually come from our, well, several renown scientists. But why is that a case? Because actually that definition didn't exist when they were trying to when they were working on this problem. And so that definition looks obvious today, but it's actually not obvious at all. And I think this algorithm came in the 70s, '78. And in '82, Goldwasser and Micali, two professors from MIT, proposed the definition for signature encryption and basically everything in cryptography. And they won another Turing Award for that. OK, so let's try to fix it. Any ideas? We do not want to change the framework. Let's still use RSA and combine it with some other primitive you have seen to try to fix it. What do we want to do? We want to break this multiplicative property. And we want to break this, this step, whatever it's called. Go ahead. Can we change the n, maybe? Change this n? Yeah. Right now, just to remind you, it's a product of two primes, OK, pq. It's a product of two primes. It's how RSA works. What's your idea? Go ahead. Before it raised to the power, we can get the hash function of it. Exactly. Let's just make a small change. So sign will be hash of m, raised to d. And verify will be-- just check whether hash of m equals signature is to e. This indeed fixes these attacks. Why? Because now you need-- well, if you do this-- hash of m and 1 times hash of m2 is not

5 going to be hash of m star because hash is supposed to be [INAUDIBLE] random. That's not going to work. And here, what the attacker needs to do is to find hash of m, such that it's sigma raised to e. It can still do this, but it does not know what this message is because of the one-wayness of hash function. If we use a good hash function there, then it indeed fixes both the attacks. But we have seen the lecture that this hash function also needs to be collision resistant. Remember that? A question? Isn't the message public? Yeah, the message-- oh, OK. Good point. Oh, no, but, OK, you are talking about this attack, right? So the attacker needs to find the public message, but all he can do is select the sigma, and raise it to e. That's going to be its hash of m. And then he cannot figure out where this m is. But what about the other way? I mean, if he has two messages, he can still get m star, and then get hash of m star. OK, so he then he gets, has of m1. He gets hash of m2. But you need to find the m star, such that its hash is the multiplication of these two. And, yeah, he does not know how to find that message. So if the hash is not multiplicative, one-way, and collision resistant, then it seems that we have fixed all the attacks we know. However, how do we know there are no other attacks? So actually, indeed, this is a good idea. We have several national standards that just use this but slightly differently. I can-- this is just for your information. So there's a standard called [? NC,?] whatever-- X93.1. It uses RSA, this word padding, so it takes the hash of the message and pad with this hex stream, and prepended and append another hex stream. Why do they do that? They don't know either, but they just think it's probably more secure than only using a hash. There's another standard that has a different steam and a difference stream here, and it doesn't matter. So that's indeed a weakness of these types of approaches. So their security is what we call ad hoc. We do not know how to break them. But we do not know how to prove they are secure either. Yet, that's what people do in practice. So, unfortunately, that's all I can tell you today, so how not to construct the digital signature. I

6 cannot tell you how to construct the secure digital signature because that's out of the scope of this class. And it's a major topic in cryptography. Any questions so far? Go ahead. The hash function here is the one way, yeah? Yes, it's one-way, collision resistance, and-- So what is the use of using the RSA? Couldn't we just use the only hash function then? OK, good question. So, OK, let's be clear what you're saying. OK, never mind. Can you answer your own question? So my question was, well, why do we have to use the RSA? Why, when we have the hatch function? You want me-- so [INAUDIBLE] couldn't create the forgery. [INAUDIBLE]. How does it create a forgery? Just answer your own question. Let everyone else know. Maybe they have the same question. So answer your own question. So my answer is so adversary can't just choose random message and hash it and [INAUDIBLE]. Yeah. What's the problem? Problem is that a hash function is a public function that everybody can compute. So the attacker just chooses a message, compute as hash, so using a hash is not a signature. But good point. I'm actually coming to that. So far we have seen three major primitives-- private key encryption, public key encryption, and digital signature. So if we categorize them a little bit-- so these two are asymmetric key. They are public key and secret key. This one is symmetric key. And these two are for secrecy. They are trying to hide the message. And this one is for integrity. Meaning, the message is what the sender sends. So you can see we are missing one primitive here. What if the two parties, they do share a secret key, and one party wants to verify the other party? The other message indeed does come from the other party. So indeed we do have a primitive for that. It's called message authentication code. So its definition is basically exactly the same as digital signature. I'm just going to change it here. Except that it has only one key. So the sign function is replaced by a MAC. And there's

7 no notion of secret key and public key. We have only one key. And how do we verify? OK, so verify function basically just becomes the other guy also recomputes the MAC of the message and checks whether that's the signature. So verifier just to recompute and compare correctness-- we also want correctness. We also want unforgeability And it's defined in exactly the same way. Now, actually, I would have asked this question here-- is hash a valid MAC? The answer is still no because MAC is a public function that everyone can compute, and it's trivial come up with a forgery. So thank you for asking that question. But the hash is actually very close. How can we get a message authentication code? So several ideas. Can we just hash the key concatenated with the message? Then some other random attacker who doesn't have the key does not know how to compute this thing. That's a reasonable idea. But, well, if we can do it this way, how about we do the message concatenated with the key. Or if you want, you can do key concatenated with message and then concatenated with the key. So it turns out this doesn't work for some very advanced reasons. And this one may or may not. For SHA1, it doesn't work, unfortunately. And for SHA3-- that's the replacement for SHA1 and SHA2-- it actually works. So the simplest MAC we can imagine is just to choose SHA3 as the hash function, and input is the key and then the message. Not the other way. It's also, just FYI, purpose. By the way, there's another reasonable thought. That is, how about we encrypt the hash? Now, everyone can compute the hash, but they don't know how to encrypt. If I use, say, a secret key encryption, this turns out to be wrong as well. That's digital signature in MAC. But one caveat here, our unforgeability is defined this way. A little bit strange, but it makes sense. But it indeed has some weakness in some applications. So imagine, say, I send you a message-- today's recitation is canceled. And it has my signature on it. So you can verify it indeed comes from me. But once I send that message, every of you has that message. So next week, one of you can send that message again, saying, today's recitation is canceled. Then you have no idea whether it's indeed me sending the message again or someone doing

10 Yeah. Exactly. We're going to create a sigma 5, which is the hash of sigma 1, concatenated with sigma 2. So we do the same thing here. And so you all know what it is, right? I don't need to write it. And keep going until we got a root hash. And now we're going to store this thing locally on the side. So what's the local storage complexity? o of 1-- we're only storing one hash locally. Now, what's the time complexity? OK, so how do I verify? Yeah. Log in-- how do I verify? I need to, so, first verify if this hash matches, and then read this hash, and verify whether this link matches, and verify whether this one matches, and then I'm done. If I want to update, I also need to update this hash, then it causes this hash to change and then that hash to change. But it's always some path in that tree. It doesn't affect anything globally. Question. But you're not storing like sigma 5? Say again. You're not storing sigma 5. I am not. I have to go ahead and read it. From where? From Google Drive or Dropbox. So are we sure that that is secure? OK, yeah, so that's the next thing we're going to do. Is this secure? Or in other words, can the adversary change one of the files, and somehow maintain the same root hash? That's your question then. Of course, we assume the hash is collision resistant. Or I should say if the hash is collision resistant, then this hash tree is collision resistant. Any intuition? Or anyone wants to prove it? Go ahead. So like if the root eventually one of the leaves will be different because it changes. Yep.

11 Now, you want the root to be the same than the other hash has to be different, but there's no collisions. I mean, it's hard to find the other hash. Correct, so I'll just repeat what you said, but I'll start with the leaf because that's easier for me to think about. So say I change this one, this block. Now, I claim this hash here will change. If it doesn't, then I have found the collision. Because this x4 prime has the same hash as the original x4. So if this sigma 4 changes, then sigma 6 will change. Otherwise, I have found the collision. Because this sigma 3 concatenate with the new sigma 4 is my collision. So same argument-- either this one changes, or I have found the collision. I repeat the argument all the way to the root. Any question about that? What if like the adversary changes like two hash options--for example, x1 and x2-- but sigma 1 and sigma 2 changes, but sigma 5 stays the same? OK, so then we have found the collision that is sigma 1 concatenated with sigma 2. That's a collision with the new sigma 1 concatenated with the new sigma 2. Make sense? If the concatenation stayed the same, like sigma 1 and sigma 2 concatenation. So if the concatenation stayed the same, that means both of them are the same. They had to make sure they are changed? So I'm not sure I understand your question. So concatenation is basically just a bunch of bits then followed by another bunch of bits. If this entire thing is the same, that means this part is the same and this part is the same. And if your sigma 1, new sigma 1 is the same as your old sigma 1, that means I have found a collision here. Because we changed it, but your sigma doesn't change. So lastly, I'm going to do a quick review of the knapsack problem because I think in the lecture, we may run out of time and I didn't mention everything. So if you recall, the knapsack cryptosystem, it says you have a knapsack problem. I'll call u1 to un. And then we're going to transform it. OK, this is a super increasing sequence. I'm going to transform into a general one by multiplying n and then mod m. So this is an easy problem, and that is a hard problem. So how do I encrypt? I'm going to take a subset sum, which is mi, Wi, where mi is the i-th bit in the

12 message. So how do I decrypt? I'll take this, transform this s back to the super increasing domain by multiplying inverse of n. So that's going to be inverse of n multiplied by this mi Wi. That's how I encrypt it. And then each Wi is n times ui. So far so good. So that gives me mi times ui sigma. Of course, every step is modulo m. So the first thing I'm going to claim is that m has to be larger than sigma ui. If that's the case, then the t-- my t is just this subset sum. So if I solve this knapsack problem, I get the same answer as solving the original, the general knapsack problem. If my m is not that large, if the m is too small, then I have a problem. Because then my t will be the subset sum minus sum multiple of m. Then it's a different problem. I do not get the same message back. OK, then we have a problem. So because we defined density to be n over the log of max ui. Does everyone remember this part? So each ui is in the range of 1 to m, or maybe 0 to m. If I have a bunch of them, then this is not super rigorous. If I have a bunch of them, chances are that some of them are very close to m. Because it's unlikely that all of them are small. So this thing is roughly n over log of m. So then we have a dilemma. If we set m to be a small number, then my density is fine, but that means all of my ui's needs to be small because m needs to be greater than the sum of them. If all the ui's are small, then I have a very limited choices of them, then actually an attacker can just guess what ui I chose by a brute force algorithm or something like that. And if I choose m to be large, or if I choose all the ui's to be large, to choose them from large range, then my m is going to be very large. And this density is low. And that's vulnerable to the low density attacks. And so how low a density is considered low? So several people proposed that based on heuristics, that if this density is less than 0.45, then it's considered low density, and it can be attacked. And this threshold had been improved. So but while most of the knapsack cryptosystems are broken, there are few that have so far stood the test of time. So they are still interesting because knapsack problems, knapsack cryptosystems will be much faster than RSA or any number theory based, because we are just adding numbers here. An RSA have this operation where m is a 1,000 bit number, and e is also 1,000 bit number.

13 And take this exponentiation is actually very slow. So knapsack cryptosystem are still interesting. However, the original motivation turned out to be unsuccessful. The original motivation is to base cryptography on the NP complete problem. That's not going to work because NP problems are hard, only in the worst case. And we need cryptography to be hard in the average case. Because if they are only hard in the worst case, that means there are several instances of this problem that are hard. So either you pick a secret key that doesn't correspond to a hard problem, or you pick a secret key that's corresponds to a hard problem. But everyone else picks the same secret key because everyone wants to be secure. That's the reason why it's unlikely to get cryptography from NP hard problems. That's all for today's recitation. And thanks everyone for the entire semester. Thank you for participation.

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