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1

The question Alex linked in comments explains why authentication works to prevent a man-in-the-middle attack on Diffie–Hellman. So, whenever you can do the key exchange in an authenticated channel, you can be sure there is no MitM attack. (Assuming DH problem remains unbroken, of course.) Now, your questions: Is one solution for both Alice and Bob ...


1

The problem about Man-in-the-Middle attack on Diffie-Hellman is that both sides are not confident about other side's public key (g^a and g^b). If they were sure that they have correct public key of their's friend Man-in-the-Middle attack wouldn't be possible, because MITM attack is based on the forgery of public keys by adversary! If for instance Bob and ...


2

Well, as it says in your link the problem is authentication. So somehow Alice and Bob must set up an authenticated channel. One way of implementing such a channel is by Alice and Bob holding each others public verification key for a signature scheme. A CA would probably not hold a secret key for Alice and Bob. However, using a CA to get an authentic copy ...


6

First, the fact that the data is "easy" to guess (in the sense that an attacker has a one-in-2^32 or a one-in-2^64 chance of guessing correctly) doesn't mean much if the attacker has no way of checking if his guess is correct. Or at least, it's not a problem with the cryptography. Second, even if he does have that ability, the problem of protecting your ...


1

No. An RSA signature is just a single number, encoded in a certain way. The number represents $x^d$, where $x$ is a padded hash of the document and $d$ the private exponent. If you know a public key $(m,e)$, you can calculate $x = (x^d)^e \mod m$, but without a document (or at least its hash) there is nothing to compare it with to verify anything. The only ...


7

The other answer is correct in general. However, if your messages are all exactly one block long (or all one block after padding), ECB is a secure MAC. A PRP looks like a PRF up to half its bit length, i.e. up to $2^{64}$ blocks for AES. A secure PRF is a secure MAC of the same size. Thus, AES ECB used on 128-bit messages is a secure MAC as long as you use ...


9

It is not secure, because an attacker can "mix and match" the output blocks from different authentication tags on different input messages, or repeat output blocks for repeated input blocks. For example, if the attacker knows the tag $F_k(m)$ for a one-block message $m$, then it can forge the correct tag $F_k(m) \mid F_k(m)$ for the two-block message $m ...


4

Cryptography studies properties of information. Information, at least in discrete form (a sequence of bits) can inherently be duplicated. Inasmuch as a document conveys meaning, it can be expressed as a sequence of bits. If someone obtains a copy of the document, they can duplicate the sequence of bits and make other copies. It is impossible to prevent the ...


0

With RSA (or any asymmetric cryptography, for that matter), the key question is "how do you know you can trust the other peer's public key?". Without authentication, you could be sending your message using the attacker's public key. Example 1: the first time you SSH to a host, you are prompted to confirm that you trust the public key. That's because if an ...


5

The attack is even more simple with RSA than with symmetric keys, because the asymmetric encryption key is assumed to be public. Let me tell you a story involving Alice, Bob and Mallory :). Alice wants to send a message to Bob using RSA. Alice encrypts the message using Bob's public key and sends it Mallory performs a Man-In-The-Middle attack, and ...



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