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22

The SSL and TLS protocols (on which HTTPS is based) are designed in a way that no attacker (neither a passive nor an active one) can read anything of the encrypted part (if the cryptographic assumptions hold - and if you don't use the NONE cipher, which does no encryption). Of course, the attacker can read the negotiation part. But this part will not ...


17

The security of Diffie-Hellman depends upon the group in which DH is used, but not upon which generator is used for this group. See note 3.53 (chapter 3, page 103) of the Handbook of Applied Cryptography. In more details: for DH, we use a subgroup of size $q$ of the integers modulo $p$ (a big prime) with the multiplication as group operation. $q$ should be ...


16

Both RSA and Diffie-Hellman work with modular exponentiation. But they work in a different way: In RSA, there are two exponentiations which invert each other, i.e. we have $e$ and $d$ such that $(x^e)^d \equiv x$ for all $x$. E.g. if $\square^e$ is the encryption, $\square^d$ is the corresponding decryption. To create this pair of $e$ and $d$ (or derive one ...


13

The really great thing about Diffie-Hellman is how light it is, network-wise: both parties send each other a single message; neither has to wait for the message from the peer before beginning to computing his own message. If you can tolerate something heavier, you can have a look at what @Paŭlo describes; with $n$ participants, it requires $n-1$ messaging ...


13

I assume you're talking about SSL/TLS or a similar protocol. In these protocols there are two reasons to use Diffie-Hellman: Your certificate only supports signing Either it is an RSA certificate restricted to signing, or it uses an algorithm that doesn't support encryption, such as DSA or ECDSA. Forward security - What happens if the server's private key ...


12

Say you encrypt a message with a key $k$. With symmetric encryption (ie. symmetric ciphers), $k$ must be secret. The sender and recipient must agree (somehow) on $k$. No-one else can be allowed to find out $k$. Anyone else who finds out $k$, can decrypt all the messages encrypted with $k$. For that reason, symmetric ciphers are often called "secret key" ...


12

The standard Diffie-Hellman key exchange algorithm (or family of algorithms) works in an cyclic group with generator $g$, and relies on $$ {y_A}^{x_B} = (g^{x_A})^{x_B} = (g^{x_B})^{x_A} = {y_B}^{x_A}, $$ where $y_A$ and $y_B$ are publicly transmitted, while $x_A$ and $x_B$ remain private. With three parties, we still have $$((g^{x_A})^{x_B})^{x_C} = ...


9

You can do key agreement with asymmetric encryption. Any asymmetric encryption algorithm (post-quantum or not) can be used for key agreement: just choose a random key and encrypt it. Password Authenticated Key Exchange looks harder, because it cannot be applied on just any key exchange or asymmetric encryption scheme. The IPAKE framework can be applied on ...


9

Well, the advantages of static-ephemeral ECDH (and, they apply to DH as well): You get one-way authentication for free. That is, if Bob has Alice's public ECDH key, and uses it to talk to someone, Bob knows that that someone is Alice, without doing any further checks. Now, Alice has no idea who she's talking to; on the other hand, for some scenarios, ...


8

Well, it depends on the which protocol is being used. For WEP and WPA, the keys used are derived directly from the pre-shared keys; that means that as long as you know the pre-shared keys, you can immediately decrypt packets as well. On the other hand, WPA2 is somewhat stronger; the two sides exchange nonces to derive the keys. Hence, unless you listen ...


7

The check $y_b^q = 1 \mod p$ is there to prevent two possible weaknesses: Suppose someone gave us (either because of a programmer error or deliberate attack) gave us a $y_b$ value of small order. If so, then someone listening in can guess the shared secret you derive. Suppose an attacker gave us a $y_b$ value with an order with a small factor $r$. Then, ...


7

CRAM-MD5 is a protocol to demonstrate knowledge of a password. In the context of email, it is sometime used by an email client to authenticate to a POP, IMAP, or/and SMTP server. Basically, the password is used as the key of HMAC-MD5 in a challenge-response protocol. Among positive things there are to say about CRAM-MD5: The password is not exchanged in ...


7

There is nothing related to passwords in AES. AES uses 128-bit keys, i.e. sequences of 128 bits. How you come up with such a key is out of scope of AES. In some contexts, you want to generate these 128 bits in a deterministic way from a password (and possibly some publicly known contextual data, like a "salt"); this is a job for password hashing. In other ...


7

What is usually meant by "group encryption" is not what you are after. Group encryption algorithms strive to achieve the following: a given message is encrypted, and may be decrypted only if sufficiently many group members collaborate. This is not what you seek; what you want is a system such that a given message can be encrypted once and every member of the ...


7

ElGamal appears to be used instead of Diffie-Hellman (or IES) in OpenPGP mostly because when that format was put together, there were some unresolved intellectual property issues surrounding both RSA and Diffie-Hellman, while ElGamal was unproblematic. This trend for ElGamal seems to stick around, mostly by force of habit, e.g. when switching to ...


6

Well, what SSL uses to negotiate the symmetric keys depends on the ciphersuite that both sides agree upon. By far, the most common method is that the client picks a random value (the premaster secret), and encrypts it with the server's RSA public key. However, it is not that unusual for the ciphersuite to specify that the client and the server agree upon a ...


6

Yes, it is. Because of the way public key crypto works, they wouldn't be able to decrypt it. First, realize that something encrypted with a public key can only be decrypted with the corresponding private key (or, depending on the algorithm, vice-versa). So lets say everyone (including the sniffer) has the server's public key. You encrypt something with it, ...


6

An attack would be trivial if the seed of the RNG was only 32 bits; just enumerate the seeds, and test which matches the intercepted messages. That's easy. However the default Java Random class uses a 48-bit state and seed (which would still be attackable, though $2^{16}$ times less easily), and there are safe subclasses, thus use of Random does not imply ...


6

Handing keys in general is known as key management. Symmetric keys should be kept secret. Secret key is often used as a synonym for symmetric key. The establishment of symmetric keys can be performed in several ways: (Authenticated) Key Agreement (KA) Sending of an (authenticated) encrypted key, also known as key wrapping Derivation from a base key using ...


6

ECDSA is a digial signature algorithm ECIES is an Intergrated Encryption scheme ECDH is a key secure key exchange algorithm. First you should understand what are the purpose of these algorithms. Digital signature algorithms are used to authenticate a digital content.A valid digital signature gives a recipient reason to believe that the message was created ...


6

So your protocol goes like this: Alice generates a key pair $(a_{priv}, a_{pub})$ and sends $a_{pub}$ to Bob. Bob generates a key pair $(b_{priv}, b_{pub})$ and sends $b_{pub}$ to Alice. Alice generates a message $m$ and sends $Enc(Sign(m, a_{priv}), b_{pub})$ (or $Sign(Enc(m, b_{pub}), a_{priv})$, I'm not sure which of both is usually used by PGP) to Bob. ...


5

The mathematical problems behind DH and RSA are similar but not known to be directly related. It is still an open question if an oracle breaking DH can be used to construct another oracle that breaks RSA (or vice versa). It is mostly believed that the two problems are not reducible to each other in poly time. However, the complexity of the fastest known DH ...


5

Diffie-Hellman and RSA are distinct and do not use the same "trick". In Diffie-Hellman, commutativity is used: $(g^a)^b = (g^b)^a$. Both Alice and Bob do two modular exponentiations each (Alice chooses $a$, computes $g^a$ and sends it to Bob, receives $g^b$ from Bob, and finally computes $(g^b)^a$). Security relies on the difficulty of discrete logarithm: ...


5

"Is to be encrypted" is not a ultimate goal. You do not encrypt data for the sake of it; you encrypt data as a way to ensure a given security property, e.g. transmitting some data between two machines, without compromising the data confidentiality with regards to attackers who may spy on the transmission line (or even alter data in transit). If: your ...


5

The standard answer in the research literature is to use information-theoretically secure message authentication codes, typically universal hashing (aka Carter-Wegman authenticators). Of course, you could use computationally-secure message authentication codes, like CMAC or HMAC, if you wanted, though that would partly defeat one of the reasons for using ...


5

Fair exchange protocols aren't new by any means, but there is a lack of layman-friendly material out there, unfortunately. I think the high prevalence of theoretical cryptography in fair exchange protocols may be partially responsible for that. At any rate, here is the basic idea behind a fair exchange protocol. Suppose you have two parties, Alice and Bob, ...


5

Most likely, this 'shared secret' was actually an IKE "preshared key"; it is used to authenticate the two sides (and, for IKEv1, is stirred into the keys). It actually isn't used as a key (and hence someone learning that key cannot use it to listen in, unless they perform an active Man-in-the-Middle attack). I suspect the password is the authentication ...


5

One observation is that if we modify the problem so that $M, A, B$ are random invertible matrices, then it is easy to prove the security of the system. In fact, we can prove that the system is informationally secure; that is, for any observed $C_1, C_2$ pair, for any possible value of $K$, there is a unique set of values of $A, B, M$ that yield that $K$ ...


5

First, I am assuming, per https://security.stackexchange.com/questions/29172/what-changed-between-tls-and-dtls, that the client handshake protocol in DTLS is not different from that in TLS over TCP. This seems a safe bet since the client/server encrypted handshake protocol in OpenVPN's UDP implementation is the same as in standard TLS over TCP. I am not ...


4

Almost all cryptographic algorithms which use groups actually work in subgroups generated by a conventional element; even if the group as a whole is non-abelian, the subgroup is cyclic, thus abelian. The Anshel-Anshel-Goldfeld protocol tries to use non-commutativity itself, and relies on "how much non-abelian" the group is. All asymmetric cryptographic ...



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