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10

mikeazo's answer clearly covers what the question asks. However, I want to go further and answer this: How much information does a perfectly secure cipher leak about the key? Exactly all the excessive information of the key that does not help in finding anything about the message. In simple words: if the key length is $k$ bits and the message is $m$ bits ...


7

Sure, it can leak something about the key as long as that doesn't leak anything about the plaintext. Consider the following cipher, I'll call it 2-OTP. 2-OTP takes as input a message $M$ and two keys $K_1$ and $K_2$. Each key must be truly random, independent of one another, and each the same length as the message $M$. Define encryption as ...


6

This expands CodesInChaos's comment into an answer. Forward Secrecy (that is, maintaining confidentiality of messages enciphered before compromise of the long term key) can be achieved in a protocol using a public-key signature scheme with a long-term public key, and a public-key encryption scheme with a per-session key; but in the case of RSA signature and ...


6

Actually, the problem with OTP isn't the storage of the pad (although secure erasure of the parts of the pad you used is trickier than it looks), and it isn't the pad generation (although, again, that's trickier than it looks), but the secure transport. After all, it's not enough for you (Alice) to have the secure pad, you also have to give a copy to the ...


5

The one-time pad is perfectly secure. It will also leak the complete key to any attacker who knows the message (and will leak some information about the key to any attacker who knows something about the message). It's important to note that there's nothing in the definition of perfect security that says that the attacker can't already know something, or ...


4

No, DHE is secure and allows to share a common secret between two parties over an insecure channel. But you cannot know, if the one you share the secret with is the one you want (DHE is vulnerable to man in the middle attacks). So DHE-RSA uses DHE to share a common secret and signs the communication with RSA to make sure, that both persons communicate with ...


3

Since this is homework, let me just give you a hint: consider the two-character messages $m_1 = \text{"aa"}$ and $m_2 = \text{"ab"}$. Given a ciphertext $c$ encrypted with a monoalphabetic substitution cipher, can you tell which of $m_1$ or $m_2$ it corresponds to, even without knowing the key? Why (not)? What does this imply about perfect secrecy?


3

One possibility for what you might be missing: normally the same key (the same matrix) is re-used to encrypt many messages. So now try counting the total entropy in $M$ length-$N$ messages, and the entropy in a $N\times N$ matrix, and compare what happens when $M$ gets large.... Another possibility you might be missing is the consequences of the fact that ...


3

To use the proper terminology: in TLS, cipher suites which include "some Diffie-Hellman" are: Anonymous Diffie-Hellman: DH_anon Static Diffie-Hellman: DH-RSA, DH-DSS... Ephemeral Diffie-Hellman: DHE-RSA, DHE-DSS... There is no "plain DHE" cipher suite in TLS; it is called "DH_anon". As the name indicates, with DH_anon, the server is "anonymous": you ...


2

Well, for perfect secrecy, we require that for all message distributions over $\mathcal{M}$, all messages $m\in\mathcal{M}$, and all (possible) ciphertexts $c$ it holds that $$Pr[M=m\ |\ C = c] = Pr[M=m]$$ In particular, that means, if we can find a single counterexample, i.e. a distribution over $\mathcal{M}$, a message $m\in\mathcal{M}$, and a ...


1

What you are describing is One-Time-Pad encryption, and yes it does have perfect secrecy. Note that for any ciphertext $y$ there is exactly one key $k'$ for each possible plaintext $x'$ so that $E_k(x') = y$. So if you choose the key uniformly at random the ciphertext gives no information on the plaintext, because any plaintext is equally likely.


1

I'll further explain the comment of @CodesInChaos and then give a simple example: Explanation When the correctness requirement is weakened the encryption scheme can omit part of the message $m$ (of length $|m|$) to be encrypted and just "loose" it in a way that the cipher (the output of the Encrypt method) is totally independent of that part. Thus the ...


1

I think they are similar but not equivalent. The first condition says that for any ciphertext $c$ all messages $m$ have equal probability to encrypt to $c$. The second says that for any message $m$ all ciphertexts $c$ have equally probability of encrypting $m$. So I think you can demonstrate non-equivalence as follows: Imagine a (somewhat contrived) ...


1

You're missing that key generation would occur inside the box (if there was one for RSA key agreement), like for $k_{pr,A}$ and $k_{pub,A}$ on page 343, rather than outside the box, as happened for $k_{pub,CA}$ on page 347.



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