Tag Info

10

The LWE assumption I think we should start from the LWE assumption. Let $n$ and $q$ be integers and let $\chi$ be a distribution over $\mathbb{Z}_q$. We often take $\chi$ as a Gaussian with small variance. (We take an error $e$ from this distribution $\chi$ and assume that $|e| \ll q$.) The LWE assumption states that any efficient adversary cannot ...

9

The ideal encryption scheme $E$ would be one that, for every ciphertext $C=E(K, M)$, if the key remains secret for the adversary, the probability of identifying $M$ is negligible. Since that is not possible in practice, the second most reasonable approach is to define constraints strong enough to satisfy some definition of security. The $IND-$ notation ...

9

Repeatedly encrypting the same message to the same ciphertext is full of practical attacks. Encryption is supposed to leak no information about the content of the message other than its length, and there are very real ways to exploit the information leakage you mention. Some of them have to do with the fact that plaintext domains are not always very large. ...

8

Yes. Such proofs are possible for El Gamal. It involves a zero knowledge proof of equality of a discrete log, together with the homomorphic property of El Gamal encryption. Recall that given $E(a)$ and $E(b)$, anyone can form $E(a/b)$ using the homomorphic property of El Gamal. Suppose $E(a/b)=(r,s)=(g^k,h^k a/b)$ (where $g$ is the generator and $h$ is ...

8

Well, whether it is a secure tweakable block cipher depends on how resistant (E,D) are to related key attacks; that's not a standard assumption for block ciphers. For example, this would not be a secure tweakable block cipher with 3DES; because every 8th bit is ignored, the attacker can effectively test the value of the 7 adjacent bits (except for the 7 ...

7

Let me try to answer your second question, and hopefully shed some light on the first one in doing so. When we encrypt a message, it's because we want to keep something about that message secret. But what is it that we actually want to protect? Let's say the message we're encrypting is AGENT DOE REPORTS 23 UNITS ON BOARD SHIP TO BASE ALPHA, DEPARTED ON ...

6

Encryption using a block cypher such as AES by passing plaintext blocks directly to the encryption function is known as Electronic Code Book mode (ECB) and is not CPA secure as (as you say in your question) it is entirely deterministic and two identical plaintext blocks will result in two identical ciphertext blocks. To prevent this an initialisation ...

6

In your formula, $n$ appears to relate to the key space, not the message space. The message space does not intervene in the definition of IND-CPA, and that's a good thing because practical message spaces consist in messages which "make sense" in a given context. There are situations where the attacker already guesses quite a lot of the attacked message, and ...

5

Cryptography is not just about confidentiality of the message, but also confidentiality of information about the message. Given the ciphertext, an attacker should not be able to determine any information about a message without knowing the key. If you can tell that message A is equal to message B, that's a leak of information. This could be useful when ...

5

Because (I assume) $g$ is a generator, it is not a square (prove this), so its Legendre symbol is $-1$. And hence, the Legendre symbols of $g^a$ and $g^b$ leak the parities or $a$ and $b$. Hence they leak the parity of $ab$, which leaks the Legendre symbol of $g^{ab}$.

4

The Caesar cipher (aka Shift cipher) has, as you said, a key space of size 26. To achieve perfect secrecy, it thus can have at most 26 plaintexts and ciphertexts. With a message space of one character (and every key only used once), it would fit the definition of perfect secrecy. For the usual use with messages longer than one character, or multiple ...

4

Not a complete answer, but since you mentioned "unmodified RSA" I feel it's relevant. Something stronger than vanilla RSA is necessary, even if it isn't semantic security. Example: What if you have a public key exponent of 3 and the symmetric key being encrypted is 16 bytes long? Using raw RSA, $m^e$ would be about $128 * 3 = 384$ bits long and thus ...

4

The proof for the perfect secrecy property of the one time pad is quite simple. It makes use of basic probabilities and it says that: $$Pr[M=m|C=c]=Pr[M=m]$$ for a probability distribution M$\{0,1\}^n$ for the message space and a probability space C for the ciphertext space. Proof: $$Pr[C=c]=\sum{Pr[C=c|M=m']\cdot Pr[M=m']} =\sum{Pr[K=m'\oplus c]}\cdot ... 4 First, on the difference between perfect security and semantic security. Both definitions concern confidentiality, so let us first define what confidentiality means. Note first that an adversary as some a priori knowledge of the message. We can capture that by e.g. having the adversary choose two messages and then flipping a fair coin to decide which one to ... 4 Spartacus: Maybe i came out with the solution, since the cryptosystem described above is not CCA-secure, an adversary A can intercept (A,B) and compute a new ciphertext$$C = 2B\bmod N = 2^er^e \bmod N$$Since he's carring out a CCA-attack he has access to a decryption oracle and since:$$C\neq B$$the oracle output$$RSA^{-1}(C) = 2^{ed}r^{ed}\bmod N = 2r\...

3

Since the keys are fixed from beginning (the sub-protocols input are ciphertexts), isn't it possible to give the secret key to the (non-uniform) distinguisher as an extra advice (the only restrictions for the advice is that its bitlength is polynomial in the security parameter), and thus allowing the distinguisher to decrypt? This is up to your security ...

3

If an attacker can choose the points $P_i$, than this system is not semantically secure. For example, they may choose $P_2=2P_1$, and the corresponding encryption $Q_2$ would be equal to $2Q_1$. If the points are chosen at random, this system is semantically secure if decisional Diffie-Hellman assumption holds for the curve. This assumption is presumed to ...

3

The initial notion of semantic security from Goldwasser and Micali has been shown to be euqivalent to what we call today indistinguishability under chosen plaintext attacks (IND-CPA). Yes that's only security against a passive adversary and actually the weakest reasonable security notion that we use today. The authors of the second paper you link seem to ...

3

Here is the proof I came up with. Please let me know if you see any problems with it... Statement to prove: If an encryption scheme is secure in the IND\$-CPA sense, then it is secure in the IND-CPA sense as well. i.e. IND\$-CPA $\Rightarrow$ IND-CPA The contrapositive is easier to prove: $\neg$IND-CPA $\Rightarrow$ $\neg$IND\$-CPA. This statement is a ... 3 Well, the obvious way to do this is: Before the protocol occurs, Alice runs the$Gen$procedure to create a public and a private key For her round, Alice sends her public key to Bob For his round, Bob selects a random symmetric key$\in \{0,1\}^n$, encrypts it with Alice's public key, and sends that encryption to Alice. Alice decrypts the message that Bob ... 3 To expand / generalize @poncho's reply, given a block cipher$(E,D)$with keylength$n$, you can make a new one$(E',D')$with key length$n+1$, which ignores the last bit of the key and just runs$(E,D)$. If$(E,D)$is a secure PRP, then so is$(E',D')$. But plugging$(E',D')$into the OP's construction does not yield a secure tweakable block cipher. To see ... 3 I'll answer question 2, leaving the first as an exercise to the reader. I'll do this on intuitive grounds, rather than using explicit conditional probabilities. The adversary is free to compute$v_1\cdot v_2$regardless of what we ask, therefore removing everything about that and$v_3$does not change the problem, which reduces to: We somewhat have ... 3 I am stuck at the point where I proved that the complexity is$O(2^\rho)$using brute-force approach. How shall I proceed? Well, a proof that assumed a specific attack strategy is of limited use, as that proof would be inapplicable if the attacker used some other strategy. Instead, what we typically do in a proof is assume that the adversary had some ... 2 You are in a twist here: semantic security (equal to IND-CPA) can only be fulfilled by probabilistic encryption schemes. You need a deterministic encryption scheme for your drop-out tolerance. As it was pointed out previously, any homomorphic encryption allows you to proof in zero knowledge the equality of two ciphertexts: known:$c_0 = E(x,r_0)\;,\;...

2

To be secure against a chosen-plaintext attack, an encryption scheme must be non-deterministic — that is, its output must include a random element, so that e.g. encrypting the same plaintext twice will result in two different ciphertexts. Indeed, if that was not the case, an attacker could easily win the IND-CPA game just by using the encryption ...

2

Why is proof-by-reduction needed? In general, reduction proofs are a very common thing in computer science. Specifically in cryptography, the reasoning goes something like this: The general cryptographic community believes that problem $X$ is very difficult to solve. I have designed a new cipher $Y$ and want to convince the community that it would be ...

2

What if it could? What does this definition mean in practice? Consider $M_0=$ attack and $M_1=$ don't attack. If the adversary can distinguish which message you are sending to your troops, they can optimize their strategy to defeat you. Another example. Say you are casting a yes ($1$) no ($0$) vote for a proposed piece of legislation. If the adversary can ...

2

Yes, semantic security (IND-CPA) is important, even for public-key cryptosystems and for hybrid cryptosystems. Let's focus on hybrid cryptosystems, where a message $m$ is encrypted by picking a random symmetric key $k$, encrypting $m$ under $k$ with a symmetric-key algorithm, and also encrypting $k$ using a public-key algorithm. In this case, it's ...

2

(Note: This answer is based on $k$ being generated by applying a pseudorandom function to a unique message-ID – counter – each time.) It depends how many times you want to encrypt with it. If you want it as a complicated OTP, then it's secure. In order to see this, just ignore the $x$ parts and note that $s_1m \bmod p$ and $s_2m \bmod p$ are to independent ...

2

Not always, it depends on the particular encryption scheme. Strictly speaking, the proofs only say that breaking indistinguishability is equivalent to breaking the hardness assumption they are based on. There are some cryptosystems, like Rabin's, where the security of the key is equivalent to the security of the ciphertexts, i.e. factoring <=> key ...

Only top voted, non community-wiki answers of a minimum length are eligible