# Tag Info

12

I'll comment only the statement referring to an AES-256 replacement with 4096-bit key: According to our engineers, this will take 23840 times longer to crack than aes256 Bob writing that is not able to correctly transcribe even the numbers that engineer Alice allegedly spelled: most likely, $23840$ is intended to be $2^{3840}$, which is the ratio $2^{... 8 A colleague gave me the following explanation that I think makes a lot of intuitive sense, so I'm reproducing it here. Skip to the last paragraph it you don't care about the proof. Suppose you're trying to track one individual user, and you're trying to figure out whether they're in the database. You have some prior knowledge about this:$\frac{P(in)}{P(out)...

7

Homomorphic Encryption on Reals In theory, homomorphic encryption can be done on real numbers. This answer describes two options you have when dealing with real numbers or operations that will result in real numbers. Kristin Lauter is doing some of the cutting edge research in this area. In a recent paper, CryptoNets: Applying Neural Networks to Encrypted ...

7

Sending information covertly when using legitimate systems is a well-studied field. As another question mentions, key words are covert channel and subliminal channel. Designing systems with no covert channels is difficult, but not impossible. The main difficulty is to ensure that if any randomness is sampled incorrectly, the legitimate recipient must detect ...

6

RSA with PKCS#1 v1.5 padding for signature genernation would work, as the algorithm is fully deterministic. The random padding is only used for RSA with PKCS#1 v1.5 padding for encryption which is different from the padding for signature generation. Note that both OAEP and PSS are not deterministic either; they both rely on a random number generator. I ...

6

Do not invent your own authenticated encryption mode. Use a standardized one, and use a well-supported library to implement it in your code. AES-GCM, AES-CCM, AES-OCB, and AES-CBC with HMAC-SHA256 over the ciphertext are all common options. Some great direction from Matt Green here: How to choose an Authenticated Encryption mode

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The definition of perfect hash functions do not have any security/cryptographic requirements. For example, the hash function that simply outputs the first n bits of any string is a perfect hash function on the set of n bit strings. Obviously this function does not hide the input at all. So the answer to your questions is that it depends on the hash ...

5

They could use 1 out of 2 oblivious transfer. Alice offers the messages $0$ and $a$ and Bob uses $b$ as his choice bit (I.e., choosing the first message if $b = 0$ and the second if $b = 1$.). It should be easy to see that Bob now receives $a \land b$ (if in doubt write down the truth-table). Now Bob can send the result to Alice (or they can do the protocol ...

4

The claims made are pretty much all nonsense or do not represent an accurate understanding of the state of the art. I'm not going to go into a point-by-point response; suffice it to say that I would not trust any advice or representations they may make about what is or isn't secure. Their system might be fine, or it might not be, but their public ...

4

$\varepsilon$-differential privacy is absolute: for any pair of databases, you cannot gain more than a small amount of probabilistic information about a single individual. When you add or remove an individual in your database, all possible outputs of your algorithm can appear with similar probability. By contrast, $(\varepsilon,\delta)$-differential privacy ...

4

Beside XTS and CBC-ESSIV, are there any other modes of encryption that are commonly used? No. Just look at software like Veracrypt (XTS only) or dmcrypt. There was a third notable scheme for this purpose, LRW, but it's deemed not secure enough: The issues of LRW were: a) An attacker can derive the LRW tweak key K2 from the ciphertext if ...

4

Any transformation of the identifier, including those in the question, is bound to have at least one drawback: after a transformation of cryptographic quality, it will be next to impossible to re-agregate identifiers that refer to the same person but have been slightly garbled (perhaps because there are two common spelling of the mother's name). Even ...

4

The general technique for tree hashing (independently of the issue of partially revealing content) is to define the hash of nodes as the hash of the concatenation of the hash of their leaves, with a suffix (or prefix) that makes the input of the hash different for leaves and nodes. Demonstrably, this inherits from the cryptographic hash it's properties of ...

4

In short, with this multiplicative definition, it could be ruled out the possibility that an individual's record would be randomly selected and published. Consider a malicious algorithm $M^*$ that picks a random individual's record from the input database (of size $n$) and outputs it. Note that this $M^*$ should not be considered secure for a good ...

3

First - please REALLY check the certificate, download it, open and have a look into details (4096 bit, 30y validity). I don't think your government wants to be insecure themselves (so other parties will act on behalf of it). If the certificate is to be used as a root (or issuing) certificate for other nationally issued certificates, you need it to be ...

3

In this case there is no need for a rainbow table, since at most the identifier has about 15 bits of information. You can cycle through all possible values (there are $26^2 \cdot 31 \cdot 2 = 41912$) and even if they were hashed with a salt you will find the preimage in a fraction of a second. So the only way to protect the information is with a secret key, ...

3

Signing and encrypting together is not secure in this method, at least in the way most would perceive security. For example, Bob would likely interpret this message as being sent from Alice to Bob. However, Alice may have sent it to Charles who decrypted and re-encrypted the signed message under Bob's public-key. In order to do this securely, you need to add ...

3

The simplest way to do this would be to have the sender randomly shuffle the elements. The receiver chooses a random element to request. That way the receiver has no idea which of the original (before the shuffle) elements he got.

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Well, it has the obvious problem that if the UA has both $d_1H(r)^{k_1}$ (from the party) and $H(r)^{S-k_1}$ (from the UA), it can compute $d_1$ directly.

3

I'll try to give some high-level intuition of the used construction paradigms in my answer below. First, note that we distinguish between one-show and multi-show anonymous credential schemes. For one-show credential schemes, you essentially have that multiple showings of the same credential can be linked to each other, while showings are unlinkable to the ...

3

The $\delta$ item is a relaxation of the $\epsilon$-differential privacy notion. The latter is a strong security notion because it requires an algorithm $\mathcal{A}$ to have very close output distributions on "neighbor" datasets $D_1,D_2$ that differ in a single record. From its formal definition $\Pr[\mathcal{A}(D_1) \in S] \leq e^{\epsilon} \times \Pr[\... 3 No, it means that the functions are chosen from some domain with some probability distribution. This is standard for randomized algorithms. For simplicity, assume there are$N$randomized functions$\mathcal{K}$possible, and one choose one uniformly with probability$1/N.$For example, if we restricted ourselves to polynomials of degree$\leq k$over$...

2

The easiest way would be to make use of a trusted third party. In this way, users can authenticate to the trusted party (maybe with the cellphone/IMEI number) which then issues them with a "ticket" or "group/blind signature" along with a pseudo-identity, similar to those used in e-voting schemes. The pseudo-identities can then be checked for duplicates. ...

2

The answer to the first question is both. TLS uses a custom PRF based on HMAC to generate symmetric and MAC keys from a shared secret. The shared secret is created during the asymmetric key exchange between client and server as part of the handshake. The PRF generates key material of a required length. That length is determined by the key sizes and the key ...

2

Instead of trying to invent your own protocol, you'd be much better off using something that is already out there. For example, you could use TLS to transport the data. Another option would be to use GnuPG and some other transport mechanism (post the file on a website to be downloaded by Bob, send it via email, etc). Now, to your question of does this ...

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Furthermore, is using AES-CBC this way any safer than using AES-ECB? Yes, using CBC mode is (almost?) always safer than using ECB mode. Is using zero IVs with AES-CBC safe[...]? No it is not safe, because if you re-use a key with that construction you will leak the common prefix between messages, at block granularity. That is if you have three messages ...

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You're looking for "key escrow", which is often a very bad idea. Even if this fits your use case, you're creating a single point of failure. I'm sure that there are more sophisticated approaches, but here is a very simple one: You need a pseudorandom function (PRF) and a public key encryption scheme. Draw a random key $k$ for the PRF and use it as master ...

2

You asked a How do I question. You probably asked a Should I question. As in "Should I release a data set that reveals the distance between two strings where everything else is anonymized?" The answer is No unless you're making a crossword puzzle. (XKCD beat me to the joke for something similar. Your scenario is much more crossword-like however.) Even if ...

2

The problem you are describing is called Secret Sharing And the common algorithm to solve the m of N is https://en.wikipedia.org/wiki/Shamir%27s_Secret_Sharing It is not only computationally difficult, it is information theoretic impossible to discover the secret without the required number of shares.

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