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10

Do the post-quantum ciphers also automag/tically address the 1st problem? Not really, however to explore that in any detail, we need to explore what the 1st problem is. If $P=NP$ is proven true, what does that practically mean? Well, it might have absolutely no practical ramifications, or it might mean that virtually all known cryptographical systems ...


6

I can't agree with bullet 3. of the question stating quantum computers are getting stronger with a good pace (82-qbit computer last year, 512-qbit computer this year), at least in a context of cryptanalysis. Even the marketing people praising the device alluded to do not pretend that it is of a kind useful for cryptanalysis: they state about it: The ...


5

No, since finding $a$ allows offline checking of passwords. $\:$ No, although I can't back this part up.


5

First of all there does exist information theoretically secure message authentication codes suitable for use with a one time pad. An HMAC is not one of those information theoretically secure. As far as I recall the first article presenting such a construction is the 1981 article by Wegman and Carter: New hash functions and their use in authentication and ...


4

The problem is that the key is essentially random data and thus cannot be meaningfully compressed. Several variations of the McEliece cryptosystem have tried modifying it to produce public keys with special structure which are compressable. However, all such systems have been broken (as far as I know), and it seems that in most cases, adding special ...


4

If you need security against quantum attacks, there aren't that many options. I would go for a lattice-based encryption like NTRU or something based on ring learning with errors. There are no "magic numbers" involved and the assumptions they are based on have been scrutinized by the academic community. NTRU has been around for a decade and has pretty good ...


3

Dinh, Moore, Russell have shown that the quantum algorithm (Quantum Fourier sampling) used to attack RSA and ElGamal does not work on McEliece-like crypto systems. (I think) this means, that there are no known algorithms on quantum computers that decrease the complexity of attacks on McEliece, and thus McEliece is just as safe post-quantum computers as it is ...


3

You can build a gigantic, enormous tree that has capacity for up to $2^{80}$ one-time signatures (say). Then, each time you want to sign something, you randomly pick a 80-bit value and use that to select which of the $2^{80}$ subtrees to use to sign the message. As long as the number of messages you intend to sign is much less than $2^{40}$ messages, a ...


3

Symmetric algorithms are secure post-quantum, only with less bits of security (usually about half). That means you only need to care about the authentication and key exchange parts of the cipher suite. Suites that don't use public key authentication or key exchange, i.e. preshared key suites, are post-quantum secure, but not useful in most usecases. There ...


3

If you look at exact security, the height matters. The reason is that it defines the number of OTS key pairs and hence the possible number of one time signatures per MSS key pair. To forge a MSS signature, it is enough to generate a forgery for 1 out of $2^h$ OTS signatures. Hence you get a reduction in the bit security of $h$ bits.


3

My understanding is that the attack only works against McEliece with algebraic geometry codes. The paper by Bernstein, Lange and Peters recommends parameters for McEliece with binary Goppa codes, so the attack does not apply against those parameters.


3

I believe that the conjugacy search problem is broken by probabilistic attacks (see chapter 7). I am not sure if this completely ends braid cryptography, however, since there are other difficult problems in braid groups that have not been studied extensively.


2

Supersingular isogenies are a rather recent attempt at post quantum security. You will have a hard time finding an efficient and secure implementation, and even if you write one yourself, the algorithms have not yet seen that much cryptanalysis. (Although that's a subjective judgement call.) If post quantum security wasn't a concern, you could choose from ...


2

In general there is no reason to use tree hash modes for Merkle trees. The reason is that a Merkle tree itself is already some kind of tree hash mode. The important thing about this kind of mode is that it allows to compute the root node given the value of one leaf and one node per tree level. The possible ambiguity of hashes is not relevant for hash-based ...


2

Properly speaking, forward secrecy is a property of a protocol. The protocol is forward secret if compromise of the long term keys does not allow an attacker to decipher any past communications. (Occasionally a distinction is made between that and perfect forward secrecy, with the latter secure when the attacker also knows e.g. all other session keys.) You ...


2

If "it's a fact (not just theory) that quantum computing will break PKI in less then 10-20 years" then "we still use, advise PKI" because there is only a small amount of currently existing evidence for that fact. Specifically, there are known algorithms which can be used for PKI that are not known to be breakable by feasible quantum adversaries.


2

That Wikipedia article is full of errors and false claims. Most importantly, FSB has not been proven to be as hard as an NP-complete problem. This is because the syndrome decoding problem is NP-hard in the worst case, but FSB uses random instances of the problem. Indeed, these random instances may be much easier to break than arbitrary instances. There is no ...


2

One algorithm that is especially suited to one-use key pars is lamport signatures. Like many (all?) other signature functions, lamport signatures first hash the message to get it down to a size that is more reasonable to sign. For this use case, if you are willing to have $n^{2}$-bit signatures and $2n^{2}$-bit keys (public and private), you can sign a ...


2

As you probably know the public key in McEliece is an $k \times n $ binary matrix, encoding a generator matrix for a randomly permuted Goppa code (i.e. $G_{\mathsf{pub}} = SGP$, where $S$ is any $k \times k$ invertible binary matrix, $G$ a $k \times n$ generator matrix for an $(n, k, t)$ binary Goppa code, and $P$ a $n \times n$ permutation matrix). ...


2

Choose some 128-bit hash function, such as RIPEMD-128, and a way of randomizing it, such as this. The private key is either 60 uniformly random 128-bit strings s00,s01,...,s58,s59 and a uniformly random short salt or a seed to regenerate those. $\:$ For each i in {00,01,...,58,59}, vi is the result of hashing si 19 times. $\:$ The public key is the ...


1

Here's something similar but completely different... A 'one-way' cryptographic hash function which is regressible when combined with the function's parsed trapdoor index. That is to say, the hash function is the file and the trapdoor table file is the key! The trapdoor table generated is approximately 60% larger than the original file but eminently ...


1

McEliece public keys need about 100 kByte to 1 MByte depending on the desired security level. 65 kB for 80 bits of security (too low, corresponds to 1024 bit RSA) 150 kB for 112 bits of security 220 kB for 128 bits of security 1000 kB for 256 bits of security The McBits paper contains the following table:


1

Firstly, I assume we are talking about classical computers Implementing a brute force attack on a RSA may not be the most sensible thing, unless of course the security parameter of your target system is small.. (160 bit numbers! ) Even then you may not want to implement a brute force here.. try using Fermat's Factoring or Pollards $\rho$ method. If you ...


1

Lets assume an adversary knows one LD signature for message $M$ as well as public key $pk$ and can generate a forgery for an arbitrary different message $M'$. Clearly, there exists at least one bit position where the messages differ, i.e. $M_i \neq M'_i$, as $M \neq M'$. So, for simplicity assume there is only one bit difference. Then the adversary can take ...


1

Why not use a stream cipher? The requirement for the generating function is not that it is a PRF but a PRG (which in many constructions is build out of a PRF). If you use the key stream of a stream cipher as output then this is a PRG.



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