# Tag Info

0

I do not believe that braid based cryptography is dead since new ways of applying braid groups to cryptography are currently being investigated and have recently been proposed. In fact, recently researchers have observed how braids could be used for reversible circuit obfuscation. A recent paper on braid-based obfuscation The 2014 paper, Partial-...

2

Those numbers for the key sizes come from the bit length of the prime chosen for the finite field $F_p$. To clarify, there are no complex numbers used at all. The elliptic curve is over a quadratic extension of $F_p$, i.e. $K = F_{p^2}$, and a root $i$ of $x^2+1$ is chosen, so that elements of $K$ can be represented as uniquely as elements of $F_p \oplus i\... 1 Bruteforce appears to work well enough. The following Sage script finds an instance quickly: from sage.libs.fplll.fplll import FP_LLL from sage.libs.fplll.fplll import gen_uniform n = 5 # dimension q = 16 # size of matrix entries while True: M = gen_uniform(n, n, q) L = M.LLL(delta=0.999) S = FP_LLL(L).shortest_vector(algorithm='proved') ... 1 I think your proof of correctness may be hampered by the fact that you are setting the ciphertext to$(c_0, c_1) = (a, as + 2e + m)$when in fact the paper you cite sets the ciphertext to$(c_0, c_1) = (as + 2e + m, -a)$. You correctly state that decryption is then computed as$c_0 + c_1s \text{ (mod } 2)$. This reduces to$(as + 2e + m) + (-a)s \text{ (mod }...

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Actually, if RSA is being used in a deterministic way (and the public exponent $e$ is relatively small), someone could recover the value $N$. We know that $P^e = C \bmod N$; that's equivalent to $P^e - C = kN$ for some integer $k$; if $e$ is small, then Shor's algorithm might be able to factor $P^e - C$; allowing you to recover $N$. Alternatively, if you ...

1

I'ld say the answer is “no”. Usually you need to factor the modulus $N$ to break RSA. Now $N$ is not available to the attacker. So with a single plain text and cipher text I'm pretty sure the attacker has too little information to retrieve N or any other key component. Your pre-condition of not having the public key and therefore the modulus $N$ available ...

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TL;DR: As of 2015 quantum computing is not an immediate threat to RSA. But it is a long-term threat (decades away). Whether you should be worried depends whether you are interested in keeping the secrets for so long (2035, if you want a pessimistic guesstimate). As the other answers already say, D-Wave machines ARE not generic quantum computers and will ...

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