10,257 reputation
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bio website github.com/CodesInChaos
location Frankfurt, Germany
age
visits member for 3 years, 5 months
seen 9 hours ago

Oct
3
comment Alternative to NSA encryption algorithm
Quantum computers can efficiently compute discrete logarithms in groups, including finite fields and elliptic curves. So Diffie-Hellman is dead once QC's capable of running Shor's algorithm appear.
Oct
2
reviewed Approve Strength of MD5 in finding duplicate files
Oct
2
comment Hardness proof?
We can also show that AES-CBC is secure as unauthenticated encryption if AES itself is a secure block cipher.
Oct
1
revised What is h in this RSA variant?
edited title
Oct
1
comment Calculate the RSA private exponent from the CRT parameters
Compute $\phi = (P-1)(Q-1)$ and then the modular multiplicative inverse of $e$ using extended euclidean.
Oct
1
comment finding collision for truncated SHA1 hash output
Is your indentation correct? Don't you need to indent the two lines after the if statement?
Oct
1
comment finding collision for truncated SHA1 hash output
@StuartP.Bentley Doesn't your program search for preimages not collisions? A 40 bit prefix collision costs about as much to find as a 20 bit prefix preimage.
Oct
1
comment Small Encryption Exponent
1) Simply running a factoring software sounds a little dull for a homework question. It's also weird that the moduli for Problem 1 are smaller. If you bothered to setup factoring software for Problem 2 you use it to solve problem 1 as well, without exploiting the same message encryption weakness. 2) the short message and small exponent approach doesn't seem to work.
Oct
1
comment Calculate the RSA private exponent from the CRT parameters
1) Do you know e? Different es result in different ds. 2) How did you end up with knowing all those values but not knowing d?
Oct
1
comment Small Encryption Exponent
If the message is shorter than the length of the modulus divided by the public exponent, you can simply compute the $e$ root, for example via binary search. For real RSA the padding ensures that the message has similar length as the modulus, but for unpadded RSA short messages can happen.
Sep
30
awarded  Explainer
Sep
29
revised Message encrypted with a LFSR based stream cipher
edited tags; edited title
Sep
28
comment Keccak and compression functions
@Dingo13 Blake and Skein are not MD constructions, but they still use a kind of compression function. Compared with a basic compression function, the most significant difference is the addition of a tweak, which is used to uniquely mark each block and to signal the end of the input message.
Sep
28
comment Question about second preimage resistance of hash function combiner
It's pretty obvious that collisions and second pre-images of the concatenation imply the same attack against both hashes and are at least as strong as the stronger. On the other hand, concatenating a pre-image resistant hash with the leading bits of the message is practically bad concerning "it's hard to recover the message from a hash", even if it might still fulfill some definitions of pre-image security.
Sep
28
comment Question about second preimage resistance of hash function combiner
For second preimages I agree fully. For first preimages it depends on the precise definition you're using.
Sep
26
comment Roots of polynomial in Shamir secret sharing
1) Since shares can have a 0 y value they can be roots, in which case the attacker trivially learns some roots. 2) What you describe in your comment has almost nothing to do with SSS, except that both use polynomials in finite fields. 3) Just to be sure, with roots you mean the points at which y=0 and not the coefficients of the polynomial, right?
Sep
26
comment Roots of polynomial in Shamir secret sharing
Why do you want to know that? Unless it's just idle curiosity, I'm pretty sure that you're asking the wrong question.
Sep
24
awarded  Autobiographer
Sep
24
revised simple encryption scheme turns out to be “somewhat homomorphic”
added 6 characters in body
Sep
24
comment Generation of N bit prime numbers -> what is the actual range?
Personally I'd approximate the $\sqrt{2}$ by 1.5, which results in the simple algorithm of setting the two most significant bits to 1.