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location Wellington, New Zealand
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visits member for 2 years, 11 months
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I am an undergraduate computer science and mathematics student in New Zealand. My fields of interest are computer graphics, in particular the physics of light transport, and to some extent cryptography, as well as programming and software development in general.


Dec
15
comment How to customize a standard encryption algorithm?
What do you mean by "customize"? Do you really want to modify the entrails of the algorithm, or do you just need a parameterized block cipher built on top of AES (with provably equal strength) to differentiate it from other uses of AES, akin to Skein's personalization string?
Dec
12
comment Crack RSA with imaginary algorithm
@AndrewPoelstra Finding multiplicative inverses is always easy, Euclid came up with a working algorithm over two millenia ago. What is hard without the factorization is finding multiplicative orders.
Dec
11
comment 2048-bit RSA Decryption
@fgrieu "problem solved" doesn't sound like breaking an honor code to me. Anyway, the question has been rolled back (thanks CodesInChaos).
Dec
11
comment 2048-bit RSA Decryption
What is the point of deleting questions? If you do that, nobody else can benefit from it later on. Please consider posting your solution as an answer for future visitors.
Dec
9
answered Fermats Little Theorem, primitive root
Nov
30
comment Is this problem based on a known hard problem?
Well for starters the set of $a_i$ can't be linearly independent otherwise this amounts to solving an $i \times n$ system $Ab = c$ for $b$ with $A$ invertible. It can be solved in the rationals and then scaled to get an integer solution. And for linearly dependent $a_i$, I would imagine some partial solution could be obtained as well (if one exists) using the same techniques. Unless I am misreading the problem...
Nov
29
comment Does SHA-1 hash have quasi commutative property?
Are you going to ask the same question for SHA-2 and SHA-3 as well? crypto.stackexchange.com/questions/20497/…
Nov
28
comment Issue about randomness : what if random looks “human” ?
The very first thing to realize is that randomness is not a property of any particular output sequence. The sequence "1234" is neither random nor predictable, it's meaningless to refer to it as random. Rather, randomness is a property of the process used to generate them. Once you understand that, things begin to seem less paradoxal.
Nov
22
comment Addition / Multiplication modulo 13
I'm not sure what you're asking. Are you wondering why $g = 2$, a generator of $Z_{13}^*$, is also a generator of $Z_{13}^+$ (it is)? If so, have you considered finding all the generators of $Z_{13}^*$, and all the generators of $Z_{13}^+$?
Nov
21
comment java.util.Random and Dice Rolls
@GuutBoy He wants to recover the state of an instance of the Java Random class given its outputs (as dice rolls) efficiently.
Nov
18
comment special public keys and modulo n
Are you talking about textbook RSA (no padding, just interpret the message as a number and raise it to an exponent) or real world RSA (with padding)?
Nov
4
revised Security of RSA with $\gcd(pq, (p-1)(q-1))\ne 1$
typo
Nov
3
comment Security of RSA with $\gcd(pq, (p-1)(q-1))\ne 1$
@fgrieu No, because we have a multiple of $p$: the modulus $n$ itself. So we get that particular prime factor "for free" by adding $n$ to the product of prime powers.
Nov
2
comment Security of RSA with $\gcd(pq, (p-1)(q-1))\ne 1$
@fgrieu As I suspected (it was fairly obvious in retrospect) we also need $k$ to be non-smooth: the trick is that $n$ automatically gives us the factor of $p$, which gets us most of the way to $q - 1$ in the context of the Pollard $p - 1$ algorithm. This doesn't really change much, though, it just means that the key generation procedure needs to be adjusted to avoid $k$ being smooth, it is still open whether we can exploit the length difference between $p$ and $q$.
Nov
2
revised Security of RSA with $\gcd(pq, (p-1)(q-1))\ne 1$
better attack, supersedes old one completely
Nov
1
reviewed Approve Security of RSA with $\gcd(pq, (p-1)(q-1))\ne 1$
Nov
1
comment Security of RSA with $\gcd(pq, (p-1)(q-1))\ne 1$
@fgrieu Thanks! Hopefully there is a way to improve on the number theory to provide a faster algorithm to find $k$ than brute force.
Nov
1
comment Security of RSA with $\gcd(pq, (p-1)(q-1))\ne 1$
@fgrieu Hmm, the square root approach seems more direct, looks like I missed the obvious! But in any case, yes, it means that subject to the restrictions the modulus has to be pretty unbalanced to offer any security, but I don't have anything better than that so far :/
Nov
1
revised Security of RSA with $\gcd(pq, (p-1)(q-1))\ne 1$
added 294 characters in body
Nov
1
answered Security of RSA with $\gcd(pq, (p-1)(q-1))\ne 1$