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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.


Jun
1
comment How Does Progressive Hashing Work?
It would be difficult to do otherwise, conceptually..
May
31
comment Correct way to read a given permutation cipher?
The first permutation just shifts each letter to the right cyclically. Look at the permutation: (1, 2, 3) => (3, 1, 2). So (1=V, 2=E, 3=N) is mapped to (3=N, 1=V, 2=E), that is, NVE.
May
31
answered RSA: Fermat's Little Theorem and the multiplicative inverse relationship between mod n and mod phi(n)
May
31
revised RSA: Fermat's Little Theorem and the multiplicative inverse relationship between mod n and mod phi(n)
cleaned latex a bit
May
30
comment Security of the iterated Hill Cipher
Note: the first two papers are behind a paywall.
May
27
comment Want to use ECC but am clueless
This is off-topic as it as about security software recommendation rather than actual cryptography. But I think you should really rethink your problem - generally end users should not concern themselves with which algorithms are being used to secure their data, just that it is, so a general purpose tool will probably serve you better (and it may or may not use elliptic curve technology), otherwise you are probably only going to find hobby, proof-of-concept tools that are probably flawed and insecure...
May
21
awarded  Enlightened
May
21
awarded  Nice Answer
May
20
comment Common password derivation function for different encryption methods
Don't both bcrypt and PBKDF2 let you choose an arbitrary output length? In that case what's wrong with simply choosing the key length of the block cipher to use next?
May
15
comment RSA private exponent primality
Specifically, $d$ is going to be an integer such that $ed - 1 = k\cdot \mathrm{lcm}(p - 1, q - 1)$ for some integer $k$, that is, $d = \frac{k \cdot \mathrm{lcm}(p - 1, q - 1) + 1}{e}$, so, really, its prime factorization could be anything, except it will always be coprime to $\mathrm{lcm}(p - 1, q - 1)$, that's about all you can say.
May
15
comment modulo operations in crypto algorithms
@user220201 This is when, e.g. using $e = 3$, we have $m < n^{1/3}$, in which case $c$ is literally $m^3$ and you can just take the cube root in the integers and decrypt the ciphertext using only public information. If $m > n^{1/3}$, then we only know that $c + kn$ is a perfect cube for some integer $k$, but this $k$ becomes very, very large on average as $m$ increases (with a very complicated distribution), making it impossible to brute force it without knowing the factors of $n$ (or so we believe).
May
14
comment RSA private exponent primality
The public exponent doesn't have to be a prime.
May
12
comment Generating unsigned, bounded random value using signed bounded random values
"All generated random values must have a normal distribution" is a non-issue, once you have a uniform distribution you can convert it to a normal distribution efficiently via inverse-transform sampling or specialized methods like Box-Muller. Focus on getting a uniform distribution (rejection sampling will probably be fast enough, except perhaps for pathological values of $x$).
May
11
comment RSA, finding p,q
If you prefer, you can use the following idea: since $ed - 1$ is a multiple of both $p - 1$ and $q - 1$, if follows that $a^{\frac{ed - 1}{2^k}} \equiv \pm 1 \pmod{p, q}$ for some small $k$. Thus trying a bunch of random $a$'s, you will quickly find an $a$ which is a quadratic residue modulo $p$ but a quadratic nonresidue modulo $q$, such that $a^{\frac{ed - 1}{2^k}} - 1$ is a multiple of $p$ but not of $q$, and you are done.
May
11
comment RSA, finding p,q
I'm not sure I follow what you mean by "2 divides p - 1 ϕ(p-1) times", but the basic idea is this: if 2 divides $p - 1$, $x$ times, and 2 divides $q - 1$, $y$ times, then 2 divides $ed - 1$ at least $\max(x, y)$ times. So if you keep dividing $ed - 1$ by 2, at some point you will end up with a number that is a multiple of $p - 1$ but not of $q - 1$ (or vice versa). Then using Fermat's little theorem can produce a factor of $n$ (there are some details but that is essentially the idea).
May
11
comment RSA, finding p,q
Correct, now how do you use this knowledge to find the factors of $p$ and $q$ efficiently? Hint: how many times can $2$ divide $p - 1$ or $q - 1$? What about $ed - 1$? With this you should be able to find an efficient way to produce a congruence of the form $a^m \equiv 1 \pmod{p}$ and $a^m \not \equiv 1 \pmod{q}$ and thus find $p$ (can you see why?)
May
11
comment RSA, finding p,q
If you have $e$ and $d$, and you know that $ed \equiv 1 \pmod{\varphi(n)}$ - or $\mathrm{lcm}{(p - 1, q - 1)}$ - what can you deduce?
May
2
revised Symmetry for finite cyclic groups (Z/pZ)∗
added 315 characters in body
May
2
answered Symmetry for finite cyclic groups (Z/pZ)∗
May
1
comment How to argue to a paranoid that RSA is safe?
And using weak prime numbers is hardly RSA's fault.