584 reputation
1312
bio website
location Wellington, New Zealand
age
visits member for 3 years, 1 month
seen 5 hours ago

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.


Jan
19
comment Is there a polynomial-time algorithm to find a prime larger than $n$?
(technically, it is probabilistic, but takes finite time and always returns a prime along with a provable primality certificate)
Jan
19
comment Is there a polynomial-time algorithm to find a prime larger than $n$?
Is this asking whether to find the next prime larger than $n$, or drawing a prime larger than $n$ following a specific distribution? Because if not, Maurer's algorithm with a suitable lower bound should get the job done.
Jan
19
comment Need help with notation — finite set of random primes
The term "random" has pretty specific meaning in some branches of mathematics. I would definitely recommend sticking to "arbitrary" if the result to follow simply holds for any finite set of primes without restrictions, IMHO.
Jan
13
comment What are some elementary results (number theory) using theorems that went long-unproven?
And, of course, as the joke goes, FLT isn't strong enough to prove the irrationality of $\sqrt{2}$ itself :)
Jan
7
comment Prove that function is constant
@orion I don't think so, not all continuous functions are constant, so here it's shown that any continuous function that satisfies the condition must be constant. It could be that there exist discontinuous functions that also satisfy the condition (actually, there aren't) so it doesn't fully answer the problem, but it's not a vacuous proof.
Dec
20
comment Quadratic formula not working?
@JoeTaxpayer Always have myself (if a symbol is used at all, anyway).
Dec
9
awarded  Caucus
Nov
23
comment How to know if a term is divisible by 10
@imallett Yes, "both of which end in", which implies you don't need to compute the entire number to look at the last digits, which is what your first comment was implicitly suggesting (and why I gave the tongue in cheek reply of submitting unimaginably large numbers to force you to consider only the last digits :p)
Nov
23
comment How to know if a term is divisible by 10
@imallett How about $4^{10^{1000000}} - 6^{10^{50000000}}$? ;)
Nov
21
comment How to evaluate $(0.9)^4$ without calculator
@Joao That equation is already solved. The process of going from $(0.9)^4$ to a decimal number (or really any simpler form) is, in fact, called evaluating an expression :)
Nov
13
comment System of nonlinear equations that leads to cubic equation
This answer, but under exam conditions: ittc.ku.edu/~fokumdt/images/humor/Hangman.jpg :p
Nov
10
comment simple 'why' question about modular arithmetic 13 mod 5
As an aside, the C operator does not really behave like the modulo operator, it is the "remainder operator", which differs for negative values. For instance, (-3) % 2 = -1, because the solution to 2q + r = -3 where q = (-3) / 2 = -1 with / being integer division (round towards zero) is r = -1. But you would probably expect -3 mod 2 to be 1. Like I mentioned elsewhere, this is a trap many people fall into, as they mistakenly think a % b will always return values in [0, b), which is terribly wrong (and dangerous) for signed types.
Nov
3
comment Rsa encryption/decryption (Updated)
@user189285 Exactly. There's nothing magical about Alice, she just has information Charlie doesn't have. As soon as that information is given to Charlie, that advantage is lost. In this case, factoring N amounts to giving Charlie the private key, and hence the ability to decrypt messages just like Alice can.
Nov
3
revised Solving $T(n)= 2T(n/2) + \sqrt{n}$ without master theorem (algebraically & recurrence tree)
added 1 character in body
Nov
3
answered Solving $T(n)= 2T(n/2) + \sqrt{n}$ without master theorem (algebraically & recurrence tree)
Nov
2
comment Rsa encryption/decryption (Updated)
If Charlie could factor $N$, then he (she?) would have the primes $p$ and $q$, so it would be easy to just follow the instructions again to get phi and $d$ (since $e$ is public). Makes sense?
Oct
30
comment Is this a demonstration or a definition?
@AlexR If $0 \not \in \mathrm{dom}(!)$ then it doesn't make sense to define $0!$ - clearly the OP wants to include zero in the domain of the factorial function... so I don't see how he "implicitly defines" it that way unless he is schizophrenic ;)
Oct
30
comment Proof related to RSA decryption
That is true in real usage of RSA, but I think it is still worth proving, after all, most textbook examples of RSA use small 2 to 4 digit moduli, so the question is bound to come up eventually (not that this is the first question ever about the correctness of RSA, but still).
Oct
30
comment Proof related to RSA decryption
You need to justify why that holds for $a$ not coprime to $m$ (e.g. $a = p$, $a = kq$, etc.. for $m = pq$). Decryption does still work, but you can't just apply Euler's theorem, which only applies if $\gcd(a, m) = 1$ (you could use the CRT, for instance)
Oct
26
comment Is a closed-form expression of this non-linear expression attainable?
If you interpret $a_n$ as a continued fraction then $a_{n + 1}$ is the sum of two continued fractions of known coefficients. So it may just be possible to find a closed form using continued fraction arithmetic, but I doubt it. Maybe someone knows a trick for this one though.