Are you sure this is what is taking a long time? I use MPI in C, not Fortran, but what is slow in a MPI program is starting all the processes. This can take up to a minute sometimes when there's a lot of them.
Try adding a Barrier before the Send loop, and after the barrier print something on...
I don't know anything about the implementation you are using so I don't know whether your code is correct, but from my experience such problems are usually the result of using different padding schemes.
Also, since it's AES, make sure you are using the same mode (ECB, CBC, CTR...) and...
Hey,
Is there a proof of Kirchoff's theorem available somewhere online? In literature I can find only proofs of Cayley's formula, and the matrix tree theorem is usually only mentioned, if that.
I want to see a proof because I'm having a hard time intuitively understanding why this works...
It's not that mathematicians are more likely to make mistakes, but people who didn't study anything math-related tend to think it's a big deal that a mathematician made such a mistake so it gets noticed more.
I didn't notice any on PF, but these days the "get a college degree over the internet" one is everywhere. It's better than the porn one which was all over the internet 5-6 ago... For example, you would search in google and open a bunch of web pages and suddenly someone would start talking about...
This reminded me of my high school math teacher, she wrote "DERIVATIVES" in big letters (well, not in english) on the board and the derivatives of elementary functions, the (f*g)' = ... and the other couple of rules underneath. Then she showed us how to calculate them. She made no mention of...
5 + 4x - x^2 = m - (x -n)^2
5 + 4x - x^2 = m - x^2 + 2nx - n^2
4x - x^2 + x^2 - 2nx = m - 5 - n^2
4x - 2nx = m - 5 - n^2
x(4 - 2n) = m - 5 - n^2
In order for this to work for all values of x, it must not depend on x. And it won't depend on x if 4 - 2n = 0, because x * 0 will be 0...
Ok I made some progress but it still isn't complete:
(a over b means the binominal coefficient)
(2n!)/[n!(n+1)!]
= (2n)!/[n!n!(n+1)]
= (2n)!(n+1-n)/[n!n!(n+1)]
= [(2n!)(n+1) - (2n!)n]/[n!n!(n+1)]
= (2n!)(n+1)/[n!n!(n+1)] - (2n!)n/[n!n!(n+1)]
= (2n!)/(n!n!) - (2n!)/[(n-1)!(n+1)!]
=...
There are two simple ways to substitute.
First, you can add 1 to both sides of the second equation:
2y + 1 = x - 1 + 1
2y + 1 = x
So now you can substitute 2y+1 instead of x in the first equation.
Another way would be to multiply both sides with 1/2:
2y * 1/2 = (x - 1) * 1/2
y =...