Follow Slashdot blog updates by subscribing to our blog RSS feed


Forgot your password?
Check out the new SourceForge HTML5 internet speed test! No Flash necessary and runs on all devices. Also, Slashdot's Facebook page has a chat bot now. Message it for stories and more. ×

Comment Re:So slashdotters (Score 4, Insightful) 293

But this probably will close the door on the 99 cases out of 100 where an IP actually does equal a bad person who needs to be caught.

I'm not sure about the 99/100 figure. However, even if that's true, I'd argue that just because something is a 99% accurate indicator of crime, it doesn't justify a forfeiture of rights for the other 1%. Is having an IP address linked to an illegal activity justification to open an investigation? Sure. Enough to break in and confiscate property of an individual who has an open WAP living in a populated area? Probably not. Keep in mind people committing internet crimes are "crafty" and know that its important to hide their own identities (often, masking them as the identities of others)

Comment Re:Both? (Score 1) 266

Normally, I'd dog on the editors for this too, but I just noticed the, the "OMG wyte ifone" submission has almost twice the comments as this'll be interesting to check back on the comment count in a few hours to see what /.'ers really care about these days!

Comment Re:Grain of salt (Score 1) 157

My thoughts exactly...

Symmetric, (Strictly) diagonally dominant matrices are great: Non-singular, real spectrum, diagonalizable...In fact, purely from the fact that the eigenvalues can be bounded away from 0, many iterative methods will have provably fast O(n^2) convergence...beating the classic O(n^3) by an order of magnitude.

I'm not up to speed in the particular tricks used for the Symmetric, DD regime, but certainly one would only "naively" try solving this using Gaussian elimination, due to the special structure. One thing I thought was interesting was that the authors mention that the "previous" fast algorithm solves in:


Well, for n 10^52 (HUUUUUUUUUUUGE!!!) n^2 is less than nlog(n)^25, so there complexity constant becomes really important!!! I can't imagine that the "previous" algorithm was useful (practically speaking!)


Slashdot Top Deals

"The Avis WIZARD decides if you get to drive a car. Your head won't touch the pillow of a Sheraton unless their computer says it's okay." -- Arthur Miller