Valhalla Legends Archive

http://ender.x86labs.org/code/misc/echoserver/

This will compile and run on unix. It's a simple client/server echo program. By the way, you should probably introduce yourself before you ask for help. What brought you here?

For Windows you need winsock... I don't know anything about Windows programming so you'll have to find a tutorial.

invert stop trying to turn this into a playground where you can masturbate on webcam

you guys, you guys, have a little respect for pms.

Quote from: brew on April 25, 2008, 07:03 PM

Quote from: Andy on April 25, 2008, 06:38 PM
According to her, she's quite rich

Yes. She makes an annual salary of $100-150k, according to this.

Yes, according to that she's also seven feet tall.

You guys are terrible.

Here we have this really lonely and depressed chic who probably has no friends IRL and can't go anywhere without nicotine gum... and instead of welcoming her and soothing her depression you give her even MORE reason to use that heroin she's trying to avoid.

I can just imagine her, biting on her cuticles, eyeing that box in the closet.

I mean REALLY, this is like 8 on 1.

You all should be ashamed of yourselves.

He just pissed me off because he stole my name. But now I feel bad. I'm going to stop posting here now.

You have no life, and your last post is a mess.

You have no life.

Oh, okay. Maybe this works.

Solution:

Let S be the set of all boats and let ~ be a relation on S. Each boat can be represented as B = (a, b, c, d) where (a,b) is the starting point and (c,d) is the velocity vector. Let |B| = (x, y) = (a + cn, b + dn). Define B_i ~ B_j iff |B_i| = |B_j|.

Let B1 be the real boat. We know the set of all boats to be Z^4 and thus countable, so on some nth try we will hit B1. Once we hit B1, construct the ~-equivalence class for B1, we'll call it C. Every boat in C must have a different velocity vector, or else they would have originated at the same point, in which case they would be the same boat. Thus no boat in C will ever meet again. And since the equivalence class C is a subset of Z^4, it will also be countable.

Thus we can "narrow in" on the equivalence class C and only target boats in C. Index every boat in C as 1, 2, 3, ... Every boat we hit in C that's not the real boat will be a miss, since no two boats in C can ever coincide again. Since C is countable, we will eventually hit the real boat on the kth try, and thus the coordinates of the real boat will be the boat in C indexed k.

Good point, K.

The boat sinks in one hit for the bonus problem as well as for the original question, right? If the boat were to be "invincible" and keep going after getting hit, then I believe the bonus problem would be much more tractable, and the algorithm 100% accurate.

But suppose the boat does sink in one hit. Is this algorithm supposed to be successful 100% of the time?

Nevermind, I think I've thought of a way to do it. Will post later if I can turn the idea into a solution.

lol, I'm insanely busy so nevermind that.

Cool riddle! May I post it on a forum I visit? (Not x86 of course, as most people there have already seen it ) I will cite your friend as the inventor, as well as you and this thread of course.

EDIT: I don't see how the current algorithm does not already yield (a,b,c,d). If you know the value of n when you hit the boat, and the rule by which you order the boats, then you can just look for the nth quadruple on your indexed list.

haha, I didn't see the post about this exact thing in this forum, sorry, you can delete this...

edit: despite the above I am posting anyways because I haven't been able to fix it...

My site, http://www.ktmem.com, looks drastically different in IE and firefox. First of all, the navigation box is clipped in IE whereas it is not in firefox -- in firefox, it spans 90% of the height as it's supposed to. Also, it seems like IE is ignoring a lot of my other CSS. Can anyone tell me why this is happening and what I can do to make IE display my page the same way that firefox currently does?

I don't want to come off as an asshole, but those boolean literals, true/false, are ugly and unnecessary. Perhaps you did it for clarity, though.

Non-blocking IO allows for far more clients than blocking IO and is "cleaner." It doesn't matter how little state the threads have; they still eat up unnecessary CPU cycles.

Posting from Linux ;-) I haven't mustered up the courage to boot Windows up and see whether my file system is fried or not. Thanks for recommending PartitionMagic -- it did the job (well, as far as I know!) and was user-friendly.

EDIT:
Windows seems fine ;-)