Solve this

[Blarg]

http://imgs.xkcd.com/comics/nerd_sniping.png

Given two adjacent vertices p and q, the resistance across the edge from p to q is the voltage drop along the edge when a 1 amp current is injected at p and withdrawn at q. Whether or not p and q are adjacent, the unit current flow from p to q can be written as the superposition of the unit current flow from p out to infnity and the unit current flow from infinity into q.

By symmetry, in the unit current flow from p to infinity, the flow out of p is distributed equally among the 2d edges going out of p, so the flow along any one of them is 1/2d amps. Similarly, in the unit current flow into q, the flow along each edge coming into q is 1/2d . If p and q are adjacent, when the two flows are superimposed the flow along the edge from p to q is 2*1/(2d) = 1/d.

But since this edge has resistance 1 ohm, the voltage drop along it is also 1/d, so the efective resistance between p and q is 1/d.

I'll watch out for semi trucks on the way home from work.

Duhhhhhh... I mean that was just obvious..

[Khamsin]

Given two adjacent vertices p and q, the resistance across the edge from p to q is the voltage drop along the edge when a 1 amp current is injected at p and withdrawn at q. Whether or not p and q are adjacent, the unit current flow from p to q can be written as the superposition of the unit current flow from p out to infnity and the unit current flow from infinity into q.

By symmetry, in the unit current flow from p to infinity, the flow out of p is distributed equally among the 2d edges going out of p, so the flow along any one of them is 1/2d amps. Similarly, in the unit current flow into q, the flow along each edge coming into q is 1/2d . If p and q are adjacent, when the two flows are superimposed the flow along the edge from p to q is 2*1/(2d) = 1/d.

But since this edge has resistance 1 ohm, the voltage drop along it is also 1/d, so the efective resistance between p and q is 1/d.

I'll watch out for semi trucks on the way home from work.

Go back to preschool! :ashira:

[Ishido]

do we get gil if we solve it? Seeing that you get 1000g in WoW if you do (I have no idea if that is a lot)

Foof isn't on Mal'ganis (at least I don't think so).

I'm glad that I don't know enough about this stuff to start solving it, I would waste so much time on it.

[Charla]

Oh, and section 8 looks a lot like Elvish of some sort.

[Vistani]

Section 8 reads something like:
mpkw pgw yp id rt kwl mt
np ydh tht kwm (?might be a b)kw fy tm
rv ykw aa (k)hwl i(k)hw

[Pyree]

HI FOOF!!

Remember me? :x

[Neosutra]

Given two adjacent vertices p and q, the resistance across the edge from p to q is the voltage drop along the edge when a 1 amp current is injected at p and withdrawn at q. Whether or not p and q are adjacent, the unit current flow from p to q can be written as the superposition of the unit current flow from p out to infnity and the unit current flow from infinity into q.

By symmetry, in the unit current flow from p to infinity, the flow out of p is distributed equally among the 2d edges going out of p, so the flow along any one of them is 1/2d amps. Similarly, in the unit current flow into q, the flow along each edge coming into q is 1/2d . If p and q are adjacent, when the two flows are superimposed the flow along the edge from p to q is 2*1/(2d) = 1/d.

But since this edge has resistance 1 ohm, the voltage drop along it is also 1/d, so the efective resistance between p and q is 1/d.

I'll watch out for semi trucks on the way home from work.

Go back to preschool! :ashira:

Bah, my grammer in posts is always lacking when I am posting at work, as I am usually typing in a hurry >.<.

At least my math is generally mostly right sometimes.

[Foof]

do we get gil if we solve it? Seeing that you get 1000g in WoW if you do (I have no idea if that is a lot)

So we are basically helping some one get 1000G, which he could possibly turn around for about $200 if he RMTs it?

Passsssssssssssssss

When I quit FFXI, I gave my gil away along with my character (I don't like the idea of RMT). I have no interest in the reward (and I'm sure the people on the boards there will come up with an answer before anyone here does). I also don't have a character on that realm, so I wouldn't even be eligible.

And unfortunately, I doubt the person who made the puzzle plays FFXI, so no gil reward equivalent Just here for fun!

[Ishido]

I hope the answer is the lyrics to Never Gonna Give You Up, would be the most elaborate rickroll ever.

[Foof]

The first one is very easy if you are familiar with this type of stuff. (Hint, look up ROT47 on the Wikipidia)

As for the rest...maybe tomorrow

So I looked it up. Here's what is had to say.

"ROT47 is a generalisation of ROT13 which, in addition to scrambling the basic letters, also treats numbers and many other characters. Instead of using the sequence A–Z as the alphabet, ROT47 uses a larger alphabet, derived from a common character encoding known as ASCII. The use of a larger alphabet is intended to produce a more thorough obfuscation than that of ROT13, but ROT47 is far less widely supported."

Found a converter here: http://netzreport.googlepages.com/onlin ... 18_47.html

And it came out to Goodluck! !kculdooG. Cool. No hints for the next sections though so it might not even be part of the real puzzle

For my own contribution, the last part is all either : or :::. I would imagine one would represent a 0 and one would represent a 1. Going to go through that later.

Also, to add, the OP of this puzzle gave up a few hints for solving.

Clue #1: The first puzzle involves several decryption operations (the second op was born in rome but lives on usenet)

Clue #2: http://i4.tinypic.com/6wzagye.jpg

Clue #3:http://i7.tinypic.com/6ks8u1k.gif

Clue #4:le chiffre indéchiffrable pour la solution

HI FOOF!!

Remember me?

Yea, your avatar gives it away =P