The question is "On this infinite grid of ideal one-ohm resistors, what is the equivalent resistance between the two marked nodes?"

n here's the picture: http://xkcd.com/356/

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 Recognitions: Homework Help Science Advisor Arghhhh don't bring that here!!!!!!! Interestingly the answer doesn't seem to be independant of the distance between the two test points. One of the odd features of conductive anti-static mats is that the resistance between any two points is the same (the matts are assumed to be infinite) - it must be because of the discrete number of paths in the resistor case
 Recognitions: Gold Member Science Advisor Staff Emeritus laurelelizabeth, Are you asking us to solve it? It's actually really easy; just use superposition. - Warren

 Quote by chroot laurelelizabeth, Are you asking us to solve it? It's actually really easy; just use superposition. - Warren

Really easy?
It seems you can solve it in 4 or 5 lines!
...I'd like to see how...

 Quote by chroot laurelelizabeth, Are you asking us to solve it? It's actually really easy; just use superposition. - Warren

Hey Warren, wake up !

 Let me try... Total resistance = 1 ohm?

 Quote by kudoushinichi88 Let me try... Total resistance = 1 ohm?
I think it is wrong, but, in fact, I don't care about the number - I just wanna know HOW TO GET IT.

Chroot had told us it could be done easily, by using superposition.

 Recognitions: Homework Help Science Advisor The case for an exact diagonal is quite easy. R(n,n) = 2/pi * Sum{k=1...n: 1/(2*k-1)} R(1,1) = 2/pi R(2,2) = 2/pi * (1 + 1/3) R(3,3) = 2/pi * (1 + 1/3 + 1/5) For a "2 accross 1, up 2" path as in the cartoon I'm not sure, but it must lie between 2/pi and 1.33*2/pi There is a page of references for the calculation - you will probbaly need an academic subscription to get the text http://www.physics.thetangentbundle....istive_lattice
 Lol~ my answer was a wild guess. Sorry. I think you'd better conduct a lab experiment and see the results...
 Recognitions: Homework Help Science Advisor You can write something in python/mathematica to try a 100,100 array and get a good approximation. Then fiddle around with 2/pi and a constant to work out the 'correct' answer.

 Quote by mgb_phys The case for an exact diagonal is quite easy. R(n,n) = 2/pi * Sum{k=1...n: 1/(2*k-1)} R(1,1) = 2/pi R(2,2) = 2/pi * (1 + 1/3) R(3,3) = 2/pi * (1 + 1/3 + 1/5) For a "2 accross 1, up 2" path as in the cartoon I'm not sure, but it must lie between 2/pi and 1.33*2/pi There is a page of references for the calculation - you will probbaly need an academic subscription to get the text http://www.physics.thetangentbundle....istive_lattice
I don't get it at all. How did you come up with that? I am not familiar with those math at all...

 Recognitions: Homework Help Science Advisor Derivation stolen from http://www.geocities.com/frooha/grid/node2.html
 Oh my... I never thought this thing is really that complicated...

 Quote by mgb_phys Derivation stolen from http://www.geocities.com/frooha/grid/node2.html
Thanks a lot, mgb_phys !!!

And it is not that easy Warren had said...

Well, who knows, maybe CHROOT could explain it in a other way, using superpostition...

 Whoosh that all went over my head :P I hope that I didn't annoy to many people by posting that there