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I searched and couldn't find it on the site, so here it is (quoted to the letter):

On this infinite grid of ideal one-ohm resistors, what's the equivalent resistance between the two marked nodes?

Nerd Sniping

With a link to the source.

I'm not really sure if there is an answer for this question. However, given my lack of expertise with basic electronics, it could even be an easy one.

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Answer: number of all spanning trees containing an edge between the marked points on the grid graph with said edge attached divided by the number of all spanning trees on the original grid graph (as a limit of grid size going to infinity). I am going to sleep and hope I'll find out in the morning that someone will have counted those trees :-) One way to do so is as determinant of a certain matrix associated with the graph or else as a certain correlator of the $q \to 0$ limit of an associated Potts model on the given graph (with and without edge between the marked points). – Marek Dec 20 '10 at 0:39
@Mark C This question could hardly have been answered accurately in a high school text. Perhaps it asked about two adjoining nodes? – Mark Eichenlaub Dec 20 '10 at 3:08
I instantly recognized the title from XKCD [Nerd Snipping is one of my favorites]. – Mateen Ulhaq Dec 20 '10 at 6:36
Discussion on meta: – Marek Dec 20 '10 at 8:35
The question @ m.SE... – user172 Dec 20 '10 at 10:49
up vote 29 down vote accepted

Nerd Sniping!

The answer is $\frac{4}{\pi} - \frac{1}{2}$.

Simple explanation:

Successive Approximation! I'll start with the simplest case (see image below) and add more and more resistors to try and approximate an infinite grid of resistors.

Simulation results

Mathematical derivation:

The well-known result from which all the other resistances are easily computed.

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+1, but it would be even better to outline the solution in the post so that people don't have to click a link to see how it's done. – David Z Dec 20 '10 at 2:12
The stuff on that math link is pretty complicated... Too much for mere inhuman lifeforms such as me. – Mateen Ulhaq Dec 20 '10 at 6:39
Yeah, it took me a few readings to figure out how it was done. (That's what makes it "fun" :-P) – David Z Dec 20 '10 at 7:22
@Sklivvz: regardless, we should have an explanation and not just a link in the answer. (For your answer as-is I think I may have been too quick to click the upvote button) – David Z Dec 20 '10 at 9:15
@kalle: Kirchhoff's law is what I mentioned in my comment above. You'll obtain an infinite-dimensional matrix and you'll have to compute it's determinant. Or you can use various dualities that connect resistor network with models in statistical physics. Nevertheless, I very much doubt any possible method is in any way easy. You'll definitely need to do Fourier transform or non-trivial integrals (as in the Sklivvz's link) at some point to obtain the results. So you are saying that you obtained something simple that can beat these established methods? I can't say I don't doubt you ;-) – Marek Dec 20 '10 at 18:05

This is a classical problem. The trick is to use symetry whenever it's possible and to connect the symmetric points since they have the same potentiel. First diagonals, etc...

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No, this is why it snipes, it isn't reducible by symmetry like the adjacent vertices resistance. – Ron Maimon Jul 23 '12 at 14:41
Don't bother trying, the value is transcendental, it can't reduce by symmetry to a finite grid, this would necessarily give a rational answer. The answer is the inverse of the lattice Laplacian on non-adjacent points, which can be done in k-space, you get a sinusoidal factor in the numerator and denominator, and only for the resistance between adjacent vertices do the two factors cancel after symmetrization. – Ron Maimon Jul 23 '12 at 16:13
@Shaktyai If you wish to delete this or any other of your own answers, use the "delete" text that sits under the left hand side of the answer text. – dmckee Jul 24 '12 at 18:01

protected by Qmechanic Jan 23 '14 at 20:46

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