What is the resistance between two points a knights move away on a infinite grid of 1-ohm resistors

On an infinite grid of ideal one-ohm resistors, what's the equivalant resistance between two nodes a knights move away?

(please fix the tags, I didn't really know where to put it)

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Does anyone else think of Mass Ave. and MIT when looking at the cartoon? –  Pete L. Clark Dec 3 '10 at 5:35
You might also awnt to ask this on physics.stackexchange.com –  Frédéric Grosshans Dec 3 '10 at 10:08
Has anyone generalized this to a triangular hex-lattice yet? –  deoxygerbe Dec 4 '10 at 23:46
It's really a combinatorial problem more than a physics problem. –  Raskolnikov Dec 21 '10 at 14:19
A mathematician named David Ingerman has published lots of things about problems related to this one. (Once he ran afoul of Wikipedia's policy forbidding original research by trying to make one of his preprints into a Wikipedia article. I met him in Massachusetts years ago.) If you like this sort of thing, google his name. ${}\qquad{}$ –  Michael Hardy Jul 12 at 2:01

After Google asked this in their aptitude test, this has become a famous problem.

You can find a nice discussion and a more general solution here: http://www.mathpages.com/home/kmath668/kmath668.htm

I believe the answer for your case is $\displaystyle \dfrac{4}{\pi} - \dfrac{1}{2}$.

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I think it's been a famous problem a lot longer than that. (-: –  Tobin Fricke Dec 3 '10 at 18:29
@TObin: I guess... Hard to debate either way :-) –  Aryabhata Dec 3 '10 at 18:41

The good answer is indeed $\frac{4}{\pi}-\frac{1}{2}$. You can find a complete solution in the book of R. Lyons and Y. Peres "Probability on trees and networks", section 4.3, p. 124-127. This mainly uses Fourier analysis and the symmetry of the grid.

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