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

I first saw this problem on the Google Labs Aptitude Test.  A professor and I filled a blackboard without getting anywhere.  Have fun.

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

  • 8
    $\begingroup$ Does anyone else think of Mass Ave. and MIT when looking at the cartoon? $\endgroup$ Commented Dec 3, 2010 at 5:35
  • $\begingroup$ You might also awnt to ask this on physics.stackexchange.com $\endgroup$ Commented Dec 3, 2010 at 10:08
  • 2
    $\begingroup$ Has anyone generalized this to a triangular hex-lattice yet? $\endgroup$ Commented Dec 4, 2010 at 23:46
  • $\begingroup$ It's really a combinatorial problem more than a physics problem. $\endgroup$ Commented Dec 21, 2010 at 14:19
  • 1
    $\begingroup$ 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. ${}\qquad{}$ $\endgroup$ Commented Jul 12, 2015 at 2:01

2 Answers 2


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 $\displaystyle \dfrac{4}{\pi} - \dfrac{1}{2}$.


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.


Not the answer you're looking for? Browse other questions tagged .