Resistor Problem

There's a question on the Google Aptitude Test about resistors:

10. On an infinite, two-dimensional, rectangular lattice of 1-ohm resistors, what is the resistance between two nodes that are a knight's move away?

I don't think that the answer is as complicated as what people are making it, since every node is the same value. Here's what I think the solution is:

A knight's move is a combination of 2 squares in one dimension, and 1 square in the other dimension. So, if you imagine all of these 1-ohm resistors in a grid:

  ! ! ! ! ! ! ! ! ! 
 -+-+-+-+-+-+-+-+-+-
  ! ! ! ! ! ! ! ! ! 
 -+-+-+-+-+-+-+-+-+-
  ! ! ! ! ! ! ! ! ! 
 -+-+-+-+-+-+-+-+-+-
  ! ! ! ! ! ! ! ! ! 
 -+-+-+-+-+-+-+-+-+-
  ! ! ! ! ! ! ! ! ! 

then you're trying to find the total resistance from one intersection to another one that is, say, 2 down and 1 to the right (i.e., resistance from A to F):

  A-B
  ! !
  C-D
  ! !
  E-F 

Despite the fact that this is an infinite grid, all resistors outside of this area will not have any effect on the system between the points A and F.

Breaking down the network:

A-B-D = 2 ohms
A-C-D = 2 ohms

The total resistance from A-D, therefore, is 1 ohm because:

Rt = 1/(1/R1 + 1/R2) = 1/(1/2 + 1/2) = 1/1 = 1

D-F = 1 ohm

So, the total resistance following path of the network A-B-C-D-F is 2 ohms.

But, there is another network in parallel: A-C-D-E-F It's value is also 2-ohms.

In all, going from point A to F is like having two 2-ohm resistors in parallel, which we showed above to be 1 ohm total.

I'll probably build this circuit and take physical measurements, but as I type this, I think that the answer is 1-ohm.

I'd gladly admit defeat if someone who actually studied this in school has a different explanation. ;-)