Is it 0.916333... ohms?
Damn, I missed dinner!
This http://arxiv.org/abs/cond-mat/9909120" [Broken] (see appendix A) gives an absolute mess.
I actually did use a superposition - I don't know why it fails! =(How did you get that? I think that to solve such problems (where a simple superposition doesn't give you the answer), a discrete fourier transform is required, and then the final integral which gives the resistance looks very tricky.
From "xkcd.com"[/URL] :
A friend of mine finished school some years ago with a BS in physics and immediately chucked it all to become a lawyer. After seeing this Nerd Sniping XKCD strip I sent it to him on a workday to see if he was truly a lawyer or if he would be sniped, along with with full disclosure of my intentions. Couple days later I checked back, sure enough, four billable hours went down the drain. :devil: Good thing he wasn't standing in the road.
BTW, engineer's solution: assume the infinite grid looks like a short to any one node except for the four resistors directly attached into the node. The parallel resistance into the node then is 1/4 ohm and therefore the resistance between any two, non-adjacent nodes is 1/2 ohm. With actual (cheap) 20% resistors thats close enough to the actual (.7xx is it?)
Yes, ok, I'll have to make clear I'm using really cheap resistors when I go for the Google interview.The resistance betweeb any two horizontally or vertically adjacent nodes is exactly 1/2 ohms. The resistance between nodes further apart than that is not. That 20% fudge on top of 0.5 ohms yields 0.6 ohms, not 0.773 ohms.
Think I'll code up a quick sim for fun.Its even worse for nodes further apart than that.
Why don't you try working out the answer for nodes separated by two diagonal hops?