# Talk:356: Nerd Sniping

Just because the problem contains an infinite series (or parallel) doesn't mean that it's unsolvable. It's tricky, certainly, and getting the "true" answer involves some rather heavy math, but it's not impossible. Indeed, Google shows that it's already been answered. 76.122.5.96 20:42, 20 September 2012 (UTC)

I've always had an issue with this problem for one simple reason. In an infinite set of resistors, there is no space to apply a charge, thus there is no resistance. Ohm's law states Resistance = Voltage / I(current). So, in a system where there is no current (creating a divide by zero error), and there is no voltage (no change in electron work capacity, because we don't have a way to excite the electrons, because there is no power) Resistance is incalculable. lcarsos (talk) 22:22, 20 September 2012 (UTC)

We live in 3 dimensions, just place a battery above the grid with wires going to the 2 points. --84.197.34.154 22:59, 24 October 2012 (UTC)

- Not everybody does... --FlatlandDweller 11:08, 15 November 2012 (UTC)

This problem is "unsolvable" only if you try to just use the basic methods for finite networks.
There is a page on this at http://mathpages.com/home/kmath668/kmath668.htm that reports that the cited points have a resistance of **4/pi - 1/2** ohms (.773234... ohms).
The 1/2 ohm resistance between adjacent nodes is actually well known.
Divad27182 (talk) 05:05, 5 October 2012 (UTC)

Solution here as well: http://mathworld.wolfram.com/news/2004-10-13/google/ Potie15 (talk) 03:50, 18 March 2013 (UTC)

Nowhere it is said that the problem is unsolvable, just that it is interesting. Of course, the sniping is more effective is the problem is also difficult to solve, because otherwise the victim would get over it quickly. Dargor17 (talk) 17:47, 16 June 2013 (UTC)

That method for parallel resistors is wrong. You don't divide resistances by the number of paths, you sum the reciprocals and then take the reciprocal of that. The method described only works if every resistor has the same value. While that's true in this problem, it's misleading to pass that off as a method that works for all cases. --173.245.55.60 03:32, 1 April 2014 (UTC)

- Good point. I made some slight alterations to clarify that we are assuming the resistors are equal. It seems a better solution than getting into the more complex version of the problem. --BlueMoonlet (talk) 12:20, 1 April 2014 (UTC)