Explain xkcd: It's 'cause you're dumb.
Nerds have a way of getting distracted easily and focusing one thing and ignoring the rest. Black Hat has decided to make this into a disturbing game of getting nerds, in this case a physicist, to stop in the middle of a street by showing a problem for the victim to solve.
The problem Black Hat shows us is an electronics engineering thought experiment to find resistance between two points. In normal wiring, a one ohm resistor would result in one ohm of resistance. Two resistors connected in a series, where electricity has to go through each has two ohms of resistance. Two resistors in parallel give the circuit only half an ohm since you average the resistance of the path (1 ohm of resistance over 2 paths).
With an infinite grid of resistors, you have an infinite number of paths to take, and for each path an infinite number of both series and parallel paths to consider, so much more advanced methods are needed. The exact answer to the question is 4/π − 1/2.
The Google Labs Aptitude Test is a collection of puzzles published by Google as a parody of tests such as the SAT. Google is known for using logic & math puzzles in their job interviews.
- [Hat Guy is sitting on a chair, the Normal Guy is standing next to him. Across the street another man is coming from a building.]
- Hat Guy: There's a certain type of brain that's easily disabled. If you show it an interesting problem, it involuntarily drops everything else to work on it.
- [The man across the street is about to enter a crosswalk]
- Hat Guy: This has led me to invent a new sport: nerd sniping. See that physicist crossing the road?
- [Hat Guy holds up a sign]
- Hat Guy: HEY!
- [There is an image of a grid with resistors on every connection, two nodes a knight's move apart are marked with red circles.]
- The sign reads: On this infinite grid of ideal one-ohm resistors, what's the equivalent resistance between the two marked nodes?
- Physicist on the street: It's... Hmm. Interesting. Maybe if you start with... No. Wait. Hmm... You could--
- [A truck is zooming past, apparently where the physicist just stood]
add a comment!
- Normal guy: I will have no part in this.
- Hat Guy: C'mon, make a sign. It's fun! Physicists are two points, mathematicians three.
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. 18.104.22.168 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. --22.214.171.124 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. --126.96.36.199 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)