356: Nerd Sniping

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

[edit] Explanation

Nerds have a way of getting distracted easily and focusing on one thing and ignoring the rest, when they feel their specific skills are challenged by an interesting problem. 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 and get crushed by a truck by showing them an interesting problem to solve.

The problem Black Hat shows is an electronics engineering thought experiment to find the 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 one-ohm 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 equal 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 ohms, or about 0.773 ohms. See [1].

Black Hat's final comment that "physicists are two points, mathematicians three" indicates that he considers a mathematician to be a more difficult target for his game than a physicist would be. It is unclear whether this is meant as a dig on physicists or on mathematicians; it might be because physicists are interested in a wider range of problems, or because mathematicians require a higher-quality problem to hold their interest.

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. Randall explained in a speech at Google five days before this comic, that he was nerd-sniped, in a way, by that problem in this test, and got quite irritated when he ultimately found that it was actually a modern physics research problem, requiring very advanced math, far more complicated than the other puzzles it figured among.

[edit] Transcript

[Black Hat is sitting on a chair, Cueball is standing next to him. Across the street another man is coming from a building.]
Black Hat: 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.]
Black Hat: This has led me to invent a new sport: nerd sniping. See that physicist crossing the road?
[Black Hat holds up a sign.]
Black Hat: 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.]
Sign: 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.]
FOOOOM
Cueball: I will have no part in this.
Black Hat: C'mon, make a sign. It's fun! Physicists are two points, mathematicians three.
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Discussion

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)
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