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 Infinite Grid of Resistors.
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.
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)
- baDumpBump! 126.96.36.199 16:22, 28 June 2018 (UTC)
- I believe the OP is referencing the issue that an infinite circuit could not hold a current. Connecting a battery would only work for a finite grid. In addition, the orientation of the battery in physical space has no relation to its behavior in a circuit, only the points of connection matter. Think about what the battery is doing to generate a current. How does electric potential apply over an infinite grid? Even moving it through a magnetic field won't work as the flux will be uniform across each cross section. You can't rotate an infinite grid either... -- Flewk (talk) (please sign your comments with ~~~~)
Just crocodidoodle the battery to the pencil lines as and where required for an infinity of varieteediddly. I used Google News BEFORE it was clickbait (talk) 18:51, 20 January 2015 (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 if 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. --188.8.131.52 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)
The real question is: why did the physicist cross the road? --Alcatraz ii (talk) 00:53, 29 September 2014 (UTC)
Amazing. From the first comment the discussion is diverted from discussing the comic, to discussing the problem presented in the comic. The commentators have been nerd sniped by a demonstration of nerd sniping. Randall is just that good. 184.108.40.206 17:55, 30 April 2014 (UTC)
"Sniping" might also be a pun or have a deliberately dual meaning in this context, referring to both a sniper and a snipe hunt (do kids still practice the latter?). The former makes sense if Black Hat's purpose is to actually rid the world of physics and math nerds (consistent with his characteristic misanthropy and cynicism), but the latter also fits the theme of merely distracting a nerd with an impossible task, which the title text suggests may have been Randall's motivation for the strip. (On a side note, the Wikipedia article reveals that the terms sniper and snipe hunt have a common origin, which makes twice in the last month it's resolved a long-standing etymological puzzle for me. The other case united the multiple, seemingly unrelated meanings of minute ["tiny" vs. time] and second [ordinal vs. time]; see sexagesimal.) 220.127.116.11 01:40, 18 June 2014 (UTC)
I've been led to believe that 'minute' means 'tiny amount of time', 'second' is 'secondary tiny amount of time', and , I quote "Real snipe (a family of shorebirds) are difficult to catch for experienced hunters, so much so that the word "sniper" is derived from it to refer to anyone skilled enough to shoot one." from the snipe hunt wiki page. 18.104.22.168 23:45, 27 September 2014 (UTC)
Why doesn't someone solder together a thousand one ohm resistors into a grid then use an ohmmeter to measure the resistance? Then repeat with smaller and smaller grids to see if there's any effect on the measurement. If the resistance does not change, or at least doesn't change until the grid size gets quite small, then the "infinite" term in the problem is a 'red herring' to mislead. Pointless, useless, irrelevant etc information in problems is a common tactic for gauging the ability to recognize and reject such data. 22.214.171.124 00:35, 18 November 2014 (UTC)
Incidentally, should this page mention that what if 113 (I don't know how to do links, sorry) contains a picture of this comic? 126.96.36.199 23:36, 24 February 2015 (UTC)
- Yes I will do so. Have just referred to another what if where he is mentioning nerd sniping. --Kynde (talk) 11:40, 16 February 2016 (UTC)
188.8.131.52 12:17, 22 May 2015 (UTC) Am I the only one concerned with the fact that this poor guy was still on a crosswalk? The truck should have stopped. 184.108.40.206 12:17, 22 May 2015 (UTC)
- No you are not, and good point --Kynde (talk) 11:40, 16 February 2016 (UTC)
When the number of parallel resistors increase, the equivalent resistance decreases. So, in an infinite grid, wouldn't it approach zero? UrubuSelvagem (talk) 03:43, 28 September 2015 (UTC)
- They are also in series. For each parallel group, there is, in fact a corresponding group in series.
It is not directly relevant to the discussion of the comic, but this needs to be posted here. Perhaps the best nerd snipe ever actually achieved and a nearly perfect match for the comic (my professor put it in the lecture notes for my group theory class):
"Coxeter came to Cambridge and he gave a lecture, then he had this problem ... I left the lecture room thinking. As I was walking through Cambridge, suddenly the idea hit me, but it hit me while I was in the middle of the road. When the idea hit me I stopped and a large truck ran into me ... So I pretended that Coxeter had calculated the difficulty of this problem so precisely that he knew that I would get the solution just in the middle of the road ... One consequence of it is that in a group if a^2=b^3=c^5= (abc)^-1, then c^610=1."
(J.H. Conway, Math. Intelligencer v. 23 no. 2 (2001))
I did a search, and the entire passage can be read here perhaps it is even possible that this event is the inspiration for this comic? The inclusion of the "large truck" is almost too perfect. 220.127.116.11 23:45, 2 October 2015 (UTC)
- I have now added this story in a new trivia section. --Kynde (talk) 11:40, 16 February 2016 (UTC)
I know a solution that use random walks. :) 18.104.22.168 (talk) (please sign your comments with ~~~~)
I really like this comic. It says a lot about Black Hat, but so much more about Randall :-) --Kynde (talk) 11:40, 16 February 2016 (UTC)
So, *that's* how they did Gaudi in! I always suspected a plot; now I see the method. 22.214.171.124 16:30, 28 June 2018 (UTC)