storing bits of the internet away for a rainy day

Nerd Sniping / Infinite Grid of Resistors

Posted by squirreling on December 22, 2007

 The problem: (comic from xkcd, problem taken from the GLAT, google labs aptitude test)

Nerd Sniping 

The solution method: (this is the solution for the diagonal resistance, but the knight’s move resistance can be calculated from their formulas at the end)

Inject a current into the grid at the two points (1A into (0,0) and -1A into (2,1)).  By superposition, consider only the 1A injected at (0,0).  By Kirchoff’s current law all of the current entering the node must leave the node through the four resistors attached at (0,0).

I(in) = I(out)

I(0,0) = I1 + I2 +I3 + I4

I(0,0) = (V(0,0)-V(0,1))/1 ohm + (V(0,0)-V(1,0))/1 ohm + (V(0,0)-V(0,-1))/1 ohm + (V(0,0)-V(-1,0))/1 ohm

I(0,0)=4*V(0,0) – V(0,1) – V(1,0) – V(0,-1) – V(-1,0)

Perform Fourier Transform on this equation, and solve.


13 Responses to “Nerd Sniping / Infinite Grid of Resistors”

  1. Joe said

    Not entirely sure what’s wrong with my approach here, but I’m taking a highly conceptual view of the problem. In every case of ideal resistances, every new path will always reduce the overall equivalent resistance. For example, two 1 Ohm resistors in parallel will be the equivalent of 1/2 Ohm, but even the combination of a 1 Ohm resistor and a 1 MOhm resistor will give an equivalent resistance below 1 Ohm, albeit a very tiny amount less than 1. However, with an infinite grid, there are infinite paths and infinite combinations, therefore their equivalent parallel resistance must asymptotically approach 0. I believe a sum of conductances will confirm this as the denominator will grow significantly faster than the numerator due to the increased resistance of the longer paths. As the denominator approaches infinity, the numerator becomes insignificant in comparison and the equivalent resistance is 0. What am I missing? Anyone wanna spice this with 100,000,000 resistors to see what happens? 🙂

  2. squirreling said

    Just because there is an infinite number of paths, doesn’t mean that the resistance must approach 0. Think in math, you can have a infinite series which converges to some number. Ie: 1 – 1/2 + 1/3 – 1/4 + … = ln 2 (convergent series.

  3. Jerome said


    two points about your analysis.

    any point on the grid can be seen as being connected by 4 x 1ohm resistors in parallel this would give you a lower bound for the result.

    Secondly, although there are an infinite number of paths between the two points… an infinite number of them will have an infinite resistance… ho ho.

    You could look at this a bit like a finite element analysis, so I suppose that rather than try to hook up 100,000,000 resistors, you could find a material with a clearly defined resistance per meter, get a largish sheet of it and test the resistance close to the center.

  4. Ozzy said

    Wouldn’t the resistance drop Asymptotically to approximate zero?

  5. dugan said

    The answer is 2/pi

    • David said

      Hmm I get that it’s Int[x^-1 *[1-((x-1)/(x+1))^4]/2Pi , x = 0 to infinity from a 2D Fourier transform argument. That integrates to -8 (2 + 3 x (1 + x))/2Pi(3 (1 + x)^3) evaluate that
      at 0 and infinity (where its 0) you get 8/3Pi. I think this is right, its close to your answer anyway

  6. Eoin said

    if the grid is truely infinite than there is no place to attach the power source so there would be no electricity for the resistors to resist.

  7. TC Breezy said

    You people have completely overloooked the fact that an infinite number of resistors would be impossible. Geeks get so wrapped up in the theory of a problem that they overlook the obvious.
    It takes a true genius to point it out, thank you!!!

  8. Holger said


    your argument is true if each path is completely separate from any of the others, each connecting the two points and not sharing the road with any of the other paths.

    In this problem on the grid, each path has to reach the point through one of the four resistors attached to it, so they are not all separate from each other. They have a lot of overlap.


  9. BlueRaja said

    @Jerome: If there are infinite number of paths with infinite resistance, it should be easy for you to find me one 🙂

  10. MaZ anggie said

    nice info,thaks.

  11. Kate said

    I actually tried this out at school (only not on a zebra crossing) and managed to confuse most of the science and maths department.
    One science teacher reckons the answer is 0, this right?

  12. TheDude said

    You all just got hit by a truck.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: