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)
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.
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?
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.
Jerome said
Joe,
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.
Ozzy said
Wouldn’t the resistance drop Asymptotically to approximate zero?
dugan said
The answer is 2/pi