Infinite square grid of 1 ohm resistors extending in all directions.?

Question

You place two electrodes at corners in the knight position; one across and two along. What resistance do you measure between them and how do you get the result? This question was asked of job applicants by Google in an aptitude test.

Answer 1

I don't have an exact answer to this yet, but here are some things I can prove: Defining Rd as the resistance between two diagonally opposite points in a square, the relation between this and the knight move resistance Rk is:

Rk = 2Rd - 1/2

also if R2 is defined as the resistance between two points 2 resistors apart (i.e. horizontally or vertically)

Rk = 3/2 - R2

Since removing resistors always increases the resistance, calculating R2 for a finite subset of resistors gives an upper limit on R2 and therefore a lower limit on Rk. For example, consider a (2 x infinite) square grid extending vertically. The resistance between two points horizontally opposite each other 2 resistors apart, is (this can be done by recurrent application of the series and parallel rules for example):

2(sqrt(5)+1)/(5+sqrt(5))

Therefore:

R2 < 2(sqrt(5)+1)/(5+sqrt(5)) = .8944
-->
Rk > 3/2 - .8944 = .61

An upper limit on the resistance between two diagonally opposite points in a square can be calculated by just considering a 3x3 grid. The resistance between the two diagonal points in the center square is 5/7, so:

Rd < 5/7
-->
Rk < 2(5/7) - 1/2 = 13/14 = .93

So Rk is bracketed by the following limits (these could be improved by adding more resistors in the above two calculations):

.61 < Rk < .93

The actual value appears to be right at the middle of these two limits based on a computer program I wrote to calculate the answer using a simulated 32 x 32 grid (about 1000 resistors). Measuring the resistance between two points ((+1,+2) apart) near the center gave Rk = .776

Answer 2

This is a classic problem but i have a beef with it.

First of all the answer is not infinite.

To solve this problem in theory you find some sort of symmetry that will allow to ignore most of the resistors. And then you can find the total resistance of the remaining resistors. But the second you apply the two electrodes you will destroy that symmetry and you the answer you measure will be different than the theoretical value.

Answer 3

To answer previous poster. Applying the probes does not break the symmetry in the system if ample time is given for the field to equilabrate (this happens at the speed of light, so no biggie).

Furthermore, asking what the resistance is from a theoretical point of view (no probes, or perfect probes), is still a valid question.


It's a clever question. Definitely not very standard, because I haven't seen it before!

-T

Answer 4

The key word in this question is "infinite."

If there are infinite resisters in parallel to the probes (and there is) the resistance will be zero, irregardless of the series resistance. No matter how much the value of the series resistance increases, an INFINITE number of parallel circuits will provide a zero ohm path between the probes.

Answer 5

Apply a known DC current across two electrodes and measure the voltage or potential difference.If you aplied one ampere and voltage measured is 0.1 v then the resistance would be 0.1 ohm.