But… has he paid?
As an aside, have we ever figured out that resistor problem? It’s easy* to prove that it’s 0.5 ohm between two adjacent nodes, and it’s easy* to prove that on an isotropic and homogeneous two-dimensional resistive surface, the resistance is independent of the distance between the endpoints, so my guess would be that the resistance between any two nodes on that network will also be 0.5 ohm… but of course, a grid isn’t isotropic, so that’s not actually a proof.
*For two different definitions of “easy”. The second is a straightforward integral, while the first takes no calculus but needs a clever trick.
Oh, you beastly rotter, I’m going to start thinking about that now!
As an easier starter: what is the resistance between the opposite vertices of a cube?
That one’s easy: 5/6 of an ohm (assuming 1 ohm resistors on every edge). Put an amp through it, and it’s easy to see by symmetry how much current is going through each individual resistor, which means it’s easy to calculate the voltage drops, and then take the total voltage drop and divide it by your 1 amp. That’s a similar technique to what you use for the adjacent nodes on the infinite grid, except that also has a superposition step.
@Chronos a couple posts up …
I’m pretty darn sure we here did not solve it. I am sure I personally never saw the solution here if there is / was one.
IIRC the nice folks on the now shut-down XKCD discussion boards had not done so either, but did have some great cites to other speculations and partial solutions. I have not checked whether current ExplainXKCD has anything on this one that predates their existence.
My own prejudice accords with yours. 0.5 ohms between any two nodes. For a “big enough” value of infinity the granularity of resistor-wire-connection-wire-resistor … just turns into the same resistive “fabric of space” as would an idealized infinite field of carbon or copper or iron atoms.
That’s pure arm-waving on my part, but I struggle to justify any contrary notion. Discrete math and continuous math are certainly not interchangeable, but they often rhyme if you’ve got the right POV on your problem.
Right. I was once asked this as an interview question, and the questioner was surprised I got it quickly.
I think he was expecting me to get bogged down in Kirchoff law calculations.
I visualised it geometrically: there are 3 logically equivalent nodes: two at each vertice, with 3 resistors from each point. And then the middle 6, which are all at the same potential by symmetry and have 6 connections between them.
I think there is a fairly elegant proof for the infinite grid case too, but it’s after dinner and I’ve had a glass or two of wine. I’ll probably remember it tomorrow?
You armchair physicists are funny. Why not just build an infinite grid of ideal one ohm resistors and test it?
We’ve put in the research proposal. Waiting for the grant decision…
The key question: might it have military applications?
Considering what happens when you assemble an infinite mass, it’s a darn good bet something is gonna get wrecked along the way. All we need is a way to aim it. 
On the OP’s original question, I failed to understand the complexity of the answers.
If 6 units take 8 hours to do something then 1 unit will take 6x8=48 hours. The dimensions or nature of the project are irrelevant. I did this in my head while reading the rest of the post.
So nine women can make a baby in one month?
If it takes one woman to construct one with Lego, then - yes.
Human
angrily
Where did you learn to talk like this?
AI
sobbing
I learned it by watching you!
/1980s public service message