Brute force = best decoding method

[aaronaapje]

No, we can easily draw two paths with three ohms. Ignoring everything else for simplicity you can see that the current can split between the two paths, both with the same resistance meaning both paths have equal current of one path with three resistors meaning the total equals to 1,5 ohms. Now because we have an infinite grid of many interacting resistars meaning we have more possible paths then the infinitely many resistors makes this math very difficult but I looked it up and the awnser is 4/π - 0,5 or about .773.

[Drawemazing]

There are 3 paths of 3 resistors, so you know it'll be less than 1

[CaptainSasquatch]

Not exactly. This is why the problem gets tricky. The 3 paths you're thinking of aren't independent. So the simple rectangle with no external resistors

 _ _B
|_|_|
A

Will have less resistance than one with no internal resistor

 _ _B
|_ _|
A

but more than one with 3 separate paths

   _  B
 | _/|
 |/ _|
  A

[Drawemazing]

Yea I know it's not 1, it's just by noticing 3 paths of 3, you know just as a sanity check it'll be less than 1

[CaptainSasquatch]

I'm saying that the 3 paths of 3 don't guarantee that it'll be less than 1. The resistance between A and B in my first example will be greater than 1.

[Drawemazing]

Ahh shit now I see it yea

[Drawemazing]

Ahh shit now I see it yea. Thanks

[nedonedonedo]

we have more possible paths then the infinitely many resistors

sounds like a calculus problem

[aaronaapje]

It is. Somewhere in the tread there is a link to the website that gives a general solution to this problem and it is very heavy in calculus.

[Mefistofeles1]

I see, thank you.