Infinite resistor network

Google "infinite resistor grid problem" It's a fairly common puzzle.

Dave.

On Aug 12, 1:17=A0pm, "David L. Jones" :

Um, sorry.

And the trick turns out to be that there's no trick - no way of looking at the problem that makes the answer obvious.

Sylvia.

You post the question on Usenet.

It's called "The Eternal September". Google it.

Tim.

On Aug 12, 2:43=A0pm, Sylvia Else

Yes, that's what I understand reading these papers :

Proper attribution and context:

Good grief! Is this the nerd's version of Fermat's Last Theorem?

I get a whole bunch of hits on the question, but it looks like NOBODY knows the answer????!?!?!!???

Or are they keeping it a secret (a la magic tricks), just to drive an infinite number of nerds crazy? ;-)

Thanks, Rich

But, has ANYBODY ever solved it? It can't be intractable, can it?

If somebody's solved it, why don't they just publish the answer, or is it some big secret?

Thanks, Rich

Good Freaking Grief!

So, has anyone ever worked through all that and actually come up with AN ANSWER?????

Thanks, Rich

It is official forum?

;-) Rich

Watch out for trucks.

-- Les Cargill

It's the one Randall links to, yes.

Back in 2006 we discussed it here and I posted a short BASIC program to perform a numerical solution. The concept underlying that program IS EASY to understand and it makes the numerical approach obvious.

The essential idea is to imagine a sufficiently large matrix (to get close to the concept of 'infinite' for the purposes at hand.) Each matrix element represents a node of that resistor grid. That much is obvious and very easy to fathom. The rest isn't hard, either.

At this point, initialize every node at the same voltage. Doesn't matter what voltage. I use 0.5V as my arbitrary number. Pick any you want. This is the 'steady-state' assumption with no imposed voltage and no currents flowing anywhere. (All nodes at the same voltage means no current.) So far, still easy.

Now, impose a voltage difference at the two points of interest (conveniently placed near the center of your grid so that the finite perimeter is roughly equally far away in all directions) and allow the other nodes' voltages to settle. The approach to doing that is very simple. I'll get to that in a moment.

Other than the two nodes of interest, process all grid nodes where the sum of the indices, i+j, is even. Then process all grid nodes where the sum i+j is odd. Then repeat, until the node voltages don't change "much" anymore.

At this point, the resistance can be easily computed from the assumption that once current flows have settled down and the node voltages are stable that the effective resistance is now obvious from the four surrounding, now-stable node voltages and the currents they produce in aggregate. Sum up the four current flows, divide that into the impressed voltage and the result must be the effective resistance.

The following obviously relies on the fact that all R values in the grid are equal, not varying all over the place (which could also be accomodated, but isn't in the following):

(a) create a matrix V[] to represent the node voltages (b) arbitrarily initialize the matrix node values to some value Q/2 (Q can be any real number, not 0.) (c) select some near-central point A, at V[Ax,Ay] (d) select the other desired near-central point B, at V[Bx,By] (e) initialize matrix element V[Ax,Ay] to 0. (f) initialize matrix element V[Bx,By] to Q. (g) DO (h) for all nodes where i+j is even, other than V[Ax,Ay] and V[Bx,By] and edge nodes (the perimeter of the grid), set the new node value to the mean of the four surrounding four nodes. (i) for all nodes where i+j is odd, other than V[Ax,Ay] and V[Bx,By] and edge nodes (the perimeter of the grid), set the new node value to the mean of the four surrounding four nodes. (j) LOOP until 'no significant change' is satisfied.

Now just add up the currents to the four nodes surrounding V[Ax,Ay] and divide this into Q.

R is arbitrarily for purposes of computing the current and the resulting resistance. The currents will be (V[Ax-1,Ay]-V[Ax,Ay])/R, (V[Ax+1,Ay]-V[Ax,Ay])/R, (V[Ax,Ay-1]-V[Ax,Ay])/R, and (V[Ax,Ay+1]-V[Ax,Ay])/R. This works out to

Isum= (V[Ax+1,Ay]+V[Ax-1,Ay]+V[Ax,Ay+1]+V[Ax,Ay-1]-4*V[Ax,Ay])/R

Dividing that into Q provides:

Reff= Q*R/(V[Ax+1,Ay]+V[Ax-1,Ay]+V[Ax,Ay+1]+V[Ax,Ay-1]-4*V[Ax,Ay])

If R=1 for the grid and if you followed the above instructions to also set V[Ax,Ay]=0V, then it's just:

Reff= Q / (V[Ax+1,Ay]+V[Ax-1,Ay]+V[Ax,Ay+1]+V[Ax,Ay-1])

Assume (Ax,Ay)=(0,0) and a grid that spans 30 or more elements in all four directions from there, you get pretty close results. For R=1 for example, I get:

(Bx,By) R True

------- ------ ------------- (1,1) .63646 2/pi (1,2) .77284 (8-pi)/(2*pi)

Larger grids will get closer.

Jon

Are you sure?

Know what diminishing returns are?

For all intents and purposes?

Try #3 on the list of google hits for the exact term I gave.

The standard question:

Dave.

All of those solutions are wrong.

Looked at the "responses" that google gives; most are useless bush beating. One idiot gave 9 ohms for the "answer"; another idiot gave zero ohms. It is obvious the correct answer MUST be less than 3 ohms as the shortest path is thru 3 one ohm resistors.. One analysis gave 0.5 ohm total by way of current into one node is split 4 ways hence 1/4 ohm and current out of the other note is also split 4 ways hence 1/4 ohm giving 0.5 ohm total. Not too bad for first approximation attempt and possibly gives the lower limit.

Unfortunately, the first reference does NOT show the same problem, and the second reference gives no solution. Back to square one.

The solution to the resistance between adjacent nodes is given here

as 1/2 (using R = 1).

The solution to the resistance between opposite nodes is given in the first of your references as 2/Pi.

Which just leaves the question of how they should be combined. Take the relevant lattice points to be

A B C

D E F

so that we want the resistance betwee D and C.

If we hold E at 0 volts, then to get some current I to flow out of it we can apply a voltage of I/2 to D.

If instead we want a current of I to flow into E, we can set the voltage at C to -I * 2/Pi volts.

If we set the voltages at D and C simultaenously, then the currents in and out of E cancel, and we see current I going into D and out of C, with the voltage between them of I/2 + I * 2/Pi.

The resistance between D and C is therefore 1/2 + 2/Pi. In otherwords, the equivalent resistances behave as if they're in series.

Sylvia.