Re: Resistor grid puzzle

"The Phantom" <phantom@xxxxxxx> wrote
On Tue, 5 Aug 2008 10:09:10 +0100, "Fleetie" <fleetie@xxxxxxxxxxxxx> wrote:

From a quick look, I make it:

R = Sigma (n=1 -> n=(N-1)) { 1/n }

N is the number of nodes along a length of the square.

So for N=4 (a 4 x 4 grid):

R = 1 + 1/2 + 1/3 = 11/6 ohms.

For N=5:

R = 1 + 1/2 + 1/3 + 1/4 = 25/12 ohms.

I got that by drawing diagonal lines; I think all points on each
diagonal line are at the same potential,

Did you try to verify this assumption?

No. :-)

When I posted that this morning, I made the assumption based
on intuition.

This evening, I looked at it again, and arrived at another
approach, which is less based on guesswork and more rigorous
(I think!).

It relies on the symmetry of the grid about the diagonal line
AB (referring back to your diagram).

Take the N=3 case (and view in Courier or similar font):

a b c

d e f

g h i

The nodes are at positions a to i, and the 1 ohm resistors
link them on a square grid.

It is obvious from symmetry that b,d are at the same potential,
and f,h, and c,g.

So we can eliminate half of the resistors and nodes b,c,f and
change the remaining resistors from 1-ohm ones to 0.5-ohm ones.
This just exploits the symmetry about the line a-e-i above.

That leaves us with:

a

d e

g h i

Now all resistors are 0.5 ohms.

Again, it is obvious that e,g are at the same potential, so
we use the same trick again on d,e,g,h and remove node e and the 2 0.5-ohm resistors d-e, and e-h, and make resistors
d-g and g-h into 0.25-ohm ones.

So now we have:

a

d

g h i

Resistors a-d, and h-i are 0.5 ohms for a subtotal of 1.0 ohms.

Resistors d-g, and g-h are 0.25 ohms for a subtotal of 0.5 ohms.

Therefore resistance a-d-g-h-i is 1.5 ohms.

This is the same value as that predicted by my earlier summation
formula, i.e. 1/1 + 1/2 = 3/2 ohms.

I have not examined higher cases this evening.

You have me interested now.


Martin

.

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