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The resistance of n*n grid

We have a grid of N*N resistors, like this:

http://zlateski.com/static/_media/6/5/1/2/hnybded53ou83bu73n475000000284m350.jpg

And we want to calculate the resistance between A and B (picture).

Each resistor is 1 ohm...

What formula for n can I use? Or what algorithm?
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Rok-Kralj
Asked:
Rok-Kralj
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1 Solution
 
ozoCommented:
It may be easier to decompose it into serial and parallel components if you connect points at the same potential,
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Rok-KraljAuthor Commented:
I'm not sure if I understood you well...
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ozoCommented:
if you know how to find the resistance of a combinations of resistors in series  or parallel,
you may be able to convert the grid into one that you can analyze more easily by
adding connections which do not change the resistance of the grid because there will be no current trough the added connections because they connect points with the same voltage.
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WaterStreetCommented:
Unless you tell us it's not homework, I'm just going to show how to get the answer, rather than giving it to you.

Take a look at http://www.geocities.com/frooha/grid/node2.html for an infinite grid of 1-ohm resistors.  The end result gives you the resistance across the diagonal of any square within the grid.  I'll call a mini-grid.

Your problem is find the total resistance across the major diagonal of the entire grid composed of all the mini-grids.

Clue:
Isn't that the sum of the mini-grid diagonals within it?  
You already know the resistance across one of the mini-diagonals.
Now you know enough to come up with a simple algorithm  for the resistance a n by n grid of 1-ohm resistors as a function of the number "n".
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Rok-KraljAuthor Commented:
Thanks anyway... Unfortunately I wasn't able to help myself much with it... It was a part of some programming competition (getting as much help as possible is allowed)... Next time it would be better to give direct answer, because I'm a programmer, not physician.
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WaterStreetCommented:
Since you're saying outside help is allowed...

R of the major diagonal for 1-ohm resistors approaches 2 n / pi where n is the number of squares in the side of a n by n grid.
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ozoCommented:
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WaterStreetCommented:
in what way?

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Rok-KraljAuthor Commented:
ozo, you are right... It is a completely different problem.
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