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Topic: Resistors (Read 669 times) |
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ThudnBlunder
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On an infinite 2D rectangular lattice of 1-ohm resistors, what is the resistance between any two nodes that are a knight's move apart?
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| « Last Edit: Oct 15th, 2004, 6:04am by ThudnBlunder » |
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TenaliRaman
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Re: Resistors
« Reply #1 on: Oct 17th, 2004, 11:23am » |
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I wonder why u put this in easy 
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assuming u mean diagonally opposite vertices
2/pi
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Grimbal
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Re: Resistors
« Reply #2 on: Oct 17th, 2004, 1:23pm » |
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Numerically, I find 0.773239545 but I might be wrong.
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ThudnBlunder
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Re: Resistors
« Reply #3 on: Oct 17th, 2004, 9:21pm » |
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Quote:| I wonder why u put this in easy |
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Because I'm too dumb to know how hard it is. 
Quote:assuming u mean diagonally opposite vertices
2/pi :: |
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But not dumb enough to believe your answer. 
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| « Last Edit: Oct 20th, 2004, 3:17am by ThudnBlunder » |
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Barukh
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Re: Resistors
« Reply #4 on: Oct 20th, 2004, 2:29am » |
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As TenaliRaman said, this is far from being easy.
[smiley=square.gif]
I’ve found the solution in rec.puzzles archive, under physics section. There is a nice argument to find the resistance between two adjacent vertices – it equals .5. The main difficulty is to find the resistance between two diagonally adjacent vertices – the answer was given by TenaliRaman. The only satisfactory proof of this result I've seen is far from being elementary – look here.
Then, using the superposition, the question posted may be answered: 4/[pi] - .5, the number numerically calculated by Grimbal.
[smiley=square.gif]
Grimbal, how did you find it?
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SWF
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Re: Resistors
« Reply #5 on: Oct 20th, 2004, 6:03pm » |
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A similar but much easier question was discussed in the Hard Forum: Infinite Resistor Network.
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Grimbal
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Re: Resistors
« Reply #6 on: Oct 21st, 2004, 6:36am » |
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I modelled a big grid a[i,j] for i, j in -399..399
I set a[0,0] = 1 and computed a[i,j] = (a[i-1][j] + a[i+1,j] + a[i,j-1] + a[i,j+1])/4 for all [i,j] except [0,0], and fixing the border cells to 0, until it converged.
(Actually, I only computed 1/8th of the grid, adjusting border formulas.)
This models a grid where the voltage is 1 in the middle and 0 on the border. I get the voltage for each cell and I can compute the current that leaves the center cell just from the voltage of the neighbours.
Then I imagine I superpose 2 such grids centered on (0,0) and (2,1), one added and the other subtracted. The current leaving from (0,0) and entering (2,0) is the same as the current computed above, and the currents nearly cancel near the border. (If you have a single entry in the middle, then the current thru the border is non-zero. With 2 entries, the outgoing and incoming currents nearly cancel out)
Then, I can easily compute the voltage difference between (0,0) and (2,1) and so, I get the equivalent resistance.
In short, It is equivalent to forcing cell (0,0) to 1 volt, forcing cell (2,1) to -1 volt, and solving all other cells to be the average of the 4 neighbours. From there, you can compute the current leaving (0,0) and the voltage difference.
I did a second run with a smaller grid to estimate how many digits are correct in the solution.
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| « Last Edit: Oct 21st, 2004, 6:50am by Grimbal » |
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Emul P Edmon
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Re: Resistors
« Reply #7 on: Oct 26th, 2004, 9:54am » |
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 A mental excersize can be used to narrow it down to somewhere between 0.5 and 1.0 ohms. How? (a) As a lower limit, look at each spray of four resistors from the two nodes in question. No matter what the circuitry is external to those, there is at least 0.25 ohms between each node and the external world. If the start node (of the knight's move) and the end node are each 0.25 ohms away from anything else then they must be at least 0.5 ohms away from each other. (b) For an upper limit, consider that each square of four resistors has a 1 ohm resistance across its diagonal (two 2-ohm resistors in parallel). Without reusing any resistor it is easy to find the following shortest paths between the two knight's move nodes: (2 ohms, 4 ohms, 4 ohms, etc.) Just using the first three in parallel results in 1.0 ohms. I just wish I had the patience to solve it correctly, the way you did, however.
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| « Last Edit: Oct 26th, 2004, 9:57am by Emul P Edmon » |
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