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- Thread starter D H
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Gokul43201

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Is it 0.916333... ohms?

Damn, I missed dinner!

Damn, I missed dinner!

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siddharth

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Is it 0.916333... ohms?

Damn, I missed dinner!

How did you get that? I think that to solve such problems (where a simple superposition doesn't give you the answer), a discrete fourier transform is required, and then the final integral which gives the resistance looks very tricky.

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D H: 4 points (or is it 2 and 1, for the engineer? :tongue:)

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At least three points, since I made Gokul miss dinner.

This http://arxiv.org/abs/cond-mat/9909120" [Broken], question 10) uses a different expression and that enables Mathematica to yield [itex](8-\pi)/(2\pi)\approx0.773[/itex] ohms.

This http://arxiv.org/abs/cond-mat/9909120" [Broken], question 10) uses a different expression and that enables Mathematica to yield [itex](8-\pi)/(2\pi)\approx0.773[/itex] ohms.

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siddharth

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This http://arxiv.org/abs/cond-mat/9909120" [Broken] (see appendix A) gives an absolute mess.

The article uses the same principle of taking the fourier transform and applying ohm's law. The integral one gets after the fourier transform is (eqn 25 in the article)

[tex] \frac{1}{4 \pi^2} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \left( \frac{1-\cos(2x+y)}{2-\cos{x} - \cos{y}} \right) dx dy [/tex].

I'm at home and don't have access to an integrator. Does anyone have access to a math package that evaluates this integral, or a way to find the value? Does it come out as [itex](8-\pi)/(2\pi)\approx0.773[/itex]?

Mathworld seems to have used some substitution in its expression.

EDIT: Nevermind. They've apparently used a contour integral and then a substitution. It's there in appendix A, which I missed the first time.

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The difference between a normal person and a scientist:

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That's actually a normal person and Homer Simpson => Homer = Scientist?? :surprised

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Gokul43201

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I actually did use a superposition - I don't know why it fails! =(How did you get that? I think that to solve such problems (where a simple superposition doesn't give you the answer), a discrete fourier transform is required, and then the final integral which gives the resistance looks very tricky.

I haven't yet looked at the article posted by D H.

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ranger

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It just proves the first sketch of the comic strip. Nerd sniping in action.

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mheslep

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From "xkcd.com"[/URL] :

[/QUOTE]

A friend of mine finished school some years ago with a BS in physics and immediately chucked it all to become a lawyer. After seeing this Nerd Sniping XKCD strip I sent it to him on a workday to see if he was truly a lawyer or if he would be sniped, along with with full disclosure of my intentions. Couple days later I checked back, sure enough, four billable hours went down the drain. :devil: Good thing he wasn't standing in the road.

BTW, engineer's solution: assume the infinite grid looks like a short to any one node except for the four resistors directly attached into the node. The parallel resistance into the node then is 1/4 ohm and therefore the resistance between any two, non-adjacent nodes is 1/2 ohm. With actual (cheap) 20% resistors thats close enough to the actual (.7xx is it?)

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Why don't you try working out the answer for nodes separated by two diagonal hops?

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mheslep

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Yes, ok, I'll have to make clear I'm using really cheap resistors when I go for the Google interview.The resistance betweeb any two horizontally or vertically adjacent nodes is exactly 1/2 ohms. The resistance between nodes further apart than that is not. That 20% fudge on top of 0.5 ohms yields 0.6 ohms, not 0.773 ohms.

Think I'll code up a quick sim for fun.Its even worse for nodes further apart than that.

Why don't you try working out the answer for nodes separated by two diagonal hops?

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