Simple Question

Chris Foster schrieb:

Very nice and tricky question :)

not sure about it, but if you start from the result for a "simple move", that is the resistance between two adjacent points, you find 1/4 Ohm as a result. This comes from the simmetry and infinity of the structure, which in turn demands for the current to split in four equal parts. Now, it seems to me that the same argument should hold for your case, because also here the generator current can't distinguish between the four directions: in every direction the resistance has the same value. Therefore, I would have answered 1/4 Ohm.

I'd say that such a lattice has the property that resistance is always R/4 irrespective the two points taken.

Maybe tomorrow I'll regret giving this answer...Did I get the job? ;-D

M

Is there an infinite number of resistors in parallel with any two points?

What is the answer?

...

1ohm |----/\\/\\/\\/\\---| | 1ohm | |----/\\/\\/\\/\\---|y | 1ohm | |----/\\/\\/\\/\\---| | 1ohm | x|----/\\/\\/\\/\\---| | 1ohm | |----/\\/\\/\\/\\---| | 1ohm | |----/\\/\\/\\/\\---| | |

x to y = 1 ohm

Tom Biasi schrieb:

[..]

I would say no. It is a lattice (I think this should be the right word in english), that is a net of resistors. Every node joins 4 resistors toghether. One could also visualize it as a "mosquito net", where the threads are chains of resistors of some unit value, crossing with perpendicular threads of the same nature.

M

Isn't that a one dimensional array? But given even that the answer must be zero

1/1 = 1

1 + 1 . . . . . continued to infinity is zero.

Hello Chris,

Maybe we should find that company a book titled "Meaningful Interviews" or something like that. If our HR folks would have asked me to pose such questions I would have chewed them out.

--
Regards, Joerg

http://www.analogconsultants.com

If I'm not slipping up, the resulting equation would be:

PI PI / / 1 | | 1 - cos( 2u + v ) R = ------ * | | ----------------------- du dv 4*PI^2 | | 2 - cos( u ) - cos( v ) / / -PI -PI

But thinking as a programmer, I'd probably just solve it numerically, not through numerical or closed integration of the above, but instead by:

(a) create a matrix V[] to represent the node voltages (b) select some central point for A, call it V[x,y] (c) arbitrarily initialize the matrix node values to 0.5 (d) initialize matrix element V[x,y] to 0.0. (e) initialize matrix element V[x+2,y+1] to 1.0. (f) divide the grid into a checkerboard arrangement with white and black (or red and black) squares (g) process all the black squares, those not V[x,y] or V[x+2,y+1], so that their new value is the mean of the surrounding four nodes. (h) process all the white squares, those not V[x,y] or V[x+2,y+1], so that their new value is the mean of the surrounding four nodes. (i) until satisfied, go back to (g). (j) add up the four node voltages adjacent to V[x,y] into 'sum' (k) print 1/(sum-4*V[x,y]) as the resistance.

Which looking at the near center of a 80x80 grid (probably close enough) comes out to 0.7728 ohms.

Jon

If the array is infinite there has to be.

___ ___ ___ |-|___|-|-|___|-| -|___|- .-. .-. .-. .-. | | | | | | | | | | | |Start| | | | '-' ___ '-' ___ '-' ___ '-' |-|___|-o-|___|-|-|___|-| | | | | .-. .-. .-. .-. | | | | | | | | | | | | | | | | '-' ___ '-' ___ '-' ___ '-' |-|___|-|-|___|-|-|___|-| | | | | .-. .-. .-. .-. | | | | | | | | | | | | | | | | '-' ____'-' ___ '-' ___ '-' |-|___|---|___|-o-|___|-|

End

(created by AACircuit v1.28.6 beta 04/19/05

Ok, my best guestimate is 2 ohm, or more close to 1.98 ohms. JTT

Off the top of my head I think it's sqrt(5) ohms.

Graham

I don't think so. In terms of an exact calculation, it is (2/pi) for an "over one, up one" move. That's exact for an infinite grid as I understand the OP's question. The "over two, up one" has slightly higher resistance. But not so high as to be sqrt(5). That figure is wrong on its face. At a distance of "over two, up two" the exact answer is (8/(3*pi)), which sets an obvious upper limit. The actual figure for the knight's move must be between (2/pi) and (8/(3*pi)) or else between 0.6399 and 0.8488. And that is what I found when I did it using an entirely independent numerical method, from scratch.

Jon

Yes, that's what I was basing my comment on. Except that you missed adding in the rest of the infinite grid.

Jon

Nope.

Jon

I just got the exact value for this knight move position. The exact number is: (8 - pi) / (2 * pi), or 0.7732395...

By the way, here is my exact QBASIC program:

between

over

resistance

the

- 1) + g(i, j + 1)) / 4

- 1) + g(i, j + 1)) / 4

- 1) + g(i, j + 1)) / 4

- 1) + g(i, j + 1)) / 4

Chris Foster wrote in news:Xns9845848B07020eddolan@66.150.105.47:

This question was posed by a technical guy, not HR. I did get the job, but I have no idea if my answer was correct. The fellow who asked the question moved to California very soon after I stared working there, and I never found out the correct solution

--
Posted via a free Usenet account from http://www.teranews.com

How did you derive the equestion please??

Jonathan Kirwan wrote in news: snipped-for-privacy@4ax.com:

--
Posted via a free Usenet account from http://www.teranews.com

Jamie wrote in news:KTTQg.4$% snipped-for-privacy@newsfe04.lga:

I'm wondering if it's intended to be a "Kobayashi Maru" type test.

Puckdropper

--
Wise is the man who attempts to answer his question before asking it.

To email me directly, send a message to puckdropper (at) fastmail.fm