For what it's worth, I did receive this. Your approach was where I wanted to go, originally, with this [except I got sidetracked before doing that].
I have also been thinking about the use of pi, as well as the relationship between manhattan distance and linear distance. But I have not felt that I had anything interesting to say about the issue. Note also that there's a related discussion going on, in the thread spawned from http://jsoftware.com/pipermail/programming/2013-February/031717.html (but, it's been purely structural in character and has no one has been talking about admittance issues). Thanks, -- Raul On Sun, Mar 3, 2013 at 4:25 PM, Keith Park <[email protected]> wrote: > A long look back, an OOPS!, and a couple of questions. > Sixty years ago when I was a post-graduate student an old professor of > Electrical Engineering asked me this same question about the resistance > between two points in an infinite grid of resistors. In his question the > resistance was across the diagonal of one small square and the resistors > were each 2 ohms. He added in his crusty manner, "...and if you write > anything down, Park, you are wrong!". I assumed then that the expected > answer was one ohm which I was never able to demonstrate. I was rather > gruntled therefore to see the elegant proof posted by Kelly. After mulling > over that result for a few days I came to the conclusion that it was wrong. > At first I attributed the failure of the proof due to the presence > of infinities in the argument, but there I too was wrong. Consider the > simpler case of the resistance between two adjacent points on the grid. > Call these points A and B. There are four resistors connect to each of A > and B and there is one resistor, call it Rab which is common to both > points. We first inject a current into point A. Because the resistance of > the grid to infinity is infinite there must be infinite voltage at point A > under this condition. However this infinity does not cause any problem > because we only use this step to determine the current in the resistor Rab. > The current must be 0.25 amperes because the one-ampere current must > divide equally amongst the four resistors connected to point A. We then > remove the current from point A and pull one ampere out of point B. Again > there must be 0.25 amperes flowing in resistor Rab. We now apply the two > currents simultaneously. The principle of superposition allows us to > determine the current in Rab by simply summing the results obtained by > applying each current separately. There must then be 0.5 amperes flowing > in Rab which will produce a voltage of 0.5 volts. The resistance from A to > B is therefore 0.5 ohms. Rab is one ohm so the rest of the infinite grid > behaves as if it were a one-ohm resistor parallel to it. > Now consider the case where the two points are on opposite corners > of a square. As before call these points A and B. (If I am belabouring the > argument it is because there seemed to be so much confusion the last time > that this subject was discussed.) Call the four resistors connected to > point A a0,a1,a2 and a3. Similarly the four resistors connected to point > B are b0,b1,b2 and b3. The names are arbitrary so we may call the four > resistors that form the square a0,a1,b0 and b1 with a0 connected to b0 and > a1 connected b1. We now inject a current of one ampere into point A with > the current flowing out at infinity. The current flowing in a0 and a1 must > be 0.25 amperes. With the current removed from A we now pull one ampere > out of point B. Again it is certainly true that there must be 0.25 amperes > flowing in each of b0 and b1. By superposition we may find the current in > each resistor when both currents are applied by adding the results obtained > from applying each current separately . That is certainly true. So the > current in a0 and b0 must be 0.25 amperes with both currents applied, but > this is false, of course. > When we apply a current of one ampere to point A there is 0.25 > amperes flowing in a0 and a1, but there is also a current flowing in b0 and > b1. Call this current OOPS! because we overlooked it. When we pull one > ampere out of point B there will be, of course, the same current OOPS! > flowing in a0 and a1. So under superposition the current in all four > resistors will be 0.25 + OOPS! amperes, so we may conclude that the > resistance between A and B is 0.5 + 2*OOPS! ohms. The value OOPS! is not > easy to calculate so superposition is of no help to solve this problem. > Don Kelly provided a reference to a solution to the problem at: > > http://www.mathpages.com/home/kmath668/kmath668.htm > > The unnamed author of the report derives an answer of 2%pi ohms for the > resistance across the diagonal. I do not understand the method or the math > involved. The answer is almost certainly true. > I have used J to build the admittance matrix for small square grids. > The resistance calculated for a small grid is too high, but it approaches > the correct value as the size of the grid is increased. So the result can > be considered as an upper bound for the true value. If one considers a > small grid for which the outer resistors are all short-circuited together > then one obtains a resistance that is too small but rises as the grid size > is increased. These results are then a lower bound for the true > resistance. For the largest grids that my computer could handle the upper > and lower bounds are 0.636945 and 0.636305 respectively. The mean of these > two bounds differs from 2 divided by pi by less than 0.001 percent. This > result may be more of a happy accident than reality as I do not know how to > properly compare the results for the "open" and the short-circuited grids. > If Roger will ever get around to implementing %. for sparse matrices I > could then perhaps squeeze out a couple more decimal places. > It would be sort of fun if someone would write an explanation of the > method used in the mathpages reference. Please keep it simple....something > even an old engineer could understand. > A couple of questions: > 0) Why does pi appear in a calculation for a SQUARE grid! > 1) Matrix division for this special matrix is therefore a way of > computing pi, albeit a very inefficient one. Is this a new way of > computing pi or is it somehow related to a known sequence? > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
