Well, summer has now mostly gone by with no posts from me on that puzzle about the

cube of resistors, so I guess I'll solve it now.

The thing to do for the initial puzzle in three dimensions is conisder the eight vertices of the cube and apply a potential difference of V between one vertex and the opposite corner. We could then find the current and the ratio would give us the resistance. The trick is then to notice that the three vertices that are one edge away from the starting vertex are an equpotential, so they can be fused into a single vertex. Then notice that the three vertices that are adjacent to the final vertex are also an equipotential. We then have only 4 vertices, the first is connected to the second by 3 resistors, the second connected to the third by 6 resistors (takes a bit of work to confirm that 6) and the third is connected to the fourth by 3 resistors. Since M 1-Ohm resistors in parallel have an effective resistance of 1/M, the equivalent resistance of this is 1/3+1/6+1/3=5/6

OK, thats fun, and should generalize pretty fast, but we need to see exactly how. In N dimensions, each vertex can be represented as a series of N zeros and ones, and this is a complete list of the 2

^{N} vertices, each vertex is connected to the N vertices next to it, so any vertex has N edges on it. So N edges lead out from the initial vertex to the first equipotential. Next, the N vertices have 1 edge leading backward, and N-1 leading forward, so there are a total of N*(N-1) edges leading to the next equipotential (This is N*

_{N-1}C

_{1}). The next

_{N}C

_{2} vertices each have N*(N-1)/

_{N}C

_{2} (which is 2) edges from beind them, so there are N-2 going forward. This means there are a total of N(N-1)/2*(N-2) vertices going forward (this is N*

_{N-1}C

_{2}. As you can guess, the number of vertices going forward is always N*

_{N-1}C

_{k}, with k going from 0 to N-1.

So, in order to find the total resistance, we simply need to add up 1/N*

_{N-1}C

_{k}, which is 1/N times the sum of the recriprocials of the N

^{th} row of Pascal's triangle. This doesn't really have much of a closed form, but its enough to let you figure out the specific number for any particular value of N. There is a bit of a question left of the infinite limit, and fortunately Ben found a paper called "Sum of the reciprocals of the binomial coefficients", which should be easy enough to find if you are interested, the main point here though is that the limit of the sum at large N is 2, therefore the effective resistance is 2/N. Not much of an answer I guess, but I'm always happy when Pascal's triangle shows up.