==> physics/resistors.s <==
1. Cube

The key idea is to observe that if you can show that two
points in a circuit must be at the same potential, then you can
connect them, and no current will flow through the connection and the
overall properties of the circuit remain unchanged.  In particular, for
the cube, there are three resistors leaving the two "connection
corners".  Since the cube is completely symmetrical with respect to the
three resistors, the far sides of the resistors may be connected
together.  And so we end up with:

	|---WWWWWW---| |---WWWWWW---| |---WWWWWW---|
        |            | |---WWWWWW---| |            |
     *--+---WWWWWW---+-+---WWWWWW---+-+---WWWWWW---+---*
	|            | |---WWWWWW---| |            |
	|---WWWWWW---| |---WWWWWW---| |---WWWWWW---|
		       |---WWWWWW---|

This circuit has resistance 5/6 times the resistance of one resistor.

2. Platonic Solids

Same idea for 8, 12 and 20, since you use the symmetry to identify
equi-potential points.  The tetrahedron is a hair more subtle:

    *---|---WWWWWW---|---*
	|\          /|
	W W        W W
	W  W      W  W
	W   W    W   W
	|    \  /    |
	 \    ||     |
          \    |    /
	   \   W   /
	    \  W  /   <-------
	     \ W /
	      \|/
	       +

By symmetry, the endpoints of the marked resistor are equi-potential.  Hence
they can be connected together, and so it becomes a simple:

    *---+---WWWWW---+----*
	|           |
	+-WWW   WWW-+
	|    |-|    |
	|-WWW   WWW-|

3. Hypercube

Think of injecting a constant current I into the start vertex.
It splits (by symmetry) into n equal currents in the n arms; the current of
I/n then splits into I/n(n-1), which then splits into I/[n(n-1)(n-1)] and so
on till the halfway point, when these currents start adding up. What is the
voltage difference between the antipodal points? V = I x R; add up the voltages
along any of the paths:
n even:                                                         (n-2)/2
     V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... +  I/(n(n-1)       )}

n odd:                                                      (n-3)/2
 V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... +  I/(n(n-1)       )}    (n-1)/2
                                                               + I/(n(n-1)     )
And R = V/I i.e. replace the Is in the above expression by 1s.

For the 3-cube: R = 2{1/3} + 1/(3x2) = 5/6 ohm
For the 4-cube: R = 2{1/4 + 1/(4x3)} = 2/3 ohm

This formula yields the resistance from root to root of
two (n-1)-ary trees of height n/2 with their end nodes identified
(-when n is even; something similar when n is odd).
Coincidentally, the 4-cube is such an animal and thus the answer
2/3 ohms is correct in that case.
However, it does not provide the solution for n >= 5, as the hypercube
does not have quite as many edges as were counted in the formula above.

4. The Infinite Plane

For an infinite lattice: First inject a constant current I at a point; figure
out the current flows (with heavy use of symmetry). Remove that current. Draw
out a current I from the other point of interest (or inject a negative current)
and figure out the flows (identical to earlier case, but displaced and in the
other direction). By the principle of superposition, if you inject a current I
into point a and take out a current I at point b at the same time, the currents
in the paths are simply the sum of the currents obtained in the earlier two
simpler cases. As in the n-cube, find the voltage between the points of
interest, divide by I and voila`!

As an illustration, in the adjacent points case: we have a current of I/4 in
each of the four resistors:

    ^                |
    |                v
 <--o-->          -->o<--
    |                ^
    v                |
 (inject)          (take out)
And adding the currents, we have I/2 in the resistor connecting the two points.
Therefore V=(1 ohm) x I/2 and effective resistance between the points = 1/2 ohm.

We do not derive it, but the equivalent resistance between two nodes k
diagonal units apart is (2/pi)(1+1/3+1/5+...+1/(2k-1)); that, plus
symmetry and the known equivalent resistance between two adjacent
nodes, is sufficient to derive all equivalent resistances in the
lattice.

5. Continuous sheet

I think the answer is (rho/dz)log(L/r)/pi where rho is the resistivity,
dz is the sheet thickness, L is the separation, r is the terminal radius.

cf. "Random Walks and Electric Networks", by Doyle and Snell, published by the
Mathematical Association of America.