Physically Incorrect

This is a solution to this puzzle.

The quickest way to solve the problem of the cubed resistor is to use symmetry. Imagine you’ve connected a voltage V to the two opposing terminals A and B, and denote by I the current that flows through the battery and into A. By symmetry, it breaks up into 3 subcurrents of equal magnitude, I/3. These in turn split into equal currents again (I/6 amps each). Similarly, by the same symmetry argument, the current leaving B must be equal to I, so the current in each edge leading into B must be I/3:

Having observed this, we can apply Kirchoff’s law which says that along any path from A to B the voltage drop must equal V, the battery’s voltage; for example, we can use the path in red in the following figure:

Here Re is the effective resistance between points A and B - that’s what we’re after. Solving immediately yields Re = (5/6)R, which is the solution.

This method of solution, while elegant, generalizes very poorly to other situations; in particular, I don’t think it can be applied easily to the second problem I’ve posed. So before moving on, let’s solve the current problem adopting a different approach. We once again make use of the symmetry involved and note that the points C, D and B are at the same potential (marked by red dots).  The points E F G are at the same potential as well (blue dots):

By the way, pardon the re-shuffling of the letters used to describe the vertices (A and B are now next to each other, while previously they were on opposite points on the cube). Using this we redraw the resistor network on 2D paper and connect by a conducting wire B, C, D, and by a different conducting wire E, F, G; we can do this because they are at the same potential:

Redrawing the problem in 2D and connecting the red (and blue) dots shows us that, in fact, this is nothing more than 3 resistors connected in parallel, connected in series to another 6 resistors in parallel, which are connected serially to another 3 parallel resistors. The equivalent resistance of the first 3 is R/3. The equivalent resistance of the middle 6 is R/6. The equivalent resistance of the final 3 is R/3. These effective resistors are connected in series, so to get the equivalent resistance of the entire circuit we need to sum them up: R/3 + R/3 + R/6 = (5/6)R. We have recovered the previous solution.

This technique can be easily applied to the second problem. Now, due to symmetry, the points that are at equal potential are slightly different, and are marked in the following drawing:

The weaker symmetry in this problem results, of course, in fewer equivalent points. For example, C and D must be at the same potential, but we can’t say anything about E this time.  Still, this offers considerable simplification when redrawing the cube in 2D (see fig. below). It then becomes a straightforward problem of adding up resistors in parallel and in series (see steps), resulting in an effective resistance of 7R/12 between A and B: