That's quite a few resistances

The compute intensive, but somewhat uninteresting solution, of course, is the brute force method of picking two opposite vertices A and Z, applying a voltage of 1V to A, and grounding Z, (Note, this doesn't imply we only have 26, vertices, you'll need to come up with creative names for the other 94 nodes :) ) and solving Kirchoff's equations at each of the other 118 nodes.

I have labeled a few nodes here for reference here:

This gives you the following set of 118 linear equations:

• $3*V_B-1-V_F-V_E = 0$ (Satisfying Kirchoff's law at node B, the 1 is since the voltage at A is 1V)
• $3*V_F-V_B-V_H-V_G=0$ (Satisfying Kirchoff's law at node F)

$\vdots$

etc.

where $V_n$ = The voltage on node $n$

Finally, once we solve the above set of linear equations, we end up with 118 voltages. Lets consider the three on the nodes next to Z and call them W, X, and Y.

The current from A to Z is the sum of the currents from W to Z, X to Z, and Y to Z, which will be $V_W/100$ , $V_X/100$ and $V_Y/100$ (since we have 100 ohm resistors on all edges).

So, $R_{eq} = 100 * (V_W + V_X + V_Y)$

Once you solve the above set of linear equations, you get:

$V_W + V_X + V_Y = 42815/21114 \approx 2.03$

So, $R_{eq} = \boxed{203}$ .

As I said, not pretty, but it gets the job done...

If anyone has a more elegant way, I'd love to see it! :-)