Resistence of the cube: Torsten Sillke, BI, 06.04.93
Replace the edges of a cube by three different resistors a, b, and c.
Parallel edges get equal resistors.
What is the total resistance between opposite vertices of the cube?
R = %%%%%%%%%%%%%%%%%%%%%%%%%
(the solution fits into the above space. allowed symbols: "abc+*/^()")

Some relations and the solution:
 vertices
Z
/\
/  \
/ B \
A / \ C
\/ \/
/\ /\
C \ / A
\ B /
\  /
\/
Z

 edges
+
/\
a b c
/ + \
+ / \ +
\a c/ << change the middle
b/c a\b << layer a, b, c > a', b', c'
+ \ / +
\ + /
c b a
\/
+

KirchhofRules: current
Z) I = I_a + I_b + I_c
A) I_a = I_b' + I_c'
B) I_b = I_a' + I_c'
C) I_c = I_a' + I_b'
I = 2 I_a' + 2 I_b' + 2 I_c'
KirchhofRules: chain
Z>A>B, Z>B>A) U_a + U_b' = U_b + U_a'
Z>A>C, Z>C>A) U_a + U_c' = U_c + U_a'
Z>B>C, Z>B>A) U_b + U_c' = U_b + U_a'
Delta := U_a  U_a' = U_b  U_b' = U_c  U_c'
Eliminating the Us:
[ a b c 0 1 ] [I_a] [1]
[ 1 1 1 1 0 ] [I_b] [0]
[ 4 0 0 1 2/a ]*[I_c] = [0]
[ 0 4 0 1 2/b ] [ I ] = [0]
[ 0 0 4 1 2/c ] [1/R] = [0]
Resistence of the Cube:
replace the edges with resistors with a, b, c Ohms. Parallel edges get
equal resistors. Compute the resistance between opposite vertices.
1
R = 1/4 (a+b+c) + 1/4 (1/a+1/b+1/c)
If you have six different resistors, the opposite ones are equal.
(equal letters are parallel resistors.)
The current enters and leaves at the vertices where a, b, c meet.
[ 1 [ a a' b b' c c'] a b c ] [1] [ ]
[  [ +  + ]  +  +   1 ] [] [1]
[ 2 [ a+a' b+b' c+c'] a+a' b+b' c+c' ] [R] [ ]
[ ]*[ ] = [ ]
[ 1 [ a' b' c' ] 1 1 1 ] [ ] [ ]
[  [ +  + ]  1  +  +  ] [D] [0]
[ 2 [ a+a' b+b' c+c'] a+a' b+b' c+c' ] [ ] [ ]

Torsten Sillke, FRA, June 1994
An old brain teaser is:
What is the resistance between two oposite vertices of a cube,
if you replace the edges by resistors of 1 ohm each?
Some time ago, I tried a generalization. Only the parrallel edges
get the same resistance. Luckily the result had a short formular.
Then Achim Flammenkamp managed to compute the 4 and 5dim Cubes
with Mathematica. From the results we got a conjecture for the
general case.
But Achim didn't want to proof it the hard way.
Does someone see an easy way?
Formulars for the first cases (dim <= 5 are proofed):
dim:
1: R = x1
2: R = 1/2 (x1 + x2)
3: R = 1/4 (x1 + x2 + x3 + (1/x1 + 1/x2 + 1/x3)^(1))
4: R = 1/8 (x1 + x2 + x3 + x4
+ (1/x1 + 1/x2 + 1/x3)^(1) + (1/x1 + 1/x2 + 1/x4)^(1)
+ (1/x1 + 1/x3 + 1/x4)^(1) + (1/x2 + 1/x3 + 1/x4)^(1))
Conjectured function for the general case:
n: R = 1/2^(n1) SUM ( SUM 1/x_i )^(1)
A subset {1..n} i in A
#A odd
Some properties of the general function:
 R(x1,x2, ..., xn) is a symmetric function
 R(1,1,...,1) = 1/n SUM 1/Binomial[n1,k]
nterms k