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
                       \|/
                        +
                        |

Kirchhof-Rules: 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'

Kirchhof-Rules: 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 5-dim 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^(n-1)     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[n-1,k]
         n-terms         k<n

  -   R(x1,x2,...,xn) = 1/2 R(x1,x2,...,xn,0) 

  They are fullfilled by our conjecture.
  For a proof of the second property see e.g. B. Sury.



References:
-----------

- Achim Flammenkamp;
  mailto:achim@Uni-Bielefeld.DE
  personal communications, June 1994

- Louis W. Shapiro;
  An Electrical Lemma,
  Mathematics Magazine, 60:1 (1987) 36-38

- David Singmaster (Problem);
  SIAM Review, 22 (1980) 504, (Solution)
  Problem 79-16, Resistances in an n-Dimensional Cube
  (equal resistances. solutions for all pairs are given.)

- B. Sury,
  Sum of the reciprocals of the binomial coefficients,
  Europ. J. Combinatorics 14 (1993), 351-353.

- David Wagner;
  The algebra of flows in graphs,
  Adv. Appl. Math. 21 (1998), 644-684.
  Proposition 5.3 shows (*)

- F. Nedermeyer, Y. Smorodinsky;
  Resistances in the multidimensional cube,
  Quantum 7:1 (Sept./Oct. 1996) 12-15, 63
  (- for edge-transitive graphs we have R_edge = 2/d (1 - 1/n) 
     with n nodes and degree d. (valid for infinite graphs (n=oo))
   - R_(1,1) = 2/Pi for the infinite graph Z*Z (square grid).
     Notation: R_(n,m) is the resistance between (0,0) and (n,m).)

- Peter G. Doyle (Dartmouth), J. Laurie Snell (Dartmouth) 
  Random Walks and Electric Networks,
  Mathematical Association of America, 1984
  http://front.math.ucdavis.edu/math.PR/0001057  {arXiv:math.PR/0001057}


One might as well state it for networks (allowing multiple edges but not loops).
Here is a relevant observation:  let  G+uv  be the network obtained from  G  by
adjoining a new edge between  u  and  v.  Let  t(H)  be the number of
spanning trees of the network  H.  Then the resistance between u and v is

                   t(G+uv)
          r(u,v) = ------- - 1.          (*)
                    t(G)


(Each edge has resistance 1.)