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What is the resistance between 2 points in a cube of 1-ohm resistors? I am asking about each of the cases:

1) The 2 points are connected by 1 edge
2) The points are 2 edges from each other
3) The points are antipodal (if such a word can be used in the case of a cube).

The Dead Legend
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mtheorylord
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    If you don't mind me saying, this question has nothing to do with sequence and series or summation. This is purely a matter of solving Kirchoff's equations. – Kenny Wong May 06 '17 at 22:53
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    Wrote a program recently to solve these kinds of diagrams -- it returns $7/5$ ohms for the one-edge case, $3/4$ ohms for the two-edge case, and $5/6$ ohms for the "antipodal case" (using Kirchhoff's equations, as Kenny mentioned). – Marcus Andrews May 06 '17 at 22:56
  • @MarcusStuhr Yes, I remember that $5/6$ for the antipodal case! I did it a long time ago... – Kenny Wong May 06 '17 at 23:02
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    @mtheorylord By the way, when using Kirchoff's laws, it helps to make full use of symmetries. This allows you to see immediately that many currents and voltages are equal, which reduces the number of simultaneous equations you have to solve. – Kenny Wong May 06 '17 at 23:04
  • Sorry, I don't understand how I would use Kirchoff's in this case. Someone care to explain in an answer? – mtheorylord May 06 '17 at 23:06
  • This answer from the physics stackexchange is pretty helpful. Same idea, different diagram. – Marcus Andrews May 06 '17 at 23:26
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    @MarcusStuhr: the one edge case must be less than $1$ ohm, because the one resistor along the edge is $1$ ohm and everything in parallel can only lower the value. – Ross Millikan May 06 '17 at 23:33
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    @RossMillikan Whoops, thanks for the catch. Typed that one incorrectly earlier (too late to edit it, unfortunately). Program returns $7/12$ for the one-edge case, not $7/5$. – Marcus Andrews May 06 '17 at 23:40

2 Answers2

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A MATLAB solution:

% Study a network of twelve 1-ohm resistor connected to form a cube.
% We want to compute the resistances between pairs of nodes.
%
% The vertices of the cube are numbered in binary from 000 to 111.
% We want to compute the resistance between nodes 001 and 000 (adjacent), 
% 011 and 000 (across a face), and finally 111 and 000 (opposite).

% The voltage of vertex 000 is taken to be 0.
% Writing the node-voltage equations for nodes 1-7 we get:
A = [ 3  0 -1  0 -1  0  0;
      0  3 -1  0  0 -1  0;
     -1 -1  3  0  0  0 -1;
      0  0  0  3 -1 -1  0;
     -1  0  0 -1  3  0 -1;
      0 -1  0 -1  0  3 -1;
      0  0 -1  0 -1 -1  3];
% The three queries result in the following right-hand sides:
B = [1 0 0; 
     0 0 0; 
     0 1 0;
     0 0 0;
     0 0 0;
     0 0 0;
     0 0 1];
% We solve symbolically to get fractions.
S = sym(A) \ sym(B);
disp(['The resistance between nodes 001 and 000 is ', char(S(1,1))])
disp(['The resistance between nodes 011 and 000 is ', char(S(3,2))])
disp(['The resistance between nodes 111 and 000 is ', char(S(7,3))])

which produces:

The resistance between nodes 001 and 000 is 7/12
The resistance between nodes 011 and 000 is 3/4
The resistance between nodes 111 and 000 is 5/6
Fabio Somenzi
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1

We can solve this problem without applying Kirchoff's laws at all, just by exploiting symmetry until we're left with series-parallel graphs.

Label the cube as follows:

cube

If we want the effective resistance between $a$ and $b$, then we know $V_c = V_e$ and $V_d = V_f$, because everything is symmetric through the plane $abgh$, so we can contract these to two points $ce$ and $df$:

adjacent-resistors

Now we can use the series-parallel rules to write the effective resistance between $a$ and $b$ as $$\frac{1}{\frac{1}{1} + \frac{1}{\frac12 + \frac1{\frac1{1/2} +\frac1{1/2+1+1/2}} + \frac12}} = \frac7{12}.$$

If we want the effective resistance between $a$ and $g$, we have to be slightly more clever. If $V_a = 0$ and $V_g = 1$, then $V_c = V_d = V_e = V_f = \frac12$ because the cube is symmetric through plane $cdef$, but these points are equidistant from $a$ and $g$. So we can contract all four of these to one point:

diagonal-resistors

Now we can use the series-parallel rules to write the effective resistance between $a$ and $g$ as $$\frac{1}{\frac{1}{1/2} + \frac{1}{1 + 1/2}} + \frac{1}{\frac{1}{1/2} + \frac{1}{1 + 1/2}} = \frac34.$$

Finally, to get the effective resistance between $a$ and $h$, we can use rotational symmetry about the diagonal $ah$ to collapse $bce$ to one point and $dfg$ to another:

opposite-resistors

Here, the effective resistance is just $\frac13 + \frac16 + \frac13 = \frac56$.

There's also a shortcut to solving the adjacent-vertices case. If $R_{vw}$ is the effective resistance between vertices $v$ and $w$, then $$\sum_{vw \in E} R_{vw} = n-1,$$ where $vw \in E$ ranges over all edges of the graph, and $n$ is the number of vertices. In this case, all $12$ edges have equal effective resistance by symmetry, and these add up to $n-1 = 7$, so the resistances must be $\frac{7}{12}$ each.

Misha Lavrov
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