-Up to-Home/Maths
-Site Map|-Text version

- Resistance-network Problems

  Fred Curtis - Feb 2000
[Last annotated November 2004]
  1. Inifinite Lattices
    1. Why I am not comfortable with superposition
  2. Finite Lattices
    1. Hypercube of Arbitrary Dimension
    2. Simplex
    3. Cross Polytope
    4. Square Lattices
  3. Further Directions
  4. Chronology
  5. References

1. Infinite Lattices

I became interested in resistance network problems from the recurring questions posted to the rec.puzzles archive.  In particular, the resistance between two adjacent nodes on an infinite 2D rectangular lattice of 1ohm resistors was given as ½ohm. The solution posted in the rec.puzzles FAQ (section "physics/resistors.p") was roughly

I was so skeptical of this reasoning that I wrote a C++ program, surl.cc, to examine the convergence in ever-larger finite lattices. Not surprisingly, the FAQ gave the correct answer.

I used the same reasoning (see References for more on the superposition principle) on some non-rectangular infinite lattices and found the following results. Simulations in ever-larger subsets of the lattices produced results which converged to the values predicted by the superposition principle:

1.1 Why I am not comfortable with superposition

I'm still not comfortable with the superpositon principle as an explanation because:

Because the superposition principle only ever makes predictions for the resistance in one edge, and because the reasoning isn't strong, it seems plausible to me that the principle is just a coincidence. That is, the principle is a just story that agrees with observed results but doesn't explain the results. Contrast this with the works listed in the References below which mathematically prove the results.

There are some experiments I could perform (see Further Directions below) that might convince me to treat the superposition principle as more than a mathematical coincidence.

Note: [2004 November] I came across József Cserti's 1999 paper [see References], which deals with infinite lattice resistance problems. The techniques used in Cserti's paper [and, I'm guessing, many of the papers cited by Cserti] involve far more integral calculus than I can understand.

2. Finite Lattices

A general method in analysing finite lattices is to locate equipotential points and join them, yielding a simpler network.

2.1 Hypercube of Arbitrary Dimension

2.1.1 Opposite vertices

When I read the rec.puzzles FAQ there was [November 2004: & still is] no general solution to the resistance between opposite corners of an arbitrary hypercube of 1-ohm resistors; solutions were only given for 3D and 4D hypercubes. This problem seemed quite tractable - just combinatorics. The same reasoning about symmetry for the cube case can be applied to the arbitrary d-dimensional case: joining the equipotential points results in d layers containing d(d-1 choose i) resistors in parallel in the i'th layer, giving an overall resistance of (sum from i=0 to d-1 of 1/(d-1 choose i)) / d. The values for the first few d are:


As d gets larger, the resistance tends toward 2/d [more accurately, 2/d + 2/(d-1)^2] ; the dense layers in the middle become relatively perfect conductors compared to the outermost layers.

2.1.2 Adjacent vertices

The resistance between adjacent vertices of a d-dimensional hypercube is (2^d - 1)/d(2^(d-1)) ohm, which has a nice geometric interpretation of (1-less-than-number-of-vertices)/(number of edges). This value, like the opposite-vertices resistance, converges to 2/d.

2.2 Simplex

A simplex is a d-dimensional analogue of the tetrahedron. It has d+1 vertices, each of which is joined to all the others by an edge. If a voltage v is applied across any two vertices of a d-dimensional simplex, the potential at each of the remaining vertices is v/2 by symmetry. Joining equipotential vertices gives a net resistance of (1/(d-1) series 1/(d-1)) parallel 1
= 2/(d-1) parallel 1
= 2/(d+1) ohms.

2.3 Cross Polytope

The cross polytope is the d-dimensional analogue of the octahedron. There are 2d vertices, two on each coordinate axis, each vertex joined to every other vertex except its opposite on the same axis.

2.3.1 Opposite vertices

If a voltage v is applied between opposite vertices, the remaining vertices are at a potential of v/2 by symmetry, yielding a resistance of 1/(2d-2) series 1/(2d-2) = 1/(d-1) ohms.

2.3.2 Adjacent vertices

If a voltage v is applied between adjacent vertices, the remaining vertices *not opposite* to the test points are at a potential of v/2 by symmetry yielding a simplified network:

 |\       /|
 | \     / |
 |  \   /  |
 |   \ /   |
 |   v/2   |
 |  /   \  |
 | /     \ |
 |/       \|

where the edge resistances are 1 ohm and the diagonal resistances are each 1/(2d-4). Further symmetry indicates the potential at B is (v-A). The resistance of the network, by tedious algebra, is (d-0.5)/(d(d-1)), which also yields the correct result when d=2.  The potential at vertex A is v(d-1)/(2d-1).

2.4 Square Lattices

Thanks to Arvind Giridhar posing the question about diagonal resistance on an NxN (2D) lattice. No exact solution to date.

A similar but more tractable model (test points at (0,N) and (N,0), all points on lines parallel to y=x connected to make them equipotential) has a diagonal resistance of (1 + 1/2 + 1/3 + ... 1/N). This underestimates the simulated diagonal resistance of an NxN lattice by about 10% for smaller N, rising to 13% for N=80.

NDiagonal resistance
on NxN lattice
(by simulation)
1 + 1/2 + ... 1/N
(From altered model with
equipotential diagonals)

3. Further Directions

Some things which would be interesting to investigate are:

4. Chronology

5. References

-This page
last changed:
31 May 2012
[Validate HTML]
-Donate free
food & land
|Feedback by email
or Web form