## Introduction to Part 5

At this point, let’s reflect upon the topics we have included in our little exploration. I have tried to place the topics into two groups:

Physics |
Mathematics |

Resistor Addition (Series and Parallel) | Combinations |

Ohm’s Law | Factorials |

Kirchoff’s Laws of Electrical Circuits | Pascal’s Triangle |

Electrical Potential Difference | Summation Notation |

You’ll notice that a great many of these topics are study material for pre-university courses of both physics and mathematics. If you are or were a UK student, these are things you might have seen when you were 17, or younger.

If you have made it thus far to Part 5, I hope you see that all the previous work, although presented in an unfamiliar way, and combined under obscure circumstances, is rather simple to understand when viewed individually. However, in any example of theoretical physics that aims not to be superficial, the combination of different mathematical and physical ideas is necessary. Indeed, that’s the whole point.

Perhaps the pattern of vertices can be used to construct the pattern of edges?

## Hasse Diagrams Revisited

The concept of *partial order* with the diagrams I introduced in Part 4 was another way of thinking about the edge connections of the n-cube. The vertices could be thought of as carrying information about the shortest route to the start vertex. Then, partial order can defined as follows:

There is a connection between one vertex and the next, if (and only if) their shortest path to the start vertex differs in size by one dimension.

This is obviously true for n-cube, without having to know about Hasse Diagrams. Yet, saying it explicitly in this way provides an insight to our solution.

To remind you, here is the Hasse Diagram for the 4-cube:

Check that the definition of edge connections we just formalized sufficiently describes the above diagram.

### Binary Representation

Thinking of the vertices carrying information about the shortest route back to the start provides another way of thinking about the total number of vertices. Instead of listing the number of dimension that have been traversed to get to a vertex, we could write the same thing in binary where the ith binary digit tells you if the ith dimension has been traversed:

This new way of thinking about the same thing makes it immediately obvious that the total number of vertices must be 2 to the power of n. There are two options for each binary digit (0 or 1) and n possible choices.

Also, we can define “partial order” by grouping vertices that have the same number of 1s. These can then be graded (from start to finish) by ordering the groups by number of 1s. Here is the same Hasse diagram as above, but in binary form:

## Number of Resistors in the Parallel Branches

When the n-cube has been reduced with equipotentials, we are left with n equipotentials. The ith equipotential has (n,i) vertices, as previously determined in Part 4. What was left to consider was how many edges connected one equipotential to the next.

Now we we have considered the Hasse Diagrams, is is clear that after traversing i dimension, there are (n-i) dimensions left to travel to the final vertex. Thus, each vertex in the ith equipotential has (n-i) edges to the next equipotential. One edge heading off in each untraversed dimension.

We could summarize this as:

We have now found the final piece we need for our solution:

[Incidentally, this is also the equation for the second outer later of Pascal's Pyramid]

## Solution for n-Cube Resistor Network

This was the scheme for a solution that derived in Part 4:

Substituting our new equation for the resistor branches, we obtain:

Inserting our equation for *combinations* gives:

Tidying up the fraction:

The n! doesn’t depend upon “i” and so can be removed from the summation (as it could be factorized out from the sum). With some further simplification of the factorials, we obtain our solution:

### Case Examples

Let’s test it with one familiar example and another unsolved case.

#### n=4

#### n=5

### First 20 Solutions

n |
Total Resistance /R(fraction) |
Total Resistance /R(10 decimal places) |

0 | 0 | 0.0000000000 |

1 | 1 | 1.0000000000 |

2 | 1 | 1.0000000000 |

3 | 5/6 | 0.8333333333 |

4 | 2/3 | 0.6666666666 |

5 | 8/15 | 0.5333333333 |

6 | 13/30 | 0.4333333333 |

7 | 151/420 | 0.3595238095 |

8 | 32/105 | 0.3047619047 |

9 | 83/315 | 0.2634920634 |

10 | 73/315 | 0.2317460317 |

11 | 1433/6930 | 0.2067821067 |

12 | 647/3465 | 0.1867243867 |

13 | 15341/90090 | 0.1702852702 |

14 | 28211/180180 | 0.1565712065 |

15 | 10447/72072 | 0.1449522699 |

16 | 1216/9009 | 0.1349761349 |

17 | 19345/153153 | 0.1263115968 |

18 | 18181/153153 | 0.1187113540 |

19 | 651745/5819814 | 0.1119872559 |

20 | 1542158/14549535 | 0.1059936206 |

Don’t pay any attention to red line, other than for it to aid your eye to see the pattern. It has no physical significance, as we are not considering fractional dimensions with our n-cube definitions (although such a thing can be defined with fractals).

Is it mysterious that the first 3 results are so different from the others? The graph looks so disjoint at the beginning for the first 3 solutions. When you reflect on the objects we were using in our inquiry, it is not so strange. Our investigation involved resistors, which are one dimensional objects. So, by definition, there are none for a 0-cube, and there is only one for the 1-cube. So, we would expect those results to be zero and 1.

As for the other results (n greater or equal to 2), the graph suggests that as the dimension increases, the total resistance of the n-cube tends to zero. I won’t prove that here, as the mathematics will be too difficult, but we can try to imagine the physical situation. What happens to the the number of resistors in each parallel component of the circuit diagrams, as n increases? Well, they increase too – check the red number of the Pascal diagram.

From Part 2, we learnt a handy calculation for parallel circuits with identical resistors: simply divide the resistance by number of resistors. Thus, as the number of resistors increases, the resistance of each parallel component will decrease to zero. The tricky thing to do would be to calculate which happens quicker: do the number of branches increase faster than the resistance in each branch decreases? It would seem not as our graph appears to tend to zero, but we would need to prove this with mathematical analysis.

### Pascal Diagram

With additional solutions, it is possible to continue the Pascal diagram from Part 4:

I have added in n=5 and n=6. Those values and all the previously obtained results can be calculated with the above equation.

## Testing the Theory

If an experiment to test this equation contradicts our predictions, then either we have made a mathematical (or logical) error, or the assumptions on which it rests are incorrect.

In our case, it is unlikely to be the latter, as the assumptions are some fundamentals about the conservation of energy and charge. There are been far more significant tests for which these ideas have endured.

So, any test will be for our own curiosity and to check if we have made an error in the analysis. I have built the first 5 n-cubes – the 0-cube was especially easy! Here are the results, with images of the measurements and the n-cubes I made.

Dimension (n) |
Ratio |
Predicted (kΩ) |
Measured (kΩ) |
Image |

0 | 0 | 0.000 | 0.000 | |

1 | 1 | 1.000 | 0.987 | |

2 | 1 | 1.000 | 0.983 | |

3 | 5/6 | 0.800 | 0.820 | |

4 | 2/3 | 0.667 | 0.658 |

I have provided the predictions to the same number of significant figures as the measuring device. Notice they do not agree fully, but to an agreeable percentage. An more thorough analysis of the acceptable error would be called for were the results more than anecdotal. For our purposes, it is sufficient to comment that the resistors used to build the n-cubes are manufactured to within an error of 10%.

The results fall very comfortable within this range. Furthermore, again if the results were more vital, we could calculate the combined error for each resistor cube (as we are combining many resistors each with a potential 10% error).

## Epilogue

The original motivation for this series or articles came from a problem I taught to a few students as an extra class in preparation for the British Physics Olympiad. I also prepared a challenge in the form of a cabinet display pictured below:

Federico

2012/01/29

Thank you for your interesting post! Recently, I’m working on the problem of unit-resistance hypercube and your post was really helps me understand this topic. BTW, do you have a closed-form solution for hypercube resistor network?

jamesthenabignumber

2012/01/31

Hello,

I’m glad I was helpful.

No, I don’t have a closed-form solution. I haven’t really worked on the problem.

If you find a solution, i’d be very interested to hear about it!