> Is there a discernible pattern?
>
> I haven't calculated it myself, so I would be guessing too.
A while back, in response to a similar discussion on the
Planet Physics discussion forum, I solved the problem for
hypercubes of arbitrary dimension. The method used is
the one described here of reducing it to a series-parallel
problem by using symmetry to identify equipotential nodes.
If we make a coordinate system with the center of the
hypercube of length 2 as origin and it edges parallel to
the coordinate axes, then the coordinates of the vertices
will all be either +1 or -1. Hence, the vertices of an
n-dimensional hypercube may be labeled by n-tuplets
of + and - and two vertices are the endpoints of an edge
if and only if their tuplets differ in exactly one slot. As the opposite
vertices between which we will consider the resistance
we may, without loss of generality, take as (+++++...) and
(-----...); put the former at a potential of 1 volt and
ground the latter. Then we note that any two vertices
whose tuplets have the same number of plus and minus
signs can be transformed into each other by a symmetry
operation which preserves the two vertices between which
we are measuring the resistance hence, by symmetry, they
will all be at the same potential.
Hence, the problem is equivalent to one in which we
collapse all the vertices at the same potential into a
single vertex. In this case, that is a problem in which
there are number of resistors between one node and the
next node in parallel. By replacing the parallel resistors
with their equivalents, all that is left is to sum series
resistors.
Since the resistors all have the same value (say 100
Ohms because that seems to be the standard value for
making resistor cubes) , we only need to count how many
resistors are between adjacent collapsed nodes --- if we
have m resistors in parallel, the resistance between those
two nodes will be 1/m hectoohms.
Before doing this in general, let me revisit the case of
3 dimensions using this notation. In three dimensions,
the vertices of our cube are labelled ---, --+, -+-, +--,
-++, +-+, ++-, +++. Arranging them by number of
plusses and writing the edges between them, we have
0 plusses: ---
3 edges: --- to --+; --- to -+-; --- to +--
1 plus: --+ -+- +--
6 edges: --+ to -++; --+ to +-+; -+- to -++;
-+- to +-+; +-- to +-+; +-- to ++-
2 plusses : -++, +-+, ++-
3 edges: -++ to +++; +-+ to +++; ++- to +++
3 plusses: +++
Hence, the total resistance is 1/3 + 1/6 + 1/3 = 5/6 hΩ as
we know well.
Now we return to the general problem of the n-dimensional
hypercube. By basic combinatorics, we know that there are
\( {n \choose k} \) ways to make a string of k plus
signs and n-k minus signs, hence we have \( {n \choose k} \)
vertices whose labels have k plus signs. Given a vertex with
k plus signs, there are k edges connecting it to vertices with
k-1 plus signs because we can make an edge by changing any
one of the k minus signs. For instance, if we look at the
vertex +-+, we can either change the first + to get the edge
from +-+ to --+ or the second + to get the edge from
+-+ to +--. Hence, there are \( {n \choose k} \)
edges connecting vertices with k plusses to vertices with
k-1 plusses. In the 3 dimensional example, we have
\( 1 \cdot {3 \choose 1} = 1 \cdot 3 = 3 \)
\( 2 \cdot {3 \choose 2} = 2 \cdot 3 = 6 \)
\( 3 \cdot {3 \choose 3} = 3 \cdot 1 = 3 \)
which agrees with the explicit enumeration above.
Making the series-parallel combination in general,
we thus have:
\( R = \sum_{k=1}^n {1 \over k {n \choose k}} \)
Expressing the binomial coefficients in terms of
factorials and tidying up, this can also be written as
\( R = {1 \over n!} \sum_{k=1}^n (n-k)! (k-1)! \)
Here is a table of values:
n R
2 1 100.0 Ω
3 5/6 ≈ 83.33 Ω
4 16/24 = 2/3 ≈ 66.67 Ω
5 64/120 = 8/15 ≈ 53.33 Ω
6 312/720 = 13/30 ≈ 43.33 Ω
7 1812/5040 = 151/420 ≈ 35.95 Ω
8 12288 /40320 = 32/105 ≈ 30.48 Ω
9 95616/362880 = 83/315 ≈ 26.35 Ω
I would have to admire the tenacity of anyone
who would hook up \(2^9\) resistors
into a hypercube arrangement
>
> I haven't calculated it myself, so I would be guessing too.
A while back, in response to a similar discussion on the
Planet Physics discussion forum, I solved the problem for
hypercubes of arbitrary dimension. The method used is
the one described here of reducing it to a series-parallel
problem by using symmetry to identify equipotential nodes.
If we make a coordinate system with the center of the
hypercube of length 2 as origin and it edges parallel to
the coordinate axes, then the coordinates of the vertices
will all be either +1 or -1. Hence, the vertices of an
n-dimensional hypercube may be labeled by n-tuplets
of + and - and two vertices are the endpoints of an edge
if and only if their tuplets differ in exactly one slot. As the opposite
vertices between which we will consider the resistance
we may, without loss of generality, take as (+++++...) and
(-----...); put the former at a potential of 1 volt and
ground the latter. Then we note that any two vertices
whose tuplets have the same number of plus and minus
signs can be transformed into each other by a symmetry
operation which preserves the two vertices between which
we are measuring the resistance hence, by symmetry, they
will all be at the same potential.
Hence, the problem is equivalent to one in which we
collapse all the vertices at the same potential into a
single vertex. In this case, that is a problem in which
there are number of resistors between one node and the
next node in parallel. By replacing the parallel resistors
with their equivalents, all that is left is to sum series
resistors.
Since the resistors all have the same value (say 100
Ohms because that seems to be the standard value for
making resistor cubes) , we only need to count how many
resistors are between adjacent collapsed nodes --- if we
have m resistors in parallel, the resistance between those
two nodes will be 1/m hectoohms.
Before doing this in general, let me revisit the case of
3 dimensions using this notation. In three dimensions,
the vertices of our cube are labelled ---, --+, -+-, +--,
-++, +-+, ++-, +++. Arranging them by number of
plusses and writing the edges between them, we have
0 plusses: ---
3 edges: --- to --+; --- to -+-; --- to +--
1 plus: --+ -+- +--
6 edges: --+ to -++; --+ to +-+; -+- to -++;
-+- to +-+; +-- to +-+; +-- to ++-
2 plusses : -++, +-+, ++-
3 edges: -++ to +++; +-+ to +++; ++- to +++
3 plusses: +++
Hence, the total resistance is 1/3 + 1/6 + 1/3 = 5/6 hΩ as
we know well.
Now we return to the general problem of the n-dimensional
hypercube. By basic combinatorics, we know that there are
\( {n \choose k} \) ways to make a string of k plus
signs and n-k minus signs, hence we have \( {n \choose k} \)
vertices whose labels have k plus signs. Given a vertex with
k plus signs, there are k edges connecting it to vertices with
k-1 plus signs because we can make an edge by changing any
one of the k minus signs. For instance, if we look at the
vertex +-+, we can either change the first + to get the edge
from +-+ to --+ or the second + to get the edge from
+-+ to +--. Hence, there are \( {n \choose k} \)
edges connecting vertices with k plusses to vertices with
k-1 plusses. In the 3 dimensional example, we have
\( 1 \cdot {3 \choose 1} = 1 \cdot 3 = 3 \)
\( 2 \cdot {3 \choose 2} = 2 \cdot 3 = 6 \)
\( 3 \cdot {3 \choose 3} = 3 \cdot 1 = 3 \)
which agrees with the explicit enumeration above.
Making the series-parallel combination in general,
we thus have:
\( R = \sum_{k=1}^n {1 \over k {n \choose k}} \)
Expressing the binomial coefficients in terms of
factorials and tidying up, this can also be written as
\( R = {1 \over n!} \sum_{k=1}^n (n-k)! (k-1)! \)
Here is a table of values:
n R
2 1 100.0 Ω
3 5/6 ≈ 83.33 Ω
4 16/24 = 2/3 ≈ 66.67 Ω
5 64/120 = 8/15 ≈ 53.33 Ω
6 312/720 = 13/30 ≈ 43.33 Ω
7 1812/5040 = 151/420 ≈ 35.95 Ω
8 12288 /40320 = 32/105 ≈ 30.48 Ω
9 95616/362880 = 83/315 ≈ 26.35 Ω
I would have to admire the tenacity of anyone
who would hook up \(2^9\) resistors
into a hypercube arrangement
Last edited:
