Find the resistance between A and B

[cheesehead]

Originally Posted by Batalov

College-bound is gramatically correct

... but so hidebound (definition 2).

[davieddy]

Originally Posted by Batalov

I've already prepared a picture, so let it not be wasted.
For the benefit of later readers.

The problem is quite elegant and has a benefit over the GLAT problem: I was able to explain it to my college-bound daughter, and had fun in the process.

Thank you, David!

Sorry for the late response.
In order:

Nice pic, but not sure what it adds, apart from 1/2 being midway
between 0 and 1.

As for elegance, I'm sure that William will testify that after
sussing the two hints I provided, the rest is somewhat tedious
and error prone.

Glad you got fun talking about it with with your daughter.

The only credit I would like to claim is in spotting a neat
problem in "Physicsforums" the day before.

That said, I have received a PM from someone who guessed
(correctly) that I taught him under the nickname of Doc Eddy
preumably prior to 1987, on the strength of my username and the problem
bearing my trademark. This might sound like one of those
junk Emails that inform you that every girl within twenty miles
of Milton Keynes wants sex with you, but I would like to think
otherwise.

There are a couple of obvious extensions to the puzzle:

What if you extend outwards?
If you are worried about the logistics of nesting triangles
of the same resistance per unit length ad infinitum, you could
always keep the resistivitiy constant and allow the diameter of
the wire to decrease by half each time.

David

[m_f_h]

Originally Posted by davieddy

Nice pic, but not sure what it adds, apart from 1/2 being midway
between 0 and 1.

The hint to consider the triangle as "black box" with resistance proportional to side length.

There are a couple of obvious extensions to the puzzle:

What if you extend outwards?

Yes, or to 3 (or more?) dimensions. (triangle => tetrahedron)

PS: also agree with contfrac([0,[1,1,4,1]]).

[davieddy]

Originally Posted by m_f_h

The hint to consider the triangle as "black box" with resistance proportional to side length.

Yes, or to 3 (or more?) dimensions. (triangle => tetrahedron)

PS: also agree with contfrac([0,[1,1,4,1]]).

I think we have already hinted enough to render the spoilers unnecessary.

I know enough (but no more) about crystallography to know
that extending patterns to 3D is non-trivial. Ditto the 17 patterns
of 2D wallpaper.

Not unrelated is the exercise of deriving the moment of inertia
of a triangle or rectangle using only the fact that it goes as
length^4 and the || axis theorem.

[davieddy]

Originally Posted by m_f_h

Yes, or to 3 (or more?) dimensions. (triangle => tetrahedron)

Can you pack all of 3D space with tetrahedrons?
You may be thinking of diamond.

[m_f_h]

Originally Posted by davieddy

Can you pack all of 3D space with tetrahedrons?
You may be thinking of diamond.

Er... I really did think of tedrahedrons, but I don't know if I can pack what you say... should I ?
Specifically, I *was* thinking of a tedrahedron, but in my mind I had +1'd all dimensions, so wires became faces...

Now there may be another possibility, keeping wires, but then the resulting object may be more complicated....
anyway, 'twas just (the beginning of) an idea...

[R.D. Silverman]

Originally Posted by m_f_h

Er... I really did think of tedrahedrons, but I don't know if I can pack what you say... should I ?
Specifically, I *was* thinking of a tedrahedron, but in my mind I had +1'd all dimensions, so wires became faces...

Now there may be another possibility, keeping wires, but then the resulting object may be more complicated....
anyway, 'twas just (the beginning of) an idea...

Something a bit more challenging....


Suppose every edge of a hypercube is a resistor with r=1.
What is the total resistance between opposite corners.

[Kees]

Start with small dimension (n=3, n=4)

For n=3 we get a resistance of

1/3 + 1/6 + 1/3 = 5/6

For n=4 we get a resistance of

1/4 + 1/12 + 1/12 + 1/4= 2/3

For n=5 we get

1/5 + 1/20 + 1/30 + 1/20+1/5= 8/15

[davieddy]

Originally Posted by R.D. Silverman

Something a bit more challenging....


Suppose every edge of a hypercube is a resistor with r=1.
What is the total resistance between opposite corners.

If Dr Silverman mentions "challenging" I would normally take this
as a warning not to touch the problem with a barge pole.
However, after a bit of thought (high time I tried that again), I find I've
intuitively made some progress. Excuse lack of spoilers, and playing
fast and loose with math terminology.

The vertices of the hypercube are n dimensional vectors with
each component +/- 1.
Each vertex is joined by an edge (resistor) to all others whose
vectors differ in the sign of one and only one component. So each
vertex has n edges, none of which are connected to vertices with the
same number of minus signs.

If we take the "opposite corners" to be the vertices with all +
and all - components, then I guess all vertices with the same number
of minuses will be at the same potential, and none of them connected
to each other.

The number of vertices with r minus signs is nCr.
Each of these vertices has n connections to vertices
with r+1 or r-1 minus signs.

What we want to find is "how many resistors between r and r+1?".

Enough for now

David

[davieddy]

If x(r) resistors join vertices with r minuses to those
with r+1 minuses, we have:

x(0) = n
x(r) = n*nCr - x(r-1)

I assume someone is up to getting an expression for x(r) from this.

The resistance between opposite corners is the sum of 1/x(r)
from r=0 to r=n-1.

[Kees]

http://arxiv.org/abs/0904.1757