Resistor grid puzzle

[The Phantom]

Imagine a rectangular grid of 1 ohm resistors, 3 on a side like this:

A ___ ___ ___
|-|___|-|-|___|-|-|___|-|
.-. .-. .-. .-.
| | | | | | | |
| | | | | | | |
'-' ___ '-' ___ '-' ___ '-'
|-|___|-|-|___|-|-|___|-|
.-. .-. .-. .-.
| | | | | | | |
| | | | | | | |
'-' ___ '-' ___ '-' ___ '-'
|-|___|-|-|___|-|-|___|-|
.-. .-. .-. .-.
| | | | | | | |
| | | | | | | |
'-' ___ '-' ___ '-' ___ '-'
|-|___|-|-|___|-|-|___|-|
B


Now expand the grid so it's 20 resistors on a side. What is the resistance
between diagonally opposite corners A and B?

Does anyone see a shortcut method?

 

[The Phantom]

On Tue, 5 Aug 2008 10:09:10 +0100, "Fleetie" <fleetie@fleetie.co.uk> wrote:

From a quick look, I make it:

R = Sigma (n=1 -> n=(N-1)) { 1/n }

N is the number of nodes along a length of the square.

So for N=4 (a 4 x 4 grid):

R = 1 + 1/2 + 1/3 = 11/6 ohms.

For N=5:

R = 1 + 1/2 + 1/3 + 1/4 = 25/12 ohms.

I got that by drawing diagonal lines; I think all points on each
diagonal line are at the same potential,

Did you try to verify this assumption?

so all points on a line can
be treated as one node.

So from the first corner to the first diagonal line (which links 2
nodes), there are 2 resistors in parallel, so the resistance is 1/2 ohm.

That is in series with the resistance between the 1st diagonal line (2 nodes)
and the second diagonal line (3 nodes), so there are 4 paralleled resistors
between those 2 diagonal lines, so another 1/4 ohm.

Continue and you get the expression above.

I think.


Martin

 

[Fleetie]

"The Phantom" <phantom@aol.com> wrote

On Tue, 5 Aug 2008 10:09:10 +0100, "Fleetie" <fleetie@fleetie.co.uk> wrote:

From a quick look, I make it:

R = Sigma (n=1 -> n=(N-1)) { 1/n }

N is the number of nodes along a length of the square.

So for N=4 (a 4 x 4 grid):

R = 1 + 1/2 + 1/3 = 11/6 ohms.

For N=5:

R = 1 + 1/2 + 1/3 + 1/4 = 25/12 ohms.

I got that by drawing diagonal lines; I think all points on each
diagonal line are at the same potential,

Did you try to verify this assumption?

No.

When I posted that this morning, I made the assumption based
on intuition.

This evening, I looked at it again, and arrived at another
approach, which is less based on guesswork and more rigorous
(I think!).

It relies on the symmetry of the grid about the diagonal line
AB (referring back to your diagram).

Take the N=3 case (and view in Courier or similar font):

a b c

d e f

g h i

The nodes are at positions a to i, and the 1 ohm resistors
link them on a square grid.

It is obvious from symmetry that b,d are at the same potential,
and f,h, and c,g.

So we can eliminate half of the resistors and nodes b,c,f and
change the remaining resistors from 1-ohm ones to 0.5-ohm ones.
This just exploits the symmetry about the line a-e-i above.

That leaves us with:

a

d e

g h i

Now all resistors are 0.5 ohms.

Again, it is obvious that e,g are at the same potential, so
we use the same trick again on d,e,g,h and remove node e and
the 2 0.5-ohm resistors d-e, and e-h, and make resistors
d-g and g-h into 0.25-ohm ones.

So now we have:

a

d

g h i

Resistors a-d, and h-i are 0.5 ohms for a subtotal of 1.0 ohms.

Resistors d-g, and g-h are 0.25 ohms for a subtotal of 0.5 ohms.

Therefore resistance a-d-g-h-i is 1.5 ohms.

This is the same value as that predicted by my earlier summation
formula, i.e. 1/1 + 1/2 = 3/2 ohms.

I have not examined higher cases this evening.

You have me interested now.


Martin

 

[The Phantom]

On Tue, 5 Aug 2008 20:21:57 +0100, "Fleetie" <fleetie@fleetie.demon.co.uk>
wrote:

"The Phantom" <phantom@aol.com> wrote
On Tue, 5 Aug 2008 10:09:10 +0100, "Fleetie" <fleetie@fleetie.co.uk> wrote:

From a quick look, I make it:

R = Sigma (n=1 -> n=(N-1)) { 1/n }

N is the number of nodes along a length of the square.

So for N=4 (a 4 x 4 grid):

R = 1 + 1/2 + 1/3 = 11/6 ohms.

For N=5:

R = 1 + 1/2 + 1/3 + 1/4 = 25/12 ohms.

I got that by drawing diagonal lines; I think all points on each
diagonal line are at the same potential,

Did you try to verify this assumption?

No.

When I posted that this morning, I made the assumption based
on intuition.

This evening, I looked at it again, and arrived at another
approach, which is less based on guesswork and more rigorous
(I think!).

It relies on the symmetry of the grid about the diagonal line
AB (referring back to your diagram).

Take the N=3 case (and view in Courier or similar font):

a b c

d e f

g h i

The nodes are at positions a to i, and the 1 ohm resistors
link them on a square grid.

It is obvious from symmetry that b,d are at the same potential,
and f,h, and c,g.

So we can eliminate half of the resistors and nodes b,c,f and
change the remaining resistors from 1-ohm ones to 0.5-ohm ones.
This just exploits the symmetry about the line a-e-i above.

That leaves us with:

a

d e

g h i

Now all resistors are 0.5 ohms.

Again, it is obvious that e,g are at the same potential, so
we use the same trick again on d,e,g,h and remove node e and
the 2 0.5-ohm resistors d-e, and e-h, and make resistors
d-g and g-h into 0.25-ohm ones.

So now we have:

a

d

g h i

Resistors a-d, and h-i are 0.5 ohms for a subtotal of 1.0 ohms.

Resistors d-g, and g-h are 0.25 ohms for a subtotal of 0.5 ohms.

Therefore resistance a-d-g-h-i is 1.5 ohms.

This is the same value as that predicted by my earlier summation
formula, i.e. 1/1 + 1/2 = 3/2 ohms.

I have not examined higher cases this evening.

How would you proceed in the n=4 case? Don't quit now; you are *this*
close to enlightenment!

You have me interested now.


Martin

 

[Fleetie]

"The Phantom" <phantom@aol.com> wrote

How would you proceed in the n=4 case? Don't quit now; you are *this*
close to enlightenment!

Hi.

Well, here's what I got today. I'm not too sure about it
though. It's a bit long.

A---B---C---D
| | | |
| | | |
E---F---G---H
| | | |
| | | |
I---J---K---L
| | | |
| | | |
M---N---O---P

Where a simple line join indicates a 1 ohm resistor, is equivalent to:

A
|
-
|
E-|-F
| |
- -
| |
I-|-J-|-K
| | |
- - -
| | |
M-|-N-|-O-|-P

Where a single-barred line ( -|- ) indicates a 1/2 ohm resistor.

Consider links A-E and O-P:

Obviously, R(A-P) == 1/2 + R(E-O) + 1/2
== R(E-O) + 1


E-|-F
| |
- -
| |
I-|-J-|-K
| | |
- - -
| | |
M-|-N-|-O + 1


Now because the entire network is symmetrical about line M-J, nodes M,J must be
at the same potential.

Then, also, because all 4 resistors in square I-J-N-M are equal, the square can
be "folded" along the diagonal I-N, so that nodes J,N end up as 1 node (J),
node M can be removed, and the remaining resistors I-J and J-N are again halved
in value:

E-|-F
| |
- -
| |
I-=-J-|-K
| |
= -
| |
N-|-O + 1

Where a double-barred link -=- indicates a 1/4 ohm resistor.

Now the resistance R(E-O) is just 2 of network E-F-J-I, in series,
so the resistance of the original grid is:

(E-|-F
| |
- -
| |
I-=-J * 2 ) + 1

R(E-J) = R(E-F-J) || R(E-I-J)
= (0.5 + 0.5) || (0.5 + 0.25)
= 1 || 0.75

Using 1/R = 1/R1 + 1/R2 :

1/R(E-J) = 1/1 + 1/0.75
= 1 + 4/3
= 7/3

So R(E-J) = 3/7 ohm

So R(E-O) above = 2*(3/7) = 6/7 ohm.

So R(A-P) above = (6/7) + 1 = 13/7 ohms.

This, IF CORRECT, does NOT equal the value predicted by my initial summation
formula, of:

1/1 + 1/2 + 1/3 = 11/6 ohms.

That in turn would seem to render invalid my assumption that all points on the
diagonal lines I mentioned are at the same potential.

However, it is not clear to me why that should be.

Also, I do not now see how to generate a formula for R for arbitrary N.

I did try the same approach for N=5, but ran into a problem,
because I managed to reduce the network to a small sub-network
that I do not know how to solve for overall resistance.

However, the approach I'm using to reduce the square grid network
can be summarised into a few steps:

1) Fold it dagonally along diagonal line AP (in top diagram above)
to remove one half of the network.

2) Change all resistors in the remaining triangular grid to
half of their previous value

3) Take off the 2 resistors at A and P because they're only connected
to E and O, and just note their total series resistance and add it back
on later.

4) Now consider all nodes on the middle diagonal going from (say) bottom
left to top right (in above example), i.e. the longest diagonal at 90 deg
to the original diagonal AB. The network is symmetrical about this new
diagonal, so all nodes on it are at the same potential. So they can all be
collapsed onto 1 node by removing nodes and resistors and reducing values
of existing resistors in that area (at least for N=4, N=5).

5) Now you end up with 2 identical subnetworks, in series (see
penultimate diagram above), so you only need to solve 1 of them
for R, and then double the result. (This caused me a problem for N=5.)

6) Add back on the series resistance removed in step 3 above.

I'm a bit stuck now, though. And I don't know for sure that my
value of 13/7 ohms for N=4 above is even correct.


Incidentally, the problematic subnetwork is:

----------- A
| |
R1 R4
|__ |
| | |
R2 | R5
| R7 |
| |___|
R3 R6
| |
----------- B

I don't know how to solve for R(A-B).

In my particular case, R1=R4=0.5, R2=R5=0.5, R3=R6=0.25,
and R7=0.5.

I wonder whether since the vertical chains are identical,
I can move the bottom of R7 to the junction of R2, R3, and
it'll still be the same in terms of R(A-B).


Martin
--
Manchester, U.K.

 

[The Phantom]

On Wed, 6 Aug 2008 17:36:00 +0100, "Fleetie" <fleetie@fleetie.demon.co.uk>
wrote:

"The Phantom" <phantom@aol.com> wrote
How would you proceed in the n=4 case? Don't quit now; you are *this*
close to enlightenment!

Hi.

Well, here's what I got today. I'm not too sure about it
though. It's a bit long.

A---B---C---D
| | | |
| | | |
E---F---G---H
| | | |
| | | |
I---J---K---L
| | | |
| | | |
M---N---O---P

Where a simple line join indicates a 1 ohm resistor, is equivalent to:

A
|
-
|
E-|-F
| |
- -
| |
I-|-J-|-K
| | |
- - -
| | |
M-|-N-|-O-|-P

Where a single-barred line ( -|- ) indicates a 1/2 ohm resistor.

Consider links A-E and O-P:

Obviously, R(A-P) == 1/2 + R(E-O) + 1/2
== R(E-O) + 1


E-|-F
| |
- -
| |
I-|-J-|-K
| | |
- - -
| | |
M-|-N-|-O + 1


Now because the entire network is symmetrical about line M-J, nodes M,J must be
at the same potential.

Then, also, because all 4 resistors in square I-J-N-M are equal, the square can
be "folded" along the diagonal I-N, so that nodes J,N end up as 1 node (J),
node M can be removed, and the remaining resistors I-J and J-N are again halved
in value:

E-|-F
| |
- -
| |
I-=-J-|-K
| |
= -
| |
N-|-O + 1

Where a double-barred link -=- indicates a 1/4 ohm resistor.

Now the resistance R(E-O) is just 2 of network E-F-J-I, in series,
so the resistance of the original grid is:

(E-|-F
| |
- -
| |
I-=-J * 2 ) + 1

R(E-J) = R(E-F-J) || R(E-I-J)
= (0.5 + 0.5) || (0.5 + 0.25)
= 1 || 0.75

Using 1/R = 1/R1 + 1/R2 :

1/R(E-J) = 1/1 + 1/0.75
= 1 + 4/3
= 7/3

So R(E-J) = 3/7 ohm

So R(E-O) above = 2*(3/7) = 6/7 ohm.

So R(A-P) above = (6/7) + 1 = 13/7 ohms.

This, IF CORRECT

This is indeed correct.

, does NOT equal the value predicted by my initial summation
formula, of:

1/1 + 1/2 + 1/3 = 11/6 ohms.

And this is incorrect.

That in turn would seem to render invalid my assumption that all points on the
diagonal lines I mentioned are at the same potential.

That assumption is incorrect for the cases where there are 3 or more
resistors (4 or more nodes) on a side. For the cases where n >= 4, some of
the diagonal lines have equipotential nodes, and some do not.

However, it is not clear to me why that should be.

Inadequate symmetry. There is the classic resistor cube problem:

http://www.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/64.42.html

People who have seen that problem and know about the method of solution
that exploits symmetry, upon seeing the finite resistor grid of this
problem, will likely think (as I did at first) that the same kind of
symmetry argument will solve it. I haven't been able to come up with a
shortcut method yet, but there may well be one. Here are a few solutions,
showing the correct value (determined by a brute force solution of the
network), and the (incorrect, if n > 3) value determined by the sum of the
reciprocals of integers (the method you initially proposed, and which I
also initially thought would work):

n correct summation method

3 3/2 3/2
4 13/7 11/6
5 47/22 25/12
6 1171/495 137/60
21 3.95433 3.59774

The n=21 case (the case with 20 resistors on a side) has an exact rational
solution, but it involves a really big numerator and denominator, so I give
the decimal version.

Also, I do not now see how to generate a formula for R for arbitrary N.

I did try the same approach for N=5, but ran into a problem,
because I managed to reduce the network to a small sub-network
that I do not know how to solve for overall resistance.

However, the approach I'm using to reduce the square grid network
can be summarised into a few steps:

1) Fold it dagonally along diagonal line AP (in top diagram above)
to remove one half of the network.

2) Change all resistors in the remaining triangular grid to
half of their previous value

3) Take off the 2 resistors at A and P because they're only connected
to E and O, and just note their total series resistance and add it back
on later.

4) Now consider all nodes on the middle diagonal going from (say) bottom
left to top right (in above example), i.e. the longest diagonal at 90 deg
to the original diagonal AB. The network is symmetrical about this new
diagonal, so all nodes on it are at the same potential. So they can all be
collapsed onto 1 node by removing nodes and resistors and reducing values
of existing resistors in that area (at least for N=4, N=5).

5) Now you end up with 2 identical subnetworks, in series (see
penultimate diagram above), so you only need to solve 1 of them
for R, and then double the result. (This caused me a problem for N=5.)

6) Add back on the series resistance removed in step 3 above.

I'm a bit stuck now, though. And I don't know for sure that my
value of 13/7 ohms for N=4 above is even correct.


Incidentally, the problematic subnetwork is:

----------- A
| |
R1 R4
|__ |
| | |
R2 | R5
| R7 |
| |___|
R3 R6
| |
----------- B

I don't know how to solve for R(A-B).

In my particular case, R1=R4=0.5, R2=R5=0.5, R3=R6=0.25,
and R7=0.5.

I wonder whether since the vertical chains are identical,
I can move the bottom of R7 to the junction of R2, R3, and
it'll still be the same in terms of R(A-B).

No, you can't do that. If you replace R4 and R5 with a single equivalent
resistor, and R2 and R3 similarly, you end up with a Wheatstone bridge
circuit:
http://en.wikipedia.org/wiki/Wheatstone_bridge

This can be solved with the Wye-Delta transformation:
http://en.wikipedia.org/wiki/Y-%CE%94_transform

See also:
http://en.wikipedia.org/wiki/Analysis_of_resistive_circuits

I get a value of 25/44 for this subnetwork. If you continue with your
analysis using that value, do you get the correct value of 47/22 for the
entire n=5 network?

Martin

 

[Jasen Betts]

On 2008-08-05, The Phantom <phantom@aol.com> wrote:

Imagine a rectangular grid of 1 ohm resistors, 3 on a side like this:

A ___ ___ ___
|-|___|-|-|___|-|-|___|-|
.-. .-. .-. .-.
| | | | | | | |
| | | | | | | |
'-' ___ '-' ___ '-' ___ '-'
|-|___|-|-|___|-|-|___|-|
.-. .-. .-. .-.
| | | | | | | |
| | | | | | | |
'-' ___ '-' ___ '-' ___ '-'
|-|___|-|-|___|-|-|___|-|
.-. .-. .-. .-.
| | | | | | | |
| | | | | | | |
'-' ___ '-' ___ '-' ___ '-'
|-|___|-|-|___|-|-|___|-|
B


Now expand the grid so it's 20 resistors on a side. What is the resistance
between diagonally opposite corners A and B?

Does anyone see a shortcut method?

analyse the problem,
derive a formula for size n
let n=20

or use SPICE

Bye.
Jasen

 

[Robert Monsen]

On Wed, 06 Aug 2008 20:10:13 -0700, The Phantom <phantom@aol.com>
wrote:

On Wed, 6 Aug 2008 17:36:00 +0100, "Fleetie" <fleetie@fleetie.demon.co.uk
wrote:

"The Phantom" <phantom@aol.com> wrote
How would you proceed in the n=4 case? Don't quit now; you are *this*
close to enlightenment!

Hi.

Well, here's what I got today. I'm not too sure about it
though. It's a bit long.

A---B---C---D
| | | |
| | | |
E---F---G---H
| | | |
| | | |
I---J---K---L
| | | |
| | | |
M---N---O---P

Where a simple line join indicates a 1 ohm resistor, is equivalent to:

A
|
-
|
E-|-F
| |
- -
| |
I-|-J-|-K
| | |
- - -
| | |
M-|-N-|-O-|-P

Where a single-barred line ( -|- ) indicates a 1/2 ohm resistor.

Consider links A-E and O-P:

Obviously, R(A-P) == 1/2 + R(E-O) + 1/2
== R(E-O) + 1


E-|-F
| |
- -
| |
I-|-J-|-K
| | |
- - -
| | |
M-|-N-|-O + 1


Now because the entire network is symmetrical about line M-J, nodes M,J must be
at the same potential.

Then, also, because all 4 resistors in square I-J-N-M are equal, the square can
be "folded" along the diagonal I-N, so that nodes J,N end up as 1 node (J),
node M can be removed, and the remaining resistors I-J and J-N are again halved
in value:

E-|-F
| |
- -
| |
I-=-J-|-K
| |
= -
| |
N-|-O + 1

Where a double-barred link -=- indicates a 1/4 ohm resistor.

Now the resistance R(E-O) is just 2 of network E-F-J-I, in series,
so the resistance of the original grid is:

(E-|-F
| |
- -
| |
I-=-J * 2 ) + 1

R(E-J) = R(E-F-J) || R(E-I-J)
= (0.5 + 0.5) || (0.5 + 0.25)
= 1 || 0.75

Using 1/R = 1/R1 + 1/R2 :

1/R(E-J) = 1/1 + 1/0.75
= 1 + 4/3
= 7/3

So R(E-J) = 3/7 ohm

So R(E-O) above = 2*(3/7) = 6/7 ohm.

So R(A-P) above = (6/7) + 1 = 13/7 ohms.

This, IF CORRECT

This is indeed correct.

, does NOT equal the value predicted by my initial summation
formula, of:

1/1 + 1/2 + 1/3 = 11/6 ohms.

And this is incorrect.


That in turn would seem to render invalid my assumption that all points on the
diagonal lines I mentioned are at the same potential.

That assumption is incorrect for the cases where there are 3 or more
resistors (4 or more nodes) on a side. For the cases where n >= 4, some of
the diagonal lines have equipotential nodes, and some do not.


However, it is not clear to me why that should be.

Inadequate symmetry. There is the classic resistor cube problem:

http://www.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/64.42.html

People who have seen that problem and know about the method of solution
that exploits symmetry, upon seeing the finite resistor grid of this
problem, will likely think (as I did at first) that the same kind of
symmetry argument will solve it. I haven't been able to come up with a
shortcut method yet, but there may well be one. Here are a few solutions,
showing the correct value (determined by a brute force solution of the
network), and the (incorrect, if n > 3) value determined by the sum of the
reciprocals of integers (the method you initially proposed, and which I
also initially thought would work):

n correct summation method

3 3/2 3/2
4 13/7 11/6
5 47/22 25/12
6 1171/495 137/60
21 3.95433 3.59774

The n=21 case (the case with 20 resistors on a side) has an exact rational
solution, but it involves a really big numerator and denominator, so I give
the decimal version.


Also, I do not now see how to generate a formula for R for arbitrary N.

I did try the same approach for N=5, but ran into a problem,
because I managed to reduce the network to a small sub-network
that I do not know how to solve for overall resistance.

However, the approach I'm using to reduce the square grid network
can be summarised into a few steps:

1) Fold it dagonally along diagonal line AP (in top diagram above)
to remove one half of the network.

2) Change all resistors in the remaining triangular grid to
half of their previous value

3) Take off the 2 resistors at A and P because they're only connected
to E and O, and just note their total series resistance and add it back
on later.

4) Now consider all nodes on the middle diagonal going from (say) bottom
left to top right (in above example), i.e. the longest diagonal at 90 deg
to the original diagonal AB. The network is symmetrical about this new
diagonal, so all nodes on it are at the same potential. So they can all be
collapsed onto 1 node by removing nodes and resistors and reducing values
of existing resistors in that area (at least for N=4, N=5).

5) Now you end up with 2 identical subnetworks, in series (see
penultimate diagram above), so you only need to solve 1 of them
for R, and then double the result. (This caused me a problem for N=5.)

6) Add back on the series resistance removed in step 3 above.

I'm a bit stuck now, though. And I don't know for sure that my
value of 13/7 ohms for N=4 above is even correct.


Incidentally, the problematic subnetwork is:

----------- A
| |
R1 R4
|__ |
| | |
R2 | R5
| R7 |
| |___|
R3 R6
| |
----------- B

I don't know how to solve for R(A-B).

In my particular case, R1=R4=0.5, R2=R5=0.5, R3=R6=0.25,
and R7=0.5.

I wonder whether since the vertical chains are identical,
I can move the bottom of R7 to the junction of R2, R3, and
it'll still be the same in terms of R(A-B).

No, you can't do that. If you replace R4 and R5 with a single equivalent
resistor, and R2 and R3 similarly, you end up with a Wheatstone bridge
circuit:
http://en.wikipedia.org/wiki/Wheatstone_bridge

This can be solved with the Wye-Delta transformation:
http://en.wikipedia.org/wiki/Y-%CE%94_transform

See also:
http://en.wikipedia.org/wiki/Analysis_of_resistive_circuits

I get a value of 25/44 for this subnetwork. If you continue with your
analysis using that value, do you get the correct value of 47/22 for the
entire n=5 network?


Martin

If you have a grid like this and want the resistance between a and p:

a b c d
e f g h
i j k l
m n o p

You can reduce it to a grid like this (by symmetry), where you are
getting the resistance between a and x:

a b c x
f x

For a grid like this:

a b c d e
f g h i j
k l m n o
p q r s t
u v w x y

You can reduce it to the equivalent problem, which is the resistance
between a and x in this network:

a b c d x
g h x
x

This makes the problem much easier. The only issue is the delta-wye
conversion that is required for Ra=agh, Rb=ch, Rc=bc. However, working
it out, you end up with the 47/22 value one might expect.

I'm sure there is a scheme that depends on adding up the total number
of different non-looping paths through the network, with the lengths
of those paths factored in somehow. I'm not sure that this is simpler
than the brute force method, though.

Regards,
Bob Monsen

 

[whit3rd]

On Aug 5, 12:59 am, The Phantom <phan...@aol.com> wrote:

Imagine a rectangular grid of 1 ohm resistors, 3 on a side like this:

A ___ ___ ___
|-|___|-|-|___|-|-|___|-|
.-. .-. .-. .-.
| | | | | | | |
| | | | | | | |
'-' ___ '-' ___ '-' ___ '-'
|-|___|-|-|___|-|-|___|-|
.-. .-. .-. .-.
| | | | | | | |
| | | | | | | |
'-' ___ '-' ___ '-' ___ '-'
|-|___|-|-|___|-|-|___|-|
.-. .-. .-. .-.
| | | | | | | |
| | | | | | | |
'-' ___ '-' ___ '-' ___ '-'
|-|___|-|-|___|-|-|___|-|
B

Now expand the grid so it's 20 resistors on a side. What is the resistance
between diagonally opposite corners A and B?

Does anyone see a shortcut method?

Well, in the continuum approximation, this is one ohm per square sheet
material. You can
solve Poisson's equation for the potential at any point in the
sheet... it's the square boundary
that is the problem in that case, an infinite sheet is.. relatively
easy.

 

[Jasen Betts]

On 2008-08-06, Fleetie <fleetie@fleetie.demon.co.uk> wrote:

"The Phantom" <phantom@aol.com> wrote
How would you proceed in the n=4 case? Don't quit now; you are *this*
close to enlightenment!

That in turn would seem to render invalid my assumption that all points on the
diagonal lines I mentioned are at the same potential.

However, it is not clear to me why that should be.

It's because some nodes have two connections to the next diagonal and
some nodes have only one. in other words some nodes aren't evenly
spaced between their adjacent diagonals.

Bye.
Jasen