I know this question might sound like it belongs in physics.SE, and I posted it there, but it doesn't seem to get much attention. What's more, the question to my mind is more mathematical in nature than physics.
I came up with this question as I tried to extend the problem of maximizing the resistance in a circle by selecting two points where to connect the electrodes. The solution for this problem is obvious - antipodal points( which can also be proved mathematically), however, naturally my other problem is more complex, and I don't know how to go about solving it.
I have a circuit which looks like this:
All the rings are connected at all intersection points and all have equal cross sectional area and density.
The problem: Where would you connect two electrodes so that the effective resistance of the circuit would be maximal?
I know you could try doing some complex mathematics with Kirchhoff laws, but is there a simpler way? Could there be a more clever approach using symmetry?
Edit: You can connect electrodes anywhere on the circuit. The centre of each circle lies in another circle.
Disclaimer
Please note that this answer doesn't provide any "simple" way to determine the maximum resistance. It is more on the side of documenting some formula I find during the process of studying this problem. I hope some of them will be useful for others facing similar problem.
Assumption/Summary
For the problem at hand, we will assume the circumference of the $3$ circles are $6$ and the resistance per unit length is $1$. The resistor network we study consists of $6$ nodes joined by $12$ resistors. $3$ resistors has resistance $3$ while the rest has resistance $1$. The maximum resistance is $\frac{63}{37}$, the endpoints are located on two resistors sharing a common node. Both endpoints are at a distance $\frac{63}{37}$ from the common node.
First, consider the case where the endpoints coincide with some nodes.
Given any resistor network with $n$ nodes $v_1, \ldots, v_n$.
Let $C = (c_{ij})$ be its conductance matrix. i.e. an $n \times n$ matrix whose off-diagonal entries $c_{ij}$ is $-\frac{1}{r_{ij}}$ when node $v_i$ is connected to node $v_j$ by a resistor with resistance $r_{ij}$ and $0$ otherwise. The diagonal entries of $C$ are defined by off-diagonal entries by the relation $c_{ii} = -\!\!\sum\limits_{j=1,\ne i}^n c_{ij}$.
Let $V = (V_i)$ be the column vector whose $i^{th}$ component $V_i$ is the potential set up at node $v_i$.
Let $I = (I_i)$ be the column vector whose $i^{th}$ component $I_i$ is the current flowing into the resistor network from environment through node $v_i$.
Let $\theta$ be the unit column vector with all entries equal to $\frac{1}{\sqrt{n}}$.
To determine the resistance between two nodes, we need to set up a current configuration where one unit of current enters one node, same amount of current leaves at another node and then measure the potential drop across these two nodes.
To setup any current configuration $I$ on the resistor network, Kirchhoff's laws tells us the potential need to satisfy $C V = I$. This prompt us to invert the matrix $C$. However, $C$ is not invertible. It has $\theta$ as an eigenvector with eigenvalue zero. To solve this problem, let $\lambda_2, \ldots, \lambda_n$ be the non-zero eigenvalues of $C$ and $\psi_i = (\psi_{ia} )_{a=1,n}$ be the unit eigenvector corresponds to $\lambda_i$. $C$ has following decomposition, $C = \sum\limits_{i=2}^n\lambda_i \psi_i \otimes \psi_i$. (For any column vector $\xi = (\xi_i), \eta = (\eta_j)$, $\xi \otimes \eta$ stands for their outer product, a $n\times n$ matrices with entries $\xi_i \eta_j$ ). We define the matrix $\Lambda$ by following formula:
$$\Lambda \stackrel{def}{=}\sum_{i=2}^n\lambda_i^{-1}\psi_i \otimes \psi_i = ( C + \theta\otimes\theta)^{-1} - \theta \otimes\theta$$
As one can see from below, this matrix $\Lambda$ do act like an inverse of $C$.
For any physical current configuration $I$, there is no net current flowing in/out of the resistor network. This means $\theta\cdot I = 0$. For such a current configuration, we have $$\begin{align} CV = I \implies & (C + \theta\otimes\theta)V = I + (\theta\cdot V)\theta\\ \implies & V = (\Lambda + \theta \otimes \theta)( I + (\theta\cdot V)\theta) = \Lambda I + (\theta\cdot V)\theta \end{align} $$ This tell us $\Lambda I$ determines $V$ up to an overall offset of constant.
In particular, let $I_{ab}$ be the current configuration where unit current enter the network at node $v_a$ and leave at $v_b$. The corresponding column vector has $+1$ at $a^{th}$ slot, $-1$ at $b^{th}$ slot and zero otherwise. The voltage drop across $v_a$ to $v_b$ and hence the resistance equals to
$$\begin{align}R_{ab} \stackrel{def}{=} V_a - V_b & = I_{ab}V = I_{ab}( \Lambda I_{ab} + (\theta\cdot V)\theta ) = I_{ab} \Lambda I_{ab}\\ &= \Lambda_{aa} + \Lambda_{bb} - 2\Lambda_{ab} \end{align}$$
This is the resistance we seek when node $v_a$ and $v_b$ are the endpoints.
Apply this to the problem at hand and label the nodes as shown in following diagram.
We find
$$ C = \frac13\begin{bmatrix} 8 & -1 & -1 & -3 & 0 & -3\\ -1 & 8 & -1 & -3 & -3 & 0\\ -1 & -1 & 8 & 0 & -3 & -3\\ -3 & -3 & 0 & 12 & -3 & -3\\ 0 & -3 & -3 & -3 & 12 & -3\\ -3 & 0 & -3 & -3 & -3 & 12 \end{bmatrix} \quad\text{and}\quad (R_{ab}) = \frac{1}{14} \begin{bmatrix} 0 & 10 & 10 & 7 & 9 & 7\\ 10 & 0 & 10 & 7 & 7 & 9\\ 10 & 10 & 0 & 9 & 7 & 7\\ 7 & 7 & 9 & 0 & 6 & 6\\ 9 & 7 & 7 & 6 & 0 & 6\\ 7 & 9 & 7 & 6 & 6 & 0 \end{bmatrix} $$ Since the largest entry in $(R_{ab})$ is $\displaystyle\;\frac{5}{7}\;$, this is the largest resistance if we restrict the endpoints to the nodes.
If we lift this restriction, there are two possibilities. We either insert the two endpoints to a single resistor or place them or two different resisters.
Case I - two endpoints on single resistor.
Let's say we place the two endpoints at relative position $0 \le \lambda_1, \lambda_2 \le 1$ on the resistor connecting nodes $v_a$ and $v_b$. (relative position $\lambda = 0$ means place the endpoint at $v_a$, $1$ means at $v_b$ and $\frac12$ at midpoint of resistor). The resistance between the two endpoints is given by the formula
$$R_{I}(\lambda_1,\lambda_2) = p((1-p)r_{ab} + pR_{ab}) \quad\text{ where }\quad p = |\lambda_1 - \lambda_2| $$ Since $r_{ab} \le 3$ and $R_{ab} \le \frac{5}{7}$, the resistance for this case is always smaller than $1$. $$R_{I}(\lambda_1,\lambda_2) \le \max_{p \in [0,1]}p\left(3(1-p) + \frac{5}{7}p\right) = \frac{63}{64}$$
Case II - endpoints on different resistors.
One key feature of the formula $R_{I}(\lambda_1,\lambda_2)$ above is that it is a quadratic polynomial in terms of the relative positions of the endpoints. This is also true when the two endpoints are located on different resistors.
Let's say we place one endpoint at relative position $p \in [0,1]$ on the resistor joining $v_a$ and $v_b$ and another endpoint at relative position $q \in [0,1]$ on the resistor joining $v_c$ and $v_d$. The resistance between the endpoints is given by the formula:
$$ \begin{align} R_{II}(p,q) = \color{blue}{R_{ac}} &+ \color{blue}{(R_{bc} - R_{ac})p} + (r_{ab}-R_{ab})p(1-p)\\ &+ \color{blue}{(R_{ad} - R_{ac})q} + (r_{cd} - R_{cd})q(1-q)\\ &+ \color{blue}{( R_{ac} + R_{bd} - R_{ad} - R_{bc})pq} \end{align}\tag{*1} $$ The four terms in blue are essentially bi-linear interpolation of the four values $R_{ac}, R_{ad},R_{bd},R_{bc}$, their contribution together is at most $\max\{R_{ac}, R_{ad},R_{bd},R_{bc}\} \le \frac{5}{7}$. For the remaining two terms, if $r_{ab} = 3$, then $r_{ab} - R_{ab} = 3 - \frac{5}{7} = \frac{16}{7}$. Otherwise, $r_{ab} - R_{ab} < r_{ab} = 1$. Same thing happens to the resistor joining $v_c$ to $v_d$.
Case IIa - $r_{ab} \ne 3$ or $r_{cd} \ne 3$, we have $$R_{IIa}(p,q) \le \frac{5}{7} + \frac{16}{7} \times \frac14 + 1 \times \frac14 = \frac{43}{28} $$
Case IIb - $r_{ab} = r_{cd} = 3$,
The two resistors has a common node. Let's say $v_a = v_c$, formula $(*1)$ reduces to $$R_{IIb}(p,q) = \frac{1}{7}( 5p + 16p(1-p) + 5q + 16q(1-q) -5pq )$$ Notice $$R_{IIb}(u+v,u-v) = \frac17\left(42u - 37u^2 - 27v^2\right) = \frac{63}{37} - \frac{37}{7}\left(u-\frac{21}{37}\right)^2 - \frac{27}{7}v^2 $$ The maximum happens at $(u,v) = (\frac{21}{37},0) \iff p = q = \frac{21}{37}$ with value $\frac{63}{37}$.
Since $\frac{63}{37} > \frac{43}{28} > \frac{63}{64} > \frac{5}{7}$, we can conclude the maximum resistance is $\frac{63}{37}$. The maximum value is achieved when two endpoints are located on two resistors of resistance $3$ sharing a common node. The distance between the endpoint and the common node is $3\times \frac{21}{37} = \frac{63}{37}$.