Recovering Conductances of Resistor Networks in a Punctured Disk

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Problem: Reconstructed Conjecture 2: for a connected resistor network in a punctured disk $\Gamma$, with medial graph $M(\Gamma)$, $\Gamma$ is irreducible by

$$Y\!-\!\Delta,\ \text{pendant removal, self-edge removal, parallel/series reductions, antenna jumping, antenna absorption}$$

iff $\Gamma$ is irreducible by the same moves except antenna absorption, $M(\Gamma)$ has three medial strands pairwise intersecting twice, and some medial strand self-intersects.

Result: The conjecture is false.

Take $\Gamma$ to be the graph $K_4$ embedded as a wheel: three boundary vertices $v_1,v_2,v_3$ on the outer boundary forming a triangle, and the unique interior boundary vertex $b$ joined to all three $v_i$. Thus

$$E(\Gamma)=\{v_1v_2,v_2v_3,v_3v_1,bv_1,bv_2,bv_3\}.$$

This is a nontrivial rnpd: $b$ is surrounded by the triangle.

All four vertices are boundary vertices, so the response matrix is just the weighted graph Laplacian. Its six off-diagonal entries are exactly the six edge conductances. Hence the response set contains a $6$-dimensional open set. Any graph with fewer than six edges has a response map depending rationally on at most five conductance parameters, so its image has Zariski dimension at most $5$, and cannot contain that open set. Therefore $\Gamma$ is not electrically equivalent to any graph with fewer edges. In particular it is irreducible by any sequence of the listed local moves, including antenna absorption.

Now compute its medial strands. Let

$$e_i=v_iv_{i+1},\qquad f_i=bv_i$$

with indices mod $3$. Tracing straight through the medial vertices gives the three strands

$$\begin{aligned} S_1&: m_{e_1},m_{f_2},m_{f_3},m_{e_3},\\ S_2&: m_{e_1},m_{f_1},m_{f_3},m_{e_2},\\ S_3&: m_{e_2},m_{f_2},m_{f_1},m_{e_3}. \end{aligned}$$

Thus each pair of strands intersects exactly twice:

$$S_1\cap S_2=\{m_{e_1},m_{f_3}\},\quad S_1\cap S_3=\{m_{f_2},m_{e_3}\},\quad S_2\cap S_3=\{m_{f_1},m_{e_2}\}.$$

But no strand self-intersects: each $S_i$ passes through four distinct medial vertices, and every medial vertex is shared by two distinct strands.

So $\Gamma$ satisfies the left-hand side of the conjecture but fails the required self-intersecting-strand condition. Hence Conjecture 2, as stated, is false.

Audit: the counterexample is connected, planar, has exactly one interior boundary vertex, is nontrivial, and irreducibility was proved in the stronger electrical sense, so no extra assumptions are being used.

Citation: Conjecture and terminology: Yulia Alexandr, Brian Burks, Sunita Chepuri, Patricia Commins, “Recovering Conductances of Resistor Networks in a Punctured Disk,” arXiv:1812.01517, Conjecture 2. Counterexample above.