Worked solution: Gladkov–Pak–Zimin's explicit counterexample disproving the bunkbed conjecture (2024)

Take Hollom's hypergraph $H$ from Step 2 and, for each of its six $3$-uniform hyperedges $\{a,b,c\}$, substitute a fresh copy of the path gadget $G_n$ from Step 5 (with $n = 1204$), identifying the gadget's $a$ with the hyperedge's transversal vertex and its $b, c$ with the other two vertices (Gladkov, Pak & Zimin 2024, §4.2). The resulting graph $G$ is planar (since $H$ is drawn planar and $G_n$ is planar) with $|V(G)| = 10 + 6 \times 1202 = 7{,}222$ vertices and $|E(G)| = 6 \times 2407 = 14{,}442$ edges, retaining the same three transversal vertices $T = \{u_2, u_7, u_9\}$ as Hollom's hypergraph.

By Step 5's verification that $G_n$'s probabilities satisfy the robust hyperedge lemma's inequality, and Step 4's lemma itself, ordinary $\tfrac12$-bond percolation on $G$ satisfies $\Pr[u_1 \leftrightarrow u_{10}] < \Pr[u_1 \leftrightarrow u_{10}']$ — this is Theorem 1.2 of Gladkov, Pak & Zimin (2024): there exists a connected planar graph on $7{,}222$ vertices and $14{,}442$ edges, a transversal set of size $3$, and two vertices $u, v$, for which $\Pr^{\text{bb}}_{1/2}[u \leftrightarrow v] < \Pr^{\text{bb}}_{1/2}[u \leftrightarrow v']$ — an explicit, fully constructive counterexample to the bunkbed conjecture. Notably, the argument establishes the inequality analytically, through the chain of lemmas built up over the previous steps, rather than by any brute-force computer search over percolation outcomes on the full graph (which would be far too large to enumerate).

The authors note that three transversal vertices is provably the minimum possible for any counterexample (fewer cannot work, by the known validity of the conjecture for one or two transversal vertices — see the final step), though the total vertex count $7{,}222$ is very unlikely to be optimal; smaller counterexamples almost certainly exist but were not the goal of this construction.