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

For $n \ge 3$ and $0 < p < 1$, define the weighted graph $G_n$ on $n+1$ vertices: a path $a = v_0, v_1, \dots, v_n$ of $n$ edges, plus one extra edge directly joining $a$ to $v_n$; write $b := v_1$ and $c := v_n$ as the two 'outer' attachment points (Gladkov, Pak & Zimin 2024, Lemma 4.1). Because $G_n$ is a simple weighted graph, ordinary bond percolation applies to it directly, side-stepping the impossibility of Step 3 by not trying to be an exact hyperedge — only an approximate one, for $n$ large.

A short recurrence computation (2024, §5, Lemma 5.1) shows that, for the complete gadget (including the direct edge) and the edge-closing parameter $p$, $\Pr_p(a \leftrightarrow v_n) = \tfrac{1 - p^{2n}}{1+p}$, which is exponentially close to $\tfrac{1}{1+p}$ for large $n$; combined with further exact computations for the remaining WZ-model probabilities of $G_n$ (relating $p_{abc}, p_{a|bc}, p_{ab|c}, p_{ac|b}, p_{a|b|c}$ to $p$ and $n$), the authors verify directly that the robust hyperedge lemma's inequality $400\, p_{a|bc} \le p_{abc}\, p_{a|b|c} - p_{ab|c}^2$ holds once $n \ge 3 \cdot 401 + 1 = 1204$ at $p = \tfrac12$.

At this specific value $n = 1204$, the gadget $G_n$ has $1205$ vertices and $2407$ edges, is planar, and satisfies exactly the conditions the robust hyperedge lemma of the previous step requires. Six independent copies of this one gadget are all that is needed to replace the six hyperedges of Hollom's hypergraph, which the next step assembles into the final counterexample.