Reconstruction threshold for the three-state symmetric tree channel
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For the three-state symmetric broadcast channel with diagonal (1+2λ)/3 and off-diagonal (1−λ)/3, is reconstruction on every regular b-ary or observed Poisson(d) tree possible exactly above dλ²=1? Here −1/2≤λ≤1 and d=b in the regular case.
Why it matters and what counts as progress
Why it matters: All-degree sharpness would extend known high-branching and selected small-degree threshold results.
What counts as progress: Nonreconstruction below and at the admissible threshold for every b≥2 and d>1; Poisson advantage averages over the observed tree without conditioning on survival.
How the problem developed
Allan Sly Proved the Kesten–Stigum reconstruction threshold for the three-state Potts broadcast model on regular trees of sufficiently large degree; all-degree sharpness remained open.
- 25 Sept 2026 · Reported claim [2]
OpenAI The preprint claims the exact dλ² > 1 threshold, including nonreconstruction at equality for both signs of λ, for every regular b-ary tree and observed Poisson tree in its stated range.
AI contribution
The exact reconstruction threshold for the three-state symmetric channel
For the three-state symmetric channel Kλ(i,i)=(1+2λ)/3 and Kλ(i,j)=(1−λ)/3 for i≠j, with −1/2≤λ≤1, reconstruction from depth-ℓ spins and the observed tree on every regular b-ary tree (integer b≥2) and observed Poisson(d) tree (d>1) occurs exactly when dλ²>1 (d=b in the regular case). At admissible equality there is nonreconstruction. Poisson advantage averages over the tree without conditioning on survival.
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Publication dates and editorial changes.
OpenAI reports the exact three-state tree reconstruction threshold