transversal.py: a STRONG transversal (n-2 faces whose Boolean assignment
single-valuedly and injectively determines the boundary) would give a
constructive proof. It exists in 0/2948 disks with k>=1 -- once there is
any interior vertex, fixing n-2 faces leaves boundary-visible completion
freedom, so the boundary is never single-valued in them. Works only at
k=0 (base case). Both elementary routes (reduction localization, direct
transversal) now closed.
Co-Authored-By: Claude Opus 4.8 <noreply@anthropic.com>
didericis
c482bc5633