Add chained-seam findings note (medial pigeonhole)

Write up the R_T coupling, the uniform-family result (feasible n=9, infeasible
n=12 via empty size-7 universal, 0001112 blocked 210/211), and the open threads.

Co-Authored-By: Claude Opus 4.8 <noreply@anthropic.com>

papers/medial_tire_decompositions_of_plane_triangulations/experiments/chained_seam_findings.md

@@ -0,0 +1,78 @@
+# Chained seams: the medial pigeonhole, located
+
+Companion to the interface-alphabet results (`kempe_interface_admissibility_probe.py`,
+`kempe_tile_overlap_probe.py`). This note pursues the paper's §6 *medial pigeonhole
+programme* — the restriction relation `R_T` between a tire's outer and inner boundary
+states, and the open *chain-pigeonhole conjecture* — at the data level.
+
+Scripts: `kempe_transfer_relation_probe.py`, `kempe_uniform_family_probe.py`.
+
+## Background
+
+A nested chain `T_0 ⊃ T_1 ⊃ …` glues into a proper 3-colouring of `M(G)` iff
+consecutive boundary states match: `inner-state(T_i) = outer-state(T_{i+1})`
+(compatible family, paper Prop "gluing criterion"). Boundary states are necklaces
+(colours permuted, boundary walk rotated/reflected). The restriction relation
+
+```
+R_T ⊆ outer-states × inner-states
+```
+
+records which (outer, inner) boundary-state pairs one Kempe-balanced colouring of `T`
+realises jointly. The chain-pigeonhole conjecture asks whether nested `R_T`'s can ever
+compose to empty.
+
+Established earlier: the realised boundary alphabet (each side) is exactly the full
+parity-admissible set, identical inner vs outer, `n`-independent (n=9, n=12, m=3..8);
+and every *pair* of tiles overlaps, so single seams never dead-end.
+
+## Finding 1 — `R_T` is genuinely coupled
+
+`R_T` is **not** a product of its projections: a tile's inner boundary necklace
+constrains its outer one. Seen at n=9 in the `(p,q) = (4,5)` and `(5,4)` size classes
+(`product? = False`), and broadly at n=12. So "every seam individually realisable"
+does **not** trivially pass through a tile.
+
+## Finding 2 — the uniform shortcut works at n=9, breaks at n=12
+
+Question: is there one boundary state `σ_m` per level-cycle *size* that threads every
+tile simultaneously (outer face and every inner face)? If so, painting every level
+cycle `σ_m` glues any tree with no pigeonhole.
+
+- **n=9: FEASIBLE.** Unique universal per size (`|D[m]| = 1`), and notably *not*
+  monochromatic — the balanced-block necklaces
+
+  ```
+  σ3 = 012   σ4 = 0011   σ5 = 00012   σ6 = 000011
+  ```
+
+  threads all 66 tiles. (Caveat: n=9 has **no** branching tiles.)
+
+- **n=12: INFEASIBLE.** 1237 tiles (1029 bite, 175 genuinely branching). The CSP fails
+  for one sharp reason: **size-7 seams admit no universal state** (`|D[7]| = 0`). The
+  211 size-7 boundaries realise all 10 admissible necklaces between them, but their
+  intersection is empty — the near-universal `0001112` lands on **210 of 211**,
+  blocked by a single tile.
+
+Per-size universal domain sizes at n=12: `3:1  4:1  5:1  6:2  7:0  8:4  9:1`.
+
+## Finding 3 — interpretation
+
+Universals vanish as the tile population grows (more boundaries of a given size → more
+sets to intersect → empty), so the "paint every seam identically" shortcut is doomed
+at scale; the per-interface pigeonhole choice is genuinely necessary. **But this is
+not an obstruction to gluing** — pairwise overlap still always holds, so chains still
+glue by choosing states per interface. We have *located why the conjecture is hard*:
+the difficulty lives in the non-uniform, branch-coupled selection across the tree,
+exactly where the paper's pigeonhole sits — not in any single seam or a uniform
+assignment.
+
+## Open threads
+
+- **Odd-size seams are the pressure point.** Uniformity first fails at size 7 (odd);
+  small odd sizes 3, 5 kept unique universals. Does every large odd size lose its
+  universal as `n` grows? (tracked below / in follow-up).
+- **The 210/211 near-miss.** A single tile blocks the size-7 universal — identifying
+  the "universal-breaker" tiles may expose the combinatorial core.
+- **Branch-tree composition.** n=12 has 175 true branching tiles; composing `R_T`
+  along actual trees (not just per-size uniformity) is the conjecture proper.