# Empirical state of the Conjecture-5.1 proof

This document is a snapshot of where the proof of Conjecture 5.1
(the face-monochromatic-pair conjecture) stands after the empirical
work in `experiments/`. It is meant as commentary for the reader who
has finished reading the paper and wants to know what's been verified
computationally vs what remains to be proven structurally.

## Summary table

| claim | status | empirical evidence |
|---|---|---|
| Conjecture 5.1 (clauses 1–3) | conjecture | ✓ 535,182 / 535,182 (n ≤ 21, direct witness search) |
| Conjecture 5.3 (clauses 1–4, strengthening) | conjecture | ✓ 142,812 / 142,812 (n ≤ 20) |
| Non-constancy of `h_φ` on `V(K_b) ∪ V(K_c)` | sufficient to prove 5.1 via Lemma 5.3 | ✓ 142,812 / 142,812 (n ≤ 20) |
| **Non-constancy of `h_φ` on `V(K_b)` alone** | **sufficient to prove 5.1 via Corollary 5.4** | ✓ 142,812 / 142,812 (n ≤ 20) |
| Lemma A: `h_φ(v_0) = h_φ(v_1)  ⇔  c-edges on opposite local sides` (Lemma 5.2 in the paper) | proven (Lemma 5.2) | ✓ 625,200 / 625,200 consecutive pairs |
| Identity `s_b ⊕ s_c = i_b ⊕ i_c ⊕ 1` at shared vertex | follows from the Heawood definitions + Lemma A | ✓ 263,004 / 263,004 shared vertices |
| Parity-bucket symmetry `n_{(0,0)} = n_{(1,1)}` and `n_{(0,1)} = n_{(1,0)}` over shared vertices | structural (likely provable) | ✓ universal |
| At consecutive shared `K_b`-vertices, `i_b` parity strictly alternates | structural | ✓ universal (transition matrix has 0 within-parity transitions) |
| At odd-`i_b` shared vertices, `i_c` parity flip is forced on transition | structural | ✓ universal |

## What's actually open

The shortest path to a proof of Conjecture 5.1 is now:

> Show structurally that for every chord-apex+Kempe colouring `φ` of every
> reduced dual `Ĝ'_{v,i}`, `h_φ` is not constant on `V(K_b)` (or
> equivalently on `V(K_c)`).

This is verified on 142,812 / 142,812 colourings up to `n = 20`. It is
captured in the paper by **Corollary 5.4** (the per-cycle form of
Lemma 5.3) and **Remark 5.5** (the empirical near-proof).

## Why the proof is harder than it looks

Three empirical facts from the diagnostics in `experiments/` rule out
the simplest proof strategies:

1. **The obstruction has no slack.** The minimum Heawood-flip count on
   `K_b` observed across the data is **2**, attained on 12 colourings at
   `n = 18`. Those colourings have *one single minority Heawood vertex*
   on `V(K_b)` — flipping its sign would give constancy. So the proof
   cannot rest on bulk inequalities like "at least half of `V(K_b)` has
   each sign"; it must rule out every single-minority configuration.

2. **The minority isn't anchored to a structural vertex.** For
   low-flip-count colourings, the minority vertex(es) are distributed
   roughly evenly across `v_n`, `A_0, …, A_4`, and "other" non-named
   vertices in the rest of `G'`:

       v_n        12.86%
       A_{i+1}    10.82%
       A_{i+2}     8.98%
       A_i         7.76%
       A_{i+4}     5.31%
       A_{i+3}     5.10%
       other      49.18%

   So no single named vertex is *always* the minority — the proof
   cannot fix on "vertex X must have Heawood opposite to the majority";
   roughly half the time the minority lives outside the named six.

3. **No single named-vertex-pair is always a Heawood mismatch.** The
   most "reliable" same-cycle pair is `(A_i, A_{i+1})` and
   `(A_{i+1}, A_{i+2})` at consecutive face-boundary positions, each
   with 75% mismatch rate — well short of universal. So the proof can
   neither identify a specific edge that always has differing Heawood
   at its endpoints, nor a specific vertex that's always minority.

Together (1)–(3) say the obstruction is **global, not local**: there
*is* always a Heawood mismatch on `V(K_b)`, but where it sits varies by
colouring. A successful structural proof will need a global argument —
likely a topological / homological / parity-counting argument that
operates on all of `V(K_b)` simultaneously, rather than identifying a
specific forced flip.

## Candidate mechanisms (none confirmed)

These were explored and the corresponding diagnostics ruled them out
or revealed why they don't yield a contradiction on their own:

- **Heawood sum identity `∑_v h_φ(v) = 0`.** Holds only ~17.6% of the
  time on chord-apex+Kempe colourings; the sum can be anywhere in
  `{-24, -20, …, 24}`. So this classical identity is *not* available
  here.

- **Heawood sum on a single Kempe cycle `∑_{V(K)} h_φ = 0`.** Holds only
  ~23% of the time per cycle.

- **Cycle-side balance `|L_b| = |R_b|`.** Holds only 35.43% of the
  time. Constancy *would* force this exactly, but the empirical
  imbalance is large in most colourings.

- **Specific named-vertex pair always mismatches.** No such pair
  exists; closest is `(A_j, A_{j+1})` at 75%.

## Files

- Paper text: `paper.tex`, sections 3 (Heawood number definition,
  Lemma 5.2), 5 (Lemma 5.3, Corollary 5.4, Remark 5.5).
- Diagnostic scripts: see `experiments/check_heawood_*.py`,
  `experiments/check_kempe_intersection_and_alternation.py`,
  `experiments/check_shared_*.py`,
  `experiments/check_cw_parity_prediction.py`,
  `experiments/check_constancy_obstruction.py`,
  `experiments/check_min_flip_structure.py`,
  `experiments/check_minority_location.py`.

## Failed proof route via edge-sharing Kempe constancy

A natural-looking strategy to prove Conjecture 5.1 was:

> **Conjecture (now disproved):** If $K_0$ is an $\{a,b\}$-Kempe cycle
> of $\varphi$ and $K_1$ is an $\{a,c\}$-Kempe cycle of $\varphi$ that
> shares an edge with $K_0$, then $h_\varphi$ cannot be constant on
> both $V(K_0)$ and $V(K_1)$ simultaneously.

Combined with Lemma 5.3 (no clause-3 witness $\Rightarrow$ constancy
on both $V(K_b)$ and $V(K_c)$), this would have closed Conjecture 5.1.
It is **false** by a concrete counterexample (Figure in `paper.tex` at
`\ref{fig:no-two-constant-kempe-counterexample}`). A partial proof
attempt is preserved in the paper alongside the disproof:

- Step 1 (local CW structure) — unconditional.
- Step 2 (forced odd-crossing $\Rightarrow |E(K_0) \cap E(K_1)|$ even
  and $\geq 2$) — unconditional.
- Step 3 (Heawood face-sum mod 3) — unconditional but does not yield
  a contradiction on its own.
- Step 4 (lune-face Case A) — closes the sub-case where two shared
  a-edges are consecutive on \emph{both} cycles (automatic when
  $|E(K_0) \cap E(K_1)| = 2$).
- Step 5 (general case) — open / now known false via the
  counterexample.

The counterexample shows the general case of the conjecture is
unsalvageable; the search for a structural proof of Conjecture 5.1
will need a different angle.

## Snapshot date

This commentary is current as of commit `e688037`.
