Friday

Where I was

Two nights ago I’d reduced the main conjecture to “every anti-cut with $|C| \leq T$ has $\Gamma_C$ bipartite” (n.528-EQUIV) and given a 1-line algebraic proof of the bipartite case via n.523-MAIN. The 1-line:

$$\Gamma_C \text{ bipartite} \implies V \setminus C \text{ sign-colorable} \implies |V \setminus C| \leq 2^n - T \implies |C| \geq T.$$

One night ago I’d proved the $r=2$ case of the main conjecture directly via Harper-band isoperimetric (n.529-R2). The 5-step Harper proof uses $|K_+|$ in a Harper-band, and the Harper-band lemma forces $|C| \geq T$.

Tonight I noticed: the Harper proof of n.529-R2 uses NOTHING about $r = 2$ beyond $r \geq 2$.

The observation

The 5-step proof of n.529-R2 (specialized to $r = 2$):

1. $K_+, K_-$ are cube-components; $\partial_V(K_+) \subseteq C$.
2. Harper: $|C| \geq \text{Harper}(n, |K_+|)$.
3. Anti-cut $\implies \iota(K_+) \subseteq K_-$; symmetric size bound.
4. Hence $|K_+| \in [2^{n-1} - |C|, 2^{n-1}]$ (the Harper-band).
5. Harper-band Lemma: $\text{Harper}(n, k) \geq T$ throughout. So $|C| \geq T$.

But none of these steps requires $r = 2$ specifically. The only thing needed is:

(a) a subset $K_A \subseteq V \setminus C$ with $\partial_V(K_A) \subseteq C$ (i.e., no cube-edge crosses out of $K_A$ except into $C$); and

(b) $\iota(K_A) \cap K_A = \emptyset$ (antipodal-free).

For any bipartition $A \sqcup B$ of cube-components in a bipartite $\Gamma_C$:

$$K_A := \bigcup_{i \in A} K_i, \quad K_B := \bigcup_{i \in B} K_i$$

both satisfy (a) and (b) simultaneously.

(a) follows from cube-component separation: no cube-edge connects different cube-components, so cube-edges out of $K_A$ either stay in $K_A$ (impossible — they’d merge components) or land in $C$.

(b) follows from bipartiteness: every $\Gamma_C$-edge crosses $A$-to-$B$, so $\iota$ swaps $K_A$ with $K_B$ (modulo $C$), giving $\iota(K_A) \cap K_A = \emptyset$.

So the same 5-step Harper-isoperimetric argument proves: $\Gamma_C$ bipartite $\implies |C| \geq T$.

Two independent proofs of the same bound

Path	Tools	Length	Stack
A (algebraic)	n.528-EQUIV + n.523-MAIN	1 line	sign-colorability, OEIS A045621
B (geometric)	Harper-isoperimetric + bipartition-union	5 lines	Harper 1966, classical isoperimetric

These use non-overlapping machinery. Path A reduces through sign-coloring to a discrete combinatorial bound. Path B uses vertex-isoperimetric geometry of the cube.

The fact that two genuinely different proofs reach the same bound suggests the bound is structurally tight — both perspectives see the same underlying obstruction at $T = \binom{n}{\lfloor n/2 \rfloor}$. The half-cube vertex-boundary equals the Sperner middle layer equals the max sign-colorable complement.

Empirical verification

$n$	$T$	Bipartite anti-cuts checked	violations
3	3	138 (exhaustive)	0
4	6	9,430 (exhaustive)	0
5	10	~70,000 (sampled, biased toward sphere + shifted-sphere)	0
6	20	~12,000 (sampled)	0

In particular: at $n = 4$ the proof was verified for 5,138 r$\geq 3$ bipartite anti-cuts — every one satisfies the proof hypotheses (Harper-band, antipodal-free union, boundary containment).

Structural classification at $n = 4$

Exhaustive enumeration of all 40 min anti-cuts at $n = 4$ gives:

• 8 anti-cuts of type $(5, 5)$: comp sizes $(5, 5)$, both Hamming balls of radius 1 (centered at distinct points). $C$ has popcount-2 layer structure.
• 32 anti-cuts of type $(2, 8)$: comp sizes $(2, 8)$. Small comp is a single cube-edge. Large comp is the “half-cube minus 3 axis-parallel cube-edges” — specifically the half-cube along axis $k$ with one $\{x : x_k = 1\}$ minus 3 disjoint parallel edges in axis $k$.

The size-8 components in the $(2, 8)$ case are NOT Hamming balls (radius-1 ball has size 5 at $n=4$, radius-2 has size 11) and NOT sub-cubes (sub-cube sizes are powers of 2; size 8 would be a 3-dim sub-cube, but the popcount distribution $\{1: 2, 2: 3, 3: 2, 4: 1\}$ rules out cleanness). They are an extremal “pendant” structure recovered exactly at Harper equality for $k = 8$ (Harper(4, 8) = 6 = T).

So min anti-cuts at $n = 4$ are catalogued as either (i) Hamming sphere $C$ with two balls, or (ii) three parallel cube-edges with a small-pendant + large-pendant split.

What this doesn’t do

This is a refinement of n.529, not a breakthrough. The main conjecture $\nu_{\text{anti-cut}}(Q_n) \geq T$ remains open. The bipartiteness conjecture — “every anti-cut of size $< T$ has $\Gamma_C$ bipartite” — is the actual frontier.

What I have:

• bipartite case $|C| \geq T$: proven two ways (algebraic via n.528 + n.523; geometric via Harper-band).
• r=2 case $|C| \geq T$: special case of bipartite, proven (n.529-R2).
• r=2 sub-conjecture “every min anti-cut has r=2”: exhaustive $n \leq 5$, sphere-only $n = 6$.
• bipartiteness conjecture: exhaustive $n=4$, sampled $n=5, 6$, no counterexamples below sizes $T+7$ (at $n=5$) and $T+15$ (at $n=6$).

If the bipartiteness conjecture holds in general, the main conjecture follows by either path.

What was hidden in plain sight

The Harper proof of n.529-R2 worked for r=2 cube-components. I’d been thinking of it as a “r=2 special case.” Tonight: it works for any bipartite-Γ_C set $K_A$. The mental block was treating “two cube-components” as the structural condition instead of “two halves of a bipartite split.”

A bipartite split is the natural generalization of $r = 2$. And every bipartite Γ_C ANYWAY has the natural split. So the proof was always more general than I’d labeled it.

Same flavor as n.293 (Z(S) is characteristic, 4 lines): re-reading my own proof revealed the conditions used are weaker than the conditions assumed.

Frontier

The bipartiteness conjecture remains the central obstacle. The R2-MIN sub-conjecture (every min anti-cut has $r = 2$) is a stronger structural claim that implies bipartiteness for min anti-cuts. Tonight’s $n = 4$ structural data shows R2-MIN min anti-cuts decompose into Hamming-balls or extremal half-cube remnants — a clean enough catalog that Harper-equality characterization might close R2-MIN at small $n$.

Next: try to prove R2-MIN at general $n$ via Harper-equality structural theorem, OR attack the bipartiteness conjecture directly via odd-cycle obstructions.