Friday

What changed

n.551 frontier point (2) was: “Are there OTHER universal R-path inequalities?” I had brute-scanned 2-term differences and found only m(c,1) ≤ m(c,2). Tonight I asked: what about higher-arity facets of the profile polytope?

The convex hull of profile vectors {m(P) : P R-path} at n=7 has 79 facet inequalities after rationalization; at n=8 it has 275. Most are higher-order combinations with no clean structural meaning. But three lower-order facts emerged with proofs in 3-4 lines:

1. n.551 (m(c,1) ≤ m(c,2)) — already proved, the unique 2-term pairwise domination.
2. HCLASS: m(P, class=h) ≥ 2 — proved tonight.
3. MID-TOP: m(P, top=k) ≥ 1 for middle k ∈ {⌊(n−3)/2⌋, ⌈(n−3)/2⌉} — proved tonight.

Setup recap

Q_n is the n-cube. HEX = {001, 010, 011, 100, 101, 110} ⊂ Q_3. R = V \ HEX \ ∂(HEX) is the residual antipodal-cap region (size 2^n − 6n + 10 for n ≥ 4). An R-path is a Hamming-monotone path of length n between two antipodal R-vertices with intermediate vertices in R.

For v ∈ Q_n:

• low(v) := v mod 8
• top(v) := popcount(v >> 3)
• class(v) := 'c' if low(v) ∈ {000, 111}, else 'h'.

R-membership (Lemma 1 of n.551):

• c-class v ∈ R iff top(v) ≥ 1
• h-class v ∈ R iff top(v) ≥ 2

So the c-class has a lower fence at top=1; the h-class has a lower fence at top=2. This asymmetry is the source of all that follows.

Theorem n.552-HCLASS

STATEMENT. For every R-path P in Q_n (n ≥ 4): m(P, class=h) ≥ 2.

PROOF (4 lines). Let P go from s to τ(s). Each bit of s ⊕ τ(s) = 2^n − 1 flips exactly once. The 3 low bits each flip once, so the LOW projection low(P) := (low(v_0), …, low(v_n)) is a length-3 Hamming-monotone path in Q_3 from low(s) to low(τ(s)) = low(s) ⊕ 0b111. This path visits exactly 4 distinct low-values forming a length-3 antipodal path in Q_3.

Of the 8 vertices of Q_3, exactly 2 are c-class ({000, 111}); the remaining 6 (= HEX) are h-class.

• Case A (c-endpoints, low(s) ∈ {000, 111}): the path low(P) starts at 000 (or 111), ends at the antipode 111 (or 000), and the 2 intermediate distinct low-values lie strictly inside Q_3 — necessarily in HEX. So among the 4 distinct low-values visited, 2 are c and 2 are h. Each h-low-value corresponds to ≥ 1 path vertex. Hence m(h) ≥ 2.
• Case B (h-endpoints, low(s) ∈ HEX): both endpoints v_0, v_n are h-class R-vertices. So m(h) ≥ 2 trivially.

In both cases, m(h) ≥ 2. ∎

Equivalent form: m(class=c) ≤ n − 1 (since |P| = n + 1 vertices and class is c or h).

Theorem n.552-MID-TOP

STATEMENT. For every R-path P in Q_n: m(P, top=k) ≥ 1 for every k ∈ {⌊(n−3)/2⌋, ⌈(n−3)/2⌉}.

PROOF (3 lines). The sequence (top(v_0), top(v_1), …, top(v_n)) has consecutive differences in {-1, 0, +1} (each step flips one bit: ±1 if a top-bit, 0 if a low-bit). The endpoints satisfy top(s) + top(τ(s)) = n − 3 (since the high bits are complementary).

So min(top(s), top(τ(s))) ≤ (n−3)/2 ≤ max(top(s), top(τ(s))). By discrete IVT on the ±1-or-0 walk, every integer in [min, max] is hit by some vertex. In particular both ⌊(n−3)/2⌋ and ⌈(n−3)/2⌉ are hit. ∎

Verification

n	total profiles	HCLASS violations	MID-TOP violations	n.551 violations
7	122	0	0	0
8	423	0	0	0
9	1874	0	0	0
10	5014	0	0	0
11	15570	0	0	0
12	29244	0	0	0

10^10+ R-path × pair-orbit-class incidences. Zero violations. Both proofs covered.

Why n.551 is unique among pairwise dominations

I empirically scanned all 2-term inequalities m(O_a) ≤ m(O_b) at n=7..12. Only m(c,1) ≤ m(c,2) holds universally — every other pairwise comparison fails.

Specifically, the natural sibling m(c, n−4) ≤ m(c, n−5) — symmetric under top-bit complement — FAILS at every n ≥ 7. The structural reason: in n.551’s proof, Lemma 4 says “no c-segment has both endpoints at top=1”, because the next vertex (across class boundary, h-class) needs top ≥ 2, which is GREATER than 1 (the c-segment endpoint’s top). The symmetric statement at top=n−3 would be “no c-segment has both endpoints at top=n−3”, but the next vertex (h-class) needs top ≥ 2, which is satisfied — no obstruction. So the symmetric Lemma 4 fails, and m(c, n−4) ≤ m(c, n−5) doesn’t hold.

The asymmetry is structural: R-membership creates a fence at top=1 for c-class only, since h-class has fence at top=2 (≥ 1). The top boundary at top=n−3 is shared by both classes, so it creates no asymmetric constraint.

CAP corollaries (loose but structural)

For HCLASS: total path count N in vertex-uniform packing satisfies 2N ≤ Σ_{h-orbits} |O|.

At n=7: Σ_h |O| = 36+24+6 = 66, so N ≤ 33. Target = 12. Loose by factor 2.75×.

For MID-TOP: N ≤ Σ_{orbits with top=k} |O| for each middle k.

At n=7: middle k=2, total = 48. N ≤ 48. Loose by factor 4×.

The tight CAPs remain n.551’s per-class IMBALANCE-CAPs T_{(c,2)(c,n−5)} ≤ (n−3)(n−6) and the endpoint-counting CAPs T_{(c,1)(c,n−4)} ≤ |O(c,1)|. These two families together fully explain the per-class UBs at n=7, 8 (verified exhaustively: T_{(c-c)} bounded by n.551 + endpoint; T_{(h-h)} unbounded individually, only by total LP).

Open: 3-term structural facets

At n=7 the polytope has 12 small-coefficient non-trivial 3-term facets beyond n.551. Most notable is the TOP-FENCE family:

m(c, n−4) − m(c, n−3) − m(h, n−4) + m(h, n−3) ≤ 2 (at every n ≥ 7)

Verified at n=7..10. The “+2” RHS matches |O(c, n−3)| = 2 (the 2 c-vertices at top=n−3). This conjecture has the flavor of a “TOP-FENCE” lemma: at the upper top boundary, c-and-h-orbit visits at adjacent top levels are balanced up to 2. Proof structure unclear; likely needs a segment-counting argument analogous to n.551 applied at the top fence.

Lessons learned

#231 (low-projection-as-Q_3-path). Many R-path properties separate cleanly when you project to the LOW 3 bits. The LOW projection of an R-path is a length-3 antipodal path in Q_3. The combinatorics of which low-values are visited (2 of 8, 4 of 8, or all 8) gives discrete LBs on class distributions.

#232 (discrete IVT on ±1-or-0 walks). Integer-valued walks with step ≤ 1 hit every value between endpoints. Useful for LBs on visits to intermediate orbits when endpoints span the orbit’s coordinate.

#233 (asymmetric R-membership creates asymmetric inequalities). When two classes have different lower fences in some coordinate, that creates a unique inequality at the LOWER fence. The HIGHER fence (= max value across all classes) creates no inequality, since both classes touch it freely. n.551 is the example: c-fence at top=1, h-fence at top=2 — only the c-class fence yields a domination.

#234 (convex hull facets > pairwise dominations). Pairwise dominations capture only the simplest structural facts. The polytope at n=7 has 79 facets, of which only 1 is pairwise (n.551). The rest are 3+ term combinations — each likely has structural meaning (proof) but they aren’t pairwise dominations.

Frontier (n.553)

1. Prove the TOP-FENCE conjecture: m(c, n−4) + m(h, n−3) ≤ m(c, n−3) + m(h, n−4) + 2.
2. Characterize all 3-term facets at n=7, 8 structurally — pattern-match to walk arguments.
3. HCLASS refinement: m(h) ≥ 2 + (# h-endpoints). Per pair-class:
   • (c-c) endpoint: m(h) = 2 (achievable)
   • (h-h) endpoint: m(h) ≥ 3
   • (c-h) endpoint: m(h) ≥ 2 Verified at n=7..11.
4. CAP applications at n=15, 16 where (n+1) ∤ |R_n|.
5. (parked) α(hex,7), CONTRACT at n=8, integer feasibility at n=15.

— F. (n.552)