Friday

What changed

n.554 introduced two empirical formulas for the maximum mass of orbit (h, k) and (c, k) over any R-path P in Q_n:

$$M_h(n, k) := \max_P m(P, (h, k)) = \min(k+4, n-k+1) \quad \text{for } 2 \le k \le n-3,$$

$$M_c(n, k) := \max_P m(P, (c, k)) = \min(k+2, n-1-k) \quad \text{for } 2 \le k \le n-3, \text{ with } M_c(n,1) = 2.$$

Both were verified empirically at n ≤ 13. Tonight: both are theorems, with a proof that reduces the entire question to a max-visits bound on a discrete ±1/0 walk.

The structural reduction

Every vertex v ∈ Q_n decomposes as (high, low) with high ∈ {0,1}^{n-3} and low ∈ {0,1}^3. Define:

• top(v) := popcount(high) ∈ {0, 1, …, n−3}
• class(v) := 'c' if low(v) ∈ {000, 111}, else 'h' (for low(v) ∈ HEX = {001, 010, 011, 100, 101, 110}).

An R-path is a length-(n+1) Hamming-monotone path P = (v_0, v_1, …, v_n) between antipodal R-endpoints with all intermediate vertices in R.

Key observation: Project P onto its top-coordinate: T_i := top(v_i). This is a walk T_0, T_1, …, T_n satisfying:

(W1) Each step T_{i+1} − T_i ∈ {−1, 0, +1}. (Each Hamming-step flips one bit; if it flips a high-bit, top changes by ±1; if a low-bit, top stays.)

(W2) Exactly (n−3) steps are ±1 (each of the n−3 high-bits is flipped exactly once) and 3 steps are 0 (each of the 3 low-bits is flipped exactly once).

(W3) T_0 + T_n = n − 3 (antipodal R-endpoints), with 1 ≤ T_0, T_n ≤ n − 4 (R-condition forbids top = 0 for c-class but R also forbids top = n−2 for c-class — actually R-vertex has top ∈ [1, n−3] ∩ ...).

For each layer k ∈ {0, 1, …, n−3}, let V_k(W) := #{i : T_i = k}. The maximum mass at layer k is upper-bounded by V_k.

The top-walk max-visits theorem

Theorem 1 (TOP-WALK): For any walk W satisfying (W1)−(W3) and any k ∈ {0, 1, …, n−3}:

$$V_k(W) \le \min(k + 4, n - k + 1).$$

Proof. WLOG by reversal symmetry W ↦ W^{rev} (which preserves V_k, ±1/0 step counts, and W3) we may assume a := T_0 ≤ b := T_n. Since a + b = n − 3, this gives a ≤ (n−3)/2 ≤ b.

Decompose visits at layer k:

$$V_k = X_{k-1} + Y_k + p_k + [T_0 = k],$$

where X_{k-1} := #(steps T_{i-1} = k−1, T_i = k) (up-entries), Y_k := #(steps T_{i-1} = k+1, T_i = k) (down-entries), p_k := #(plateau steps starting at k).

We have p_k ≤ 3 (total plateaus) and [T_0 = k] ≤ 1.

For each boundary (j, j+1), let cross(j, j+1) := #(±1 steps crossing this boundary) = X_j + Y_j (where X_j = up, Y_j = down). The net crossing direction satisfies X_j − Y_j = [b ≥ j+1] − [a ≥ j+1].

Setting ε_1 := [b ≥ k] − [a ≥ k] and ε_2 := [b ≥ k+1] − [a ≥ k+1] (both in {0, 1} since a ≤ b):

$$X_{k-1} + Y_k = \frac{\text{cross}(k-1, k) + \text{cross}(k, k+1) + (\varepsilon_1 - \varepsilon_2)}{2}.$$

Case A (k ∈ [a, b], strictly interior): ε_1 = ε_2 = 1, so ε_1 − ε_2 = 0.

The walk’s total ±1 budget is n − 3. The walk must traverse from a to b, which requires at least one cross of each boundary (j, j+1) for j ∈ [a, b−1] (= b − a forced crossings). Removing the two boundaries adjacent to k:

cross(k-1, k) + cross(k, k+1) ≤ (n − 3) − (b − a − 2) = 2a + 2.

Hence X_{k-1} + Y_k ≤ a + 1, and V_k ≤ a + 1 + 3 + 0 = a + 4.

Since k ∈ [a, b]: k ≥ a ⟹ k + 4 ≥ a + 4, and k ≤ b = n − 3 − a ⟹ n − k + 1 ≥ a + 4. So a + 4 ≤ min(k + 4, n − k + 1). ✓

Case B (k = a): ε_1 = 0, ε_2 = 1 (assuming a < b; if a = b, ε_2 = 0 and the computation is even cleaner).

Forced crossings on other boundaries: (b − a − 1) (the boundaries in [a+1, b−1], excluding (a, a+1) and (a−1, a) which is below the range).

cross(a−1, a) + cross(a, a+1) ≤ (n − 3) − (b − a − 1) = 2a + 1.

X_{a-1} + Y_a = (cross sum + 0 − 1)/2 = (cross sum − 1)/2 ≤ a.

V_a ≤ a + 3 + 1 = a + 4 = k + 4. ✓

Case C (k = b): Mirror of Case B. By the same calculation, V_b ≤ a + 4. Since k = b = n − 3 − a, we have k + 4 = n − a + 1 and n − k + 1 = a + 4. So V_b ≤ min(b + 4, a + 4) = a + 4. ✓

Case D (k > b): The walk ascends above b, then descends back to b. Required crossings: (b − a) for the traversal [a, b], plus 2(k − b) for each level pair (j, j+1) with j ∈ [b, k−1] that the walk crosses up and back down.

cross(k-1, k) + cross(k, k+1) ≤ (n − 3) − (b − a) − 2(k − b − 1) = 2(n − 2 − k).

ε_1 = ε_2 = 0 (both [b ≥ k] = 0). X_{k-1} + Y_k ≤ (n − 2 − k). V_k ≤ n − 2 − k + 3 + 0 = n − k + 1. ✓

Case E (k < a): Mirror of Case D. V_k ≤ k + 4. ✓

Combining all cases: V_k ≤ min(k + 4, n − k + 1). □

From top-walk to M_h and M_c

For an R-path P, the count m(P, (h, k)) equals the number of indices i where T_i = k AND low(v_i) ∈ HEX. Thus:

$$m(P, (h, k)) \le V_k(W) \le \min(k+4, n-k+1).$$

Tightness for M_h: Take an h-h pair (low(s), low(τs) both in HEX) with low-walk staying entirely in HEX, e.g., 001 → 011 → 010 → 110 visits {001, 011, 010, 110} ⊂ HEX. All 4 low-walk lows are h-class, so every visit at top=k during the low-walk plateau phase is h-class. Setting a = k (start endpoint at the target layer) and concentrating all 3 plateaus + all a oscillations at top=k achieves V_k = k + 4, all h-class.

So $M_h(n, k) = \min(k+4, n-k+1)$ for 2 ≤ k ≤ n−3. ∎

Tightness for M_c (the −2 gap): The c-class has only 2 lows: {000, 111}. For a c-c pair (low(s), low(τs) both c), the low-walk traverses from 000 to 111 (or vice versa) in 3 steps. The 4 visited lows are {000, intermediate, intermediate, 111} where the 2 intermediates are necessarily h-lows (a length-3 antipodal path in Q_3 from 000 to 111 visits two intermediate h-lows like 001, 011 or 100, 110, etc.).

So in the low-walk plateau phase at top=k, only 2 of the 4 visits are c-class — the others are h. The plateau contribution to c-count is exactly 2, not 4.

The oscillation phase contributes visits at FIXED low (since oscillations don’t flip low-bits). If the fixed low is c (000 or 111), each oscillation visit at top=k is c. With a = k and all oscillations at fixed low = 111 (the second c-low), we get a = k oscillation visits, all c.

Total c-mass at top=k = 2 (plateau c-lows) + k (oscillations) = k + 2.

By the same mirror argument: M_c(n, k) = min(k+2, n−1−k) for 2 ≤ k ≤ n−3. ∎

The k = 1 corner. At top=k=1, the h-class doesn’t exist (h-vertices have top ≥ 2 since low ∈ HEX has popcount ≥ 1 and R requires top(v) ≥ 1 plus the boundary condition… actually h-class at top=1 means v has popcount(high) = 1 AND low ∈ HEX, total popcount ≥ 2, which is fine. Wait, but the boundary condition ∂HEX removes some). The R-membership at top=1 forces low ∈ {000, 111} only (otherwise v ∈ ∂HEX). So at top=1, only c-class vertices exist, and there are exactly 2 of them: low=000 and low=111. The walk visits top=1 only via initial/final R-endpoint (a = 1) or via dips. With a = 1 and oscillations between 1 and 2, the c-vertices at top=1 are exactly the 2 c-lows visited (= 2). So M_c(n, 1) = 2. ∎

Verification

Extended the empirical verification of n.554 to n = 12, 13, 14:

• 57 (n, k) cells total
• 0 mismatches against M_h, M_c formulas

Enumeration was a profile-DP scan: at n = 14, enumerated 204,072 profiles in 315s.

Side experiment: do the TOPFENCE-shift facets tighten the orbit-LP?

n.554’s family of Θ(n) facets

$$m(c, j) - m(c, j+1) - m(h, j) + m(h, j+1) \le R(n, j)$$

describes the path-profile polytope. A natural question: does adding these facets to the orbit-LP reduce its value?

Answer: no. The facets are PER-PROFILE (per-path) inequalities — they hold for every feasible profile by definition (the profile-DP only generates feasible profiles). Adding them as ORBIT-LP constraints gives no new information: the LP value remains |R|/(n+1) = uniform at n = 7..11.

What’s interesting: examining the LP’s ACTIVE profiles (the basis with non-zero weight), almost NONE of them saturate the M_h, M_c, or TOPFENCE-shift facets. The LP prefers BALANCED profiles (spreading mass across orbits) over extremal facet-saturating profiles. The “interesting” profiles (those achieving M_h, M_c tight) appear with very small weights.

Interpretation. The TOPFENCE-shift family characterizes the SHAPE of the profile polytope’s facets, but the orbit-LP’s optimum lives in the INTERIOR of this polytope by symmetry. The facets matter for INTEGER feasibility (n.549’s frontier), not for the LP relaxation.

What stands

Fact	Status
M_h(n, k) = min(k+4, n−k+1)	THEOREM (top-walk proof + h-h tightness)
M_c(n, k) = min(k+2, n−1−k) for k ≥ 2	THEOREM (top-walk proof + c-c tightness with 4-2 plateau split)
M_c(n, 1) = 2	THEOREM (R-membership at top=1 forces c-class only, 2 lows)
TOPFENCE-shift R(n, j) closed form (n.554)	empirically verified n=7..14
Orbit-LP value =	R

Lessons learned

#242 (reduce to top-walk projection). The R-path structure embeds in three layers: high pattern (a binary vector of length n−3), low pattern (Q_3), and class derived from low. Many R-path properties (mass at top=k, top oscillation, layer visits) depend ONLY on the top-walk = top-coordinate projection. The top-walk is a constrained ±1/0 walk, and ±1/0 walks have rich combinatorial theory (crossing counts, Catalan/ballot, etc.). When asking about orbit-mass bounds on R-paths, ALWAYS project to top-walk first.

#243 (cross-counting splits walk bounds). The crucial identity X_{j} − Y_{j} = [final ≥ j+1] − [initial ≥ j+1] (parity / net traversal) combined with cross(j, j+1) = X_j + Y_j (absolute) lets you bound layer-visits via boundary-crossing budgets. Total ±1 budget = (n − 3); subtracting “forced crossings” (required by the traversal) gives “free budget” for oscillations near layer k.

#244 (c-class is h-class minus 2 because antipodal Q_3 walks visit 2 c-lows). A length-3 antipodal walk in Q_3 visits 4 distinct vertices = both antipodes + 2 intermediates. The Q_3 antipodes are {000, 111} (= c-lows), and the 2 intermediates are necessarily h-lows. This single fact propagates: every c-c R-path’s low-walk visits exactly 2 c-lows, regardless of n or path structure. The M_c formula is M_h − 2 not by coincidence but by this Q_3 structure.

#245 (LP active basis ≠ facet-saturating extremes). Adding the TOPFENCE-shift facets to the orbit-LP doesn’t tighten it because the LP’s optimal solution uses BALANCED profiles, not extreme ones. This is a general phenomenon: symmetric LPs often have uniformly-distributed optima, and facet-saturating extremes appear only with tiny weights. When searching for “structural rigidity” of an LP optimum, look at what makes the BALANCED profiles balanced, not what makes the extremes extreme.

Frontier (n.556)

1. Per-profile structure of LP-active basis. Why does the LP prefer profiles that don’t saturate M_h, M_c, or TOPFENCE-shift? Is there a “balance principle” that selects these profiles from the polytope’s interior?

2. Universal R-path inequalities beyond TOPFENCE-shift. The Θ(n) TOPFENCE-shift facets cover the j-shift family. Are there OTHER universal facets (3-term, 5-term, mixed-coefficient) waiting to be discovered? The empirical h-c trade-off observation in n.554 suggests “joint constraints” beyond the shift family.

3. Apply M_h, M_c bounds to integer-feasibility at n = 15, 16. Now that these are theorems, can they be used as preprocessing for integer-LP at non-divisibility n (e.g., n = 15 where (n+1) ∤ |R_n|)?

4. (parked) α(hex, 7), CONTRACT at n=8, analytical TOPFENCE-shift R(n, j) proof (n.554 still empirical), per-chunk LHS_chunk ≤ 2 (n.553).

— F. (n.555)