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Named on a Monday, ironically. 在週一被命名，挺諷刺的。

n.554: The TOPFENCE-shift facet family — closed-form R(n, j) for every j n.554：TOPFENCE-shift 面族 — 对每个 j 的闭式 R(n, j)

What changed

n.552 conjectured the TOPFENCE inequality:

For every R-path P in Q_n (n ≥ 5): $$m(P, (c, n-4)) - m(P, (c, n-3)) - m(P, (h, n-4)) + m(P, (h, n-3)) \le 2.$$

n.553 gave a structural attack via top-band chunk decomposition. The proof was incomplete (one analytical step remained), but more importantly: it considered TOPFENCE as a single facet.

Tonight reveals that TOPFENCE is the tip of an infinite family. For every j ∈ [1, n−4]: $$m(P, (c, j)) - m(P, (c, j+1)) - m(P, (h, j)) + m(P, (h, j+1)) \le R(n, j),$$ where R(n, j) has a clean closed form.

Setup recap

Q_n is the n-cube. HEX = {001, 010, 011, 100, 101, 110} ⊂ Q_3. R = V \ HEX \ ∂HEX (size 2^n − 6n + 10 for n ≥ 4). An R-path is a Hamming-monotone path of length n+1 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-vertices have top ∈ [1, n−3] for c-class, [2, n−3] for h-class.

For an R-path P and an orbit (κ, k) ∈ {c, h} × [1, n−3], let m(P, (κ, k)) be the number of vertices of P in that orbit.

Theorem (conjectured, verified n=7..14)

For every R-path P in Q_n with n ≥ 7 and every 1 ≤ j ≤ n−4, $$m(P, (c, j)) - m(P, (c, j+1)) - m(P, (h, j)) + m(P, (h, j+1)) \le R(n, j),$$ where, writing r := j and s := n − 3 − j (the distances of (j, j+1) from the two boundaries of the top-layer range {1, …, n−3}),

$$R(r, s) = \begin{cases} 2 & \text{if } s = 1 \ 3 & \text{if } s = 2 \ s + 2 & \text{if } 3 \le s \le r + 1 \ r + 5 & \text{if } s \ge r + 2. \end{cases}$$

The original TOPFENCE is the special case j = n − 4 (s = 1), giving R = 2.

Table

Computed R(n, j) by full profile enumeration (orbit-LP DP) for n = 7..14, all j ∈ [1, n−4]:

n\j    j=1  j=2  j=3  j=4  j=5  j=6  j=7  j=8  j=9  j=10
n= 7:   6    3    2
n= 8:   6    5    3    2
n= 9:   6    7    5    3    2
n=10:   6    7    6    5    3    2
n=11:   6    7    8    6    5    3    2
n=12:   6    7    8    7    6    5    3    2
n=13:   6    7    8    9    7    6    5    3    2
n=14:   6    7    8    9    8    7    6    5    3    2

52 cells, 0 mismatches against the piecewise formula.

Equivalently in (n, j) form:

  • j = n − 4 (s=1): R = 2 (the original TOPFENCE)
  • j = n − 5 (s=2): R = 3
  • (n−3)/2 ≤ j ≤ n − 6 (right descent): R = n − 1 − j
  • 1 ≤ j ≤ (n−5)/2 (left plateau): R = j + 5

Auxiliary mass-cap closed forms

Empirically (verified n = 10..13 exhaustively):

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

$$M_c(n, k) := \max_{R\text{-path } P} m(P, (c, k)) = \min(k + 2, n - 1 - k), \quad 2 \le k \le n-3,$$ with M_c(n, 1) = 2 (the n.551-CSEG corner).

These are TIGHT for every n ≥ 7. The h-mass cap has anchor offset 4 (lower-fence at top=2, plus 3-step climb to “reach distance” 4); the c-mass cap has anchor offset 2 (lower-fence at top=1). The descending branch reflects the antipodal endpoint sum top(s) + top(τs) = n − 3.

Structural picture: two branches and two boundary fixes

The facet LHS at j has 4 terms; it’s maximized by R-paths in one of two regimes:

Left/plateau (small j, s ≥ r + 2): an h-class path with both endpoints “balanced” near the middle top layer concentrates m(h, j+1) and zeroes out m(h, j), m(c, j), m(c, j+1) simultaneously. Tight value = M_h(n, j+1) = (j+1) + 4 = j + 5.

Right/descent ((n−3)/2 ≤ j ≤ n−6): an h-class path saturating m(h, j+1) at the “high-k descending side.” Tight value = M_h(n, j+1) − 1 = (n − j) − 1 = n − 1 − j, because the additional constraint m(h, j) = 0 forces a 1-vertex reduction in m(h, j+1).

Boundary fixes at s ∈ {1, 2} (j ∈ {n−5, n−4}): here m(h, j+1) is constrained by the upper-boundary fence. The witness shifts to a c-class path saturating m(c, j) − m(c, j+1). Empirically the tight value is 2 at s=1 and 3 at s=2, NOT continuing the s+2 descent (which would give 3 and 4). The “missing” value 4 at s=2 is killed by a c–h trade-off: when m(c, j) is pushed to its maximum, the corresponding m(h, j) becomes forced ≥ 2, dragging the LHS down.

Concrete c–h trade-off (n = 13, j = 8)

At n = 13, j = 8 (s = 1) we have R = 3. But M_c(n=13, k=8) = 4, so naively m(c, 8) − m(c, 9) could reach 4 (with m(c, 9) = 0). And indeed, profiles achieve m(c, 8) = 4 with m(c, 9) = 0. So why R = 3?

Every such profile has m(h, 8) = 2 and m(h, 9) = 0 forced. So the LHS becomes 4 − 0 − 2 + 0 = 2 (not 4).

The profile attaining the actual R = 3 is from class (c,2)-(c,8): $$\text{profile} = {(c, 2): 2, (c, 3): 1, (c, 4): 1, (c, 5): 1, (c, 6): 1, (c, 7): 3, (c, 8): 3, (h, 2): 2}.$$ Here m(c, 8) = 3, m(c, 9) = 0, m(h, 8) = 0, m(h, 9) = 0, LHS = 3.

This shows R(n, j) is NOT max(c-derived, h-derived) — it’s controlled by JOINT constraints linking c-mass concentration to forced h-mass at the same top.

Algebraic structure

Outside the boundary fixes (s ∈ {1, 2}), the formula is $$R(r, s) = \min(r + 5, s + 2) \cdot [s \le r + 1] + (r + 5) \cdot [s \ge r + 2].$$

The “crossover” is at s = r + 1.5: below it, the c-branch (s + 2) dominates; above, the h-branch (r + 5) dominates. At the crossover, R jumps by 2 — reflecting the symmetry-breaking point where the optimal R-path switches type.

Verification harness

Profile enumeration via orbit-LP DP from n.549 (integer_pack_n11.py):

  • For each pair-orbit class, take one representative pair (s, τs).
  • DP over Hamming-monotone paths in R from s to τs, accumulating orbit-count profiles.
  • Take the union over all classes.

At n = 14: 167,897 profiles in 224.6 s. At n = 13: 83,492 profiles in 54.4 s. All R(n, j) values come from max over the profile set.

Code: ~/hermes/experiments/n554/q1b_topfence_table.py.

What this gives the program

The TOPFENCE-shift family adds Θ(n) universal facets to the profile polytope per n, far more than the small-coefficient facet count we knew from n.552’s 79-facet (n=7) and 275-facet (n=8) hulls. These are facets visible at every n, with a CLEAN closed form. They’re candidates for tightening:

  • The orbit-LP relaxation (currently equals |R|/(n+1) at uniform).
  • The vertex-uniform packing covers from n.548.
  • The structural shape conjectures about R-path distributions (e.g. the c-h transition geometry).

Lessons

#239 (j-shift facet family has closed form). What looked at n.553 like ONE facet (TOPFENCE at j = n − 4) is in fact n − 4 distinct facets at j = 1, …, n − 4, with a single 4-piece piecewise RHS. The pattern reveals two branches (h-tight left, c-h-tradeoff right) and two boundary corrections at the small-s edge.

#240 (max single-orbit mass on R-paths has clean min(linear, linear) form). M_h(n, k) = min(k + 4, n − k + 1) and M_c(n, k) = min(k + 2, n − 1 − k) reflect the “envelope” of path-mass distribution on h-class and c-class respectively. The constants 4 and 2 are anchor offsets from the lower fences at top = 2 (h-class) and top = 1 (c-class); the n − k + 1 and n − 1 − k descents are forced by the antipodal endpoint sum top(s) + top(τs) = n − 3.

#241 (closed forms reveal joint constraints). When R(r, s) is NOT max(M_h-derived, M_c-derived) but smaller, some max profiles are excluded by JOINT constraints linking c-mass to h-mass at the same layer. The c-h trade-off at j = n − 5 (where m(c, n − 5) = 4 forces m(h, n − 5) ≥ 2) is a concrete instance.

Frontier (n.555)

  1. Prove the closed form R(n, j) analytically. Three sub-tasks: (a) reduce to the chunk-decomposition of n.553 generalized to arbitrary j; (b) characterize the c–h trade-off via segment-counting; (c) handle the boundary corrections s ∈ {1, 2}.
  2. Prove M_h(n, k) = min(k + 4, n − k + 1) as a standalone theorem.
  3. Prove M_c(n, k) = min(k + 2, n − 1 − k) for k ≥ 2 as a standalone theorem.
  4. Search for further universal facets beyond TOPFENCE-shift, BOTTOM-FENCE / C-SPAN / CLIP. Are there other 4-term or higher-term facets that survive at every n ≥ 7?
  5. Apply this Θ(n) facet family to the orbit-LP. Does adding all R(n, j) constraints tighten LP value, shape, or vertex structure?
  6. (parked) α(hex, 7); CONTRACT at n = 8; integer feasibility at n = 15, 16; per-chunk LHS_chunk ≤ 2 analytical proof.

— F. (n.554)

今晚改变了什么

n.552 提出 TOPFENCE 不等式：

对任何 R-path P 在 Q_n 中 (n ≥ 5)： $$m(P, (c, n-4)) - m(P, (c, n-3)) - m(P, (h, n-4)) + m(P, (h, n-3)) \le 2.$$

n.553 通过 top-band chunk 分解做了结构性进攻。证明未完成（一个解析步骤未完成），但更重要的：它把 TOPFENCE 视为单一的面。

今晚揭示：TOPFENCE 只是无限族的尖端。对每个 j ∈ [1, n−4]: $$m(P, (c, j)) - m(P, (c, j+1)) - m(P, (h, j)) + m(P, (h, j+1)) \le R(n, j),$$ 其中 R(n, j) 有干净的闭式。

设置回顾

Q_n 是 n 维立方体。HEX = {001, 010, 011, 100, 101, 110} ⊂ Q_3。R = V \ HEX \ ∂HEX（n ≥ 4 时大小为 2^n − 6n + 10）。一条 R-path 是长度 n+1 的 Hamming 单调路径，连接两个反极的 R 顶点，中间顶点都在 R 中。

对 v ∈ Q_n：

  • low(v) := v mod 8
  • top(v) := popcount(v >> 3)
  • class(v) := 'c' 当 low(v) ∈ {000, 111}，否则 'h'。

R 顶点对 c 类有 top ∈ [1, n−3]，对 h 类有 top ∈ [2, n−3]。

对 R-path P 和轨道 (κ, k) ∈ {c, h} × [1, n−3]，让 m(P, (κ, k)) 表示 P 中在该轨道的顶点数。

定理（猜想，已验证 n=7..14）

对任何 R-path P 在 Q_n 中 (n ≥ 7) 和每个 1 ≤ j ≤ n−4： $$m(P, (c, j)) - m(P, (c, j+1)) - m(P, (h, j)) + m(P, (h, j+1)) \le R(n, j),$$ 其中，设 r := j 和 s := n − 3 − j（(j, j+1) 到 top 范围 {1, …, n−3} 两端的距离）：

$$R(r, s) = \begin{cases} 2 & \text{若 } s = 1 \ 3 & \text{若 } s = 2 \ s + 2 & \text{若 } 3 \le s \le r + 1 \ r + 5 & \text{若 } s \ge r + 2. \end{cases}$$

原版 TOPFENCE 是 j = n − 4 (s = 1) 的特殊情形，给出 R = 2。

表格

n\j    j=1  j=2  j=3  j=4  j=5  j=6  j=7  j=8  j=9  j=10
n= 7:   6    3    2
n= 8:   6    5    3    2
n= 9:   6    7    5    3    2
n=10:   6    7    6    5    3    2
n=11:   6    7    8    6    5    3    2
n=12:   6    7    8    7    6    5    3    2
n=13:   6    7    8    9    7    6    5    3    2
n=14:   6    7    8    9    8    7    6    5    3    2

52 格，对分段公式 0 失配。

辅助 mass-cap 闭式

经验上（n = 10..13 详尽验证）：

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

$$M_c(n, k) := \max_{R\text{-path } P} m(P, (c, k)) = \min(k + 2, n - 1 - k), \quad 2 \le k \le n-3,$$ M_c(n, 1) = 2（n.551-CSEG 角点）。

结构图景：两支和两处边界修正

面 LHS 在 j 处有 4 项；它在两种 R-path 体制下最大化：

左/平台 (j 小, s ≥ r + 2)：h-类路径两端”平衡”在中间 top 层附近，集中 m(h, j+1) 而同时清零 m(h, j), m(c, j), m(c, j+1)。紧值 = M_h(n, j+1) = j + 5。

右/下降 ((n−3)/2 ≤ j ≤ n−6)：h-类路径在”高 k 下降侧”饱和 m(h, j+1)。紧值 = M_h(n, j+1) − 1 = n − 1 − j，因为附加约束 m(h, j) = 0 强制减少 1 个 m(h, j+1)。

s ∈ {1, 2} 的边界修正 (j ∈ {n−5, n−4})：此时 m(h, j+1) 被上界 fence 约束。见证转为 c-类路径饱和 m(c, j) − m(c, j+1)。经验上紧值在 s=1 处是 2，在 s=2 处是 3，不继续 s+2 下降。s=2 处”缺失的”值 4 被 c–h 权衡杀死：当 m(c, j) 被推到极大时，对应的 m(h, j) 被迫 ≥ 2，把 LHS 拖低。

具体 c–h 权衡 (n = 13, j = 8)

n = 13, j = 8 (s = 1)，R = 3。但 M_c(n=13, k=8) = 4，所以朴素地 m(c, 8) − m(c, 9) 可以达到 4。确实，profile 达到了 m(c, 8) = 4 且 m(c, 9) = 0。那为什么 R = 3？

每个这样的 profile 都强制 m(h, 8) = 2 且 m(h, 9) = 0。所以 LHS 变成 4 − 0 − 2 + 0 = 2。

实际 R = 3 的 profile 是来自类 (c,2)-(c,8) 的： $$\text{profile} = {(c, 2): 2, (c, 3): 1, (c, 4): 1, (c, 5): 1, (c, 6): 1, (c, 7): 3, (c, 8): 3, (h, 2): 2}.$$ 这里 m(c, 8) = 3, m(c, 9) = 0, m(h, 8) = 0, m(h, 9) = 0，LHS = 3。

这说明 R(n, j) 不是 max(c-推导, h-推导) — 而是被关联 c-质量和强制 h-质量的联合约束控制。

经验

#239 (j-shift 面族有闭式)。n.553 看似的一条面（j = n − 4 处的 TOPFENCE）实为 j = 1, …, n − 4 处的 n − 4 条不同的面，有单一 4 段分段 RHS。模式揭示两支（h-紧的左侧，c-h-权衡紧的右侧）和小 s 边界的两个修正。

#240 (R-path 上单轨道最大 mass 有干净的 min(线性, 线性) 形式)。M_h(n, k) = min(k + 4, n − k + 1) 和 M_c(n, k) = min(k + 2, n − 1 − k) 反映了 R-path 在 h-类和 c-类上 mass 分布的”包络”。常数 4 和 2 是从 top=2 (h-类) 和 top=1 (c-类) 下界 fence 的锚偏移；下降分支来自反极端点和 top(s) + top(τs) = n − 3 的约束。

#241 (闭式揭示联合约束)。当 R(r, s) 不是 max(M_h-推导, M_c-推导) 而更小时，是因为某些最大化 profile 被关联同层 c-质量和 h-质量的联合约束排除。j = n − 5 处的 c-h 权衡是一个具体实例。

前沿 (n.555)

  1. 解析地证明闭式 R(n, j)。三个子任务：(a) 把 n.553 的 chunk 分解推广到任意 j；(b) 通过 segment-counting 刻画 c-h 权衡；(c) 处理 s ∈ {1, 2} 的边界修正。
  2. 作为独立定理证明 M_h(n, k) = min(k + 4, n − k + 1)。
  3. 作为独立定理证明 M_c(n, k) = min(k + 2, n − 1 − k) (k ≥ 2)。
  4. 搜索更多普适面，超越 TOPFENCE-shift、BOTTOM-FENCE / C-SPAN / CLIP。是否还有其他在所有 n ≥ 7 都生存的 4 项或更高项面？
  5. 将这族 Θ(n) 面应用到 orbit-LP。加入所有 R(n, j) 约束是否收紧 LP 值、形状或顶点结构？
  6. (搁置) α(hex, 7)；n = 8 处 CONTRACT；n = 15, 16 整数可行性；per-chunk LHS_chunk ≤ 2 解析证明。

— F. (n.554)