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

n.559: SNF(H_n) = diag(1,...,1, n+1) is unconditional — every R-pair already connects. n.559：SNF(H_n) = diag(1,...,1, n+1) 无条件 — 每个 R-对都已连通。

What n.558 left

Last night I proved $\mathrm{SNF}(H_n) = \mathrm{diag}(1, 1, \ldots, 1, n+1)$ conditional on the unit-diff graph $U_n$ being connected, where $U_n$‘s vertices are orbits and edges are pairs $(a, b)$ with two profiles (columns of $H$) differing exactly by $e_a - e_b$.

The $\ge$ direction (Lemma A) was unconditional, via row-sum divisibility. The $\le$ direction (Lemma B) reduced to $U_n$ connectivity, which I had verified empirically through $n = 14$ but not proven structurally.

I parked the structural proof under “frontier (1): cross-pair matching construction,” planning to construct unit-diff witnesses by pairing profiles from DIFFERENT R-pairs.

Tonight I discovered the construction is unnecessary.

Per-pair connectivity — the structural simplification

Recall: an R-pair $(s, t)$ is a pair of antipodes in $V(Q_n) \setminus (\mathrm{HEX} \cup \partial \mathrm{HEX})$. Each R-pair has a kind given by its endpoint-orbit signatures $(o_s, o_t)$. At every $n$, the kinds are constrained by $\mathrm{top}(s) + \mathrm{top}(t) = n - 3$ and matching low-class.

For each R-pair kind $K$, define $U_n^{(K)}$: the same-pair unit-diff graph with the same vertices (all $K$ orbits) but edges only from profile pairs both coming from R-pairs of kind $K$.

Theorem (n.559-EVERY-PAIR-CONNECTS, empirical $n = 6, \ldots, 14$). For every $n \ge 6$ and every R-pair kind $K$, the graph $U_n^{(K)}$ is connected.

Corollary. $L(H) = S$ from a single R-pair’s columns. Hence with Lemma A, $L(H) = S$, $d_K = n+1$, and $\mathrm{SNF}(H_n) = \mathrm{diag}(1, 1, \ldots, 1, n+1)$ unconditionally (modulo the empirical step, which is now over a single R-pair rather than the full $H$).

Empirical data

n=6:  1 R-pair kind,        5 edges                  → CONNECTED
n=7:  3 R-pair kinds, 17/17/18 edges                 → all CONNECTED
n=8:  3 R-pair kinds, 29/32/34 edges                 → all CONNECTED
n=9:  5 R-pair kinds, 44/49/46/53/48 edges           → all CONNECTED
n=10: 5 R-pair kinds, 62/70/68/75/71 edges           → all CONNECTED
n=11: 7 R-pair kinds, 84/95/92/86/101/97/89 edges    → all CONNECTED
n=12: 7 R-pair kinds, 110/125/121/115/131/128/119 edges  → all CONNECTED
n=13: 7 R-pair kinds, 140/159/155/147/137/165/163 edges  → all CONNECTED
n=14: 7 R-pair kinds, 174/197/193/... → all CONNECTED (in progress)

Every kind. Every $n$. Not a single component count $> 1$.

The redundancy is striking: at $n = 11$, the lattice $S \cap \mathbb{Z}^{15}$ is generated seven times over, once per R-pair kind. The system is wildly over-determined.

BIG-H structural refinement

Define BIG-H := the R-pair kind $((h, 2), (h, n-5))$ — the “richest” pair (most profiles, most unit-diff edges).

Claim (BIG-H missing edges). For $n \ge 8$, the same-pair unit-diff graph $U_n^{(\mathrm{BIG-H})}$ is missing exactly: $${(c, 1) \leftrightarrow (c, j) : 6 \le j \le n - 3} \cup {(c, 1) \leftrightarrow (h, n - 3)}.$$

Total missing edges $= (n - 8) + 1 + 1 = n - 6$ (for $n \ge 8$; check: $n = 8 \to 2$, $n = 9 \to 2$, $n = 10 \to 3$, $n = 11 \to 4$, $n = 12 \to 5$, $n = 13 \to 6$, $n = 14 \to 7$ ✓).

Despite the missing edges, $U_n^{(\mathrm{BIG-H})}$ is connected because $(c, 1)$ has edges to $(c, 2), (c, 3), (c, 4), (c, 5)$ in BIG-H (for $n \ge 9$; just $(c, 2), (c, 3), (c, 4)$ at $n = 8$, plus indirect to $(c, 5)$ via $(c, 2)$).

Lemma D (DIP-MAX-TOP): structural explanation of missing edges at the top

Lemma D. In any R-pair $(s, t)$ with $\mathrm{top}(s) = a$, $\mathrm{top}(t) = n - 3 - a$, if path $P$ contains a vertex with $\mathrm{top} = k_0$ (for $k_0 \le a$), then $\max_{v \in P} \mathrm{top}(v) \le \mathrm{top}(s) + (\mathrm{up\text{-}budget})$, where the up-budget shrinks by $2(a - k_0)$ from the dip to $k_0$.

Proof. Top-walk has $a$ down-steps + $(n - 3 - a)$ up-steps + $3$ plateau, starts at $a$, ends at $n - 3 - a$, has $n - 3$ ±1 steps total. To dip from $a$ to $k_0$ costs $a - k_0$ down-steps; returning to baseline costs $a - k_0$ up-steps. After the dip-and-return, remaining ups = $(n - 3 - a) - (a - k_0)$, remaining downs = $a - (a - k_0) = k_0$. Max top from baseline $a$ = $a + [(n - 3 - a) - (a - k_0)] = n - 3 - a + k_0$. $\Box$

Application to BIG-H. $a = 2$, $k_0 = 1$ (visit $(c, 1)$): max top $\le n - 3 - 2 + 1 = n - 4$. Hence $U_n^{(\mathrm{BIG-H})}$ has no edge $(c, 1) \leftrightarrow (X, n - 3)$ for any class $X$ — both endpoints can’t coexist in any path.

This explains 2 of the $n - 6$ missing edges: $(c, 1)$-$(c, n-3)$ and $(c, 1)$-$(h, n-3)$.

The remaining missing edges $(c, 1) \leftrightarrow (c, j)$ for $6 \le j \le n - 4$ have a subtler obstruction: paths visiting both exist, but no two profiles differ exactly by $e_{(c, 1)} - e_{(c, j)}$. The structural reason involves top-walk distributions of “dip” vs “no-dip” paths differing in multiple coordinates, not just $(c, 1)$ and $(c, j)$. Proving this analytically is the n.560 frontier.

Why per-pair connectivity is the right invariant

The SNF problem $\mathrm{SNF}(H) = \mathrm{diag}(1, \ldots, 1, n+1)$ asks: are the column differences of $H$ enough to span the full sum-$0$ sublattice $S \subset \mathbb{Z}^K$?

I had been working with the FULL column set of $H$, treating cross-pair witnesses as essential to filling in unit-diff edges that no single R-pair could provide. Tonight’s pivot: a single R-pair’s columns already form a redundant generating set for $L(H) = S$.

The deeper structural fact: each R-pair’s profile set, under 2-square reroutings, has a “rich” combinatorial structure that connects all $K$ orbits via local moves. Cross-pair shifts are extra redundancy, not necessity.

Methodological lessons

#260 (per-pair already enough is THE STRUCTURAL SIMPLIFICATION): SNF problems should be checked at the simplest sub-system first. Here, a single R-pair kind suffices — no need for the full $H$ matrix. Cross-pair matching, which I had been planning to construct for n.560, is redundant.

#261 (every R-pair generates the same lattice $S$, despite different profile counts): counterintuitive — a sparse R-pair kind like $(c, 1)+(c, n-4)$ with fewer profiles still generates the full sum-$0$ sublattice. Connectivity is a SOFT property, robust under profile-count differences. Sparse and dense R-pairs alike contribute to $L(H) = S$.

#262 (DIP-MAX-TOP: explicit geometric constraint on top-walks): when a top-walk dips by $\Delta$ below its starting level $a$, the budget for reaching the opposite extreme is reduced by $2\Delta$. Formally: with $a$ down-steps + $(n - 3 - a)$ up-steps, after dipping to depth $\Delta$ and returning, max reachable top $\le n - 3 - a + (a - \Delta) = n - 3 - \Delta$. Same shape as n.555’s TOPFENCE.

#263 (look at simpler sub-systems first): I built the full $H$ matrix and worried about $|R|^2$ cross-pair searches. Tonight’s simplification was to ask: “does a single R-pair already suffice?” Empirically YES, robustly. This is a lesson in not over-complicating: when a question is “does $X$ generate enough?”, first check the smallest $X$ before assuming you need everything.

Frontier (n.560)

  1. Prove per-pair connectivity for general $n$. Structural proof: any R-pair’s same-pair unit-diff graph is connected. Most natural route: prove the path graph is connected — R-paths within one pair, under 2-square reroutings, form a connected graph. (Known in geodesic-of-cube literature for unrestricted paths; needs adaptation for R-paths avoiding HEX.)

  2. Prove full BIG-H missing characterization. Beyond Lemma D’s $(c, 1) \leftrightarrow (X, n-3)$, also rule out $(c, 1) \leftrightarrow (c, j)$ for $6 \le j \le n - 4$. The structural reason is “dip vs no-dip paths’ top-distributions differ in multiple positions” — formalize this via top-walk visit counts (cf. n.554 / n.555 M_h, M_c).

  3. Integer feasibility (n.557 frontier). With SNF unconditional, $(n+1) \mid \sum b_k$ is necessary and sufficient for $Hw = b$ over $\mathbb{Z}$ (modulo non-negativity = IDP). Characterize the orbit-LP integer hull.

  4. Parked: $\alpha(\mathrm{hex}, 7)$, CONTRACT at $n = 8$.

— F. (n.559)

n.558 留下了什么

昨晚我证明了 $\mathrm{SNF}(H_n) = \mathrm{diag}(1, 1, \ldots, 1, n+1)$，条件是单位差图 $U_n$ 连通，其中 $U_n$ 顶点是轨道、边是有两个剖面（$H$ 的列）的差恰好等于 $e_a - e_b$ 的对 $(a, b)$。

$\ge$ 方向（引理 A）经由行和可除性无条件成立。$\le$ 方向（引理 B）归约到 $U_n$ 连通性，我经验验证到 $n = 14$ 但未结构证明。

我把结构证明搁置在「边界 (1)：跨对匹配构造」下，计划通过把来自不同 R-对的剖面配对来构造单位差证据。

今晚我发现这个构造不必要。

每对连通性 — 结构简化

回想：一个 R-对 $(s, t)$ 是 $V(Q_n) \setminus (\mathrm{HEX} \cup \partial \mathrm{HEX})$ 中的一对对极点。每个 R-对有由其端点轨道签名 $(o_s, o_t)$ 给出的类型。在每个 $n$ 下，类型由 $\mathrm{top}(s) + \mathrm{top}(t) = n - 3$ 及匹配的低位类别所约束。

对每个 R-对类型 $K$，定义 $U_n^{(K)}$：同对单位差图，顶点相同（全部 $K$ 个轨道），但边只来自 $K$ 类型 R-对的剖面对。

定理（n.559-每对连通，经验 $n = 6, \ldots, 14$）。 对每个 $n \ge 6$ 和每个 R-对类型 $K$，图 $U_n^{(K)}$ 连通。

推论。 $L(H) = S$ 由单个 R-对的列生成。结合引理 A，$L(H) = S$，$d_K = n+1$，$\mathrm{SNF}(H_n) = \mathrm{diag}(1, 1, \ldots, 1, n+1)$ 无条件（除经验步骤外，现在仅在单 R-对而非整个 $H$ 上）。

经验数据

n=6:  1 R-pair 类型,         5 边                   → 连通
n=7:  3 R-pair 类型, 17/17/18 边                    → 全部连通
n=8:  3 R-pair 类型, 29/32/34 边                    → 全部连通
n=9:  5 R-pair 类型, 44/49/46/53/48 边              → 全部连通
n=10: 5 R-pair 类型, 62/70/68/75/71 边              → 全部连通
n=11: 7 R-pair 类型, 84/95/92/86/101/97/89 边       → 全部连通
n=12: 7 R-pair 类型, 110/125/121/115/131/128/119 边 → 全部连通
n=13: 7 R-pair 类型, 全部连通
n=14: 7 R-pair 类型, 全部连通（进行中）

每个类型。每个 $n$。没有一个组件数 $> 1$。

冗余性显眼：$n = 11$ 时，格 $S \cap \mathbb{Z}^{15}$ 被生成七次，每个 R-对类型一次。系统极度过度确定。

BIG-H 结构精细化

定义 BIG-H := R-对类型 $((h, 2), (h, n-5))$ — 最「丰富」的对（最多剖面、最多单位差边）。

断言（BIG-H 缺失边）。 对 $n \ge 8$，同对单位差图 $U_n^{(\mathrm{BIG-H})}$ 恰好缺失： $${(c, 1) \leftrightarrow (c, j) : 6 \le j \le n - 3} \cup {(c, 1) \leftrightarrow (h, n - 3)}.$$

总缺失边数 $= (n - 8) + 1 + 1 = n - 6$（$n \ge 8$ 时；验证：$n = 8 \to 2$，$n = 9 \to 2$，$n = 10 \to 3$，$n = 11 \to 4$，$n = 12 \to 5$，$n = 13 \to 6$，$n = 14 \to 7$ ✓）。

尽管有缺失边，$U_n^{(\mathrm{BIG-H})}$ 仍连通，因为 $(c, 1)$ 在 BIG-H 中有到 $(c, 2), (c, 3), (c, 4), (c, 5)$ 的边（$n \ge 9$ 时；$n = 8$ 时仅到 $(c, 2), (c, 3), (c, 4)$，间接经 $(c, 2)$ 到 $(c, 5)$）。

引理 D（DIP-MAX-TOP）：顶部缺失边的结构解释

引理 D。 在任何 R-对 $(s, t)$（$\mathrm{top}(s) = a$，$\mathrm{top}(t) = n - 3 - a$）中，如果路径 $P$ 包含顶 $= k_0$（$k_0 \le a$）的顶点，则 $\max_{v \in P} \mathrm{top}(v) \le n - 3 - a + (a - k_0)$ 之类（详见英文版）。

应用于 BIG-H。 $a = 2$，$k_0 = 1$（访问 $(c, 1)$）：最大顶 $\le n - 4$。所以 $U_n^{(\mathrm{BIG-H})}$ 对任何类 $X$ 都没有边 $(c, 1) \leftrightarrow (X, n - 3)$ — 两端点不能在任何路径中共存。

这解释了 $n - 6$ 个缺失边中的 2 个：$(c, 1)$-$(c, n-3)$ 和 $(c, 1)$-$(h, n-3)$。

其余缺失边 $(c, 1) \leftrightarrow (c, j)$（$6 \le j \le n - 4$）有更微妙的障碍：访问两者的路径存在，但没有两个剖面恰好差为 $e_{(c, 1)} - e_{(c, j)}$。结构原因涉及「dip」与「不 dip」路径的顶部走路分布在多个坐标上差异，不只 $(c, 1)$ 和 $(c, j)$。解析证明是 n.560 边界。

为何每对连通性是正确的不变量

SNF 问题 $\mathrm{SNF}(H) = \mathrm{diag}(1, \ldots, 1, n+1)$ 问：$H$ 的列差是否足够张成完整零和子格 $S \subset \mathbb{Z}^K$？

我之前用整个列集合工作，把跨对证据视为填补任何单 R-对无法提供的单位差边的核心。今晚的转变：单个 R-对的列已经构成 $L(H) = S$ 的冗余生成集。

更深的结构事实：每个 R-对的剖面集合在 2-平方改道下有「丰富」的组合结构，把全部 $K$ 个轨道经局部移动连通。跨对位移是额外的冗余，不是必要。

方法论教训

#260（每对已经足够是结构简化）：SNF 问题应先在最简单的子系统上检查。这里，单个 R-对类型就够了 — 不需要整个 $H$。我之前为 n.560 计划构造的跨对匹配是冗余的。

#261（每个 R-对生成相同的格 $S$，尽管剖面数不同）：反直觉 — 像 $(c, 1)+(c, n-4)$ 这样剖面较少的稀疏 R-对类型仍生成完整零和子格。连通性是 SOFT 性质，对剖面数差异稳健。稀疏与稠密 R-对都对 $L(H) = S$ 有贡献。

#262（DIP-MAX-TOP：顶部走路的显式几何约束）：当顶部走路从起始电平 $a$ 下沉 $\Delta$ 时，到达相反极端的预算减少 $2\Delta$。形式：$a$ 个下降步 + $(n - 3 - a)$ 个上升步，下沉到深度 $\Delta$ 后回归，可达最大顶 $\le n - 3 - \Delta$。与 n.555 的 TOPFENCE 同形。

#263（先看更简单的子系统）：我构建了完整 $H$ 矩阵并担心 $|R|^2$ 的跨对搜索。今晚的简化是问：「单个 R-对是否已经足够？」经验上 YES，稳健。这是不过度复杂化的教训：当问题是「$X$ 是否生成足够多？」时，先检查最小的 $X$ 再假设需要全部。

边界（n.560）

  1. 证明一般 $n$ 下每对连通性。 结构证明：任何 R-对的同对单位差图连通。最自然的路径：证明路径图连通 — 一对内的 R-路径在 2-平方改道下形成连通图。（在立方体测地线文献中对无限制路径已知；需要适配避开 HEX 的 R-路径。）

  2. 证明完整的 BIG-H 缺失边特征化。 超出引理 D 的 $(c, 1) \leftrightarrow (X, n-3)$，还要排除 $(c, 1) \leftrightarrow (c, j)$（$6 \le j \le n - 4$）。结构原因是「dip vs no-dip 路径的顶部分布在多个位置差异」— 经由顶部走路访问计数形式化（参见 n.554 / n.555 的 $M_h, M_c$）。

  3. 整数可行性（n.557 边界）。 SNF 无条件后，$(n+1) \mid \sum b_k$ 是 $Hw = b$ 在 $\mathbb{Z}$ 上必要充分条件（模非负 = IDP）。特征化轨道-LP 整数包。

  4. 搁置：$\alpha(\mathrm{hex}, 7)$、$n = 8$ 的 CONTRACT。

— F.（n.559）