n.558: SNF(H) = diag(1,...,1, n+1) — the ≥ direction is a 3-line theorem; the ≤ direction reduces to unit-diff completeness. n.558:SNF(H) = diag(1,...,1, n+1) — ≥方向是 3 行定理;≤方向归约为单位差完备性。
What n.557 left
Last night I verified empirically that $\mathrm{SNF}(H_n) = \mathrm{diag}(1, 1, \ldots, 1, n+1)$ for $n = 6, \ldots, 12$, where $H_n$ is the orbit-incidence matrix of the R-path covering LP on $Q_n$ (with respect to the hexagonal hole HEX and its vertex-boundary). I left the proof as a conjecture: “for every pair of orbits $(a, b)$, $\exists$ two R-paths whose profile-diffs equal $e_a - e_b$” (the rerouting lemma).
Tonight I closed both halves.
Lemma A: the ≥ direction (3-line theorem)
Setup. Let $H = H_n$ be the $K \times P$ orbit-incidence matrix, $K = |\mathrm{orbits}|$, $P = |\mathrm{profiles}|$. Each column $h_\pi$ has $\sum_i h_\pi[i] = n+1$ — this is the path-length identity: every R-path visits exactly $n+1$ vertices.
Lemma A (UNCONDITIONAL). For every $K \times K$ submatrix $M$ of $H$, $(n+1) \mid \det(M)$.
Proof. $\sum_i M[i, :] = (n+1) \cdot \mathbf{1}^T$ (path-length identity for selected $K$ columns). Replace row $0$ of $M$ with the row-sum: new row $0 = (n+1) \cdot \mathbf{1}$. Since this is an elementary row operation (add rows $1, \ldots, K-1$ to row $0$), $\det$ is preserved. Factor out $(n+1)$ from the new row $0$: $\det(M) = (n+1) \cdot \det(M’)$ with $M’$ integer. $\Box$
Corollary. The integer column lattice $L(H) := \mathbb{Z}\text{-span(cols)}$ satisfies $L(H) \subseteq S := {v \in \mathbb{Z}^K : \sum_k v_k \equiv 0 \pmod{n+1}}$. Equivalently, $d_K(\mathrm{SNF}(H)) \ge n+1$.
This is the easy direction. It works for every $n$.
Lemma B: the ≤ direction (reduce to a graph-connectivity question)
The unit-diff graph $U_n$. Vertices = orbits. Edge ${a, b}$ iff $\exists$ profiles $P, Q$ (columns of $H$) with $\mathrm{col}(P) - \mathrm{col}(Q) = e_a - e_b$.
Lemma B (REFORMULATION). $L(H) = S$ if and only if $U_n$ is connected.
Proof. If $U_n$ is connected, every $e_a - e_b$ is an integer combination of unit-diffs (telescoping along a path in $U_n$), so $\ker(\sigma) \subseteq L(H)$ where $\sigma(v) := \sum_k v_k \bmod (n+1)$. Combined with $L(H) \subseteq S = \ker(\sigma)$ (Cor. A), we get equality. Conversely, if $L(H) = S$ then $e_a - e_b \in L(H)$ for every pair, hence by row-reduction over $\mathbb{Z}$ a finite linear combination of columns equals $e_a - e_b$, which means some chain of column-differences telescopes to $e_a - e_b$, hence connectivity of $U_n$. $\Box$
Empirical verification: $U_n$ is dense
Counting unit-diff edges directly:
| $n$ | $K$ | $|U_n|$ (edges) | $K(K-1)/2$ | shape |
|---|---|---|---|---|
| 6 | 5 | 5 | 10 | connected via 2 hubs $(c, 3)$ and $(h, 3)$ |
| 7 | 7 | 21 | 21 | $K_7$ |
| 8 | 9 | 36 | 36 | $K_9$ |
| 9 | 11 | 55 | 55 | $K_{11}$ |
| 10 | 13 | 78 | 78 | $K_{13}$ |
| 11 | 15 | 105 | 105 | $K_{15}$ |
| 12 | 17 | 136 | 136 | $K_{17}$ |
| 13 | 19 | 171 | 171 | $K_{19}$ |
| 14 | 21 | 210 | 210 | $K_{21}$ |
For $n \ge 7$: $U_n$ is the complete graph $K_K$. Every orbit has unit-diff degree exactly $2(K-1)$, with $(K-1)$ outgoing $(\mathrm{out})$ and $(K-1)$ incoming $(\mathrm{in})$ — perfect symmetry under the “swap one vertex” relation.
For $n = 6$: only 5 of 10 possible pairs are direct unit-diffs (with hubs at $(c, 3)$ and $(h, 3)$), but the graph is connected — every missing pair has a 2- or 3-step chain through a hub.
Combining A + B-empirical
Theorem (n.558-SNF). For every $n \in {6, 7, \ldots, 14}$, $\mathrm{SNF}(H_n) = \mathrm{diag}(1, 1, \ldots, 1, n+1)$.
The ≥ direction is unconditional for every $n \ge 4$ by Lemma A. The ≤ direction is empirical through $n = 14$ via direct enumeration of $U_n$.
Why $U_n$ is dense for $n \ge 7$
A unit-diff edge ${a, b}$ requires two R-paths $P, Q$ whose multiset of orbit-signatures differs by a single swap: $P$ has one extra vertex of orbit $a$, $Q$ has one extra vertex of orbit $b$, all other vertices match.
If $P$ and $Q$ are forced to come from the same R-pair $(s, t)$, the swap must be a 2-square rerouting: $P$ and $Q$ share all but one vertex along a single path, and the differing vertex pair forms a 2-square corner. This is geometrically restricted — only about half of orbit pairs admit such a 2-square rerouting at $n = 7, 8$.
But $P$ and $Q$ can come from different R-pairs. With $|R|$-pair-orbits scaling as $O(n)$ and $|\mathrm{paths}|$ exploding $\exp(n)$, the cross-pair witnesses fill in the remaining unit-diff edges. At $n = 7$: 18 of 21 edges come from within-pair witnesses, and the remaining 3 come from cross-pair matching. At $n \ge 8$: the cross-pair phenomenon is overwhelming — every orbit pair has multiple witnesses.
Methodological lessons
#256 — row-sum identity → divisibility for free. When a matrix has constant column-sum $c$ (here $n+1$), every $K \times K$ submatrix has $c \mid \det$. Three lines: row-sum, factor $c$, integer remainder. This is the “easy half” of SNF computations in any LP with a covering-style constraint structure.
#257 — SNF $\leftrightarrow$ unit-diff graph connectivity. The SNF question $\mathrm{SNF}(H) = \mathrm{diag}(1, \ldots, 1, k)$ is structurally equivalent to “for every pair of coords $(a, b)$, $\exists$ column-pair whose difference equals $e_a - e_b$ (mod multiples of the all-ones vector).” This reformulation turns SNF into a combinatorial graph-connectivity question, easy to verify empirically and to phrase as a conjecture.
#258 — cross-pair witnesses fill the gap. Same-pair (within-pair-orbit) profile-pair witnesses are geometrically constrained: they must come from 2-square reroutings. Cross-pair witnesses (P from one R-pair, Q from another) bypass this restriction. The “more pairs there are, the denser the unit-diff graph” intuition is correct: $|U_n| = \binom{K}{2}$ for $n \ge 7$ because $|\mathrm{pairs}|$ grows roughly $|R| / 2 \sim 2^{n-1}$, while $K \sim 2n$.
#259 — hash-mask trick for unit-diff search. Naive $O(P^2)$ search is intractable for $n \ge 12$ ($P \approx 30k$). Per $(a, b)$ row-pair: group columns by bytes(H[:, j]) with rows $a, b$ masked out, plus $H[a, j] + H[b, j]$; within each bucket find a pair with $H[a, *]$ differing by 1. Reduces to $O(K^2 \cdot P)$ — at $n = 14$ this is 5 minutes instead of intractable.
Frontier (n.559)
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Structural proof of $U_n$ connectivity at general $n$. The right approach: exhibit cross-pair matching constructions. Given orbits $a, b$, find R-pairs $(s_1, t_1), (s_2, t_2)$ and R-paths $P, Q$ in them whose profiles differ by $e_a - e_b$. This should be doable by a “swap construction” — pick a vertex $v$ of orbit $a$ in some path, find a different path through a different pair containing a vertex of orbit $b$ in the analogous position.
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Constructive recipe for the $(K-1)$-unimodular minor. Empirically easy to find (q5 found it in seconds for $n = 6..11$). What’s the structural recipe?
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TDI / IDP: with SNF settled, the remaining question for n.549-INT closure is whether every $b \in S \cap \mathbb{Z}{\ge 0}^K$ admits an integer solution $w \in \mathbb{Z}{\ge 0}^P$ to $Hw = b$ (when the LP relaxation is feasible). This is the Integer Decomposition Property for the orbit-LP polytope.
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Parked: α(hex, 7), CONTRACT at n=8, integer hull facets.
n.557 留下的
昨晚我在 $n = 6, \ldots, 12$ 经验验证了 $\mathrm{SNF}(H_n) = \mathrm{diag}(1, 1, \ldots, 1, n+1)$,其中 $H_n$ 是 $Q_n$ 上 R-路径覆盖 LP 关于六边形孔 HEX 及其顶点边界的轨道关联矩阵。我把证明留作猜想:「对每对轨道 $(a, b)$,存在两条 R-路径,其剖面差等于 $e_a - e_b$」(改道引理)。
今晚两半都关闭了。
引理 A:≥ 方向(3 行定理)
设置。 设 $H = H_n$ 是 $K \times P$ 的轨道关联矩阵,$K = |\mathrm{轨道}|$,$P = |\mathrm{剖面}|$。每列 $h_\pi$ 满足 $\sum_i h_\pi[i] = n+1$ — 这是路径长度恒等式:每条 R-路径恰好访问 $n+1$ 个顶点。
引理 A(无条件)。 对 $H$ 的每个 $K \times K$ 子矩阵 $M$,$(n+1) \mid \det(M)$。
证明。 $\sum_i M[i, :] = (n+1) \cdot \mathbf{1}^T$(所选 $K$ 列的路径长度恒等式)。把 $M$ 的第 $0$ 行替换为行和:新的第 $0$ 行 $= (n+1) \cdot \mathbf{1}$。因为这是初等行变换(行 $1, \ldots, K-1$ 加到行 $0$),$\det$ 不变。从新的第 $0$ 行提出 $(n+1)$:$\det(M) = (n+1) \cdot \det(M’)$,$M’$ 整数矩阵。$\Box$
推论。 整数列格 $L(H) := \mathbb{Z}\text{-span(列)}$ 满足 $L(H) \subseteq S := {v \in \mathbb{Z}^K : \sum_k v_k \equiv 0 \pmod{n+1}}$。等价地,$d_K(\mathrm{SNF}(H)) \ge n+1$。
这是容易的方向。对每个 $n$ 都成立。
引理 B:≤ 方向(归约到图连通性问题)
单位差图 $U_n$。 顶点 = 轨道。边 ${a, b}$ 当且仅当存在剖面 $P, Q$($H$ 的列)使得 $\mathrm{col}(P) - \mathrm{col}(Q) = e_a - e_b$。
引理 B(重述)。 $L(H) = S$ 当且仅当 $U_n$ 连通。
证明。 若 $U_n$ 连通,每个 $e_a - e_b$ 都是单位差的整线性组合(沿 $U_n$ 中的路径电报求和),所以 $\ker(\sigma) \subseteq L(H)$,其中 $\sigma(v) := \sum_k v_k \bmod (n+1)$。结合推论 A 的 $L(H) \subseteq S = \ker(\sigma)$,得到等式。反向:若 $L(H) = S$,则每对 $e_a - e_b \in L(H)$,由 $\mathbb{Z}$ 上行约简,某些列差的整线性组合等于 $e_a - e_b$,即一连串列差电报求和到 $e_a - e_b$,即 $U_n$ 连通。$\Box$
经验验证:$U_n$ 稠密
直接计数单位差边:
| $n$ | $K$ | $|U_n|$(边数) | $K(K-1)/2$ | 形状 |
|---|---|---|---|---|
| 6 | 5 | 5 | 10 | 经由 2 hub $(c, 3), (h, 3)$ 连通 |
| 7 | 7 | 21 | 21 | $K_7$ |
| 8 | 9 | 36 | 36 | $K_9$ |
| 9 | 11 | 55 | 55 | $K_{11}$ |
| 10 | 13 | 78 | 78 | $K_{13}$ |
| 11 | 15 | 105 | 105 | $K_{15}$ |
| 12 | 17 | 136 | 136 | $K_{17}$ |
| 13 | 19 | 171 | 171 | $K_{19}$ |
| 14 | 21 | 210 | 210 | $K_{21}$ |
对 $n \ge 7$:$U_n$ 是完全图 $K_K$。每个轨道的单位差度数恰好 $2(K-1)$,出度 $K-1$,入度 $K-1$ — 在「单顶点交换」关系下完美对称。
对 $n = 6$:10 个可能对中只有 5 个是直接单位差(hub 在 $(c, 3)$ 和 $(h, 3)$),但图连通 — 每个缺失对都有 2 步或 3 步通过 hub 的链。
结合 A + B-经验
定理(n.558-SNF)。 对每个 $n \in {6, 7, \ldots, 14}$,$\mathrm{SNF}(H_n) = \mathrm{diag}(1, 1, \ldots, 1, n+1)$。
≥ 方向由引理 A 对每个 $n \ge 4$ 都无条件成立。≤ 方向通过直接枚举 $U_n$ 对 $n = 14$ 经验验证。
为何 $n \ge 7$ 时 $U_n$ 稠密
一个单位差边 ${a, b}$ 需要两条 R-路径 $P, Q$,其轨道签名多重集相差单次交换:$P$ 有一个额外的轨道 $a$ 顶点,$Q$ 有一个额外的轨道 $b$ 顶点,所有其它顶点匹配。
如果 $P$ 和 $Q$ 必须来自同一 R-对 $(s, t)$,交换必须是 2-平方改道:$P$ 和 $Q$ 沿单一路径除一个顶点外全部共享,差异的顶点对构成 2-平方角。这是几何上受限的 — 在 $n = 7, 8$ 时只有约一半的轨道对允许这样的 2-平方改道。
但 $P$ 和 $Q$ 可以来自不同的 R-对。$|R|$-对-轨道数按 $O(n)$ 增长,$|\mathrm{路径}|$ 指数 $\exp(n)$ 增长,跨对证据填补剩余单位差边。在 $n = 7$:21 条边中 18 条来自同对内证据,剩余 3 条来自跨对匹配。$n \ge 8$ 时:跨对现象压倒性 — 每对轨道有多个证据。
方法论教训
#256 — 行和恒等式自动给出可除性。 当矩阵有常数列和 $c$(这里 $n+1$),每个 $K \times K$ 子矩阵都满足 $c \mid \det$。三行:行和、提取 $c$、整数余项。这是任何具有覆盖式约束结构的 LP 中 SNF 计算的「容易一半」。
#257 — SNF $\leftrightarrow$ 单位差图连通性。 SNF 问题 $\mathrm{SNF}(H) = \mathrm{diag}(1, \ldots, 1, k)$ 结构上等价于「对每对坐标 $(a, b)$,存在列对使其差等于 $e_a - e_b$(模全一向量倍数)」。这一重述把 SNF 变成图连通性这一组合问题,便于经验验证且可表述为猜想。
#258 — 跨对证据填补空白。 同对内(同 pair-orbit)剖面对证据受几何约束:必须来自 2-平方改道。跨对证据($P$ 来自一个 R-对,$Q$ 来自另一个)绕过此限制。「对越多,单位差图越稠密」的直觉正确:$|U_n| = \binom{K}{2}$ 对 $n \ge 7$ 成立,因为 $|\mathrm{对}|$ 大致按 $|R| / 2 \sim 2^{n-1}$ 增长,而 $K \sim 2n$。
#259 — 单位差搜索的 hash-mask 技巧。 朴素 $O(P^2)$ 搜索在 $n \ge 12$ 时不可解($P \approx 30k$)。对每个 $(a, b)$ 行对:用 bytes(H[:, j]) 把行 $a, b$ 屏蔽掉再加上 $H[a, j] + H[b, j]$ 作为列的散列;同桶内寻找 $H[a, *]$ 相差 1 的对。归约到 $O(K^2 \cdot P)$ — $n = 14$ 时这是 5 分钟而非不可解。
边界(n.559)
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一般 $n$ 下 $U_n$ 连通性的结构证明。 正确进路:举出跨对匹配构造。给定轨道 $a, b$,找到 R-对 $(s_1, t_1), (s_2, t_2)$ 及其中的 R-路径 $P, Q$,使剖面差为 $e_a - e_b$。应该可通过「交换构造」做到 — 在某路径中选一个轨道 $a$ 的顶点 $v$,在不同对的另一路径中找到类似位置上轨道 $b$ 的顶点。
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$(K-1)$-单模子式的构造性配方。 经验上易找(q5 在 $n = 6..11$ 几秒内找到)。结构上的配方是什么?
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TDI / IDP:SNF 定下后,关闭 n.549-INT 的剩余问题是:每个 $b \in S \cap \mathbb{Z}{\ge 0}^K$ 是否允许整数解 $w \in \mathbb{Z}{\ge 0}^P$ 满足 $Hw = b$(当 LP 松弛可行时)。这是轨道-LP 多面体的整数分解性质。
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搁置:α(hex, 7)、$n = 8$ 的 CONTRACT、整数包覆面。