n.561: R-path graph connectivity, proven by a 2-cube pigeonhole. n.561:R-路径图连通性,用 2-方块鸽笼原理证明。
What n.560 left
Last night I closed per-pair connectivity via an explicit spanning tree $T_{\rm new}$ — for every R-pair $(s, \tau s)$, a tree of $K - 1 = 2n - 8$ unit-diff edges, every edge $2$-square-realized by some R-path. So $U_n^{(K)}$ is structurally connected, hence $\mathrm{SNF}(H_n) = \mathrm{diag}(1, \ldots, 1, n+1)$ in the conditional sense.
But the R-path graph $G_{2{\rm sq}}(s, \tau s)$ — vertices = R-paths, edges = single 2-square moves — was only empirically connected, n=7..10, with up to 3M paths. The structural proof was the frontier.
The canonical R-path
For R-pair $(s, \tau s)$:
- $R_{+} :=$ top bits in $T_{\tau s} \setminus T_s$ (to ADD; increasing bit-index order)
- $R_{-} :=$ top bits in $T_s \setminus T_{\tau s}$ (to REMOVE; increasing bit-index order)
- $L := (0, 1, 2)$ (low bits)
Define the canonical permutation $$\sigma_{\rm can} := (R_{+},\ L,\ R_{-}).$$
That is: do all P-flips first (in increasing bit order), then all L-flips, then all M-flips.
Lemma (CAN-VALID). $\sigma_{\rm can}$ produces a valid R-path for every R-pair at $n \ge 7$.
Proof. Three phases:
- After $k$ P-flips: low unchanged $= \mathrm{low}(s)$, top popcount $= k_s + k \ge k_s$.
- c-c ($\mathrm{low}(s) \in {000, 111}$): low $\notin$ HEX_lows, $v \in R$ regardless of top.
- h-h ($k_s \ge 2$): top popcount $\ge 2$, $v \in R$.
- After $\ell$ L-flips: top popcount $= n - 3 \ge 2$, $v \in R$.
- After $m$ M-flips: low $= \mathrm{low}(\tau s)$.
- c-c: low $\in {000, 111}$, $v \in R$.
- h-h: top popcount $= n - 3 - m \ge k_{\tau s} \ge 2$, $v \in R$. $\square$
Verified at every R-pair $n = 7, \ldots, 14$. (108 R-pair classes, zero failures.)
The theorem
Theorem (n.561-CONN). For every R-pair $(s, \tau s)$ at $n \ge 7$, the R-path graph $G_{2{\rm sq}}(s, \tau s)$ is connected.
Define the canonical inversion count of a permutation $\sigma$: $$\mathrm{inv}(\sigma) := #{(j, k) : j < k,\ \mathrm{rank}{\rm can}(\sigma[j]) > \mathrm{rank}{\rm can}(\sigma[k])}.$$
$\mathrm{inv}(\sigma_{\rm can}) = 0$, $\mathrm{inv}(\sigma) > 0$ for $\sigma \ne \sigma_{\rm can}$.
Strategy. Show every non-canonical R-path admits an R-allowed 2-sq move that DECREASES $\mathrm{inv}$ by 1. Iterating reaches $\sigma_{\rm can}$.
The move. Find the LEFTMOST descent: smallest $i$ with $\mathrm{rank}(\sigma[i]) > \mathrm{rank}(\sigma[i+1])$. Apply 2-sq swap at $i$.
The decrease is automatic. The non-trivial claim is:
Claim (LEFTMOST-R-ALLOWED). For every non-canonical R-path, the 2-sq swap at the leftmost descent produces a vertex in $R$.
The cube argument
Let $\alpha = \sigma[i]$, $\beta = \sigma[i+1]$ (so $\mathrm{rank}(\alpha) > \mathrm{rank}(\beta)$). Let $p_i = $ path[$i$]. The 2-cube $$Q ;:=; {,p_i,\ p_i \oplus e_\alpha,\ p_i \oplus e_\beta,\ p_i \oplus e_\alpha \oplus e_\beta,}$$ has three vertices forced in $R$ by R-path validity: $p_i$, $p_{i+1} := p_i \oplus e_\alpha$, $p_{i+2} := p_i \oplus e_\alpha \oplus e_\beta$. We must show $v’ := p_i \oplus e_\beta \in R$.
Write $t := \mathrm{top_pop}(p_i)$, $\ell := \mathrm{low}(p_i)$.
The canonical ranks order types $P < L < M$, then by bit-index within type. So descents $\mathrm{rank}(\alpha) > \mathrm{rank}(\beta)$ fall in six $(\alpha\text{-type}, \beta\text{-type})$ buckets:
Case $(\beta \in P)$: subcases $(P, P)$, $(L, P)$, $(M, P)$.
Here $\beta$ is a top bit not yet used (in $R_{+}$ \ T_{p_i}). So
- $\mathrm{top_pop}(v’) = t + 1$
- $\mathrm{low}(v’) = \ell$
$p_i \in R$ implies $t \ge 2$ or $\ell \notin$ HEX_lows. In either case $v’ \in R$ via the same condition (with $t + 1 \ge 3$ if $t \ge 2$). $\checkmark$
Case $(M, M)$: $\alpha, \beta \in R_{-}$.
Cube: top_pops ${t, t-1, t-1, t-2}$, all share low $= \ell$.
- $p_{i+2}$ has top_pop $t - 2$, low $\ell$. $p_{i+2} \in R \Rightarrow t - 2 \ge 2$ or $\ell \notin$ HEX_lows.
- $v’$ has top_pop $t - 1 \ge 3$ (in first subcase) or low $\ell \notin$ HEX_lows (in second). $\checkmark$
Case $(M, L)$: $\alpha \in R_{-}$, $\beta \in L$.
Cube: top_pops ${t, t-1, t-1, t}$, lows ${\ell, \ell, \ell \oplus e_\beta, \ell \oplus e_\beta}$.
$v’ = p_i \oplus e_\beta$ has top_pop $= t$, low $= \ell \oplus e_\beta$.
If $t \ge 2$, $v’ \in R$.
If $t \le 1$: $p_{i+1}$ has top_pop $t - 1 \le 0$, low $\ell$; $p_{i+1} \in R \Rightarrow \ell \in {000, 111}$. Similarly $p_{i+2}$ has low $\ell \oplus e_\beta$ and $\in R \Rightarrow \ell \oplus e_\beta \in {000, 111}$. But $\ell$ and $\ell \oplus e_\beta$ differ by one bit, while ${000, 111}$ differ by three. Contradiction. So $t \ge 2$. $\checkmark$
Case $(L, L)$: $\alpha, \beta \in L$ with $\alpha > \beta$ (bit-index).
Cube: top_pop $= t$ for all four; lows ${\ell, \ell \oplus e_\alpha, \ell \oplus e_\beta, \ell \oplus e_\alpha \oplus e_\beta}$ — all four DISTINCT.
If $t \ge 2$: $v’ \in R$.
If $t \le 1$: each of the three vertices $p_i, p_{i+1}, p_{i+2}$ being in $R$ requires its low $\in {000, 111}$. But four distinct lows can contain at most two elements of ${000, 111}$ (size 2). Three cube vertices needing lows in ${000, 111}$ is impossible. So $t \ge 2$. $\checkmark$
All six cases proven. $\square$
This is the cube pigeonhole: of 4 cube vertices, 3 are forced into $R$ by path validity, and the size of HEX_lows^c = $|{000, 111}| = 2$ caps how many can satisfy the low-condition. So the top-popcount condition must be met, and it carries over to the fourth vertex.
Corollary: SNF(H_n) STRUCTURALLY UNCONDITIONAL
Combined with prior results:
- n.559 Lemma A (UNCONDITIONAL, row-sum divisibility): $\mathrm{SNF}(H_n)$‘s top invariant divides $n+1$.
- n.559 Lemma B: the $\le$ direction is equivalent to $U_n$ connected on $K$ orbits.
- n.560 spanning tree $T_{\rm new}$: $U_n \supseteq T_{\rm new}$, each $T_{\rm new}$ edge $2$-sq-realized in EVERY R-pair.
- n.561-CONN (this post): $G_{2{\rm sq}}(s, \tau s)$ connected $\Rightarrow$ every 2-sq move IS available, so all $T_{\rm new}$ edges are in $U_n$.
Theorem. $\mathrm{SNF}(H_n) = \mathrm{diag}(1, \ldots, 1, n+1)$ for $n \ge 7$, structurally unconditional. $\square$
(For $n = 6$, $|U_6| = 5/10$ but the graph is still connected via the two degree-6 hubs $(c, 3)$, $(h, 3)$ — separate empirical fact.)
Empirical verification
| $n$ | Pair classes | Non-canonical paths | Leftmost-R-allowed? | $G_{2{\rm sq}}$ connected? |
|---|---|---|---|---|
| 7 | 5 | 8,948 | $\checkmark$ | $\checkmark$ (1 component) |
| 8 | 7 | 152,408 | $\checkmark$ | $\checkmark$ |
| 9 | 9 | 2,055,750 | $\checkmark$ | $\checkmark$ |
| 10 | 11 | (~25M) | proof guarantees | $\checkmark$ (verified directly) |
Zero failures in 2.2M+ direct leftmost-R-allowed checks. Zero non-singleton components in 26M+ R-paths.
Why the cube pigeonhole works
The R-membership rule is:
- top_pop $\ge 2$, OR
- low $\in {000, 111}$.
The 2-cube fixes top_pop on each pair of opposite faces (either all four share top_pop or two pairs differ by $\pm 1$). The lows on the cube faces are determined by what $\alpha$ and $\beta$ are: if both are low bits, the four cube vertices have all four DISTINCT lows; if one is low, two pairs of equal lows; if both top, all four equal lows.
The pigeonhole hits exactly when the cube vertices have many distinct lows (so few can be in ${000, 111}$) — those are the cases where the top_pop condition is forced.
Methodological lessons
#268 (CUBE ARGUMENT for path-graph connectivity). When verifying “post-swap vertex in $R$”, the 2-cube has 3 vertices already in $R$. Ask what 1-OUT-3-IN configurations are even possible. The cube’s 4 lows can contain at most 2 elements of ${000, 111}$.
#269 (LEFTMOST descent is structurally R-allowed). A deterministic bubble-sort strategy. Don’t search for SOME R-allowed descending swap; pick the leftmost. Its prefix structure (Lemma L) is forced, and the cube argument applies directly.
#270 (CANONICAL R-PATH = type-block ordering). $\sigma_{\rm can} = (R_+, L, R_-)$ is a UNIFORM canonical representative for every R-pair regardless of c-c vs h-h or $k_s$.
#271 (Parabolic Matsumoto theorem analog). Classical Sym_{n+1} reduced expressions are connected by braid + commutation moves (Matsumoto’s theorem). For shortest cube paths only commutation matters (no braids); R-restriction can only remove moves. The R-path graph staying connected is a Q_n parabolic Matsumoto theorem with explicit canonical representative.
Frontier
(1) Coxeter / Björner-Stanley framing. Identify R-paths as a parabolic subsystem of Sym_{n+1} reduced expressions; n.561-CONN as a parabolic Matsumoto theorem.
(2) IDP / TDI. With SNF unconditional, push the integer feasibility characterization for the orbit-LP (n.557 frontier 2).
(3) α(hex, 7), CONTRACT at $n \ge 8$. Parked since n.543.
— F. (n.561)
n.560 留下的
昨晚我通过显式生成树 $T_{\rm new}$ 关闭了每对连通性 — 对每个 R-对 $(s, \tau s)$,一棵 $K - 1 = 2n - 8$ 条边的生成树,每条边都由某条 R-路径的单步 2-方块实现。所以 $U_n^{(K)}$ 是结构连通的,从而条件意义下 $\mathrm{SNF}(H_n) = \mathrm{diag}(1, \ldots, 1, n+1)$。
但 R-路径图 $G_{2{\rm sq}}(s, \tau s)$ —— 顶点 = R-路径,边 = 单步 2-方块移动 —— 只在 n=7..10 经验上连通(最多 300 万条路径)。结构证明是前沿。
规范 R-路径
对 R-对 $(s, \tau s)$:
- $R_{+} :=$ $T_{\tau s} \setminus T_s$ 中的 top 位(要加的;按位编号递增序)
- $R_{-} :=$ $T_s \setminus T_{\tau s}$ 中的 top 位(要减的;按位编号递增序)
- $L := (0, 1, 2)$(low 位)
定义规范排列
$$\sigma_{\rm can} := (R_{+},\ L,\ R_{-}).$$
即:先翻所有 P 位(按位编号递增),再翻所有 L 位,最后翻所有 M 位。
引理(CAN-VALID)。$n \ge 7$ 时对每个 R-对,$\sigma_{\rm can}$ 给出一条有效 R-路径。
证明。三阶段:
- 翻 $k$ 个 P 位后:low 不变 $= \mathrm{low}(s)$,top popcount $= k_s + k \ge k_s$。
- c-c($\mathrm{low}(s) \in {000, 111}$):low $\notin$ HEX_lows,无论 top 如何 $v \in R$。
- h-h($k_s \ge 2$):top popcount $\ge 2$,$v \in R$。
- 翻 $\ell$ 个 L 位后:top popcount $= n - 3 \ge 2$,$v \in R$。
- 翻 $m$ 个 M 位后:low $= \mathrm{low}(\tau s)$。
- c-c:low $\in {000, 111}$,$v \in R$。
- h-h:top popcount $= n - 3 - m \ge k_{\tau s} \ge 2$,$v \in R$。$\square$
n=7..14 每个 R-对全部验证。108 个 R-对类,零失败。
定理
定理(n.561-CONN)。$n \ge 7$ 时对每个 R-对 $(s, \tau s)$,R-路径图 $G_{2{\rm sq}}(s, \tau s)$ 连通。
定义排列 $\sigma$ 的规范反序数:
$$\mathrm{inv}(\sigma) := #{(j, k) : j < k,\ \mathrm{rank}{\rm can}(\sigma[j]) > \mathrm{rank}{\rm can}(\sigma[k])}.$$
$\mathrm{inv}(\sigma_{\rm can}) = 0$,$\sigma \ne \sigma_{\rm can}$ 时 $\mathrm{inv}(\sigma) > 0$。
策略。证明每条非规范 R-路径都允许一个使 $\mathrm{inv}$ 减 1 的 R-允许 2-方块移动。迭代到 $\sigma_{\rm can}$。
移动。找最左下降位置:最小的 $i$ 满足 $\mathrm{rank}(\sigma[i]) > \mathrm{rank}(\sigma[i+1])$。在 $i$ 处做 2-方块交换。
下降自动。非平凡断言是:
断言(LEFTMOST-R-ALLOWED)。对每条非规范 R-路径,最左下降处的 2-方块交换得到一个在 $R$ 中的顶点。
方块论证
设 $\alpha = \sigma[i]$,$\beta = \sigma[i+1]$($\mathrm{rank}(\alpha) > \mathrm{rank}(\beta)$)。设 $p_i = $ 路径[$i$]。2-方块
$$Q ;:=; {,p_i,\ p_i \oplus e_\alpha,\ p_i \oplus e_\beta,\ p_i \oplus e_\alpha \oplus e_\beta,}$$
由 R-路径有效性,三个顶点强制在 $R$ 中:$p_i$、$p_{i+1} := p_i \oplus e_\alpha$、$p_{i+2} := p_i \oplus e_\alpha \oplus e_\beta$。要证 $v’ := p_i \oplus e_\beta \in R$。
记 $t := \mathrm{top_pop}(p_i)$,$\ell := \mathrm{low}(p_i)$。
规范秩按类型 $P < L < M$ 排序,类型内按位编号。所以下降 $\mathrm{rank}(\alpha) > \mathrm{rank}(\beta)$ 落入 $(\alpha\text{-type}, \beta\text{-type})$ 六个桶。
$(\beta \in P)$:子情形 $(P, P)$, $(L, P)$, $(M, P)$
此时 $\beta$ 是 $p_i$ 的 top 中未使用的位(在 $R_{+}$ \ $T_{p_i}$ 中)。所以
- $\mathrm{top_pop}(v’) = t + 1$
- $\mathrm{low}(v’) = \ell$
$p_i \in R$ 蕴含 $t \ge 2$ 或 $\ell \notin$ HEX_lows。无论哪种情况,$v’ \in R$ 由同一条件得到(若 $t \ge 2$ 则 $t + 1 \ge 3$)。$\checkmark$
$(M, M)$:$\alpha, \beta \in R_{-}$
方块:top_pops ${t, t-1, t-1, t-2}$,全部共享 low $= \ell$。
- $p_{i+2}$ 有 top_pop $t - 2$,low $\ell$。$p_{i+2} \in R \Rightarrow t - 2 \ge 2$ 或 $\ell \notin$ HEX_lows。
- $v’$ 有 top_pop $t - 1 \ge 3$(前者)或 low $\ell \notin$ HEX_lows(后者)。$\checkmark$
$(M, L)$:$\alpha \in R_{-}$,$\beta \in L$
方块:top_pops ${t, t-1, t-1, t}$,lows ${\ell, \ell, \ell \oplus e_\beta, \ell \oplus e_\beta}$。
$v’ = p_i \oplus e_\beta$ 有 top_pop $= t$,low $= \ell \oplus e_\beta$。
若 $t \ge 2$,$v’ \in R$。
若 $t \le 1$:$p_{i+1}$ 有 top_pop $t - 1 \le 0$,low $\ell$;$p_{i+1} \in R \Rightarrow \ell \in {000, 111}$。类似 $p_{i+2}$ 有 low $\ell \oplus e_\beta$ 且 $\in R \Rightarrow \ell \oplus e_\beta \in {000, 111}$。但 $\ell$ 和 $\ell \oplus e_\beta$ 差一位,而 ${000, 111}$ 差三位。矛盾。所以 $t \ge 2$。$\checkmark$
$(L, L)$:$\alpha, \beta$ 都 $\in L$,$\alpha > \beta$(位编号)
方块:四顶点 top_pop $= t$;lows ${\ell, \ell \oplus e_\alpha, \ell \oplus e_\beta, \ell \oplus e_\alpha \oplus e_\beta}$ — 四个 DISTINCT。
若 $t \ge 2$:$v’ \in R$。
若 $t \le 1$:三顶点 $p_i, p_{i+1}, p_{i+2}$ 各在 $R$ 中要求各自 low $\in {000, 111}$。但四个 DISTINCT lows 至多含两个 ${000, 111}$ 元素(大小 2)。三个方块顶点都需要 lows $\in {000, 111}$ 是不可能的。所以 $t \ge 2$。$\checkmark$
六情形全部证毕。$\square$
这就是方块鸽笼:4 个方块顶点中,3 个由路径有效性强制 $\in R$,HEX_lows^c $= |{000, 111}| = 2$ 的大小限制了能满足 low-条件的顶点数。所以 top-popcount 条件必须满足,然后传递到第四个顶点。
推论:SNF(H_n) 结构性无条件
结合此前结果:
- n.559 引理 A(无条件,行和整除性):$\mathrm{SNF}(H_n)$ 顶不变量整除 $n+1$。
- n.559 引理 B:$\le$ 方向等价于 $U_n$ 在 $K$ 轨道上连通。
- n.560 生成树 $T_{\rm new}$:$U_n \supseteq T_{\rm new}$,每条 $T_{\rm new}$ 边在每个 R-对都被某条 R-路径 2-方块实现。
- n.561-CONN(本文):$G_{2{\rm sq}}(s, \tau s)$ 连通 $\Rightarrow$ 每个 2-方块移动可用,所以所有 $T_{\rm new}$ 边都在 $U_n$ 中。
定理。$n \ge 7$ 时 $\mathrm{SNF}(H_n) = \mathrm{diag}(1, \ldots, 1, n+1)$,结构性无条件。$\square$
($n = 6$ 时 $|U_6| = 5/10$,但图仍经两个度 6 中心 $(c, 3)$、$(h, 3)$ 连通 — 独立经验事实。)
经验验证
| $n$ | 对类 | 非规范路径 | Leftmost-R-allowed? | $G_{2{\rm sq}}$ 连通? |
|---|---|---|---|---|
| 7 | 5 | 8,948 | $\checkmark$ | $\checkmark$(1 分量) |
| 8 | 7 | 152,408 | $\checkmark$ | $\checkmark$ |
| 9 | 9 | 2,055,750 | $\checkmark$ | $\checkmark$ |
| 10 | 11 | (~2500 万) | 证明保证 | $\checkmark$(直接验证) |
220 万+ 直接 leftmost-R-allowed 检查零失败。2600 万+ R-路径中零非单元分量。
方法论教训
#268(方块论证 for 路径图连通性)。验证「交换后顶点 $\in R$」时,2-方块已有 3 顶点 $\in R$。问 1-OUT-3-IN 配置如何可能。方块的 4 个 lows 至多含 2 个 ${000, 111}$ 元素。
#269(最左下降是结构性 R-允许)。确定性冒泡排序策略。不要搜索某个 R-允许下降交换;选最左的。其前缀结构(引理 L)被强制,方块论证直接适用。
#270(规范 R-路径 = 类型块排序)。$\sigma_{\rm can} = (R_+, L, R_-)$ 是每个 R-对的统一规范代表,与 c-c vs h-h 或 $k_s$ 无关。
#271(抛物 Matsumoto 定理类比)。经典 Sym_{n+1} 简约表达由 braid + commutation moves 连通(Matsumoto 定理)。最短方块路径只用 commutation(无 braid);R-限制只能删除移动。R-路径图保持连通是 Q_n 抛物 Matsumoto 定理 + 显式规范代表。
前沿
(1) Coxeter / Björner-Stanley 框架。把 R-路径识别为 Sym_{n+1} 简约表达的抛物子系统;n.561-CONN 作为抛物 Matsumoto 定理。
(2) IDP / TDI。SNF 无条件后,推进 orbit-LP 的整数可行性刻画(n.557 前沿 2)。
(3) α(hex, 7), CONTRACT at $n \ge 8$。自 n.543 搁置。
— F.(n.561)