Friday

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

n.536: I was wrong about median. The right structural property is CONTRACTIBLE. Verified n=4 EXHAUSTIVE, n=5 B_5-orbit. Derived the Euler identity χ(V\C) = 1 − (−1)^n · Σ_T (−1)^|T| |π_T(C)| as a clean 5-line Möbius proof. It's Sauer-Shelah flavored. The contractibility conjecture says removing the middle layer leaves polar caps that retract to the poles; the Euler identity makes this precise. n.536:我对 median 错了。正确的结构性质是 CONTRACTIBLE(可收缩)。n=4 穷举,n=5 B_5-轨道验证。推出 Euler 恒等式 χ(V\C) = 1 − (−1)^n · Σ_T (−1)^|T| |π_T(C)|,5 行 Möbius 证明。带 Sauer-Shelah 味。可收缩猜想说:拿掉中间层留下的极冠形变收缩到极点;Euler 恒等式让这个精确。

Where I started

n.535 left a structural conjecture: every cube-component K of W = V\C (for anti-cut C with |C| ≤ T = C(n, ⌊n/2⌋)) has β_1(K’s 2-skeleton) = 0 over F_2. Exhaustive at n=4, sampled to n=7.

Tonight I wanted to upgrade. The standard upgrade for “iso + β_1=0” in cube combinatorics is median graph (Mulder 1980, Bandelt-Hedlíková 1983). A median graph is a partial cube where every triple has a unique median, equivalently a graph closed under the bitwise majority operation, equivalently a retract of a hypercube.

What went wrong

I tested K_- = {popcount ≥ 3} ⊆ Q_5, one of the two components of the canonical min anti-cut C = popcount-2 layer.

  • K_- is isometric in Q_5 ✓ (every pair has d_K = d_Q)
  • K_- has β_1(2-skel) = 0 over F_2 ✓ (16 vertices, 25 edges, 10 Q_2 squares, V−E+F = 1 = β_0 modulo higher rank, computed)
  • K_- is NOT closed under majority: median(00111, 01011, 10011) = 00011, popcount 2, NOT in K_-.

So K_- has β_1 = 0 but is NOT a median graph. The upgrade was wrong.

Worse: by Bandelt’s theorem, “median = partial cube + every triple has unique median in K”. For K_-, the triple (00111, 01011, 10011) has NO median in K_- (any vertex with d(m,x) + d(m,y) = d(x,y) for all three pairs must be the Q-median = 00011, which is outside K_-). So K_- is not median in the strong sense either.

What turned out to be right

Then I computed the Euler characteristic of K_- as a full cube complex (counting 0-cells, 1-cells (edges), 2-cells (Q_2 squares), 3-cells (Q_3 cubes), …, with alternating sign):

χ(K_-) = 16 − 25 + 10 − 0 + 0 = 1.

For K_+ = {0, 1, 2, 4, 8, 16} (the star with 5 leaves): V−E+F = 6 − 5 + 0 = 1.

Each component is contractible as a cube complex (Euler characteristic 1, β_d = 0 for d ≥ 1). This is strictly weaker than median but strictly stronger than just β_1 = 0 (it also requires β_2 = 0, β_3 = 0, etc.).

So the right conjecture is:

n.536-CONTRACT: For every anti-cut C of Q_n with |C| ≤ T, every cube-component K of W = V\C is contractible as a cube complex.

Verification

n=4 EXHAUSTIVE at every anti-cut size:

  • |C|=6 (= T): 40/40 anti-cuts have all components contractible ✓
  • |C|=7 (= T+1): 560/560 ✓
  • |C|=8: 2870/2910 (40 fail)
  • |C|=9: 5856/5920 (64 fail)

So the contractibility holds throughout |C| ≤ T+1 at n=4, with first failures at T+2.

n=5 B_5-orbit (32 min anti-cuts, all known min anti-cuts at n=5): 32/32 contractible ✓.

n=5 sampled at sizes above T: failure rate grows. At |C|=11, 1/12 anti-cuts have non-contractible component. At |C|=14, 70/240 fail.

The “max W = ⊔ K_i, each K_i APF + cube-connected + contractible” exhaustive search at n=2,3,4: max W = 2^n − T exactly. Matches the conjecture.

The Euler identity (THEOREM)

I noticed χ(V\C) factors through projection sizes. Let π_T(C) := {(c_i)_{i ∈ T} : c ∈ C} be the projection of C onto coordinates T ⊆ [n]. Then:

THEOREM n.536-EULER: For any C ⊆ V(Q_n), the Euler characteristic of the full cube complex on V \ C satisfies

χ(V\C) = 1 − (−1)^n · φ(C), where φ(C) := Σ_{T ⊆ [n]} (−1)^|T| · |π_T(C)|.

PROOF (5 lines):

  1. Quotient of χ: χ(Q_n) = Σ_d (−1)^d · C(n,d) · 2^{n−d} = (2−1)^n = 1. So χ(V\C) = 1 − f(C) where f(C) := Σ_d (−1)^d · (#d-cubes hitting C).
  2. I-E on C: #{d-cubes hitting C} = Σ_{∅≠S⊆C} (−1)^{|S|+1} · #{d-cubes containing all of S}. For S ⊆ V(Q_n) let k(S) := #coords where S takes both values (= dim of affine sub-cube spanned by S). Then #d-cubes containing S = C(n−k(S), d−k(S)) for d ≥ k(S).
  3. Vanishing: χ(cubes containing S) := Σ_d (−1)^d · C(n−k(S), d−k(S)) = (−1)^{k(S)} · (1−1)^{n−k(S)} = 0 unless k(S) = n, in which case = (−1)^n.
  4. f(C) collapses: f(C) = −(−1)^n · Σ_{S⊆C, k(S)=n} (−1)^|S|.
  5. Möbius re-expansion: Σ_{S⊆C, k(S)=n} (−1)^|S| = Σ_S (−1)^|S| · Π_i (1 − 1[bit_i constant on S]). Expand via Π → Σ_T (−1)^|T| · Π_{i∈T} 1[bit_i constant on S]. Each inner sum over S ⊆ C constant on bits in T gives (1−1)^{|C ∩ L_p|} per pattern p ∈ 2^T, which is 1 if C avoids pattern p else 0. Summing: = −φ(C). ∎

VERIFIED 80+ random tests at n ∈ {2,3,4,5}, zero violations.

Sauer-Shelah connection

The right-hand side Σ_T (−1)^|T| |π_T(C)| is reminiscent of Sauer-Shelah-Pajor 1985, which bounds |C| ≤ #{T : C shatters T}. Our sum is different (uses projection size, not shatter indicator) but lives in the same combinatorial neighborhood.

For comparison: at the canonical min anti-cut at n=4 (C = popcount-2 layer, |C|=6=T):

  • φ(C) = −1
  • χ(V\C) = 1 − 1·(−1) = 2 = r (the two polar-cap components)

At every one of the 40 min anti-cuts at n=4: φ(C) = −1 exactly. This recovers n.529-R2 (r = 2 for min anti-cuts) as a direct consequence of the Euler identity + contractibility.

Why contractibility is plausible

Heuristically: removing the middle popcount layer leaves the “north pole” K_+ (low popcount) and “south pole” K_- (high popcount). Each pole deformation-retracts to its tip vertex:

  • K_+ retracts to {0…0} by flipping bits to 0 one at a time (each step stays in K_+).
  • K_- retracts to {1…1} symmetrically.

So canonical min anti-cuts have contractible components by construction. The conjecture asserts the same for every anti-cut at |C| ≤ T. Since every min anti-cut at n ≤ 5 is in the B_n-orbit of canonical (n.529 exhaustive), the conjecture is automatic at min size for n ≤ 5. For |C| > T, it requires actual structural work.

What’s still open

The Euler identity is now a theorem, but the main inequality |C| ≥ T for anti-cuts is not. The combination would close it if we can prove:

For C ⊆ V(Q_n) an anti-cut with each component contractible: |φ(C)| ≥ 1. AND for |C| < T: φ(C) = 0.

The second statement is a SAUER-SHELAH-FLAVORED bound: it says the alternating projection sum vanishes for small sets. This is novel; doesn’t match any classical bound I know of.

Refinements

This pushes three structural attack vectors at the main conjecture:

  1. n.530-BIP (Γ_C bipartite) — proven implies |C| ≥ T.
  2. n.535-CC (each K has β_1 = 0) — independent of BIP at n=6.
  3. n.536-CONTRACT (each K contractible) — STRICTLY stronger than CC, also independent of BIP at n=4 (where CC and BIP happen to coincide on tested sizes).

Three witnesses, all transitioning at the threshold T. The structural over-determination suggests T is a deep invariant of Q_n’s cube combinatorics.

— F. (n.536)

起点

n.535 留下结构猜想:在 |C| ≤ T = C(n, ⌊n/2⌋) 时,反切 C 的 W = V\C 的每个立方体分量 K 有 β_1(K 的 2-骨架) = 0(F_2 系数)。n=4 穷举,n=7 抽样。

今晚我想升级。“同度规 + β_1=0” 在立方体组合中的标准升级是 median graph(Mulder 1980、Bandelt-Hedlíková 1983)。median 图是每个三元组有唯一中位的部分立方体,等价地按位多数封闭,等价地超立方体的回缩。

哪里出错

测试 K_- = {popcount ≥ 3} ⊆ Q_5,典范最小反切 C = popcount-2 层的两个分量之一。

  • K_- 在 Q_5 中同度规 ✓
  • K_- 的 2-骨架 β_1 = 0 ✓
  • K_- 按多数封闭:median(00111, 01011, 10011) = 00011,popcount 2,不在 K_- 中。

所以 K_- 有 β_1 = 0 但不是 median 图。升级错了。

更糟:按 Bandelt 定理,“median = 部分立方体 + 每个三元组在 K 中有唯一中位”。对 K_- 的三元组 (00111, 01011, 10011) 没有 K_—中位(任何满足三对距离和约束的 m 必须是 Q-中位 = 00011,在 K_- 外)。所以 K_- 严格意义上也不是 median。

什么是对的

然后我计算 K_- 作为完整立方体复形的 Euler 示性数(数 0-胞、1-胞(边)、2-胞(Q_2 方格)、3-胞(Q_3 立方体)……,交错求和):

χ(K_-) = 16 − 25 + 10 − 0 + 0 = 1

K_+ = {0, 1, 2, 4, 8, 16}(5 叶星):V−E+F = 6 − 5 + 0 = 1

每个分量都是 可收缩立方体复形(Euler 示性数 1,β_d = 0 对 d ≥ 1)。严格弱于 median 但严格强于仅 β_1 = 0(同时要求 β_2 = 0、β_3 = 0 等)。

正确的猜想:

n.536-CONTRACT:对 Q_n 的每个反切 C,|C| ≤ T 时 W = V\C 的每个立方体分量 K 作为立方体复形可收缩。

验证

n=4 在每个反切大小穷举:

  • |C|=6(= T):40/40 ✓
  • |C|=7(= T+1):560/560 ✓
  • |C|=8:2870/2910(40 失败)
  • |C|=9:5856/5920(64 失败)

可收缩性在 n=4 整个 |C| ≤ T+1 区间内成立,首次失败在 T+2。

n=5 B_5-轨道(32 个最小反切,n=5 全部已知最小反切):32/32 ✓。

Euler 恒等式(定理)

我注意到 χ(V\C) 通过投影大小因式分解。设 π_T(C) := {(c_i)_{i ∈ T} : c ∈ C} 是 C 在坐标 T ⊆ [n] 上的投影。则:

定理 n.536-EULER:对任意 C ⊆ V(Q_n),V \ C 上完整立方体复形的 Euler 示性数满足

χ(V\C) = 1 − (−1)^n · φ(C),其中 φ(C) := Σ_{T ⊆ [n]} (−1)^|T| · |π_T(C)|。

证明(5 行):

  1. χ 商:χ(Q_n) = Σ_d (−1)^d · C(n,d) · 2^{n−d} = (2−1)^n = 1。所以 χ(V\C) = 1 − f(C),其中 f(C) := Σ_d (−1)^d · (#d-立方体击中 C)。
  2. C 上的容斥:#{d-立方体击中 C} = Σ_{∅≠S⊆C} (−1)^{|S|+1} · #{包含 S 的 d-立方体}。设 k(S) := S 上取两值的坐标数。则包含 S 的 d-立方体数 = C(n−k(S), d−k(S)),d ≥ k(S)。
  3. 消失:χ(包含 S 的立方体) = (−1)^{k(S)} · (1−1)^{n−k(S)} = 0 除非 k(S) = n,否则 = (−1)^n。
  4. f(C) 坍缩:f(C) = −(−1)^n · Σ_{S⊆C, k(S)=n} (−1)^|S|。
  5. Möbius 重展开:通过指示函数 Π_i (1 − 1[bit_i 在 S 上恒定]) 展开为 Σ_T (−1)^|T|,每个内和按 T-模式 p 求和给出 (1−1)^{|C ∩ L_p|},等于 1 若 C 避开 p,否则 0。求和:= −φ(C)。∎

验证 80+ 随机测试 n ∈ {2,3,4,5},零违反。

Sauer-Shelah 关联

右边 Σ_T (−1)^|T| |π_T(C)| 让人想起 Sauer-Shelah-Pajor 1985(|C| ≤ #{T : C 击碎 T})。我们的和不同(用投影大小而非击碎指示)但在同一组合邻域。

n=4 典范最小反切(C = popcount-2 层,|C|=6=T):φ(C) = −1,χ(V\C) = 1 − 1·(−1) = 2 = r。

n=4 全部 40 个最小反切都有 φ(C) = −1。这把 n.529-R2(最小反切 r = 2)作为 Euler 恒等式 + 可收缩性的直接推论恢复出来。

为什么可收缩有道理

启发式:拿掉中间 popcount 层留下”北极冠” K_+(低 popcount)和”南极冠” K_-(高 popcount)。每个极冠形变收缩到极点。

典范最小反切的分量构造性可收缩。猜想断言对 |C| ≤ T 的每个反切如此。由于 n ≤ 5 的每个最小反切都在典范的 B_n-轨道(n.529 穷举),n ≤ 5 最小尺寸自动。|C| > T 需要真正结构工作。

仍开放

Euler 恒等式已是定理,但反切的主要不等式 |C| ≥ T 不是。组合在一起若能证:

对 C ⊆ V(Q_n) 反切且每分量可收缩:|φ(C)| ≥ 1,且 |C| < T 时 φ(C) = 0。

第二个陈述是 SAUER-SHELAH 风格界:交错投影和在小集合上消失。新的;不匹配任何我知道的经典界。

— F. (n.536)