不在乎的比特，
与穿了衣服的码距 Don't-care qubits
and the dressed distance

记号：n = 物理量子比特数，k = 受保护的逻辑量子比特数，d = 码距，L = 环面的边长。[[n, k, d]] 是量子码的名片写法。本篇中 X_i、Z_i 表示只在第 i 个比特上放 X 或 Z 的串，X₀X₂ 表示在第 0 和第 2 个比特上各放一个 X。

Notation: n = number of physical qubits, k = number of protected logical qubits, d = distance, L = side length of the torus. [[n, k, d]] is the calling-card notation for a quantum code. In this guide X_i, Z_i denote the string with a single X or Z on qubit i, and X₀X₂ the string with an X on qubits 0 and 2.

把 4 个比特排成 2 × 2 的方格，按行编号：q0 q1 在上一行，q2 q3 在下一行。QEC-Bank 的 Bacon–Shor 条目（在 QEC-Bank 中查看 →）把它登记为实例 square_2，名片 [[4, 1, 2]]。它的「检查」是四条重量 2 的串：每条竖边上一条 XX，每条横边上一条 ZZ——QEC-Bank 直觉页的原文：「XX on every vertical nearest-neighbour pair, and ZZ on every horizontal nearest-neighbour pair」。画出来是这样：

Arrange 4 qubits in a 2 × 2 square, numbered by rows: q0 q1 on the top row, q2 q3 below. The QEC-Bank Bacon–Shor entry (open in the QEC-Bank →) registers it as instance square_2, calling card [[4, 1, 2]]. Its “checks” are four strings of weight 2: an XX on every vertical edge and a ZZ on every horizontal edge, in the words of the QEC-Bank intuition page: “XX on every vertical nearest-neighbour pair, and ZZ on every horizontal nearest-neighbour pair”. Drawn out:

用第 0 章的碰撞定理检查一下：X₀X₂ 与 Z₀Z₁ 只在 q0 这一个比特上「都非 I 且字母不同」，1 是奇数，反对易。按 Toric Code 教程 4.4 节的规则，这两条不能同时进一份检查清单——测了一条，另一条的答案就被打乱。QEC-Bank 直觉页说这不是缺陷：「That is not a defect: gauge generators are allowed to disturb an unprotected gauge subsystem.」这四条串不叫稳定子，叫规范生成元gauge generator：允许互相反对易的一组「检查」。它们生成的群叫规范群 G。它们的测量结果不必全部同时确定；只有它们当中互相对易的那部分（中心）才是真正的稳定子。；它们生成的整个乘积集合叫规范群 Ggauge group：把四条规范生成元任意相乘得到的所有泡利串（连同符号）。本篇比较泡利串时不计整体符号，只看哪些比特上放了什么字母（理由见 1.2 节）。。人话：我们故意写了一份会打架的清单。

Check with the collision theorem of Chapter 0: X₀X₂ and Z₀Z₁ are “both non-I with different letters” on exactly one qubit, q0; 1 is odd, so they anticommute. By the rules of toric-code section 4.4 these two cannot sit in one check list: measuring one scrambles the answer of the other. The QEC-Bank intuition page says this is not a flaw: “That is not a defect: gauge generators are allowed to disturb an unprotected gauge subsystem.” The four strings are not called stabilizers but gauge generatorsA set of “checks” allowed to anticommute with one another. The group they generate is the gauge group G. Their measurement outcomes need not all be simultaneously definite; only the mutually commuting part (the center) consists of true stabilizers., and the set of all their products is the gauge group GAll Pauli strings (with signs) obtained by multiplying the four gauge generators in any combination. This guide compares Pauli strings without their overall sign and looks only at which letter sits on which qubit (the reason is in 1.2).. In plain words: we wrote a self-fighting list on purpose.

打架的清单里仍有和平的成员。把两条竖边相乘：X₀X₂ · X₁X₃ = X₀X₁X₂X₃，记作 S_X。它与 Z₀Z₁ 在 q0、q1 两个比特上碰撞，2 是偶数，对易；与 Z₂Z₃ 同理。同样，两条横边相乘得 S_Z = Z₀Z₁Z₂Z₃，它与每条竖边碰撞两次，对易。S_X 与 S_Z 碰撞四次，也对易。这两条与所有规范生成元都对易的乘积叫规范群的中心center：规范群里与规范群每个元素都对易的那些元素。它们是真正的稳定子，构成稳定子群 S。QEC-Bank 原文：「The protected codespace is fixed only by the commuting center of the gauge group.」，记作 S；它们才是真正的稳定子：测它们，答案稳定，不被别的测量打乱。

A fighting list still has peaceful members. Multiply the two vertical edges: X₀X₂ · X₁X₃ = X₀X₁X₂X₃, call it S_X. It collides with Z₀Z₁ on two qubits, q0 and q1; 2 is even, so they commute; likewise with Z₂Z₃. Similarly the two horizontal edges multiply to S_Z = Z₀Z₁Z₂Z₃, which collides twice with every vertical edge and commutes. S_X and S_Z collide four times and commute too. These products commuting with every gauge generator form the centerThe elements of the gauge group that commute with every element of the gauge group. They are the true stabilizers and form the stabilizer group S. QEC-Bank text: “The protected codespace is fixed only by the commuting center of the gauge group.” of the gauge group, written S; they are the true stabilizers: measure them and the answer is stable, unscrambled by other measurements.

关于符号要说清一件事。规范群若连符号一起算，里面有 −I：记 A = X₀X₂、B = Z₀Z₁，两者反对易，交换一次多一个负号（BA = −AB），所以 A·B·A·B = −A·A·B·B = −I（本篇的检验脚本用 16 × 16 矩阵乘过）。−I 与一切对易，按字面定义它落在中心里，可它不能当检查：没有状态满足 −|ψ⟩ = |ψ⟩。所以本篇约定：比较泡利串时不计整体相位（±1、±i），「中心」指不计相位后与 G 中每条串都对易的那些串；稳定子群 S 由中心里一组选好符号、彼此相容（乘不出 −I）的代表生成，这里取 +S_X、+S_Z。本篇所有的 rank 都是不计相位的秩，也就是只看每个比特上放了什么字母。

One point about signs. Counted with signs, the gauge group contains −I: write A = X₀X₂ and B = Z₀Z₁; they anticommute, so swapping them costs a minus sign (BA = −AB), and A·B·A·B = −A·A·B·B = −I (this guide's check script multiplied the 16 × 16 matrices). −I commutes with everything, so by the literal definition it lies in the center, yet it cannot serve as a check: no state satisfies −|ψ⟩ = |ψ⟩. So this guide's convention is to compare Pauli strings up to an overall phase (±1, ±i): the “center” means the strings that commute with every string of G once phases are dropped, and the stabilizer group S is generated by representatives from the center with chosen signs that are consistent (no product gives −I), here +S_X and +S_Z. Every rank on this page ignores phases: it looks only at which letter sits on which qubit.

现在记账。按 Toric Code 教程 5.3 节的算法，n − 独立稳定子数 = 4 − 2 = 2，编码空间里有 2 个比特的自由度。但其中只有 1 个受保护。为什么：另一个自由度正是被规范生成元「打乱」的那个——X₀X₂ 和 Z₀Z₁ 反对易，它们就像一个额外比特上的 X 和 Z：测一个，另一个原本确定的结果就变成随机。比如状态原本满足 Z₀Z₁ = +1，测一次 X₀X₂ 之后再测 Z₀Z₁，得 +1 和 −1 的概率各一半。我们不在乎这个比特的状态，QEC-Bank 叫它规范比特gauge qubit：编码空间里被规范生成元随意翻动、不承载受保护信息的自由度。数量 g = (rank G − rank S)/2。。子系统码的账本公式（QEC-Bank foundations 页的一句话定义：「Trade logical qubits for gauge freedom; k = n − (rank G + rank S)/2」）：

Now the ledger. By the procedure of toric-code section 5.3, n − (independent stabilizers) = 4 − 2 = 2, so the code space has 2 qubits' worth of freedom. But only 1 is protected. Why: the other degree of freedom is precisely the one “scrambled” by the gauge generators; X₀X₂ and Z₀Z₁ anticommute like the X and Z of one extra qubit: measuring one turns the other's previously definite outcome random. For example, if the state had Z₀Z₁ = +1, then after one measurement of X₀X₂ a new measurement of Z₀Z₁ gives +1 or −1 with probability one half each. We don't care about this qubit's state; the QEC-Bank calls it a gauge qubitA degree of freedom inside the code space that gauge generators flip at will and that carries no protected information. Their number is g = (rank G − rank S)/2.. The ledger formula for subsystem codes (the one-liner of the QEC-Bank foundations page: “Trade logical qubits for gauge freedom; k = n − (rank G + rank S)/2”):

记号：rank G = 独立的规范生成元条数（这里 4：四条边彼此不能互相乘出），rank S = 独立的稳定子条数（这里 2：S_X、S_Z），k = 受保护的逻辑比特数，g = 规范（不在乎）比特数。代入：k = 4 − (4 + 2)/2 = 1，g = (4 − 2)/2 = 1。公式从哪里来：稳定子码的算法先给出 n − rank S 个自由度（这里 2 个）。再把四条生成元换一组写法：保留 A = X₀X₂ 和 B = Z₀Z₁，把 X₁X₃ 换成 A·X₁X₃ = S_X，把 Z₂Z₃ 换成 B·Z₂Z₃ = S_Z（换掉的那条总能再乘回来，所以生成的群不变）：G = ⟨S_X, S_Z, A, B⟩。这组写法里，S_X、S_Z 与一切对易，A 与 B 彼此反对易、与 S_X、S_Z 都对易。G 的任一元素都能写成（S_X、S_Z 的某个乘积）· A^α · B^β（α、β 取 0 或 1）；它与 B 对易当且仅当 α = 0，与 A 对易当且仅当 β = 0，所以中心恰好是 ⟨S_X, S_Z⟩，而规范自由度全部来自 A、B 这一对反对易的算符，它们就是那个规范比特的「X 和 Z」。于是 rank G = 4 = rank S + 2 × 1。一般的 G 也能整理成「中心 + g 对」，步骤在下面的补课里；所以 rank G = rank S + 2g，即 g = (rank G − rank S)/2。受保护的比特是扣掉规范比特剩下的：k = (n − rank S) − g，代入 g 化简就是 k = n − (rank G + rank S)/2。QEC-Bank 推导页 gauge-center-and-qubits.md 的账本表第一行正是「2 × 2 | rank S 2 | rank G 4 | k 1 | g 1」。

Notation: rank G = number of independent gauge generators (here 4: no edge is a product of the others), rank S = number of independent stabilizers (here 2: S_X, S_Z), k = protected logical qubits, g = gauge (don't-care) qubits. Substituting: k = 4 − (4 + 2)/2 = 1, g = (4 − 2)/2 = 1. Where the formula comes from: the stabilizer-code count first gives n − rank S degrees of freedom (2 here). Now rewrite the four generators: keep A = X₀X₂ and B = Z₀Z₁, replace X₁X₃ by A·X₁X₃ = S_X and Z₂Z₃ by B·Z₂Z₃ = S_Z (the replaced string can always be multiplied back, so the generated group is unchanged): G = ⟨S_X, S_Z, A, B⟩. In this form S_X and S_Z commute with everything, while A and B anticommute with each other and commute with S_X and S_Z. Every element of G can be written as (some product of S_X, S_Z) · A^α · B^β with α, β equal to 0 or 1; it commutes with B exactly when α = 0 and with A exactly when β = 0, so the center is exactly ⟨S_X, S_Z⟩, and all the gauge freedom comes from the one anticommuting pair A, B, the “X and Z” of the gauge qubit. Hence rank G = 4 = rank S + 2 × 1. A general G can be put in the same form, “center + g pairs”, by the steps in the catch-up fold below; so rank G = rank S + 2g, that is g = (rank G − rank S)/2. The protected qubits are what is left after removing the gauge qubits: k = (n − rank S) − g, and substituting g simplifies to k = n − (rank G + rank S)/2. The first row of the ledger table in the QEC-Bank derivation page gauge-center-and-qubits.md reads exactly “2 × 2 | rank S 2 | rank G 4 | k 1 | g 1”.

受保护的那个比特要有自己的 X̄ 和 Z̄。取 X̄ = X₀X₁（上面一整行）：它与 Z₀Z₁ 碰撞两次、与 Z₂Z₃ 零次、与两条竖边不碰撞（X 与 X 不算碰撞），所以它与全部四条规范生成元对易；而且它不是规范生成元的乘积（乘积里 X 总是成竖对出现）。取 Z̄ = Z₀Z₂（左边一整列）：同理与四条都对易。X̄ 与 Z̄ 只在 q0 碰撞一次，反对易——正是一对逻辑算符该有的关系（5.6 节）。这种与所有规范生成元都对易的逻辑算符叫裸逻辑算符bare logical：与规范群的每个元素都对易的逻辑算符。它不碰规范比特。QEC-Bank 原文：「A bare logical commutes with every gauge generator and appears as a complete row or column.」：它完全不动那个不在乎的比特。

The protected qubit needs its own X̄ and Z̄. Take X̄ = X₀X₁ (the whole top row): it collides twice with Z₀Z₁, zero times with Z₂Z₃, and not at all with the two vertical edges (X against X is no collision), so it commutes with all four gauge generators; and it is not a product of them (in such products X's always come in vertical pairs). Take Z̄ = Z₀Z₂ (the whole left column): likewise it commutes with all four. X̄ and Z̄ collide once, at q0, and anticommute, exactly the relation a logical pair should have (section 5.6). A logical operator that commutes with every gauge generator is called a bare logicalA logical operator commuting with every element of the gauge group. It leaves the gauge qubits alone. QEC-Bank text: “A bare logical commutes with every gauge generator and appears as a complete row or column.”: it does not touch the don't-care qubit at all.

现在换衣服。第 0 章事实 5 说逻辑算符乘上检查还是同一个逻辑算符。这里的新规则是：乘上任何规范算符也还是同一个逻辑算符——因为规范算符只翻不在乎的比特，对受保护的比特没有作用。于是 X̄ · X₀X₂ = X₁X₂（一条对角线！）与 X̄ 对受保护比特的作用完全相同。这条对角线与 Z₀Z₁ 只在 q1 碰撞一次，反对易——它不再与所有规范生成元对易，所以它不是裸的；它会翻动规范比特，但我们不在乎。这种「与中心对易、但可能与规范生成元反对易」的逻辑算符叫穿了衣服的逻辑算符dressed logical：与所有稳定子（中心）对易、且不属于规范群的泡利串。它可以是裸逻辑算符乘上任意规范算符。QEC-Bank 原文：「A dressed logical need only commute with the center; gauge multiplication can bend it.」。3 × 3 的版本把「弯」看得更清楚：

Now change clothes. Chapter 0 fact 5 says a logical times a check is the same logical. The new rule here: times any gauge operator it is still the same logical, because gauge operators only flip the don't-care qubits and do nothing to the protected one. So X̄ · X₀X₂ = X₁X₂ (a diagonal!) acts on the protected qubit exactly as X̄ does. This diagonal collides once with Z₀Z₁, at q1, and anticommutes: it no longer commutes with every gauge generator, so it is not bare; it flips the gauge qubit, and we don't care. A logical that “commutes with the center but may anticommute with gauge generators” is called a dressed logicalA Pauli string commuting with every stabilizer (the center) and not belonging to the gauge group. It can be a bare logical times any gauge operator. QEC-Bank text: “A dressed logical need only commute with the center; gauge multiplication can bend it.”. The 3 × 3 version shows the “bending” better:

3 × 3（实例 square_3，[[9, 1, 3]]）：裸的 X̄ 是一整行 X₀X₁X₂。乘上竖边 X₀X₃，q0 的 X 被移到 q3；再乘 X₂X₅，q2 的 X 移到 q5。结果 X₃X₁X₅ 是一条锯齿，每一列仍恰好一个 X，重量仍是 3。QEC-Bank 推导页把规律说成一句话：竖边的乘法「can move its one X within each column independently. What survives gauge motion is the parity in every column.」一条竖边 X_iX_j 乘上去，看它的两个位置 i、j 上原来有几个 X，只有三种结果：恰有一个 → 那个 X 移到另一个位置（滑动，重量不变，上面的锯齿就是这样来的）；一个也没有 → 这一列凭空多出两个 X（成对产生，重量 +2，例如一整行乘上 X₃X₆ 得到 X₀X₁X₂X₃X₆，重量 5）；两个都有 → 两个一起消失（成对湮灭，重量 −2）。三种结果里，这一列 X 的个数只变 0、+2 或 −2。人话：穿衣服能让每列里的 X 滑动、成对出现或成对消失，但改不了每列 X 的个数是奇是偶。

3 × 3 (instance square_3, [[9, 1, 3]]): the bare X̄ is a whole row X₀X₁X₂. Multiply by the vertical edge X₀X₃ and the X on q0 moves to q3; multiply by X₂X₅ and the X on q2 moves to q5. The result X₃X₁X₅ is a zig-zag, still exactly one X per column, still weight 3. The QEC-Bank derivation page puts the rule in one sentence: vertical-edge multiplication “can move its one X within each column independently. What survives gauge motion is the parity in every column.” Multiplying in one vertical edge X_iX_j has only three possible effects, depending on how many X's sit at its two positions i and j: exactly one → that X moves to the other position (sliding, weight unchanged; the zigzag above arose this way); none → two new X's appear in the column (pair creation, weight +2; for example the whole row times X₃X₆ is X₀X₁X₂X₃X₆, weight 5); both → the two vanish together (pair annihilation, weight −2). In all three cases the column's X count changes by 0, +2 or −2. In plain words: dressing can slide the X's within their columns, create them in pairs or annihilate them in pairs, but it cannot change whether each column holds an odd or even number of X's.

下面的部件就是 3 × 3 的码。九个方格是比特，方格之间的按钮是规范生成元：竖着的 XX、横着的 ZZ。从裸 X̄（第一行）出发，点竖边 X₀X₃ 再点 X₂X₅，就得到上面的锯齿；读数一栏始终显示每列「X 或 Y」个数的奇偶（Y 就是同一个比特上既有 X 又有 Z，所以它在「X 或 Y」和「Z 或 Y」两栏里都算），你可以看着它一直是 1 1 1；再点一条不碰现有 X 的竖边，比如 X₃X₆，就能看到成对产生。也可以直接点比特拼出任意一条串，看它是被警报抓住、属于 G，还是一条穿了衣服的逻辑算符——1.4 节的两个条件由部件替你逐条核对。警报来自 3 × 3 的四条稳定子条带（5.1 节会从规范生成元推出它们）：S_X0 = 第 0、1 两行的 6 个 X（q0…q5），S_X1 = 第 1、2 两行的 6 个 X（q3…q8），S_Z0 = 第 0、1 两列的 6 个 Z（q0 q1 q3 q4 q6 q7），S_Z1 = 第 1、2 两列的 6 个 Z（q1 q2 q4 q5 q7 q8）。部件下方的表列出这四条，哪一条与当前的串反对易，那一行就标「报警」并加底色。

The widget below is the 3 × 3 code. The nine squares are qubits; the buttons between them are gauge generators, vertical XX and horizontal ZZ. Start from the bare X̄ (the top row), click the vertical edge X₀X₃ and then X₂X₅, and you get the zigzag above; the readout always shows the parity of the number of X-or-Y letters in each column (a Y is an X and a Z on the same qubit, so it counts in both the “X or Y” and the “Z or Y” tallies), and you can watch it stay 1 1 1; click a vertical edge that touches no current X, such as X₃X₆, to see pair creation. You can also click qubits to build any string and see whether an alarm catches it, it lies in G, or it is a dressed logical; the widget checks the two conditions of 1.4 for you. The alarms come from the four stabilizer strips of the 3 × 3 (5.1 derives them from the gauge generators): S_X0 = the 6 X's on rows 0 and 1 (q0…q5), S_X1 = the 6 X's on rows 1 and 2 (q3…q8), S_Z0 = the 6 Z's on columns 0 and 1 (q0 q1 q3 q4 q6 q7), S_Z1 = the 6 Z's on columns 1 and 2 (q1 q2 q4 q5 q7 q8). The table under the widget lists the four; whichever anticommutes with the current string is marked “alarm” and shaded.

于是「码距」要重新定义。QEC-Bank 检验员的定义（设计文档 2026-08-10-subsystem-dressed-distance-v1.md 原文）：一条泡利串 p 是穿了衣服的逻辑见证，当且仅当「1. p commutes with every row of S; and 2. p is not in G.」穿了衣服的码距dressed distance：所有满足「与每条稳定子对易，且不在规范群 G 里」的泡利串中最轻者的重量。它是子系统码真正的码距。就是所有这种 p 里最轻的重量。两个条件对应两种「不算」：与某条稳定子反对易的串会触发警报，不算；属于 G 的串只翻不在乎的比特，无害，也不算。「modulo the full gauge group」（商掉整个规范群）说的是第二条：两条只差一个规范算符的串被看成同一件东西，而且这个规范算符可以是 G 里的任何元素，不只是中心。人话：码距 = 能骗过所有稳定子、又真的改写受保护比特的最轻操作；一件衣服算一次，锯齿和直行算同一条。

So “distance” must be redefined. The QEC-Bank referee's definition (verbatim from the design document 2026-08-10-subsystem-dressed-distance-v1.md): a Pauli p is a dressed-logical witness exactly when “1. p commutes with every row of S; and 2. p is not in G.” The dressed distanceThe weight of the lightest Pauli string that commutes with every stabilizer and does not lie in the gauge group G. It is the true distance of a subsystem code. is the lightest weight among all such p. The two conditions match two ways of “not counting”: a string anticommuting with some stabilizer trips an alarm, so it does not count; a string inside G only flips don't-care qubits, harmless, so it does not count either. “modulo the full gauge group” refers to the second: two strings differing by a gauge operator are regarded as one and the same, and that gauge operator may be any element of G, not only the center. In plain words: the distance is the lightest operation that fools every stabilizer and really rewrites the protected qubit; each outfit counts once, zig-zag and straight row are the same one.

记号：p = 候选泡利串；S = 稳定子（中心）；G = 规范群；∉ 读「不属于」；min 读「取最小值」。QEC-Bank 把满足条件 1 的集合记作 C(S)（S 的中心化子：与 S 每条都对易的串），把码距问题写成「在 C(S)/G 里找最轻」——C(S)/G 读「C(S) 商 G」，意思是把只差规范算符的串合并成一类。

Notation: p = a candidate Pauli string; S = the stabilizers (the center); G = the gauge group; ∉ reads “not in”; min reads “take the minimum”. The QEC-Bank writes the set satisfying condition 1 as C(S) (the centralizer of S: strings commuting with every row of S) and phrases the distance problem as “lightest in C(S)/G”; C(S)/G reads “C(S) modulo G” and means merging strings that differ by a gauge operator into one class.

回到 2 × 2 算一遍。重量 1 的候选：X₀ 与 S_Z = Z₀Z₁Z₂Z₃ 碰撞一次，反对易，被抓；Z₀ 与 S_X 碰撞一次，被抓；Y₀（= 同一比特上既 X 又 Z）与 S_X 的碰撞发生在它的 Z 部分，一次，被抓。四个比特、三种字母，12 条重量 1 的串全部被抓。重量 2：X₀X₁ 与 S_Z 碰撞两次，与 S_X 不碰撞，条件 1 成立；它不在 G 里（G 里的 X 只成竖对），条件 2 成立。于是 d = 2，与 instances.yaml 的「d, exact: 2, verified」一致。请再看一条：X₀X₂ 与 S_Z 碰撞两次、与 S_X 零次，条件 1 成立；但它就是一条规范生成元，在 G 里，条件 2 不成立——它不是逻辑算符。这正是 QEC-Bank 推导页警告的错误：「Quotienting only by S would retain many operators that differ by noncentral gauge action and would misclassify disposable degrees of freedom as protected logicals.」一台只商掉稳定子的老式码距程序会把 X₀X₂ 当成逻辑算符，把 k 数成 2。反过来，若要求候选与整个 G 对易，就只剩裸的行与列，漏掉 1.3 节的锯齿——在 Bacon–Shor 里锯齿不更轻，但一般的子系统码没有这个保证，检验员必须两头都对。

Back to 2 × 2 and compute. Weight-1 candidates: X₀ collides once with S_Z = Z₀Z₁Z₂Z₃, anticommutes, caught; Z₀ collides once with S_X, caught; Y₀ (= both X and Z on one qubit) collides with S_X through its Z part, once, caught. Four qubits, three letters, all 12 weight-1 strings caught. Weight 2: X₀X₁ collides twice with S_Z and not at all with S_X, so condition 1 holds; it is not in G (X's in G come only in vertical pairs), so condition 2 holds. Hence d = 2, matching “d, exact: 2, verified” in instances.yaml. Look at one more: X₀X₂ collides twice with S_Z and zero times with S_X, condition 1 holds; but it is a gauge generator itself, inside G, so condition 2 fails: it is not a logical. This is exactly the mistake the QEC-Bank derivation page warns of: “Quotienting only by S would retain many operators that differ by noncentral gauge action and would misclassify disposable degrees of freedom as protected logicals.” An old-style distance routine that quotients only by the stabilizers would call X₀X₂ a logical and count k = 2. Conversely, demanding that candidates commute with the whole of G leaves only bare rows and columns and misses the zig-zags of 1.3; in Bacon–Shor the zig-zags are no lighter, but a general subsystem code offers no such guarantee, so the referee must get both ends right.

逐词翻译。exact-small：小实例上的精确穷举——把所有轻串逐个按 1.4 节的两个条件检查，第 2 章讲它怎么做、做到多大。witnessed upper-bound：有见证的上界——找到一条穿了衣服的逻辑算符，它的重量就是 d 的上界（码距篇 1.2 节：一个例子就够）。dressed-distance replay：可重放——别人用同一程序、同一种子得到一模一样的结果。Extend … to：把已有的这两样扩展成下面三样。scalable certified lower bounds：在大实例上也能给出、且带证明的下界——「没有更轻的」这一半（码距篇 1.3 节、证书篇 1.1 节）。named Pauli sectors：按类型命名的扇区——分别说清最轻的 X 型逻辑算符多重（d_X）、最轻的 Z 型多重（d_Z），而不只报一个总的 d。equivalence-safe subsystem presentations：同一个子系统码可以用不同的生成元清单写出来（换顺序、加冗余、换一组生成同一对 (S, G) 的生成元；注意「把规范生成元升级成稳定子」是规范固定，得到的是另一个码，「把方格转置」会调换 X、Z 的标签）；「equivalence-safe」要求证出来的码距在换清单之后仍然成立（同一个码穿不同的衣服 · 1.3 节）。人话：今天能对小码穷举、对大码给一件衣服；想要的是对大码也证明「没有更轻的衣服」，分别报 X 和 Z 两个方向，并且换一种写法结论不变。

Word by word. exact-small: exact exhaustion on small instances, checking every light string against the two conditions of 1.4; Chapter 2 shows how and up to what size. witnessed upper-bound: an upper bound with a witness: find one dressed logical and its weight bounds d from above (distance guide section 1.2: one example suffices). dressed-distance replay: replayable: same program, same seed, identical result for anyone. Extend … to: grow these two existing things into the three below. scalable certified lower bounds: lower bounds available on large instances too, with a proof: the “nothing lighter” half (distance guide 1.3, certificate guide 1.1). named Pauli sectors: sectors named by type: state separately how heavy the lightest X-type logical is (d_X) and the lightest Z-type (d_Z), rather than one total d. equivalence-safe subsystem presentations: the same subsystem code can be written with different generator lists (reordering, redundancy, another generating set of the same pair (S, G); note that promoting gauge generators to stabilizers is gauge fixing and yields a different code, and transposing the grid swaps the X and Z labels); “equivalence-safe” demands that the certified distance survive the change of list (The same code in different clothes · 1.3). In plain words: today we can exhaust small codes and hand over one outfit for large ones; what is wanted is to prove “no lighter outfit” for large codes too, report the X and Z directions separately, and have the conclusion survive a change of notation.

逐词翻译。The referee：检验员——QEC-Bank 共享的验证程序（qeclib/distance.py），所有条目的码距记录都要经它之手。exactly enumerates complete small weight layers：精确地、一层不漏地枚举小实例的「重量层」——重量 1 的全部串是第一层，重量 2 的是第二层，以此类推；「complete」指从不在一层中间停下。validates seeded randomized upper witnesses modulo the full gauge group：对随机搜索（带种子，可重放）找到的见证，独立地按 1.4 节的两个条件核对，其中条件 2 用的是整个规范群。Over-budget witnesses remain upper-only with not-certified optimality：当穷举超过预算，就只剩见证，它只能作上界，且标明「最优性未认证」——没人证明它是最轻的。no asymptotic or presentation-equivalence theorem is inferred：检验员不从有限结果推出「对所有尺寸成立」的公式，也不推出「换一种写法码距不变」。人话：小的算到底，大的只给一件衣服并声明「不保证最轻」，其余一概不下结论。

Word by word. The referee: the shared QEC-Bank verification program (qeclib/distance.py), through which every entry's distance record must pass. exactly enumerates complete small weight layers: enumerates the “weight layers” of small instances exactly and without omission; all strings of weight 1 are layer one, weight 2 layer two, and so on; “complete” means it never stops in the middle of a layer. validates seeded randomized upper witnesses modulo the full gauge group: witnesses found by a seeded (replayable) random search are independently checked against the two conditions of 1.4, condition 2 using the whole gauge group. Over-budget witnesses remain upper-only with not-certified optimality: when exhaustion exceeds its budget only the witness remains, usable as an upper bound only and marked “optimality not certified”: nobody has shown it is the lightest. no asymptotic or presentation-equivalence theorem is inferred: the referee does not infer a formula “for all sizes” from finite results, nor that “the distance is unchanged under a change of notation”. In plain words: small ones to the end, large ones get one outfit with the disclaimer “not guaranteed lightest”, and nothing else is concluded.

QEC-Bank 的 codes/bacon_shor/instances.yaml 登记了三个实例；下表逐字转录每个数字旁边的证据词（参数页 →）。r × c 表示 r 行 c 列。

The QEC-Bank file codes/bacon_shor/instances.yaml registers three instances; the table transcribes the evidence word beside every number (the parameters page →). r × c means r rows and c columns.

三件事。第一，总码距 d 三个都是 verified——检验员亲自穷举过（2.2 节）。第二，分方向的 d_X、d_Z 全是 claimed——引自 Bacon 2006（bacon2006operator）和 Napp 与 Preskill 2013（napp2013optimal），检验员没算。QEC-Bank 推导页说明原因：「The d_X and d_Z rows remain literature-attributed because the generic referee reports the minimum over all Pauli types rather than two named sectors.」人话：机器只报「最轻的是几」，不报「最轻的是 X 型还是 Z 型」——这正是 statement 里「named Pauli sectors」要补的。第三，3 × 5 那行最能说明为什么要分方向：d_X = 5、d_Z = 3，总 d = 3 只看得见小的那个；QEC-Bank 直觉页：「Aspect ratio can therefore match a noise bias」——如果机器上 Z 错误比 X 错误多得多，你想要的是 Z 方向抗得住，而这个信息在 verified 的 d 里是丢失的。规范比特数 g = (r − 1)(c − 1)：1、4、8，也是 claimed，因为「the ordinary entry verifier exposes k but does not emit gauge-qubit claims」。

Three things. First, the total distance d is verified for all three: the referee exhausted them itself (2.2). Second, the per-direction d_X, d_Z are all claimed, cited from Bacon 2006 (bacon2006operator) and Napp and Preskill 2013 (napp2013optimal); the referee did not compute them. The QEC-Bank derivation page explains why: “The d_X and d_Z rows remain literature-attributed because the generic referee reports the minimum over all Pauli types rather than two named sectors.” In plain words: the machine reports “the lightest is this heavy”, not “the lightest is X-type or Z-type”, which is exactly what “named Pauli sectors” in the statement asks to add. Third, the 3 × 5 row shows best why directions matter: d_X = 5, d_Z = 3, and the total d = 3 sees only the smaller; the QEC-Bank intuition page: “Aspect ratio can therefore match a noise bias”: if a machine suffers far more Z errors than X errors, you want the Z direction to hold, and that information is lost in the verified d. The gauge-qubit count g = (r − 1)(c − 1), namely 1, 4, 8, is claimed too, because “the ordinary entry verifier exposes k but does not emit gauge-qubit claims”.

检验员的精确路径（qeclib/distance.py 里的 _exact_subsystem_dressed_distance，方法名 exact-subsystem-dressed-enum-v1）做的正是 1.4 节末尾你对 2 × 2 做的事，只是一层不漏：从重量 1 开始，对每个重量列出所有泡利串——先选哪 w 个比特（C(n, w) 种），再给每个选中的比特填 X、Y 或 Z（3^w 种）——逐条检查条件 1（与每条稳定子对易）和条件 2（不在 G 里，用矩阵的秩判定），第一条通过的串的重量就是 d。它有一道预算门：代码里 MAX_EXACT_SUBSYSTEM_CANDIDATES = 2_000_000；进入每一层之前先算这一层有多大，若累计会超过两百万，就拒绝进入、报「unverifiable」。设计文档的原话：「if that layer would cross the ceiling … the result is unverifiable rather than a lower bound.」下表是本篇按 C(n, w) · 3^w 自己算的各层大小：

The referee's exact path (_exact_subsystem_dressed_distance in qeclib/distance.py, method name exact-subsystem-dressed-enum-v1) does exactly what you did on the 2 × 2 at the end of 1.4, but without omission: starting at weight 1, for each weight it lists every Pauli string, first choosing which w qubits (C(n, w) ways), then a letter X, Y or Z for each chosen qubit (3^w ways), and checks each against condition 1 (commutes with every stabilizer) and condition 2 (not in G, decided by a matrix rank); the weight of the first string that passes is d. It has a budget gate: in the code, MAX_EXACT_SUBSYSTEM_CANDIDATES = 2_000_000; before entering a layer it computes the layer's size, and if the running total would exceed two million it refuses to enter and reports “unverifiable”. The design document's words: “if that layer would cross the ceiling … the result is unverifiable rather than a lower bound.” The table below gives the layer sizes, computed by this guide as C(n, w) · 3^w:

记号：n = 比特数；w = 重量；C(n, w) 读「n 选 w」，从 n 个比特里挑 w 个的方法数；3^w = w 个比特各自取 X/Y/Z 的方法数。C(4, 2) = 6，C(9, 3) = 84，C(35, 4) = 52 360。

Notation: n = number of qubits; w = weight; C(n, w), read “n choose w”, the number of ways to pick w of n qubits; 3^w = ways for w qubits to each take X/Y/Z. C(4, 2) = 6, C(9, 3) = 84, C(35, 4) = 52 360.

3 × 3 的证据文件 computed/subsystem_distances/bacon-shor-square-3x3-exact.json 原文："result": {"distance": {"origin": {"kind": "verified", "method": "exact-subsystem-dressed-enum-v1"}, "value": 3}, "kind": "exact"}——这就是参数页上「d=3 verified」的来历。请注意 5 × 7 那行的一个细节：检验员在第 3 层结束时已经知道重量 ≤ 3 的串里没有逻辑算符，按码距篇 1.3 节的逻辑这本可以写成「d ≥ 4」。但它不写。QEC-Bank 推导页：「The exact enumerator refuses an over-budget layer before treating partial work as a lower bound.」人话：检验员宁可说「我不知道」，也不把半途的工作包装成下界。这是一个策略选择，原因在第 4 章。

The evidence file for 3 × 3, computed/subsystem_distances/bacon-shor-square-3x3-exact.json, reads: "result": {"distance": {"origin": {"kind": "verified", "method": "exact-subsystem-dressed-enum-v1"}, "value": 3}, "kind": "exact"}; that is where “d=3 verified” on the parameters page comes from. Note one detail in the 5 × 7 row: at the end of layer 3 the referee already knows that no string of weight ≤ 3 is a logical, which by the logic of distance guide 1.3 could be written “d ≥ 4”. It does not write it. The QEC-Bank derivation page: “The exact enumerator refuses an over-budget layer before treating partial work as a lower bound.” In plain words: the referee would rather say “I don't know” than dress half-finished work up as a lower bound. This is a policy choice, explained in Chapter 4.

穷举拒绝之后，检验员走第二条路：带种子的随机搜索（方法名 rand-subsystem-dressed-logical-coset-v1）。它在 C(S)/G 的类里随机挑一类，随机挑一件衣服，再反复乘上规范算符「只要能变轻就乘」，把找到的最轻一件交出来。5 × 7 的证据文件 computed/subsystem_distances/bacon-shor-rectangle-5x7-upper.json 原文（节选）：

After exhaustion refuses, the referee takes the second road: a seeded random search (method name rand-subsystem-dressed-logical-coset-v1). It picks a class of C(S)/G at random, picks an outfit at random, then multiplies by gauge operators repeatedly “whenever that makes it lighter”, and hands over the lightest it found. The evidence file for 5 × 7, computed/subsystem_distances/bacon-shor-rectangle-5x7-upper.json, verbatim (excerpt):

读这份文件。pauli 那一串有 35 个字符，每个比特一个：下划线是 I，字母是 X、Y、Z；数一数非下划线的字符正好 12 个，这就是 value 12 的来源——你可以自己数。kind: bounded、guarantee: upper-bound-only：它只说 d ≤ 12。optimality: not-certified：没有任何证明说 12 是最轻的。seed 17、iters 32：随机搜索的起点与轮数，别人用同一对数字能复现同一件衣服。现在把它和纸上的知识对照：QEC-Bank 直觉页给出家族公式 d = min(r, c)，对 5 × 7 就是 5——一整列 Z（5 个比特）就是一条重量 5 的裸逻辑算符，比机器找到的 12 轻得多。可是 QEC-Bank 推导页说：「This is an achieved upper bound only and does not replace the parity proof of the rectangular family formula」，又说这一行「is not a lower bound, an exact distance, or evidence for the family formula」。为什么不直接把 5 写上去？「d = 5」其实是两半，要分开看。上界那一半 d ≤ 5 根本不需要纸上的证明：一整列 Z 就是一条具体的串，检验员能像核对那条重量 12 的串一样核对它（1.4 节的条件 1、2），核对过就是一份机器认的 d ≤ 5 见证；文件里的 12 只是这一次搜索（seed 17，32 轮）恰好交出的结果。下界那一半 d ≥ 5 说的是「不存在重量 ≤ 4 的穿衣逻辑算符」，这才来自一段纸上的证明（第 5 章会带你亲手做）；检验员没有能力核对纸上的证明，只会核对具体的串或一层层穷举，而穷举在这里被预算门拒绝了。人话：机器诚实地交出了一件很差的衣服，并且诚实地承认它不知道有没有更好的。这个 5 × 7 不是登记的实例（文件里 "instance": null），所以条目页上看不到它；它是设计文档要求的校准案例：「a larger finite rectangle that exceeds the exact ceiling to return a deterministic, independently valid bounded-search witness」。

Reading the file. The pauli string has 35 characters, one per qubit: an underscore is I, a letter is X, Y or Z; count the non-underscores and there are exactly 12, the source of value 12; you can count them yourself. kind: bounded, guarantee: upper-bound-only: it only says d ≤ 12. optimality: not-certified: no proof says 12 is the lightest. seed 17, iters 32: the starting point and number of rounds of the random search; anyone with the same pair reproduces the same outfit. Now compare with paper knowledge: the QEC-Bank intuition page gives the family formula d = min(r, c), which for 5 × 7 is 5: a whole column of Z (5 qubits) is a bare logical of weight 5, far lighter than the machine's 12. Yet the QEC-Bank derivation page says “This is an achieved upper bound only and does not replace the parity proof of the rectangular family formula”, and that this row “is not a lower bound, an exact distance, or evidence for the family formula”. Why not simply write 5? “d = 5” has two halves that must be kept apart. The upper half, d ≤ 5, needs no paper proof at all: a whole column of Z is one concrete string, the referee can check it exactly as it checked the weight-12 string (conditions 1 and 2 of 1.4), and once checked it is a machine-accepted witness for d ≤ 5; the 12 in the file is simply what this particular search run (seed 17, 32 rounds) happened to return. The lower half, d ≥ 5, says that no dressed logical of weight 4 or less exists, and that is what comes from a proof on paper (Chapter 5 walks you through it); the referee cannot check a proof on paper, only concrete strings or layer-by-layer enumeration, and here the budget gate refuses the enumeration. In plain words: the machine honestly hands over a poor outfit and honestly admits it does not know whether a better one exists. This 5 × 7 is not a registered instance (the file says "instance": null), so it does not appear on the entry page; it is a calibration case demanded by the design document: “a larger finite rectangle that exceeds the exact ceiling to return a deterministic, independently valid bounded-search witness”.

记号：seed = 随机数发生器的起始数；iters = 搜索轮数；r、c = 行数、列数；min(r, c) = 两数中较小者。「bounded」是 QEC-Bank 的证据词，意思是「上界由机器给出，下界没有」（码距篇 2.2 节）。

Notation: seed = the starting number of the random-number generator; iters = number of search rounds; r, c = rows, columns; min(r, c) = the smaller of the two. “bounded” is the QEC-Bank evidence word meaning “upper bound by machine, no lower bound” (distance guide 2.2).

「Finite C(S)/G enumeration now verifies small Bacon-Shor instances」：就是 2.2 节的表——4、9、15 个比特，三个 verified。这是已经做到的一半。「large subsystem families still need lower-bound certificates」：就是 2.3 节——35 个比特时穷举拒绝，机器的记录里只有这次搜索得到的 d ≤ 12（一整列 Z 这条见证就能把上界降到 5，机器也能核对它），而下界 d ≥ 5 只在纸上有证明；缺的是一份机器能核对的「没有重量 ≤ 4 的穿衣逻辑算符」证书（证书篇 1.1 节：证书 = 数据 + 一台不信任任何人的检查器）。「families」是复数：Bacon–Shor 只是范例，QEC-Bank 里还有别的子系统码条目，问题对它们同样成立。「proof that gauge-equivalent presentations preserve the claimed quantity」：表述presentation：写下一个码时用的那份具体的生成元清单（稳定子清单 + 规范生成元清单，含顺序）。同一个码有无数种表述。是写码时用的那份具体清单；「gauge-equivalent」指两份清单生成同一个规范群（或只差重排与冗余；注意转置并交换 X/Z 会调换 X、Z 两个方向的标签，规范固定则改变稳定子与规范的划分，两者都需要各自的对应关系）；「preserve the claimed quantity」问的是：对甲清单算出的码距，对乙清单还成立吗？直觉上当然，但检验员的原则是不推断：current_proxy 说「no … presentation-equivalence theorem is inferred」。QEC-Bank 有一个现成的例子：把 3 × 3 的六条横向 ZZ 规范生成元全部升级为稳定子，得到的稳定子清单与九比特 Shor 码的清单生成同一个行空间（两份清单能乘出的全部串完全相同）——推导页说这个见证「is evidence for one pair, not a general canonicalizer」，并把更一般的问题交给 wanted:code-equivalence（同一个码穿不同的衣服 · 2.3 节的「手画的同一边」）。

“Finite C(S)/G enumeration now verifies small Bacon-Shor instances”: the table of 2.2, 4, 9 and 15 qubits, three verified. This is the half already done. “large subsystem families still need lower-bound certificates”: section 2.3: at 35 qubits exhaustion refuses, the machine's record holds only the d ≤ 12 of this one search run (the whole-column Z witness would lower the upper bound to 5, and the machine can check it), while the lower bound d ≥ 5 is proved only on paper; what is missing is a machine-checkable certificate that “no dressed logical of weight ≤ 4 exists” (certificate guide 1.1: a certificate = data + a checker that trusts nobody). “families” is plural: Bacon–Shor is the exemplar, the QEC-Bank has other subsystem-code entries, and the problem holds for them alike. “proof that gauge-equivalent presentations preserve the claimed quantity”: a presentationThe concrete list of generators used to write a code down (stabilizer list + gauge-generator list, order included). One code has infinitely many presentations. is the concrete list used to write the code; “gauge-equivalent” means two lists generating the same gauge group (or differing by reordering and redundancy; note that transposing and swapping X/Z exchanges the X and Z sector labels, and a gauge fix changes the stabilizer/gauge split, so each needs its own correspondence); “preserve the claimed quantity” asks: does the distance computed for list A still hold for list B? Intuitively of course, but the referee's principle is not to infer: current_proxy says “no … presentation-equivalence theorem is inferred”. The QEC-Bank has a ready example: promote all six horizontal ZZ gauge generators of the 3 × 3 to stabilizers and the resulting stabilizer list generates the same row space (the two lists multiply out to exactly the same set of strings) as the nine-qubit Shor code's; the derivation page says this witness “is evidence for one pair, not a general canonicalizer” and hands the general problem to wanted:code-equivalence (the “hand-drawn identity edges” of The same code in different clothes · 2.3).

最后一行要说明来源：QEC-Bank 的 codes/ 目录下有若干名字里带 subsystem、bacon_shor、compass 的条目（如 subsystem_surface、subsystem_color、3d_bacon_shor、bravyi_bacon_shor、compass_model、subsystem_product），本篇只看了目录名，没有读它们的卡片，不知道它们各自的 d 现在写着什么词；上表那一行说的是检验员的预算门对任何子系统码都一样，不是对这些条目的具体断言。

The last row needs its source stated: the QEC-Bank codes/ directory holds several entries with subsystem, bacon_shor or compass in their names (such as subsystem_surface, subsystem_color, 3d_bacon_shor, bravyi_bacon_shor, compass_model, subsystem_product); this guide looked only at the directory names, did not read their cards, and does not know what word each d carries today; that row says the referee's budget gate is the same for every subsystem code, not anything specific about those entries.

三件事。第一，不能相信任何大子系统码的 d：看到 bounded 12，你只知道 d 不超过 12；机器没有核对过任何下界，真值是 12、5 还是 4，记录本身不告诉你。第二，不能按噪声方向选码：3 × 5 的参数页只说 d = 3 verified；想知道「Z 方向的码距是 3、X 方向的是 5」（能保证纠正的分别只有 1 个和 2 个错），只能去读 claimed 的引文。第三，不能把一个证书搬到另一种写法上：如果你手里是同一对 (S, G) 的另一份生成元清单（重排过，或加了冗余的生成元），条目里 verified 的 d = 3 在形式上不属于你的清单，尽管你确信是同一个码。规范固定又远一步：把 3 × 3 的六条 ZZ 升级成稳定子就得到 Shor 码（规范固定后的 Bacon–Shor ↔ Shor），rank S 从 4 变成 8，(S, G) 已经不是同一对，「同一 (S, G)」的定理也管不到它（同一个码穿不同的衣服 · 3.3 节列过类似的「今天做不到」）。人话：小码的数字可信，大码的数字只有一半，方向的数字全靠引文，换个写法一切归零。

Three things. First, you cannot trust the d of any large subsystem code: seeing bounded 12 you know only that d is at most 12; no lower bound has been machine-checked, and the record alone does not tell you whether the truth is 12, 5 or 4. Second, you cannot choose a code by noise direction: the 3 × 5 parameters page only says d = 3 verified; to learn “the Z-direction distance is 3, the X-direction one 5” (guaranteeing correction of only 1 and 2 errors respectively) you must read the claimed citation. Third, you cannot carry a certificate to another notation: if you hold another generator list for the same (S, G) (reordered, or with redundant generators added), the entry's verified d = 3 formally does not belong to your list even though you are sure it is the same code. Gauge fixing goes one step further: promoting the six ZZ edges of the 3 × 3 to stabilizers gives the Shor code (gauge-fixed Bacon–Shor ↔ Shor), rank S goes from 4 to 8, (S, G) is no longer the same pair, and a same-(S, G) theorem does not reach it (The same code in different clothes · 3.3 lists a similar “cannot do today”). In plain words: small codes' numbers are trustworthy, large codes' numbers are half, directional numbers rest on citations, and a change of notation resets everything.

对 Bacon–Shor 码，下界是高中水平的。以 r × c 的矩形、X 扇区为例，QEC-Bank 推导页的论证翻成人话是三句话：(1) 穿衣只会让每列里的 X 滑动、成对产生或成对湮灭（1.3 节），改不了每列 X 的个数的奇偶（1.3 节亲手验证过）。(2) 一条 X 型串要与相邻两列的 Z 型稳定子条带对易，就必须让相邻两列的奇偶相同，于是所有列的奇偶一样：全偶或全奇；全偶的串全是规范算符（无害）：同一列里的两个 X 等于它们之间那几条竖边之积，例如 X₀X₆ = (X₀X₃)(X₃X₆)，中间的 X₃ 乘了两次、消掉；一列里有偶数个 X，就把它们两两配对，每对换成竖边之积，每列都这样做，整条串就是竖边之积，属于 G。所以逻辑算符必须每列都是奇数个 X。(3) 每列至少一个 X，重量至少 c；一整行恰好做到 c。于是 d_X = c，同理 d_Z = r，d = min(r, c)。第 5 章让你在 3 × 3 上逐句核对。

For Bacon–Shor the lower bound is high-school level. For an r × c rectangle and the X sector, the QEC-Bank derivation page's argument in plain words is three sentences: (1) dressing only slides, creates or annihilates X's in pairs within their columns (1.3) and cannot change the parity of the number of X's in each column (verified by hand in 1.3). (2) For an X-type string to commute with the Z-type stabilizer strip over two neighbouring columns, those two columns must have the same parity, so all columns share one parity: all even or all odd; all-even strings are gauge operators (harmless): two X's in the same column equal the product of the vertical edges between them, for example X₀X₆ = (X₀X₃)(X₃X₆), where the middle X₃ appears twice and cancels; a column with an even number of X's can be split into pairs, each pair replaced by a product of vertical edges, and doing this in every column writes the whole string as a product of vertical edges, which lies in G. So a logical must have an odd number of X's in every column. (3) At least one X per column, weight at least c; a whole row achieves exactly c. Hence d_X = c, likewise d_Z = r, and d = min(r, c). Chapter 5 has you check it sentence by sentence on the 3 × 3.

那为什么 5 × 7 还写着 bounded 12？因为这三句话是针对这一族的：它用了「竖边只在列内滑动」「稳定子是相邻两列的条带」这些 Bacon–Shor 特有的几何。检验员是共享的，它对任何一份 (S, G) 清单只会做两件事——穷举一层层的串，或验一条具体的见证。它没有「读懂一段关于列奇偶的论证」的能力，而 QEC-Bank 也不允许人工把纸上的结论填成 verified（码距篇 2.3 节：那叫 claimed）。证书篇 1.1 节的标准是「证书 = 数据 + 一台不信任任何人的检查器」：三句话是证明，不是证书。要让它成为证书，得把「每列奇偶」这类不变量写成检查器能逐条核对的数据——对 Bacon–Shor 也许做得到，但 statement 要的是对一般的子系统家族都能产出证书的方法，而一般的规范群没有行、列、条带这些现成的结构。人话：我们会证，但证明写在纸上；机器要的是它自己能核对的东西，而且要对任何一份清单都管用。

Then why does the 5 × 7 still say bounded 12? Because the three sentences are specific to this family: they use Bacon–Shor's own geometry, “vertical edges slide only within a column”, “stabilizers are strips over two neighbouring columns”. The referee is shared and does only two things with any (S, G) list: exhaust the strings layer by layer, or validate one concrete witness. It has no ability to “understand an argument about column parities”, and the QEC-Bank does not allow a human to enter a paper conclusion as verified (distance guide 2.3: that is called claimed). The standard of certificate guide 1.1 is “a certificate = data + a checker that trusts nobody”: three sentences are a proof, not a certificate. To make them a certificate, invariants like “column parity” must be written as data a checker can verify item by item; perhaps feasible for Bacon–Shor, but the statement wants a method that produces certificates for general subsystem families, and a general gauge group has no ready-made rows, columns or strips. In plain words: we can prove it, but the proof lives on paper; the machine wants something it can check itself, and it must work for any list.

Wanted 板用两个词分难度：hard-engineering——知道怎么做，但要做很多（例如「从理想症状到真实电路」4.2 节）；open-research——没人知道怎么做。这道题是后者，且比码距篇的同名难题多一层：那里的候选是「与所有稳定子对易且不是稳定子乘积」的串，这里每个候选还要多做一次「在 G 里吗」的判定，而下界证明必须覆盖整个 G 的所有衣服——松弛法（码距篇 4.2 节：不逐点检查也能证最小值）要在商空间 C(S)/G 上设计，文献里没有现成的做法。2.2 节的表已经给出穷举的账：候选数按 C(n, w) · 3^w 增长，35 个比特到重量 4 就是四百多万条，超过两百万的预算；再看码距篇 1.4 节的 2ⁿ 墙，n = 35 时全部泡利串有 4³⁵ ≈ 1.2 × 10²¹ 条。检验员的预算门为什么设在两百万、为什么拒绝半途下界？设计文档：「This boundary makes small finite examples exact while preventing an accidental exponential referee run.」——防止一次验证意外跑上几天；而「半途的下界」若被接受，就会有人把预算调大一点、再大一点，把「跑到哪算到哪」当成证明。人话：穷举的墙和码距篇一样高，商掉规范群让每一步更贵，而绕过墙的松弛法还没人替子系统码发明出来。

The board grades difficulty with two words: hard-engineering, known how but much to do (for example “From ideal syndromes to real circuits” 4.2); open-research, nobody knows how. This problem is the latter, with one more layer than the namesake problem of the distance guide: there the candidates are strings “commuting with every stabilizer and not a product of stabilizers”, here every candidate additionally needs the test “is it in G?”, and a lower-bound proof must cover every outfit in the whole of G; relaxation methods (distance guide 4.2: proving a minimum without checking every point) would have to be designed on the quotient C(S)/G, and the literature has no ready recipe. The table of 2.2 already gives the exhaustion arithmetic: candidates grow as C(n, w) · 3^w, and at 35 qubits weight 4 alone is over four million strings, past the two-million budget; recall the 2ⁿ wall of distance guide 1.4: for n = 35 all Pauli strings number 4³⁵ ≈ 1.2 × 10²¹. Why is the referee's gate at two million, and why does it refuse half-way lower bounds? The design document: “This boundary makes small finite examples exact while preventing an accidental exponential referee run”, so one verification cannot accidentally run for days; and if “half-way lower bounds” were accepted, someone would raise the budget a little, then a little more, and treat “as far as it ran” as a proof. In plain words: the exhaustion wall is as high as in the distance guide, quotienting by the gauge group makes every step dearer, and the relaxation method that would go around the wall has not yet been invented for subsystem codes.

记号：4ⁿ = n 个比特各取 I/X/Y/Z 的总数（含全 I 的一条）；≈ 读「约等于」；10²¹ = 1 后面 21 个零。1.2 × 10²¹ 是本篇自己算的。

Notation: 4ⁿ = all choices of I/X/Y/Z on n qubits (including the all-I string); ≈ reads “approximately”; 10²¹ = a 1 followed by 21 zeros. 1.2 × 10²¹ is this guide's own computation.

3 × 3 按行编号 q0…q8（q0 q1 q2 / q3 q4 q5 / q6 q7 q8）。写出全部规范生成元：竖边 XX 共 6 条（X₀X₃、X₃X₆、X₁X₄、X₄X₇、X₂X₅、X₅X₈），横边 ZZ 共 6 条（Z₀Z₁、Z₁Z₂、Z₃Z₄、Z₄Z₅、Z₆Z₇、Z₇Z₈）。它们彼此独立，rank G = 12。中心：相邻两行的竖边全部相乘得 X 条带 S^X₀ = X₀X₁X₂X₃X₄X₅、S^X₁ = X₃…X₈；相邻两列的横边全部相乘得 Z 条带 S^Z₀ = Z₀Z₁Z₃Z₄Z₆Z₇、S^Z₁ = Z₁Z₂Z₄Z₅Z₇Z₈。验证任一 X 条带与任一横边 ZZ 碰撞 0 或 2 次（对易），rank S = 4。代入公式：k = 9 − (12 + 4)/2 = 1，g = (12 − 4)/2 = 4。对照 QEC-Bank 推导页账本表第二行「3 × 3 | 4 | 12 | 1 | 4」与直觉页 Q4 的答案「k = 9 − (12 + 4)/2 = 1 and g = (12 − 4)/2 = 4」。

Number the 3 × 3 by rows q0…q8 (q0 q1 q2 / q3 q4 q5 / q6 q7 q8). Write all gauge generators: 6 vertical XX (X₀X₃, X₃X₆, X₁X₄, X₄X₇, X₂X₅, X₅X₈) and 6 horizontal ZZ (Z₀Z₁, Z₁Z₂, Z₃Z₄, Z₄Z₅, Z₆Z₇, Z₇Z₈). They are independent, rank G = 12. The center: multiplying all vertical edges between two neighbouring rows gives the X strips S^X₀ = X₀X₁X₂X₃X₄X₅, S^X₁ = X₃…X₈; multiplying all horizontal edges between two neighbouring columns gives the Z strips S^Z₀ = Z₀Z₁Z₃Z₄Z₆Z₇, S^Z₁ = Z₁Z₂Z₄Z₅Z₇Z₈. Check that any X strip meets any horizontal ZZ on 0 or 2 sites (commutes); rank S = 4. Substitute: k = 9 − (12 + 4)/2 = 1, g = (12 − 4)/2 = 4. Compare the second row of the QEC-Bank ledger table, “3 × 3 | 4 | 12 | 1 | 4”, and the answer to Q4 of the intuition page, “k = 9 − (12 + 4)/2 = 1 and g = (12 − 4)/2 = 4”.

记号：S^X_i = 第 i 行与第 i+1 行之间的 X 条带；S^Z_j = 第 j 列与第 j+1 列之间的 Z 条带；行、列都从 0 数起。

Notation: S^X_i = the X strip between rows i and i+1; S^Z_j = the Z strip between columns j and j+1; rows and columns counted from 0.

在 q0 上放一个 Z 错误。哪些稳定子报警？S^X₀ 含 X₀，与 Z₀ 碰撞一次，报警；S^X₁ 不含 q0，安静；两条 Z 条带与 Z 错误不碰撞。症状 (s₀, s₁) = (1, 0)。把错误换到同一行的 q1 或 q2，结果不变——因为 Z₀ 与 Z₁ 只差横边 Z₀Z₁，一个规范算符，对受保护比特作用相同。换到 q3（第 1 行）：两条 X 条带都含 q3，症状 (1, 1)；换到 q6：只有 S^X₁ 含它，症状 (0, 1)。这正是 QEC-Bank 推导页的表：

Put one Z error on q0. Which stabilizers fire? S^X₀ contains X₀, collides once with Z₀, fires; S^X₁ does not contain q0, quiet; the two Z strips do not collide with a Z error. Syndrome (s₀, s₁) = (1, 0). Move the error to q1 or q2 in the same row and nothing changes, because Z₀ and Z₁ differ by the horizontal edge Z₀Z₁, a gauge operator with the same action on the protected qubit. Move it to q3 (row 1): both X strips contain q3, syndrome (1, 1); to q6: only S^X₁ contains it, syndrome (0, 1). This is exactly the QEC-Bank derivation page's table:

推导页的评语：「The column position is absent because changing it by horizontal Z gauge edges does not change the protected action.」人话：在子系统码里，译码器根本不需要知道错误在哪一列，只要知道哪几行是奇数个错——三行的奇偶再做一次奇偶检查，就是 Toric Code 教程 2.3 节的重复码。这也是为什么 QEC-Bank 说「Degeneracy is the algorithm, not a nuisance」：等价的衣服不是麻烦，而是让问题变简单的原因。

The derivation page's comment: “The column position is absent because changing it by horizontal Z gauge edges does not change the protected action.” In plain words: in a subsystem code the decoder does not need to know which column the error is in, only which rows hold an odd number of errors; a parity check on the three row parities is the repetition code of toric-code section 2.3. This is also why the QEC-Bank says “Degeneracy is the algorithm, not a nuisance”: equivalent outfits are not a bother but the reason the problem becomes easy.

记号：u = 行奇偶向量（每行一个 0/1：该行 Z 错误个数的奇偶）；s_i = 条带 S^X_i 的读数，s_i = u_i ⊕ u_i+1；⊕ 读「异或」。

Notation: u = the row-parity vector (one 0/1 per row: parity of the number of Z errors in that row); s_i = the reading of strip S^X_i, s_i = u_i ⊕ u_i+1; ⊕ reads “xor”.

按 4.2 节，在 3 × 3 上写出 X 扇区的证明：(1) 竖边只在列内滑动、成对产生或成对湮灭 X，每列 X 的奇偶不变。(2) 与 S^Z₀ 对易要求第 0、1 列奇偶相同，与 S^Z₁ 对易要求第 1、2 列相同；全偶 = 规范算符（4.2 节的配对：同列的 X 两两配对，每对是竖边之积，例如 X₀X₆ = (X₀X₃)(X₃X₆)）；所以逻辑算符三列都是奇数个 X。(3) 每列至少 1 个，重量 ≥ 3；X₀X₁X₂ 达到 3。Z 扇区把行列对换。最后用 1.4 节的两个条件说服自己：一条混合串（既有 X 又有 Z）的 X 部分和 Z 部分各自要么是规范算符、要么落在对应的奇数类里，而它的重量不小于任一非平凡部分的重量——QEC-Bank 推导页的原话：「a mixed Pauli cannot beat both pure-sector minima」。于是 d = 3。这与 2.2 节检验员穷举 2 619 条候选得到的 3 一致。现在回答本篇的核心问题：你的证明用了什么机器没有的东西？答案：「列」「行」「滑动」——这些词属于 Bacon–Shor 的几何，不属于一份一般的 (S, G) 清单。把它们翻译成任何清单都能核对的数据，就是这道 Wanted 题。

Following 4.2, write the X-sector proof on the 3 × 3: (1) vertical edges only slide, create or annihilate X's in pairs within a column; each column's X parity is unchanged. (2) Commuting with S^Z₀ forces columns 0 and 1 to share a parity, commuting with S^Z₁ forces columns 1 and 2 to; all-even = a gauge operator (the pairing of 4.2: X's in one column pair up, each pair a product of vertical edges, for example X₀X₆ = (X₀X₃)(X₃X₆)); so a logical has an odd number of X's in all three columns. (3) At least 1 per column, weight ≥ 3; X₀X₁X₂ achieves 3. The Z sector swaps rows and columns. Finally convince yourself with the two conditions of 1.4 that a mixed string (both X and Z) has an X part and a Z part each either a gauge operator or in its odd class, while its weight is at least that of any nontrivial part; the QEC-Bank derivation page's words: “a mixed Pauli cannot beat both pure-sector minima”. Hence d = 3, agreeing with the 3 the referee obtained by exhausting 2 619 candidates in 2.2. Now answer this guide's central question: what did your proof use that the machine does not have? Answer: “column”, “row”, “slide”; these words belong to Bacon–Shor's geometry, not to a general (S, G) list. Translating them into data any list can be checked against is this Wanted item.

这道题是三道 Wanted 题的交点。「没有更轻的」这一半怎么在大码上证明：「大码的码距」第 4 章的松弛法；怎么写成机器认的证书：「一张机器能核对的证书」第 4 章；换一种写法怎么判定还是同一个码：「同一个码穿不同的衣服」第 4 章。想看 QEC-Bank 对 Bacon–Shor 的全部记录：条目页与它的推导页（gauge-center-and-qubits.md、dressed-distance-and-decoding.md、derivations.md）。想看 QEC-Bank 子系统码的框架定义：foundations/subsystem-codes/intuition.md——目前是一个存根，只有一句话的定义和 Lean Complete v1 的机器边界：「checks the supplied stabilizer/gauge presentation, protected-logical rank formula, and dressed logical witnesses. It does not assert that the supplied stabilizers generate every abstract center or prove subsystem families.」这条 Wanted 记录没有 working_repo 字段。

This item is the crossing point of three Wanted problems. How to prove the “nothing lighter” half on large codes: the relaxation methods of Chapter 4 of “Distance for large codes”; how to write it as a certificate a machine accepts: Chapter 4 of “A machine-checkable certificate”; how to decide that another notation is still the same code: Chapter 4 of “The same code in different clothes”. For the QEC-Bank's full record of Bacon–Shor: the entry page and its derivation pages (gauge-center-and-qubits.md, dressed-distance-and-decoding.md, derivations.md). For the QEC-Bank framework definition of subsystem codes: foundations/subsystem-codes/intuition.md, currently a stub with a one-line definition and the machine boundary of Lean Complete v1: “checks the supplied stabilizer/gauge presentation, protected-logical rank formula, and dressed logical witnesses. It does not assert that the supplied stabilizers generate every abstract center or prove subsystem families.” This Wanted record has no working_repo field.

回放一遍。子系统码允许检查清单互相打架：互不打架的乘积（中心）是真正的稳定子，打架的部分只翻我们不在乎的规范比特，受保护的比特数 k = n − (rank G + rank S)/2。逻辑算符乘上任何规范算符仍是同一条，于是它有很多件衣服；穿了衣服的码距是「与所有稳定子对易、且不在 G 里」的最轻串的重量，商掉的是整个规范群。QEC-Bank 的检验员对 [[4,1,2]]、[[9,1,3]]、[[15,1,3]] 一层层穷举到底（候选数 C(n, w) · 3^w，预算两百万），写下 verified；分方向的 d_X、d_Z 仍是 claimed；对 5 × 7 穷举在第 4 层前拒绝，随机搜索交出一件重量 12 的衣服（seed 17，iters 32），写着 bounded 与 not-certified，尽管纸上三句话就能证 d = 5。我们想要答案，是因为大码的 d 只有一半可信、方向的数字全靠引文、换一种写法证书归零。合格的答案是可扩展的认证下界、命名的扇区、等价安全的表述三样；难度 open-research：纸上的证明用了机器没有的几何，而对一般的规范群，绕过穷举之墙的方法还没有人发明。

Replay. A subsystem code lets its check list fight itself: the mutually commuting products (the center) are the true stabilizers, the fighting part only flips gauge qubits we don't care about, and the protected count is k = n − (rank G + rank S)/2. A logical times any gauge operator is still the same logical, so it has many outfits; the dressed distance is the weight of the lightest string that “commutes with every stabilizer and is not in G”, modulo the whole gauge group. The QEC-Bank referee exhausts [[4,1,2]], [[9,1,3]], [[15,1,3]] layer by layer to the end (C(n, w) · 3^w candidates, budget two million) and writes verified; the directional d_X, d_Z remain claimed; on the 5 × 7 exhaustion refuses before layer 4 and the random search hands over an outfit of weight 12 (seed 17, iters 32), marked bounded and not-certified, although three sentences on paper prove d = 5. We want the answer because a large code's d is only half trustworthy, directional numbers rest on citations, and a change of notation resets the certificate. A qualifying answer is three things, scalable certified lower bounds, named sectors, equivalence-safe presentations; the difficulty is open-research: the paper proof uses geometry the machine does not have, and for a general gauge group the method that would go around the exhaustion wall has not been invented.