十二个逻辑比特挤在一块，
门从哪里来 Twelve logical qubits in one block:
where do the gates come from

《格点手术 · 高中数学版》1.1 节讲过门集：任何量子运算都能拆成单比特门加 CNOT；纠错架构里只用有限的一组——H、S、CNOT 合称 Clifford 门把泡利串映成泡利串的门（H、S、CNOT 及其组合）。在稳定子语言里可以完全追踪；单靠它们不能通用计算。，再加一个 T 门才通用。表面码那一篇的结论是：平面上局域、常深度的操作最多做到 Clifford（Bravyi–König 定理），T 门要靠魔法态。BB 码的情形要重新问一遍：那条定理管的是检查作用范围远小于 √d（d 是码距）的二维码族，一块具体的 gross 不满足这个前提，定理对它不给结论（5.2 节说明）——但先别高兴：第 5 章会看到，QEC-Bank 与文献扫描的结论是「no source supplies a universal transversal gate set for the original gross block」，非 Clifford 的边界在 BB 上同样存在，只是理由不同。

Section 1.1 of “Lattice Surgery · High-School Edition” covered gate sets: any quantum computation decomposes into single-qubit gates plus CNOT; error-corrected architectures use a finite set, H, S and CNOT together called the Clifford gatesGates that map Pauli strings to Pauli strings (H, S, CNOT and their compositions). Fully trackable in the stabilizer language; not universal on their own., plus one T gate for universality. The surface-code guide concluded that local constant-depth operations in the plane reach at most Clifford (the Bravyi–König theorem) and the T gate needs magic states. For BB the question must be asked afresh: that theorem is about 2D code families whose check range is much smaller than √d (d the code distance); one concrete gross block does not meet that premise, so the theorem says nothing about it (section 5.2 explains); but hold the cheer: Chapter 5 shows that the QEC-Bank and the scout report conclude “no source supplies a universal transversal gate set for the original gross block”, so the non-Clifford boundary exists on BB as well, for a different reason.

BB 还多一层要求。表面码一块贴片一个逻辑比特，「对这块贴片做 H」就是「对这个逻辑比特做 H」。gross 一块 144 比特的码里住着 12 个逻辑比特；对整块做一个物理操作，效果落在 12 个逻辑比特的某种组合上。QEC-Bank 原文：「The packed value k = 12 makes addressability especially important. A collective operation on all twelve logical qubits is not a substitute for a compiler that selects the desired logical combination while preserving the others.」人话：一个「同时对 12 个都做点什么」的操作，不等于「只对第 3 个做 H」；后者叫可寻址addressability：能对指定的一个或几个逻辑比特做门，而不扰动同一块里的其余逻辑比特。k = 12 的块必须回答这个问题，k = 1 的贴片不必。。intuition.md 把它列为失败模式之一：「Twelve logical qubits in one block are useful only if an architecture can prepare, measure, couple, and route the desired logical combinations without erasing the memory advantage.」

BB adds one more requirement. One surface-code patch holds one logical qubit, so “apply H to this patch” is “apply H to this logical qubit”. One 144-qubit gross block houses 12 logical qubits; a physical operation on the whole block acts on some combination of the 12. The QEC-Bank text: “The packed value k = 12 makes addressability especially important. A collective operation on all twelve logical qubits is not a substitute for a compiler that selects the desired logical combination while preserving the others.” Plainly: an operation that “does something to all 12 at once” is not “H on logical 3 only”; the latter is called addressabilityThe ability to apply a gate to one or a few chosen logical qubits without disturbing the rest in the same block. A k = 12 block must answer this; a k = 1 patch need not.. intuition.md lists it among the failure modes: “Twelve logical qubits in one block are useful only if an architecture can prepare, measure, couple, and route the desired logical combinations without erasing the memory advantage.”

logical-operations.md 的第二句给出判定标准：「The relevant question is not ‘does the code have symmetries?’ but ‘which physical process preserves the code, what logical symplectic action does it induce, and what fault-tolerance and routing resources certify that action?’」拆成三问。① 哪个物理过程保持码：具体的一层物理门或一次置换，作用后每条稳定子仍是稳定子（可能变成另一条）。② 它诱导什么逻辑辛作用：辛作用symplectic action：一个 Clifford 操作把每个逻辑 Pauli 算符（X̄₁…X̄_k、Z̄₁…Z̄_k）送到哪个逻辑 Pauli 算符，写成一个 2k × 2k 的 0/1 矩阵。k = 12 时是 24 × 24。「辛」指它保持对易/反对易关系。是这个操作对 2k 个逻辑 Pauli 生成元各做了什么，写成一个 2k × 2k 的 0/1 矩阵——gross 是 24 × 24。人话：知道「码空间被映回自身」还不够，要知道里面的 12 个比特各被怎样搬动了。③ 什么容错与布线资源认证它：这个过程在真实噪声下是否仍是容错的，以及硬件上要怎样连线才能做。容错的意思是：过程中出的故障只要不超过允许的数目（最基本的要求：任意一处出一个物理故障），每个输出块里留下的错误都还在该块译码器能纠正的范围内；哪怕只造成一个逻辑错误，也算失败。三问答完，QEC-Bank 才写一条记录。在 QEC-Bank 里，第 3 章的平移停在第 ② 问：QEC-Bank 没有为 gross 登记它们的逻辑作用（文献里有答案，见 3.2.2 节）。

The second sentence of logical-operations.md gives the criterion: “The relevant question is not ‘does the code have symmetries?’ but ‘which physical process preserves the code, what logical symplectic action does it induce, and what fault-tolerance and routing resources certify that action?’” Three questions. (1) Which physical process preserves the code: a concrete layer of physical gates or a permutation after which every stabilizer is still a stabilizer (possibly a different one). (2) What logical symplectic action it induces: the symplectic actionWhere a Clifford operation sends each logical Pauli operator (X̄₁…X̄_k, Z̄₁…Z̄_k), written as a 2k × 2k binary matrix; 24 × 24 for k = 12. “Symplectic” means it preserves commutation relations. is what the operation does to each of the 2k logical Pauli generators, written as a 2k × 2k binary matrix, 24 × 24 for gross. Plainly: knowing “the code space maps to itself” is not enough; you must know how each of the 12 qubits inside was moved. (3) What fault-tolerance and routing resources certify it: whether the process stays fault tolerant under real noise, and how the hardware must be wired to perform it. Fault tolerant means: as long as the faults during the process stay within the allowed number (the basic requirement: any single physical fault anywhere), every output block is left with errors that its decoder can still correct; causing even one logical error counts as failure. Only when all three are answered does the QEC-Bank write a record. In the QEC-Bank, the translations of Chapter 3 stop at question (2): the QEC-Bank records no logical action for them on gross (the literature has answers, see section 3.2.2).

下表逐字转录 card.yaml 的 operations 字段（五条断言 atlas-assertion:bivariate_bicycle:operations:0…4）。每条有四个字段：gate（门）、mechanism（机制）、level（层级）、cost（成本），以及 origin（证据词 + 引文）。五条的 level 全是 Clifford，origin 全是 claimed——引自文献、QEC-Bank 的机器没有重算任何一条的逻辑作用；applies_to 全是 eczoo-entity:qcga#collection，即描述 BB 这一族而非 gross 一块。成本一栏是本篇最要紧的原文，后面各章逐条展开。

The table transcribes the operations field of card.yaml verbatim (the five assertions atlas-assertion:bivariate_bicycle:operations:0…4). Each has four fields, gate, mechanism, level and cost, plus origin (evidence word + citation). All five have level Clifford and origin claimed: cited from the literature, with the QEC-Bank machine recomputing the logical action of none of them; applies_to is eczoo-entity:qcga#collection throughout, i.e. the BB family rather than the single gross block. The cost column is the most important text in this guide, and the following chapters unfold it row by row.

QEC-Bank 原文的设定：「Take two identical BB blocks, labelled control c and target t, and apply physical CNOT from qubit i of c to qubit i of t for every i.」CNOT 是 Clifford 门，它把 Pauli 算符变成 Pauli 算符（《格点手术 · 高中数学版》5.2 节的 Heisenberg 图景：不跟踪态，只跟踪错误和算符怎样穿过门）。「共轭」就是问：门前的一个 Pauli，挪到门后变成什么。规则只有四条（补课 2.A 在基态上推一遍）：

The QEC-Bank set-up: “Take two identical BB blocks, labelled control c and target t, and apply physical CNOT from qubit i of c to qubit i of t for every i.” CNOT is a Clifford gate and maps Pauli operators to Pauli operators (the Heisenberg picture in section 5.2 of “Lattice Surgery · High-School Edition”: instead of following the state, follow how errors and operators pass through the gates). “Conjugation” asks: what does a Pauli in front of the gate become once moved behind it? There are only four rules (catch-up 2.A derives them on basis states):

现在把一条 gross 的 X 检查放进去。控制块的 X 检查 0 是 X 作用在比特 {1, 2, 18, 75, 78, 84} 上（参数导读 1.3 节转录的 reproduce.py 断言）。这六个比特各自与目标块的同号比特做了一次 CNOT，按第一条规则，每个 X_c 都变成 X_cX_t：结果是「控制块 {1, 2, 18, 75, 78, 84} 上的 X」乘「目标块 {1, 2, 18, 75, 78, 84} 上的 X」——也就是控制块的 X 检查 0 乘以目标块的 X 检查 0，两者都是稳定子，乘积仍是稳定子。目标块的 X 检查 0 按第二条规则不动。Z 一侧对称：控制块的 Z 检查 0（比特 {3, 60, 66, 76, 77, 126}）按第三条规则不动；目标块的 Z 检查 0 按第四条规则变成自己乘控制块的 Z 检查 0。QEC-Bank 的原话就是这四句：「an X stabilizer on the control block acquires the matching X stabilizer on the target; an X stabilizer on the target stays fixed. Dually, a target Z stabilizer acquires the matching control Z stabilizer, while a control Z stabilizer stays fixed. Every image remains in the two-block stabilizer group.」这里用到了「两块相同」：目标块的 X 检查 0 恰好也落在 {1, 2, 18, 75, 78, 84} 上，复印过去的 X 才刚好是一条现成的检查。两块不同，复印过去的图形一般不是目标块的稳定子，码就被破坏了。

Now feed in one gross X check. X check 0 of the control block is X on qubits {1, 2, 18, 75, 78, 84} (the reproduce.py assertion transcribed in section 1.3 of the parameters guide). Each of these six qubits did one CNOT with the same-numbered target qubit, so by the first rule every X_c becomes X_cX_t: the result is “X on control {1, 2, 18, 75, 78, 84}” times “X on target {1, 2, 18, 75, 78, 84}”, that is control X check 0 times target X check 0, both stabilizers, so the product is a stabilizer. Target X check 0 stays fixed by the second rule. The Z side mirrors it: control Z check 0 (qubits {3, 60, 66, 76, 77, 126}) stays fixed by the third rule; target Z check 0 becomes itself times control Z check 0 by the fourth. The QEC-Bank says exactly these four things: “an X stabilizer on the control block acquires the matching X stabilizer on the target; an X stabilizer on the target stays fixed. Dually, a target Z stabilizer acquires the matching control Z stabilizer, while a control Z stabilizer stays fixed. Every image remains in the two-block stabilizer group.” This used “two identical blocks”: target X check 0 also lies exactly on {1, 2, 18, 75, 78, 84}, so the copied X is precisely an existing check. With two different blocks the copied pattern is in general not a stabilizer of the target, and the code is broken.

逻辑算符同理。QEC-Bank：「The same conjugation applies to corresponding logical Pauli pairs, giving the parallel logical CNOT action.」取控制块第 i 个逻辑比特的 X̄_i（某个与全部 Z 检查对易、又不是 X 检查乘积的 X 串，BB 教程 4.4 节），第一条规则把它变成 X̄_i^cX̄_i^t；Z̄_i^t 变成 Z̄_i^cZ̄_i^t。这正是「逻辑比特 i（控制）对逻辑比特 i（目标）做 CNOT」的四条特征变换（《格点手术》5.2 节列过），对 i = 1 到 12 同时成立：一层 144 个物理 CNOT = 12 个并行的逻辑 CNOT。这就是 《格点手术》1.2 节说「横向 CNOT 对任何两块相同的 CSS 码成立」的证明——CSS 保证 X 检查只含 X、Z 检查只含 Z，四条规则才能分别套用。

Logical operators follow the same way. QEC-Bank: “The same conjugation applies to corresponding logical Pauli pairs, giving the parallel logical CNOT action.” Take the X̄_i of logical qubit i on the control block (some X string commuting with every Z check yet not a product of X checks, BB guide section 4.4); the first rule turns it into X̄_i^cX̄_i^t, and Z̄_i^t becomes Z̄_i^cZ̄_i^t. These are exactly the four characteristic transformations of “CNOT from logical i (control) to logical i (target)” (listed in section 5.2 of “Lattice Surgery”), holding for i = 1 to 12 simultaneously: one layer of 144 physical CNOTs = 12 parallel logical CNOTs. This is the proof behind the statement in section 1.2 of “Lattice Surgery” that “transversal CNOT works for any two identical CSS codes”: CSS guarantees that X checks contain only X and Z checks only Z, so the four rules can be applied separately.

记录 0 的成本原文：「one aligned pairwise physical-CNOT layer; module transport, alignment, and correlated decoding are additional architecture costs」。分号前是门本身：一层对齐的、逐对的物理 CNOT。每个门只碰控制块的一个比特和目标块的一个比特，这就是「横向」；这 144 个门两两不共用比特，可以同时做，所以这个构造的深度是 1（横向本身并不要求深度 1）。分号后是三笔另算的账。模块搬运与对齐：144 对比特要同时两两相邻或相连。表面码那篇1.2 节说过，可移动的中性原子阵列能把两块码叠到一起，固定布线的超导芯片做不到；gross 的两块各 144 个数据比特，还各带 144 个帮手比特，对齐的工程量随之放大。相关译码：CNOT 把控制块的 X 错误复印到目标块（规则一），把目标块的 Z 错误复印回控制块（规则四）——门做完，两块的错误不再独立，译码器必须把两块 288 个数据比特的症状一起译（译码器导读 4.4 节操作合同里「whether X/Z correlations … are present」这一项在此变成「块间相关」）。QEC-Bank 的总结：「This is a code-space statement. A hardware implementation still needs aligned pairs, a fault model for correlated two-block errors, and a decoder that handles them.」人话：2.1 节证的是「码空间层面」的事实；在真机上，它还要一份两块联动的错误模型和一个会处理这种联动的译码器。引文 yoder2025tour（据文献扫描报告：Yoder 等，“Tour de Gross: A Modular Quantum Computer Based on Bivariate Bicycle Codes”，2025，arXiv 2506.03094，登记待补）「embeds this primitive inside a gross-family modular architecture」——把这个原语嵌进一个以 gross 家族为单元的模块化架构里。

Record 0's cost, verbatim: “one aligned pairwise physical-CNOT layer; module transport, alignment, and correlated decoding are additional architecture costs”. Before the semicolon is the gate itself: one layer of aligned, pairwise physical CNOTs. Each gate touches one qubit of the control block and one of the target block, which is what makes it transversal; the 144 gates share no qubits, so they can run at once, and this construction has depth 1 (transversality by itself does not demand depth 1). After the semicolon are three separately billed items. Module transport and alignment: 144 pairs of qubits must be simultaneously adjacent or connected. Section 1.2 of the surface-code guide noted that movable neutral-atom arrays can stack two code blocks while fixed-wiring superconducting chips cannot; two gross blocks have 144 data qubits each plus 144 helper qubits each, and the alignment engineering scales with them. Correlated decoding: the CNOT copies control-block X errors onto the target (rule one) and target-block Z errors back onto the control (rule four); after the gate the errors of the two blocks are no longer independent, and the decoder must decode the syndromes of both blocks' 288 data qubits jointly (the operational-contract item “whether X/Z correlations … are present” of section 4.4 of the decoders guide becomes “inter-block correlations” here). The QEC-Bank sums up: “This is a code-space statement. A hardware implementation still needs aligned pairs, a fault model for correlated two-block errors, and a decoder that handles them.” Plainly: section 2.1 proved a fact “at the code-space level”; on a real machine it still needs a two-block error model and a decoder that handles it. The citation yoder2025tour (per the scout report: Yoder et al., “Tour de Gross: A Modular Quantum Computer Based on Bivariate Bicycle Codes”, 2025, arXiv 2506.03094, registration pending) “embeds this primitive inside a gross-family modular architecture”.

横向 CNOT 的效果是「第 i 个逻辑比特（控制块）→ 第 i 个逻辑比特（目标块），i = 1…12，同时」。它做不到三件事。第一，块内的门：控制块第 3 个与第 7 个逻辑比特之间的 CNOT，它没有任何机制。第二，挑选：只对第 3 对做 CNOT、其余 11 对不做——144 个物理 CNOT 是一起做的，拆不开。第三，任何单比特逻辑门：H、S 都不在这条记录里。QEC-Bank 的卡片承诺表（第 5 章 5.4 节）把记录 0 的边界写成：承诺「aligned pairwise CNOT preserves two BB blocks」，不可推出「zero routing or decoder cost」。所以横向 CNOT 是 BB 门集里的第一块砖，不是全部；第 3、4 章讨论的自同构与手术，正是为了补「块内」与「挑选」这两个缺口。

The transversal CNOT's effect is “logical i (control block) → logical i (target block), for i = 1…12, simultaneously”. Three things it cannot do. First, gates within a block: a CNOT between logical 3 and logical 7 of the control block has no mechanism here. Second, selection: a CNOT on pair 3 only, leaving the other 11 pairs alone; the 144 physical CNOTs are applied together and cannot be separated. Third, any single-qubit logical gate: neither H nor S is in this record. The card's promise table (section 5.4 in Chapter 5) states record 0's boundary as: promise “aligned pairwise CNOT preserves two BB blocks”, excluded inference “zero routing or decoder cost”. So the transversal CNOT is the first brick of the BB gate set, not the whole of it; the automorphisms and surgery of Chapters 3 and 4 exist precisely to fill the “within-block” and “selection” gaps.

BB 教程 2.2 节的两个平移 x、y 是 72 × 72 的置换矩阵；A、B、Aᵀ、Bᵀ 都是 x、y 的多项式，所以都与 x、y 对易。QEC-Bank 原文：「Let P be either torus translation x or y, and apply P to both data blocks. Since A, B, Aᵀ, Bᵀ all commute with P, H_X diag(P,P) = P H_X, H_Z diag(P,P) = P H_Z. Left multiplication by P merely permutes check rows. Thus the physical qubit permutation maps the stabilizer group to itself. reproduce.py checks these exact matrix identities for the gross instance.」人话：右边乘 diag(P,P) 是「把左层和右层的比特都按 P 重新编号」，左边乘 P 是「把检查的行重新编号」；两者相等，说明比特挪过之后得到的检查集合，只是原来的检查集合换了个顺序。reproduce.py 第 116–120 行对 x 和 y 各断言了这两条恒等式（模 2）。

The two translations x, y of BB guide section 2.2 are 72 × 72 permutation matrices; A, B, Aᵀ, Bᵀ are polynomials in x, y and therefore commute with them. The QEC-Bank text: “Let P be either torus translation x or y, and apply P to both data blocks. Since A, B, Aᵀ, Bᵀ all commute with P, H_X diag(P,P) = P H_X, H_Z diag(P,P) = P H_Z. Left multiplication by P merely permutes check rows. Thus the physical qubit permutation maps the stabilizer group to itself. reproduce.py checks these exact matrix identities for the gross instance.” Plainly: multiplying by diag(P,P) on the right “renumbers the qubits of both layers by P”, multiplying by P on the left “renumbers the check rows”; their equality says the set of checks obtained after moving the qubits is the original set in a different order. Lines 116–120 of reproduce.py assert both identities (mod 2) for x and for y.

到实验台上看这件事。预设 gross，点格子 (0,0) 的 X 检查（圆点），记下六条腿的位置；再点右边相邻格子 (1,0) 的 X 检查：同一个六腿图形整体右移一格，读数里的比特编号各加 6。往上点 (0,1) 的检查是 y 的效果，编号各加 1。x 的 12 个幂与 y 的 6 个幂组合出 12 × 6 = 72 个平移（这是 BB 教程说的 x¹² = y⁶ = 1 的直接推论），每一个都把 72 条 X 检查置换成 72 条 X 检查、72 条 Z 检查置换成 72 条 Z 检查。

See it on the bench. Preset gross, click the X check (circle) of site (0,0) and note the positions of its six legs; then click the X check of the neighbouring site (1,0) to the right: the same six-leg figure shifted one site right, with every qubit index in the readout increased by 6. Clicking the check of (0,1) above shows the effect of y, indices increased by 1. The 12 powers of x combined with the 6 powers of y give 12 × 6 = 72 translations (a direct consequence of x¹² = y⁶ = 1 in the BB guide), each permuting the 72 X checks among themselves and the 72 Z checks among themselves.

QEC-Bank 紧接着说：「This proves code-space preservation, not usefulness.」保持码空间只回答了第 1 章三问里的第 ①问。要认证一个逻辑门，QEC-Bank 列出五步，原文照录：

The QEC-Bank continues: “This proves code-space preservation, not usefulness.” Preserving the code space answers only question (1) of Chapter 1's three. To certify a logical gate the QEC-Bank lists five steps, verbatim:

做完第 4 步，可能得到三种结果，QEC-Bank 原文：「The action may be trivial, may permute logicals, or may implement a coupled Clifford.」人话：平移对 12 个逻辑比特可能什么也没做（24 × 24 是单位矩阵——逻辑信息对平移「看不见」；平移不产生符号，见补课 1.A，所以这里单位矩阵就是真的什么也没做）；可能把逻辑比特换了位置（第 3 个变成第 7 个之类）；也可能是一个把几个逻辑比特缠在一起的 Clifford 门。哪一种，要算了才知道。结论一句：「The card does not call translations logical gates because the gross action has not been certified there.」intuition.md 的第五道自测题问的就是这个：「If translations preserve the stabilizers, are they already twelve addressable logical gates?」答：「No. Stabilizer preservation proves only that the code space maps to itself. One must compute the induced action on a logical Pauli basis; it may be trivial, may mix several logical qubits, and may be expensive to route. The card records only operations with a cited certificate and explicit applicability.」这就是「自同构 ≠ 门」的全部内容：对 gross，QEC-Bank 只登记了第 ① 问的答案；第 ②③ 问的答案在文献里（3.2.2 节），QEC-Bank 没有复现，本篇也只复现了第 ② 问。

After step 4 there are three possible outcomes, in the QEC-Bank's words: “The action may be trivial, may permute logicals, or may implement a coupled Clifford.” Plainly: the translation may do nothing at all to the 12 logical qubits (the 24 × 24 matrix is the identity; the logical information “does not see” the shift; translations produce no signs, see Catch-up 1.A, so here the identity matrix really means nothing happens); it may swap logical qubits around (logical 3 becomes logical 7, say); or it may be a Clifford gate entangling several logical qubits. Which one, you only know after computing. The conclusion in one sentence: “The card does not call translations logical gates because the gross action has not been certified there.” The fifth litmus question of intuition.md asks exactly this: “If translations preserve the stabilizers, are they already twelve addressable logical gates?” Answer: “No. Stabilizer preservation proves only that the code space maps to itself. One must compute the induced action on a logical Pauli basis; it may be trivial, may mix several logical qubits, and may be expensive to route. The card records only operations with a cited certificate and explicit applicability.” That is the whole content of “automorphism ≠ gate”: for gross the QEC-Bank records only the answer to question (1); answers to questions (2) and (3) are in the literature (section 3.2.2), which the QEC-Bank has not reproduced, and this guide reproduces only question (2).

前四步可以在最小的 BB 实例上完全手算：toric_L3（A = 1 + x，B = 1 + y，3 × 3 环面，18 个比特，k = 2）。编号同实验台：左层比特 L(i, j) 的编号是 i·3 + j，右层比特 R(i, j) 的编号是 9 + i·3 + j。先把检查的腿读出来（实验台选 toric_L3 可以逐条点出来核对；坐标都模 3）：X 检查 (i, j) 碰 L(i, j)、L(i+1, j)、R(i, j)、R(i, j+1)；Z 检查 (i, j) 碰 L(i, j)、L(i, j−1)、R(i, j)、R(i−1, j)。

The first four steps can be done entirely by hand on the smallest BB instance: toric_L3 (A = 1 + x, B = 1 + y, 3 × 3 torus, 18 qubits, k = 2). Numbering as in the lab bench: the left-layer qubit L(i, j) has index i·3 + j, the right-layer qubit R(i, j) has index 9 + i·3 + j. First read off the legs of the checks (choose toric_L3 in the lab bench and click them one by one to check; all coordinates mod 3): X check (i, j) touches L(i, j), L(i+1, j), R(i, j), R(i, j+1); Z check (i, j) touches L(i, j), L(i, j−1), R(i, j), R(i−1, j).

第 1 步，选逻辑基。k = 2，要两对代表元。X̄₁ = 左层竖列 L(0, 0)、L(0, 1)、L(0, 2) 上的 X（比特 {0, 1, 2}）；X̄₂ = 右层横行 R(0, 0)、R(1, 0)、R(2, 0) 上的 X（比特 {9, 12, 15}）；Z̄₁ = 左层横行 L(0, 0)、L(1, 0)、L(2, 0) 上的 Z（比特 {0, 3, 6}）；Z̄₂ = 右层竖列 R(0, 0)、R(0, 1)、R(0, 2) 上的 Z（比特 {9, 10, 11}）。它们要满足三件事，每件都能数出来。(a) 骗过对方的检查：Z 检查读的是它与 X 串重叠个数的奇偶。Z 检查 (0, j) 碰 L(0, j) 和 L(0, j−1)，两个都在 X̄₁ 里，重叠 2；其余 Z 检查不碰 X̄₁，重叠 0；所以 X̄₁ 让全部 Z 检查读 0。另外三个照同样的办法数。(b) 配成对：X̄₁ 与 Z̄₁ 只共用比特 0（1 个，奇数，反对易），X̄₂ 与 Z̄₂ 只共用比特 9；X̄₁ 与 Z̄₂、X̄₂ 与 Z̄₁ 不共用比特（对易）。(c) 不是检查的乘积：Z̄₁ 骗过全部 X 检查，即与每条 X 检查重叠偶数个比特，于是与任何几条 X 检查的乘积也重叠偶数个；X̄₁ 与 Z̄₁ 重叠 1 个，所以 X̄₁ 不可能是 X 检查的乘积。X̄₂ 用 Z̄₂ 作证，Z̄₁、Z̄₂ 分别用 X̄₁、X̄₂ 作证，道理相同。

Step 1, choose a logical basis. k = 2, so two pairs of representatives are needed. X̄₁ = X on the left-layer column L(0, 0), L(0, 1), L(0, 2) (qubits {0, 1, 2}); X̄₂ = X on the right-layer row R(0, 0), R(1, 0), R(2, 0) (qubits {9, 12, 15}); Z̄₁ = Z on the left-layer row L(0, 0), L(1, 0), L(2, 0) (qubits {0, 3, 6}); Z̄₂ = Z on the right-layer column R(0, 0), R(0, 1), R(0, 2) (qubits {9, 10, 11}). They must satisfy three things, each of which can be counted. (a) Pass the other kind of check: a Z check reads the parity of its overlap with an X string. Z check (0, j) touches L(0, j) and L(0, j−1), both in X̄₁, overlap 2; the other Z checks miss X̄₁, overlap 0; so X̄₁ makes every Z check read 0. Count the other three the same way. (b) Pair up: X̄₁ and Z̄₁ share only qubit 0 (one, odd, anticommute), X̄₂ and Z̄₂ share only qubit 9; X̄₁ and Z̄₂, and X̄₂ and Z̄₁, share no qubit (commute). (c) Not a product of checks: Z̄₁ passes every X check, i.e. overlaps each X check in an even number of qubits, hence overlaps any product of X checks evenly too; X̄₁ overlaps Z̄₁ in one qubit, so X̄₁ cannot be a product of X checks. X̄₂ has Z̄₂ as its witness, and Z̄₁, Z̄₂ have X̄₁, X̄₂, by the same argument.

第 2 步，用 x 共轭。x 把两层的每个格子 (i, j) 都挪到 (i+1, j)，编号加 3（左层在 0…8 里、右层在 9…17 里循环）。第 3 步，模稳定子约化。把平移后的串与原串相加（模 2，两边都有的比特消掉），看这个差能不能写成检查的乘积。四个代表元的结果：

Step 2, conjugate by x. x moves every site (i, j) of both layers to (i+1, j), adding 3 to the index (cycling within 0…8 on the left and 9…17 on the right). Step 3, reduce modulo stabilizers. Add the shifted string to the original (mod 2: qubits in both cancel) and ask whether the difference is a product of checks. The four representatives give:

所以 x·X̄₁ = X̄₁ ×（X 检查 0、1、2），x·Z̄₂ = Z̄₂ ×（Z 检查 3、4、5），另外两个的像就是它们自己：四个像都只与原来的代表元差一个稳定子，代表同一个逻辑算符。第 4 步，拼矩阵。把每个像按 (X̄₁ X̄₂ | Z̄₁ Z̄₂) 写成一行：(1 0 | 0 0)、(0 1 | 0 0)、(0 0 | 1 0)、(0 0 | 0 1)，是 4 × 4 单位矩阵；平移不带符号（补课 1.A），所以 x 在 toric_L3 上真的什么也没做。图上的理由：沿 x 挪一个绕 y 方向的圈（X̄₁、Z̄₂），它还绕着同一个方向，只差几条检查；沿 x 挪一个绕 x 方向的圈（X̄₂、Z̄₁），它原地不动。

So x·X̄₁ = X̄₁ × (X checks 0, 1, 2), x·Z̄₂ = Z̄₂ × (Z checks 3, 4, 5), and the other two images are the representatives themselves: all four images differ from the originals by a stabilizer and represent the same logical operators. Step 4, assemble the matrix. Write each image as a row in the order (X̄₁ X̄₂ | Z̄₁ Z̄₂): (1 0 | 0 0), (0 1 | 0 0), (0 0 | 1 0), (0 0 | 0 1), the 4 × 4 identity; translations carry no signs (Catch-up 1.A), so on toric_L3 x really does nothing. The picture behind it: shift a loop that winds in the y direction (X̄₁, Z̄₂) along x and it still winds the same way, differing by a few checks; shift a loop that winds in the x direction (X̄₂, Z̄₁) along x and it stays where it was.

本篇用一段脚本对 gross 跑了同样的前四步（notes/operations/verify.py；逻辑基由脚本选取，下面的结论与怎么选无关）。结果与 toric_L3 截然不同：x 的 24 × 24 辛作用不是单位矩阵；没有任何一个非平凡的逻辑 Pauli 被 x 保持不变（矩阵减去单位矩阵后秩为 24，满秩）；x 连做 6 次，逻辑作用才回到单位，尽管 x 作为比特的挪动要做 12 次才回到原位。y 的结果相同：不是单位，没有不动的逻辑 Pauli，6 次回到单位（y 本身挪 6 次就回原位）。把 72 个平移逐个算过，只有「不动」和 x⁶ 两个在逻辑上什么也不做，所以 72 个平移恰好给出 36 种不同的逻辑作用。平移只把 X 串搬成 X 串、Z 串搬成 Z 串，所以每一种都是由逻辑 CNOT 组成的线路，不含 H、S。

This guide ran the same first four steps on gross with a script (notes/operations/verify.py; the script picks the logical basis, and the conclusions below do not depend on that choice). The result is unlike toric_L3: the 24 × 24 symplectic action of x is not the identity; no non-trivial logical Pauli is left unchanged by x (the matrix minus the identity has rank 24, full rank); and x must be applied 6 times before its logical action returns to the identity, although as a movement of qubits it takes 12 applications to return. y gives the same: not the identity, no fixed logical Pauli, identity after 6 applications (y itself returns after 6 moves). Checking all 72 translations one by one, only “stay” and x⁶ do nothing logically, so the 72 translations give exactly 36 distinct logical actions. A translation carries X strings to X strings and Z strings to Z strings, so each of these actions is a circuit of logical CNOTs, with no H or S.

第 5 步，文献已经做了。Bravyi 等（arXiv 2308.07915，第 9.2 节与表 8）给出了容错线路：只用症状测量本来就要的连线，把数据比特经过帮手比特搬一圈，一次实现形如 A_jA_kᵀ 或 B_jB_kᵀ 的平移（A_j、B_k 是 A、B 的单项式，也就是一条腿对应的那一步），CNOT 深度 4，途中把帮手比特重置到 |0⟩，错误不会在数据比特之间传开。论文指出，只要 Tanner 图连通，这些平移组合起来就得到全部平移（gross 的 72 个）。他们的用途是寻址：平移把逻辑算符的整张网格挪动，配合论文第 9.4 节的辅助测量系统，就能选定测哪一个逻辑比特；作为门，他们写道这些 CNOT 线路「怎样用于有用的计算并不清楚」。所以本章说「平移不是门」，准确的范围是：QEC-Bank 没有为 gross 登记平移的逻辑作用和容错线路，本篇也没有复现表 8 的线路；并不是说没人做过。它每次都搬动全部 12 个逻辑比特，正是 1.1 节说的「集体操作不等于可寻址」；在 Bravyi 等的方案里，它是寻址的工具，不是单独作用在某个逻辑比特上的门。

Step 5 has been done in the literature. Bravyi et al. (arXiv 2308.07915, section 9.2 and Table 8) give fault-tolerant circuits: using only the connections that syndrome measurement already needs, they move the data qubits round a loop through the helper qubits and implement in one go a translation of the form A_jA_kᵀ or B_jB_kᵀ (A_j, B_k are monomials of A and B, i.e. the step behind one leg), with CNOT depth 4, resetting the helper qubits to |0⟩ along the way so that errors cannot spread between data qubits. The paper notes that whenever the Tanner graph is connected these translations combine into all translations (the 72 of gross). Their use is addressing: a translation shifts the whole grid of logical operators, and together with the ancilla measurement system of the paper's section 9.4 it selects which logical qubit gets measured; as gates, they write that “it is not clear how to use these CNOT circuits to facilitate useful computations”. So when this chapter says “a translation is not a gate”, the exact scope is: the QEC-Bank records no logical action and no fault-tolerant circuit for gross translations, and this guide has not reproduced the circuits of Table 8; it does not mean nobody has done it. Each translation moves all 12 logical qubits at once, which is exactly the “a collective operation is not addressability” of section 1.1; in the scheme of Bravyi et al. it is a tool for addressing, not a gate acting on one chosen logical qubit.

有一处与文献对不上：Bravyi 等第 9.2 节写道，除了「不动」，每个平移的逻辑作用都不是恒等（“A transformation of this form cannot act like the logical identity”）；本篇的脚本却算得 x⁶ 在 gross 上的逻辑作用是恒等（12 个 X 型、12 个 Z 型基元素逐一核对：平移后与原来相同，或只差若干条检查）。本篇没有找到原因，记为待查；上面其余结论不依赖这一点。本节第一段和这一段的数字是本篇自己算的，不是 QEC-Bank 的证书；sayginel2025automorphisms 报告过 gross 的逻辑作用群，本篇没有与它逐项对照。

One point disagrees with the paper: section 9.2 of Bravyi et al. says that apart from “stay”, no translation acts as the logical identity (“A transformation of this form cannot act like the logical identity”); this guide's script finds that x⁶ acts as the logical identity on gross (checked element by element for the 12 X-type and 12 Z-type basis elements: after the shift each equals the original, or differs from it by a few checks). This guide has not found the reason and lists it as open; the other conclusions above do not depend on it. The numbers in this section's first paragraph and in this one are this guide's own computation, not a QEC-Bank certificate; sayginel2025automorphisms reports the logical action group of gross, and this guide has not compared the two entry by entry.

记录 1 的门叫「selected logical Clifford generators」，机制 fold-transversal，引文 eberhardt2024fold（据文献扫描报告：Eberhardt 与 Steffan，“Logical Operators and Fold-Transversal Gates of Bivariate Bicycle Codes”，2024，arXiv 2407.03973，登记待补）。QEC-Bank 的解释：「A fold pairs qubits related by a reflection or other involution so that constant-depth physical Cliffords implement a logical operation.」人话：找一个把环面「对折」的对称（反射之类，做两次回到原处的那种），把折线两侧对应的比特配成对，对每一对做同一个两比特 Clifford 门（或对折线上的比特做单比特门）——深度仍是常数，但门不再是「第 i 对第 i」，而是「第 i 对它的镜像」。文献为对称的 BB 例子构造了显式的逻辑基与这样的门，QEC-Bank 点名两个：[[98,6,12]] 与 [[162,8,12]]——注意这两个都不是 QEC-Bank 的 BB 实例，也不在 Table 3 里。

Record 1's gate is “selected logical Clifford generators”, mechanism fold-transversal, citation eberhardt2024fold (per the scout report: Eberhardt and Steffan, “Logical Operators and Fold-Transversal Gates of Bivariate Bicycle Codes”, 2024, arXiv 2407.03973, registration pending). The QEC-Bank explanation: “A fold pairs qubits related by a reflection or other involution so that constant-depth physical Cliffords implement a logical operation.” Plainly: find a symmetry that “folds” the torus (a reflection or similar, one that returns you home when done twice), pair the qubits mirrored across the fold line, and apply the same two-qubit Clifford to every pair (or a single-qubit gate on the qubits on the line); the depth is still constant, but the gate is no longer “i with i” but “i with its mirror image”. The paper constructs explicit logical bases and such gates for symmetric BB examples, of which the QEC-Bank names two: [[98,6,12]] and [[162,8,12]]; note that neither is a QEC-Bank BB instance, nor in Table 3.

证书有三部分，QEC-Bank 原文：「the physical permutation is an automorphism of the chosen check spaces; its induced logical Pauli action is computed; the folded interaction pattern is available or routable.」正是第 1 章的三问。成本原文：「a fold/permutation plus constant-depth physical Clifford layers; only symmetric BB instances with a resolved logical action qualify」。分号后是这条记录的边界：只有具备那种对称、且逻辑作用已经算出的 BB 实例才算。QEC-Bank 把它说成一句方法论：「The extra symmetry is a code-selection condition. These examples do not imply that the original [[144,12,12]] gross block has the same fold gate.」人话：这不是「BB 码有折叠门」，而是「挑出有某种对称的 BB 码，它们有折叠门」；gross 有没有，要对 gross 单独做那五步，文献没有替它做。

The certificate has three parts, in the QEC-Bank's words: “the physical permutation is an automorphism of the chosen check spaces; its induced logical Pauli action is computed; the folded interaction pattern is available or routable.” Exactly Chapter 1's three questions. The cost, verbatim: “a fold/permutation plus constant-depth physical Clifford layers; only symmetric BB instances with a resolved logical action qualify”. After the semicolon is this record's boundary: only BB instances that possess that symmetry and whose logical action has been computed qualify. The QEC-Bank puts it as a rule of method: “The extra symmetry is a code-selection condition. These examples do not imply that the original [[144,12,12]] gross block has the same fold gate.” Plainly: this is not “BB codes have fold gates” but “select BB codes with a certain symmetry, and those have fold gates”; whether gross does requires running the five steps on gross itself, which the paper did not do for it.

记录 2 的门叫「logical Clifford operations through automorphisms」，机制在卡片上同样标为 fold-transversal，引文 sayginel2025automorphisms（Sayginel、Koutsioumpas、Webster、Rajput、Browne，“Fault-Tolerant Logical Clifford Gates from Code Automorphisms”，2025，arXiv 2409.18175，已登记完整元数据）。QEC-Bank：「sayginel2025automorphisms generalizes the workflow: find Tanner-graph automorphisms, resolve their logical symplectic actions, and compile useful Cliffords. Again, graph preservation without the logical action is only the first step.」人话：不限于反射和平移，把 Tanner 图的所有自同构（把比特换比特、检查换检查而保持连线的置换）都搜出来，对每一个做 3.2 节的五步算出 24 × 24 的辛作用，再从这些作用里拼出有用的 Clifford 门。这是把「对称 → 门」变成一条可以对任何码执行的流水线。同一组作者还用自同构做译码（译码器导读 3.5 节的 koutsioumpas2025autdec）：同一批对称，一边用来搬逻辑比特，一边用来换译码帧。

Record 2's gate is “logical Clifford operations through automorphisms”, its mechanism likewise labelled fold-transversal on the card, citation sayginel2025automorphisms (Sayginel, Koutsioumpas, Webster, Rajput, Browne, “Fault-Tolerant Logical Clifford Gates from Code Automorphisms”, 2025, arXiv 2409.18175, full metadata registered). QEC-Bank: “sayginel2025automorphisms generalizes the workflow: find Tanner-graph automorphisms, resolve their logical symplectic actions, and compile useful Cliffords. Again, graph preservation without the logical action is only the first step.” Plainly: not just reflections and translations; search out every automorphism of the Tanner graph (permutations swapping qubits with qubits and checks with checks while preserving the edges), run the five steps of section 3.2 on each to get its 24 × 24 symplectic action, and then compile useful Clifford gates from those actions. This turns “symmetry → gate” into a pipeline that can be run on any code. The same group also uses automorphisms for decoding (koutsioumpas2025autdec in section 3.5 of the decoders guide): the same symmetries, used once to move logical qubits and once to change decoding frames.

成本原文：「a certified Tanner-graph permutation and its physical routing; the induced symplectic action must be computed for the chosen block」。「for the chosen block」四个字是重点：流程是一般的，结果是逐块的——对 gross 跑一遍才有 gross 的门表；QEC-Bank 没有替 gross 认证这一步，卡片的用词是「未在此认证」，不是「无人算过」（sayginel2025automorphisms 就报告了 gross 的逻辑作用群）。卡片承诺表把记录 2 的边界写成：承诺「a resolved logical action can implement Cliffords」，不可推出「every graph symmetry is useful」（第 4 步可能算出单位矩阵）。

The cost, verbatim: “a certified Tanner-graph permutation and its physical routing; the induced symplectic action must be computed for the chosen block”. The words “for the chosen block” are the point: the workflow is general, the results are per block; only running it on gross yields gross's gate table; the QEC-Bank has not certified that step for gross — the card says “not certified there”, not “never computed” (sayginel2025automorphisms does report gross's logical-action group). The card's promise table states record 2's boundary as: promise “a resolved logical action can implement Cliffords”, excluded inference “every graph symmetry is useful” (step 4 may return the identity matrix).

《格点手术 · 高中数学版》第 2 章把核心动作算透了：测 Z̄_AZ̄_B 只区分「两个逻辑比特相同 / 不同」两组振幅，组内比例原样保留（2.1 节的四个振幅）；第 3 章用一排局域的缝合检查把这个非局域算符测出来（3.2 节：缝合比特成对抵消，乘积恰好是 Z̄_AZ̄_B）；第 5 章证明两次这样的测量加一次读出等于一个 CNOT，读数记入 Pauli 帐；第 6 章说明为什么要维持 d 轮。这些代数一个字不用改。先在两比特实验台上重做一次「测 ZZ」，把「学到奇偶、不学到内容」找回手感：

Chapter 2 of “Lattice Surgery · High-School Edition” worked the core move out in full: measuring Z̄_AZ̄_B distinguishes only the “two logicals equal / different” groups of amplitudes and preserves the ratios within each group (the four amplitudes of section 2.1); Chapter 3 measures this non-local operator with a row of local seam checks (section 3.2: seam qubits cancel in pairs, the product is exactly Z̄_AZ̄_B); Chapter 5 proves that two such measurements plus one readout equal a CNOT, with outcomes booked in the Pauli frame; Chapter 6 explains why it must be held for d rounds. None of that algebra changes by a letter. First redo “measure ZZ” on the two-qubit bench to recover the feel of “learn the parity, not the content”:

BB 上的版本，QEC-Bank 原文：「Long-range surgery uses an ancilla code or patch to measure joint logical Paulis between data blocks. Repeated syndrome extraction makes the parity measurement fault tolerant; classical feed-forward completes teleportation or Clifford gadgets.」三句对应三件事：辅助码（表面码那篇的缝合线与辅助贴片，在这里是一块单独的辅助码或贴片）、重复测量（维持若干轮，让读数容错）、经典前馈（按读数记账或补一个 Pauli，完成隐形传态或 Clifford 门）。「long-range」按原文说的是辅助系统在「data blocks」之间测量：表面码的两块贴片沿边界相邻，缝合线是局域的；BB 架构里要连起来的是两块各自独立的码块，可能是相距很远的模块，辅助系统要跨过这段距离。文献：cohen2022longrange（据文献扫描报告：Cohen、Kim、Bartlett、Brown，“Low-Overhead Fault-Tolerant Quantum Computing Using Long-Range Connectivity”，2022，arXiv 2110.10794，登记待补）「develops the general infrastructure for high-rate codes」；yoder2025tour「supplies a gross-family instruction set, compilation path, and resource estimates for distance-12 and distance-18 modules」——为 d = 12（gross）与 d = 18（[[288,12,18]]）两种模块给出指令集、编译路径与资源估计。

The BB version, in the QEC-Bank's words: “Long-range surgery uses an ancilla code or patch to measure joint logical Paulis between data blocks. Repeated syndrome extraction makes the parity measurement fault tolerant; classical feed-forward completes teleportation or Clifford gadgets.” Three sentences, three things: an ancilla code (the seam and ancilla patch of the surface-code guide, here a separate ancilla code or patch), repeated measurement (held for several rounds so the outcome is fault tolerant), classical feed-forward (book the outcome or apply a Pauli, completing a teleportation or a Clifford gadget). “Long-range”, as the quote says, is about measuring “between data blocks”: two surface-code patches sit side by side along a boundary and the seam is local; in a BB architecture the two things to join are separate code blocks, possibly modules far apart, and the ancilla system has to bridge that distance. Literature: cohen2022longrange (per the scout report: Cohen, Kim, Bartlett, Brown, “Low-Overhead Fault-Tolerant Quantum Computing Using Long-Range Connectivity”, 2022, arXiv 2110.10794, registration pending) “develops the general infrastructure for high-rate codes”; yoder2025tour “supplies a gross-family instruction set, compilation path, and resource estimates for distance-12 and distance-18 modules”, i.e. instruction set, compilation path and resource estimates for the d = 12 (gross) and d = 18 ([[288,12,18]]) modules.

记录 3 的成本原文：「long-range ancilla patches, repeated measurements, and a joint decoder; the resource count belongs to the selected modular protocol」。logical-operations.md 把它展开成五项，原文照录：

Record 3's cost, verbatim: “long-range ancilla patches, repeated measurements, and a joint decoder; the resource count belongs to the selected modular protocol”. logical-operations.md expands it into five items, verbatim:

QEC-Bank 给了标签的理由：「These operations are not transversal properties of one BB check matrix. … The card labels the mechanism surgery so those resources are not hidden under ‘logical Clifford.’」人话：如果只写「Clifford 门，claimed」，读者会以为它像横向 CNOT 那样是一层门；写成 surgery，就提醒你上面五项都要另算。成本字段末句「the resource count belongs to the selected modular protocol」——具体数字（多少辅助比特、多少轮、多少耦合器）属于所选的那个模块化方案（yoder2025tour 给了它的一套），不属于 gross 的矩阵；QEC-Bank 没有登记任何数字，本篇也不给。

The QEC-Bank gives the reason for the label: “These operations are not transversal properties of one BB check matrix. … The card labels the mechanism surgery so those resources are not hidden under ‘logical Clifford.’” Plainly: had it written only “Clifford gate, claimed”, a reader would assume a single gate layer like the transversal CNOT; writing surgery reminds you that all five items above are billed separately. The cost field's last clause, “the resource count belongs to the selected modular protocol”: the concrete numbers (how many ancilla qubits, how many rounds, how many couplers) belong to the chosen modular scheme (yoder2025tour gives one such set), not to gross's matrix; the QEC-Bank registers no number, and this guide supplies none.

表面码一块贴片只有一对 X̄、Z̄，「测 Z̄_AZ̄_B」没有歧义。gross 一块有 12 对，要对第 3 个逻辑比特和另一块的第 7 个做门，先得回答：第 3 个逻辑比特的 Z̄₃ 是哪一串 Z？这正是 3.2 节五步的第 1 步「choose a logical Pauli basis」——静态卡片没有固定它，是架构方案的选择。选定之后，有两件事可以确定地说。

A surface-code patch has one pair X̄, Z̄, so “measure Z̄_AZ̄_B” is unambiguous. A gross block has 12 pairs; to apply a gate between its logical 3 and another block's logical 7 you must first answer: which string of Z's is Z̄₃? This is exactly step 1 of section 3.2, “choose a logical Pauli basis”, which the static card does not fix and the architecture must choose. Once it is chosen, two things can be said for certain.

重量有下界。任何 Z 型逻辑算符的代表元至少碰 12 个比特：码距 d = 12 的意思就是，比 12 个比特更少的 Z 串要么被某条 X 检查发现，要么是 Z 检查的乘积、不改变逻辑信息（参数导读 2.4 节）。所以测 Z̄₃ 的辅助系统至少要接到 12 个数据比特上。

The weight has a lower bound. Every representative of a Z-type logical operator touches at least 12 qubits: distance d = 12 means that a Z string on fewer than 12 qubits is either caught by some X check or is a product of Z checks that leaves the logical information alone (section 2.4 of the parameters guide). So the ancilla system that measures Z̄₃ must connect to at least 12 data qubits.

接口在哪里，由选中的代表元决定。码距和平移对称都不能推出「代表元散布在整个环面上」。图 4.3 是一个具体的例子：Z 作用在 12 个比特 {0, 2, 3, 5, 72, 74, 75, 77, 138, 140, 141, 143} 上。本篇的脚本核对了三件事：它与全部 72 条 X 检查对易（与每条的重叠都是偶数），它不是 Z 检查的乘积，重量 12 等于码距。它只占左层 L(0, j)、右层 R(0, j) 和右层 R(11, j)（j = 0、2、3、5），也就是环面上相邻的两列 i = 11 与 i = 0（i 模 12，11 的右边就是 0）。选它当代表元，接口就是这两列里的 12 个比特：辅助系统的新检查只需碰到它们。至于成本里的「long-range」，按 4.1 节的原文，来自辅助系统要在两块不同的数据块之间测量；两块摆在哪里、接口上要用哪些耦合器，属于所选模块化方案的几何，QEC-Bank 没有登记，本篇也不推。

Where the interface sits is decided by the chosen representative. Neither the distance nor the translation symmetry implies that “representatives are spread over the whole torus”. Fig. 4.3 is a concrete example: Z acting on the 12 qubits {0, 2, 3, 5, 72, 74, 75, 77, 138, 140, 141, 143}. This guide's script checked three things: it commutes with all 72 X checks (even overlap with each), it is not a product of Z checks, and its weight 12 equals the distance. It occupies only the left-layer L(0, j), the right-layer R(0, j) and the right-layer R(11, j) (j = 0, 2, 3, 5), that is, the two adjacent columns i = 11 and i = 0 of the torus (i is taken mod 12, so 0 is the right-hand neighbour of 11). With this representative, the interface is these 12 qubits in two columns: the ancilla's new checks only need to touch them. As for “long-range” in the cost, the quote of section 4.1 places it in measuring between two different data blocks; where the two blocks sit and which couplers the interface uses belong to the geometry of the chosen modular scheme, which the QEC-Bank does not record and this guide does not infer.

同时另外 11 个逻辑比特住在同一批 144 个比特里：手术期间这一整块不能被搬走，它们只能等待，并承受「idle and routing exposure」。以上是本篇对成本字段的解读；QEC-Bank 的原话是第 1 章那句：「A collective operation on all twelve logical qubits is not a substitute for a compiler that selects the desired logical combination while preserving the others.」——挑出想要的组合、同时保住其余的，是编译器的活。卡片承诺表把记录 3 写成：承诺「cited architecture provides joint operations」，不可推出「surgery is native to the static matrix」。

Meanwhile the other 11 logicals live in the same 144 qubits: the whole block cannot be moved away during the surgery, so they can only wait and bear the “idle and routing exposure”. The above is this guide's reading of the cost field; the QEC-Bank's own words are Chapter 1's: “A collective operation on all twelve logical qubits is not a substitute for a compiler that selects the desired logical combination while preserving the others.” Selecting the wanted combination while preserving the rest is the compiler's job. The card's promise table writes record 3 as: promise “cited architecture provides joint operations”, excluded inference “surgery is native to the static matrix”.

记录 4 的门叫「H / S / CNOT in the self-dual BB subfamily」，机制 transversal，引文 liang2025selfdual（据文献扫描报告：Liang 与 Chen，“Self-Dual Bivariate Bicycle Codes with Transversal Clifford Gates”，2025，arXiv 2510.05211，登记待补）。QEC-Bank：「liang2025selfdual imposes extra self-duality and twisted-torus conditions, obtaining weight-eight BB blocks with transversal CNOT, H, and S. Those conditions change the selected family.」人话：在 BB 的构造上再加两个条件——自对偶self-dual（CSS 意义下）：X 检查与 Z 检查有同样的形状，于是把每个比特上的 X、Z 互换（横向 H）会把 X 检查变成 Z 检查、Z 检查变成 X 检查，码空间不变。最简单的例子是 BB 教程 2.4 节的 bicycle 码 H_X = H_Z。liang2025selfdual 的具体条件以原文为准。（X 检查与 Z 检查同形，于是逐比特做 H 把两族检查互换，码不变）和「扭曲环面」（环面的粘合方式带一个扭转，细节见原文）——就得到检查重量 8（不是 gross 的 6）的 BB 块，它们能横向做 CNOT、H、S。成本原文：「constant-depth physical layers on blocks satisfying the extra self-duality and twisted-torus conditions; not a gross-code claim」。gross 不满足自对偶这一条（本篇算过：把 H_X 与 H_Z 叠成一张 144 × 144 的矩阵，秩是 112 而不是 66，两族检查张成的空间不同，逐比特做 H 不会把码映回自身；它的检查重量也是 6 而非 8），所以这条记录明说不是关于 gross 的。文献扫描报告的措辞更直接：「These gates do not transfer to the ordinary gross code merely because both are called BB.」

Record 4's gate is “H / S / CNOT in the self-dual BB subfamily”, mechanism transversal, citation liang2025selfdual (per the scout report: Liang and Chen, “Self-Dual Bivariate Bicycle Codes with Transversal Clifford Gates”, 2025, arXiv 2510.05211, registration pending). QEC-Bank: “liang2025selfdual imposes extra self-duality and twisted-torus conditions, obtaining weight-eight BB blocks with transversal CNOT, H, and S. Those conditions change the selected family.” Plainly: add two conditions on top of the BB construction, self-dualitySelf-dual (in the CSS sense): X checks and Z checks have the same shape, so swapping X and Z on every qubit (transversal H) turns X checks into Z checks and vice versa, leaving the code space unchanged. The simplest example is the bicycle code of BB guide section 2.4 with H_X = H_Z. The exact conditions of liang2025selfdual are as in the paper. (X and Z checks of the same shape, so a qubit-wise H swaps the two families and leaves the code unchanged) and a “twisted torus” (the torus glued with a twist, details in the paper), and you obtain BB blocks of check weight 8 (not gross's 6) that admit transversal CNOT, H and S. The cost, verbatim: “constant-depth physical layers on blocks satisfying the extra self-duality and twisted-torus conditions; not a gross-code claim”. Gross fails the self-duality condition (this guide checked: stacking H_X on H_Z gives a 144 × 144 matrix of rank 112, not 66, so the two check families span different spaces and a qubit-wise H does not map the code to itself; its check weight is also 6, not 8), so the record states explicitly that it is not about gross. The scout report puts it more bluntly: “These gates do not transfer to the ordinary gross code merely because both are called BB.”

文献扫描里还有两条相近的路线，QEC-Bank 记在推导页、没有写成卡片记录。breuckmann2026cups（据扫描报告：Breuckmann、Davydova、Eberhardt、Tantivasadakarn，“Cups and Gates I: Cohomology Invariants and Logical Quantum Operations”，2026，arXiv 2410.16250，登记待补）：「uses cohomology and cup-product invariants to certify constant-depth logical operations on codes with the required structure, including selected BB examples. The cup product is the gate certificate; a generic bivariate polynomial pair does not automatically supply it.」人话：用一种叫杯积cup product：上同调里把两个「类」乘成一个更高阶「类」的运算；在这条路线里，它的非平凡性就是「这个码有某个常深度逻辑门」的证书。属于代数拓扑，本篇不展开。的代数不变量当门的证书——不变量非零的码才有那个门，随便一对多项式不保证。xu2025batched（据扫描报告：Xu 等，“Batched High-Rate Logical Operations for Quantum LDPC Codes”，2025，arXiv 2510.06159，登记待补）：「develops batched error correction, surgery, switching, and Clifford gadgets for CSS qLDPC architectures and specializes a proposal to self-dual BB blocks. ‘Batched’ changes the resource schedule and shared ancilla accounting, not the static stabilizer parameters.」——「批处理」改的是资源的排期与共享辅助系统的记账方式，不改静态的 [[n,k,d]]。

The literature scan holds two nearby routes that the QEC-Bank records on the derivation page without turning them into card rows. breuckmann2026cups (per the scout report: Breuckmann, Davydova, Eberhardt, Tantivasadakarn, “Cups and Gates I: Cohomology Invariants and Logical Quantum Operations”, 2026, arXiv 2410.16250, registration pending): “uses cohomology and cup-product invariants to certify constant-depth logical operations on codes with the required structure, including selected BB examples. The cup product is the gate certificate; a generic bivariate polynomial pair does not automatically supply it.” Plainly: an algebraic invariant called the cup productIn cohomology, the operation multiplying two “classes” into a higher-degree “class”; on this route its non-triviality is the certificate that “this code has a certain constant-depth logical gate”. Algebraic topology, not unfolded here. serves as the gate certificate; only codes where the invariant is non-zero have the gate, and an arbitrary polynomial pair is no guarantee. xu2025batched (per the scout report: Xu et al., “Batched High-Rate Logical Operations for Quantum LDPC Codes”, 2025, arXiv 2510.06159, registration pending): “develops batched error correction, surgery, switching, and Clifford gadgets for CSS qLDPC architectures and specializes a proposal to self-dual BB blocks. ‘Batched’ changes the resource schedule and shared ancilla accounting, not the static stabilizer parameters.” “Batched” changes the resource schedule and the accounting of shared ancillas, not the static [[n,k,d]].

QEC-Bank 第 6 节的标题叫「The non-Clifford boundary」，开头一句：「No source in the six-lens sweep supplies a universal transversal gate set for the original gross block. This is the honest boundary」，随后四条原文：「transversal CNOT is available between aligned identical CSS blocks; selected symmetric or self-dual blocks gain more Cliffords; surgery and automorphisms enlarge the instruction set with extra resources; non-Clifford completion still needs magic-state injection/cultivation, switching, measurement-based resources, or another certified mechanism.」文献扫描报告的「Negative result」用几乎相同的话：「No fetched source provides a universal transversal gate set for the original [[144,12,12]] gross code. Architecture-level universality adds magic states, code switching, surgery, or another block family.」

Section 6 of the QEC-Bank file is titled “The non-Clifford boundary” and opens: “No source in the six-lens sweep supplies a universal transversal gate set for the original gross block. This is the honest boundary”, followed by four items verbatim: “transversal CNOT is available between aligned identical CSS blocks; selected symmetric or self-dual blocks gain more Cliffords; surgery and automorphisms enlarge the instruction set with extra resources; non-Clifford completion still needs magic-state injection/cultivation, switching, measurement-based resources, or another certified mechanism.” The scout report's “Negative result” uses almost the same words: “No fetched source provides a universal transversal gate set for the original [[144,12,12]] gross code. Architecture-level universality adds magic states, code switching, surgery, or another block family.”

这条边界背后其实有两条不同的定理，一条对 gross 给不出结论，一条给得出，值得分开说。Bravyi–König 定理（Bravyi 与 König，arXiv 1206.1609）说：把稳定子码摆在二维格子上，如果每条检查只碰一小块区域（直径不超过 ξ 格，叫作用范围有界），并且码距 d 相对这个范围足够大（原文的条件是 ξ 与 h·r 都远小于 √d，h 是电路深度，r 是每个门能跨的距离），那么任何保持码空间的常深度局域电路，最多实现 Clifford 门。这是表面码那条边界的来由：表面码族的检查只碰相邻的比特，ξ 固定，d 可以任意大，条件总能满足。gross 呢？把它画在 12 × 6 的环面上，A、B 里最高的项是 x³、y³，一条检查的腿最远相隔 3 步，而 √12 ≈ 3.5：「远小于」不成立，定理对这一块码不给结论，既不说它没有非 Clifford 的常深度门，也不说它有。定理要说话，需要一族码：检查的作用范围固定不变，d 不断增大。「检查绕过环面」本身不是理由，环面上的表面码检查也绕过环面。Eastin–Knill 定理（Eastin 与 Knill，“Restrictions on Transversal Encoded Quantum Gate Sets”，arXiv 0811.4262）说：只要一个量子码能探测任意单个物理比特上的任意错误（对量子比特码，就是 d ≥ 2），它就不可能有一组既通用、又全部横向的逻辑门。这条定理不要求二维、不要求局域；gross 的 d = 12 ≥ 2，满足它的条件。所以「gross 没有普适的横向门集」不只是扫描的结果，而是定理的结论。扫描真正补充的是另一层：在已有文献里，gross 这一块具体能拿到哪些 Clifford 门、各要付什么代价（本篇第 2 至 4 章）。读到「BB 码有横向 Clifford 门」这类话，先问：哪个子族、哪块码、哪个证书。

Two different theorems stand behind this boundary, one that gives no verdict on gross and one that does, and they are worth separating. The Bravyi–König theorem (Bravyi and König, arXiv 1206.1609) says: lay a stabilizer code out on a two-dimensional lattice; if every check touches only a small region (diameter at most ξ sites, a bounded interaction range) and the distance d is large compared with that range (the paper's condition is that ξ and h·r are both much smaller than √d, where h is the circuit depth and r the distance one gate can span), then every constant-depth local circuit that preserves the code space implements at most a Clifford gate. That is where the surface code's boundary comes from: the surface-code family's checks touch only neighbouring qubits, ξ is fixed and d can be as large as you like, so the condition is always met. And gross? Drawn on the 12 × 6 torus, the highest terms of A and B are x³ and y³, so the legs of one check lie up to 3 steps apart, while √12 ≈ 3.5: “much smaller” fails, and the theorem gives no verdict on this one block, neither that it lacks non-Clifford constant-depth gates nor that it has them. For the theorem to speak you need a family of codes whose check range stays fixed while d grows. “The checks wrap round the torus” is not a reason by itself; the checks of the surface code on a torus wrap round too. The Eastin–Knill theorem (Eastin and Knill, “Restrictions on Transversal Encoded Quantum Gate Sets”, arXiv 0811.4262) says that a quantum code able to detect an arbitrary error on any single physical subsystem (for qubit codes, d ≥ 2) cannot have a gate set that is both universal and entirely transversal. It needs neither two dimensions nor locality; gross has d = 12 ≥ 2 and meets its condition. So “gross has no universal transversal gate set” is not merely a result of the scan but a consequence of the theorem. What the scan adds is the other layer: which Clifford gates the existing literature actually supplies for the gross block, and at what cost (Chapters 2 to 4 of this guide). When you read “BB codes have transversal Clifford gates”, first ask: which subfamily, which block, which certificate.

五条记录的 level 全是 Clifford。要通用，还差一个 T 门（或等价的非 Clifford 门）。QEC-Bank 列出四条路：「magic-state injection/cultivation, switching, measurement-based resources, or another certified mechanism」。第一条是表面码架构的主流：先制备带噪的魔法态magic state：一个非稳定子态（例如 T 门作用在 |+⟩ 上得到的态）。有了它，用 Clifford 操作加测量就能实现 T 门；制备与提纯（蒸馏或培育）是逻辑计算开销的主项。，用 Clifford 操作提纯（蒸馏，或近年更省的「培育」cultivation），再通过门传送（用 Clifford 门和测量把魔法态「送进」电路，效果等于做了一个 T 门）消耗它实现 T——《格点手术》1.3 节介绍过蒸馏。第二条是码切换（在有 T 门的码与 BB 码之间来回切）。第三条是基于测量的资源。文献扫描里 xu2025batched 的方案正是把「parallel magic-state cultivation」配给自对偶 BB 块——仍是魔法态路线，只是并行化了。intuition.md 的总结：「Non-Clifford completion still needs magic states, switching, surgery, or another certified resource.」人话：BB 让存储便宜了很多，却没有让 T 门变便宜。存储这一侧的数字出自 Bravyi 等（arXiv 2308.07915，第 3 节与第 8 节），是电路级噪声下的数值模拟比较：每个操作（CNOT、制备、测量、空等）以概率 p = 0.1% 独立出错的标准去极化电路噪声；gross 一侧用 MIN-SUM 置信传播（上限 10,000 轮迭代）加 combination-sweep 版本的 OSD 译码，288 个物理比特（144 数据 + 144 帮手），估计每轮逻辑错误率约 2 × 10⁻⁷，论文由此估计能把 12 个逻辑比特保存近一百万轮症状测量；对照组是 12 块独立的旋转表面码贴片（d = 9 到 15，标准症状测量电路，同样的噪声模型，数据来自作者之一与 Vargo 的模拟软件，论文此处没有写明表面码一侧的译码器），要把每轮错误率从 10⁻³ 压到 10⁻⁶ 需要近 3000 个物理比特。论文据此说节省超过 10 倍；注意两边的目标错误率并不相同（约 2 × 10⁻⁷ 对 10⁻⁶），「10 倍」是近似的比较。在任何 BB 架构的总账里，T 门那一项仍要单列。

All five records have level Clifford. Universality still needs a T gate (or an equivalent non-Clifford gate). The QEC-Bank lists four roads: “magic-state injection/cultivation, switching, measurement-based resources, or another certified mechanism”. The first is the mainstream of surface-code architectures: prepare a noisy magic stateA non-stabilizer state (for example T applied to |+⟩). With it, Clifford operations plus measurement implement the T gate; preparing and purifying it (distillation, or cultivation) is the dominant cost of logical computation., purify it with Clifford operations (distillation, or the cheaper recent “cultivation”), then consume it by gate teleportation (Clifford gates and a measurement “feed” the magic state into the circuit, with the effect of a T gate) to realise T; section 1.3 of “Lattice Surgery” introduced distillation. The second is code switching (moving back and forth between a code with a T gate and the BB code). The third is measurement-based resources. In the literature scan, the proposal of xu2025batched pairs “parallel magic-state cultivation” with self-dual BB blocks, still the magic-state road, only parallelised. The summary of intuition.md: “Non-Clifford completion still needs magic states, switching, surgery, or another certified resource.” Plainly: BB made storage much cheaper, but not the T gate. The storage numbers come from Bravyi et al. (arXiv 2308.07915, sections 3 and 8) and are a numerical comparison under circuit-level noise: the standard depolarizing circuit noise in which every operation (CNOT, preparation, measurement, idling) fails independently with probability p = 0.1%; on the gross side, MIN-SUM belief propagation (at most 10,000 iterations) with the combination-sweep version of OSD as decoder, 288 physical qubits (144 data + 144 helpers), an estimated logical error rate of about 2 × 10⁻⁷ per cycle, from which the paper estimates that 12 logical qubits survive nearly one million syndrome cycles; the comparator is 12 separate rotated surface-code patches (d = 9 to 15, the standard syndrome circuit, the same noise model, data from simulation software by one of the authors and Vargo; the paper does not name the surface-code decoder at this point), which need nearly 3000 physical qubits to push the per-cycle error rate from 10⁻³ down to 10⁻⁶. On this basis the paper claims a saving of more than 10 times; note that the two target error rates differ (about 2 × 10⁻⁷ versus 10⁻⁶), so “10 times” is an approximate comparison. In the balance sheet of any BB architecture the T gate remains a separate line.

logical-operations.md 第 7 节「What the card promises」用一张表收尾，原文照录。它是读这一侧面的钥匙：每条记录承诺的是左边那句，右边那句是读者最容易多推出来、而 QEC-Bank 明确不背书的话。

Section 7 of logical-operations.md, “What the card promises”, closes with a table, reproduced verbatim. It is the key to reading this facet: what each record promises is the left sentence, and the right sentence is the one readers most easily infer in addition, which the QEC-Bank explicitly does not endorse.

表后一句说明了为什么要这样分开写：「This separation lets later work add a gross-specific logical-action certificate without rewriting family-wide folklore as fact.」人话：有一天若有人对 gross 算出了某个自同构的 24 × 24 辛作用并认证了布线，QEC-Bank 只需新增一条针对 gross 的记录，而不必把「BB 家族有这些门」这种口耳相传的说法改写成事实。这也是为什么五条记录的 applies_to 全是 qcga#collection（这一族）而不是 gross 实例：它们目前都是「族里有人做到过」，不是「gross 做到了」。

The sentence after the table explains why it is written apart like this: “This separation lets later work add a gross-specific logical-action certificate without rewriting family-wide folklore as fact.” Plainly: should someone one day compute the 24 × 24 symplectic action of some gross automorphism and certify its routing, the QEC-Bank need only add a gross-specific record, rather than rewrite the hearsay “the BB family has these gates” as fact. It is also why all five records have applies_to = qcga#collection (the family) rather than the gross instance: each currently says “someone in the family has done it”, not “gross has done it”.

回放一遍。仓库变计算机要答三问：哪个物理过程保持码、它对 12 个逻辑比特的 24 × 24 辛作用是什么、什么容错与布线资源认证它；k = 12 再加一问——可寻址；QEC-Bank 的 5 条记录全是 Clifford、全是 claimed、全对这一族而非 gross 说（第 1 章）。横向 CNOT：四条共轭规则把控制块的 X 检查复印成「自己乘目标块同号检查」，两块相同且 CSS 时全部留在稳定子群里，144 个物理 CNOT = 12 个并行逻辑 CNOT；成本另算搬运对齐与块间相关译码；它不能做块内门、不能挑对（第 2 章）。平移是自同构：H_X diag(P,P) = P H_X，检查 0 → 检查 6，72 个平移，reproduce.py 已验证；但门要再走五步，gross 的辛作用 QEC-Bank 没有登记（本篇算得是 36 种逻辑 CNOT 线路之一，Bravyi 等给出了实现平移的容错线路，用于寻址），一般而言可能是单位、置换或缠绕的 Clifford；记录 1、2 是文献对选定实例走完五步的结果，对称是选码条件（第 3 章）。手术：借辅助系统在块与块之间测乘积算符，五项资源，标签 surgery 防止它们被藏在「Clifford 门」下；k = 12 让「测哪个算符」成为编译器的问题（第 4 章）。自对偶子族靠换一族码得到横向 H/S/CNOT，not a gross-code claim；没有文献给 gross 普适横向门集，Eastin–Knill 定理也禁止它；T 门仍靠魔法态（第 5 章）。

Replay. Turning the warehouse into a computer means answering three questions: which physical process preserves the code, what 24 × 24 symplectic action it induces on the 12 logicals, what fault-tolerance and routing resources certify it; k = 12 adds a fourth, addressability; the 5 QEC-Bank records are all Clifford, all claimed, all about the family rather than gross (Chapter 1). Transversal CNOT: four conjugation rules copy a control X check into “itself times the same-numbered target check”, everything stays in the stabilizer group when the blocks are identical and CSS, 144 physical CNOTs = 12 parallel logical CNOTs; transport, alignment and inter-block correlated decoding are billed separately; it cannot do within-block gates or select pairs (Chapter 2). Translations are automorphisms: H_X diag(P,P) = P H_X, check 0 → check 6, 72 translations, verified by reproduce.py; but a gate needs five more steps; the QEC-Bank records no symplectic action for gross (this guide finds one of 36 logical CNOT circuits, and Bravyi et al. give fault-tolerant circuits for the translations, used for addressing), and in general it could be the identity, a permutation or an entangling Clifford; records 1 and 2 are the literature's completion of those steps for selected instances, symmetry being a code-selection condition (Chapter 3). Surgery: measure product operators between blocks through an ancilla system, five resources, the label surgery keeping them from hiding under “Clifford gate”; k = 12 makes “which operator to measure” a compiler question (Chapter 4). The self-dual subfamily gets transversal H/S/CNOT by changing the family, not a gross-code claim; no paper gives gross a universal transversal gate set, and Eastin–Knill forbids one; the T gate still rests on magic states (Chapter 5).