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"ca16ef9acbdde8ddea54f128413f701edf1ea4059d6e5f83a09b5d2576192ede" }, "title": "把中间结果留在计算组内，减少接力搬运", "mechanism": "逐步生成时，前馈网络的第一次矩阵变换、激活函数和第二次矩阵变换在同组内接续，最后合并各组的输出部分和。整段输入常用16行×32列、每轮累加32项的分块。", "reason": "中间向量在小存储里直接被下一道计算使用，能省掉写到外部内存再读回的路径；部分和的汇总仍有成本。", "boundary": "结构沿用旧快照的同一份程序，可核对汇编；本次提交没有搜索过程，不能补写作者怎样想到它。", "family": "无共享暂存", "process": "旧页保留此最终构造分析；无附带过程。", "reading": { "rank": 1, "question": "一段计算中，哪些中间结果可以留在原地，哪些小运算可以交给矩阵硬件批量完成？", "sections": [ { "title": "把统计也当成可以批量处理的计算", "paragraphs": [ "模型在矩阵投影前要对输入做归一化：先求均值和方差，再调整各元素的尺度。方案01把一批输入排成矩阵，与全1向量相乘来求和；平方后的输入也可用同样方式求平方和。这里的矩阵阵列既做通常的乘法，也承担统计。好处是把许多零碎操作合在一起，代价是统计与主计算会共享矩阵和寄存器资源。", "整段输入程序常见的分块是16行×32列、每轮累加32项，以及16行×16列、每轮累加64项。“分块”描述一次软件指令处理的数据，不是芯片上阵列的物理尺寸。整段程序包含80次工作组创建，不能把它读成80组同时驻留。" ] }, { "title": "前馈中间结果为什么不用完整写到外存", "paragraphs": [ "前馈网络可以理解成三步：W1矩阵把向量变宽，GELU激活函数逐元素变换，W2矩阵再把它压回原宽度。生成程序把中间的512个通道分给16个任务，每个任务持有32个通道对应的W1列和W2行。它算完这32个值、做完激活，立即在寄存器小存储里接着乘W2。", "每个任务得到的是128维输出的一份部分和，16份相加才是完整结果。这省掉了把完整中间向量写出再读回的路径，但仍有部分和传输与汇总。因而“融合”在这里有具体含义，不能理解成所有通信都消失。" ] }, { "title": "注意力还需要另一种分工", "paragraphs": [ "注意力用当前输入的查询向量Q，与历史位置的键K比较，再按比较结果汇总值V。该方案把每一路历史注意力分成四片，各片计算局部统计和加权结果，最后重新缩放并合并。两层权重与原始历史仅首步加载，后续步骤继续复用；每一步的新输入仍需读取。", "程序还存在按事件控制加载时刻的安排，说明“何时搬数据”也是成品结构的一部分。没有附带搜索记录，不能补写作者怎样选择事件间隔或怎样逐次获得这些改进。" ] } ], "lesson": "可迁移的观察角度是重新划定数据交接边界：中间结果的生产者能否直接完成下一步？同时检查新增的部分和汇总成本。静态结构说明这种机会存在，独立收益仍需对照实验。", "process": [], "sources": [ "旧版最终构造分析：final-solutions.html 中方案一；与本快照对应同一份程序。", "最终汇编中归一化统计、四片注意力与32通道前馈部分和路径。" ] }, "label": "方案01", "trace_file_count": 0, "trace_hashes": [], "area": 23.28176530432, "programs": { "M1_P1": { "cycles": 210399, "peak_power_w": 19.217624187188445, "hbm_read_bytes": 25334528, "hbm_write_bytes": 5381632, "instruction_count": 68304, "sha256": "b3a9c6a5b865c90240ad70add4cf443f413ae67d30d3d04f324bfc172106d69a", "ops": { "WG.BEGIN": 80, "VEC": 20905, "ST": 1711, "LD": 13858, "MMA.ACC": 22848, "WAIT": 7921, "SFU": 780, "WG.END": 80, 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"该方案有32条向量通道。随包记录说明，增加通道后寄存器的物理分区变小，原来的大操作数需要重新排布或拆成更细指令。最终整段输入最常见的是16行×16列、每轮32项的指令，不能直接把生成器里的名义大块当成实际执行块。", "生成阶段还把只由固定归一化参数和权重决定的计算保留下来供后续步骤复用。第一步把真实输入与两份常量投影合批处理，因此会出现3行矩阵指令；新输入相关的均值、方差和投影仍须逐步计算。" ] }, { "title": "删掉最后一段多余搬运", "paragraphs": [ "最终输出若先进入临时区，再由控制任务逐行复制到规定地址，会额外经过一次写、读、写。记录中的更新让最后一层直接写最终输出，并同步处理批次步长和提交时刻。另一些更新把归一化参数接入已有加载流，减少单独准备与等待。", "这些改动共同把数据到达时间提前。它们也可能使活动更加集中，因此整段输入的写回又按任务组分批发出。快与合法需要一起检查，不能只看某个阶段缩短。" ] } ], "lesson": "这条过程的经验是把“大操作之间的顺序”细化成“具体哪块数据何时可用”，再检验更细的依赖是否真正提前了下游执行。", "process": [ "记录先暴露了物理分区与名义布局的不匹配。agent调整寄存器排布、拆分指令，并用生成结果与完整评估确认变换有效。", "随后围绕数据就绪时刻改调度：查询先生成，历史注意力先工作；已完成的输出列向下一阶段流动。每次缩小等待都要保留对应生产者依赖。", "已保存的同硬件摘要从74,092.80分推进到78,142.69分，期间含历史保留、缓冲调整、逐块消费和直接输出等多项改动，不能把整段涨分归给一个技巧。", "参数加载调整后达到80,005.43分，再合并已验证的整段写回安排达到80,098.00分。历史完整大报告未随包保存，这段过程由阶段摘要与计划支撑。" ], "sources": [ "project/copt80_runs/README.md 的阶段记录；四份历史摘要与配置见本页真实更新图。" ] }, "label": "方案02", "trace_file_count": 6, "trace_hashes": [ 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"canonical_hardware": "6b7899e16aa11ee47c064d1f506294eb7dc15e86e63cfcc8b2205d7b774e2f4d" }, "title": "16列分块、64项累加，整段与逐步生成分开组织", "mechanism": "最终整段输入以16行×16列、每轮累加64项的矩阵指令为主，包含79个工作组生命周期；逐步生成有64个工作组，权重和原始历史在第一步读取后继续复用。", "reason": "较长的求和分块减少相同输出的累加轮次；分阶段任务可承担不同工作，但79个生命周期不等于79组同时驻留。", "boundary": "与方案5的硬件规范相同，成品差异来自程序；无法仅凭指令形状确定某项优化贡献。没有搜索记录。", "family": "无共享暂存", "process": "无附带过程；可与硬件完全相同的方案5对照程序。", "reading": { "rank": 3, "question": "相同硬件下，较长的求和块与不同任务划分，怎样改变数据复用和等待？", "sections": [ { "title": "先把“每轮累加64项”讲清楚", "paragraphs": [ "一个输出元素需要把输入与权重逐项相乘再累加。每轮处理64项，表示一条指令完成这64项的部分和；剩余项由后续轮次继续累加。方案03整段输入最常见的是16行×16列×64项。较长的求和块有机会减少发射与换块次数，同时占用更多输入、权重和临时空间。", "最终整段程序有79个工作组生命周期，生成程序有64个。生命周期统计包含不同阶段的创建与结束，不能据此确定同时有多少任务运行。需要时间线才能进一步回答哪些任务真正重叠。" ] }, { "title": "生成阶段先避免重复搬权重", "paragraphs": [ "静态读取记录显示，两层权重共1.5 MiB、原始历史共256 KiB，集中在生成首步读取；后面七步不再读这些原始对象。新步骤仍读取新输入并更新状态。这先解决了重复加载，剩余差别来自计算分块、数据交接与等待。", "生成中常见1行×16列×32项和1行×16列×64项，也有较窄的输出块。它说明作者没有给所有计算强行用同一种粒度，但仅凭形状统计还不能确定每一种块对应哪条优化假设。" ] }, { "title": 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"“没有共享暂存”不等于没有数据复用", "paragraphs": [ "该硬件不配额外共享暂存和缓存，计算仍有寄存器小存储。64个生成工作组生命周期配合首步权重和历史加载，后续步骤不重复读取这批原始数据。应从数据真正保存在哪里理解复用，不能只看缓存容量。", "该方案整段输入外存写出约3.80 MB，少于03的约5.21 MB；但整段耗时273,192周期，高于03的252,277周期。写得少是事实，等待、读取和计算组织仍可能抵消收益。" ] }, { "title": "能从成品学到的边界", "paragraphs": [ "值得带走的是“按实际形状与空间选分块”这个分析视角。若要验证某个不整齐块是否有用，需要固定其它布局，检查总覆盖、局部容量、边界处理和最终周期。", "这份提交没有搜索记录，因而正文中的容量解释是对结构的分析。它没有证实作者先做了容量推导再选择这些块，也没有给出56项比64项独立快多少。" ] } ], "lesson": "将不规则分块当作值得追查的证据，不因它看起来特殊就认定是秘诀。真正要核对的是覆盖了什么、节省了什么、又引入什么代价。", "process": [], "sources": [] }, "label": "方案04", "trace_file_count": 0, "trace_hashes": [], "area": 23.48656530432, "programs": { "M1_P1": { "cycles": 273192, "peak_power_w": 19.834769876015006, "hbm_read_bytes": 34145792, "hbm_write_bytes": 3799040, "instruction_count": 148873, "sha256": "46bcedb8f76f8bef15a2183d2f071d25d0cf49d8ae9af0c5e3955c8a8adf0021", "ops": { "WG.BEGIN": 64, "VEC": 52432, "LD": 16640, "MMA.ACC": 24048, "ST": 1376, "WAIT": 9032, "SFU": 41664, "REDUCE": 3552, "STEP.COMMIT": 1, 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"整段程序混用16行×16列×64项和32行×32列×32项等形状。生成最常见1行×16列×16项，求和块比03的一些常用块更短：单次存储需求较小，却需要更多累加轮次。" ] }, { "title": "三行指令究竟说明什么", "paragraphs": [ "生成中能看到3行×16列×8项的指令，说明程序确实把三行数据合到一次矩阵操作。这里未沿每个操作数地址重建三行的含义，因此不能照搬其它方案的解释，说它们必然是当前输入和两份常量。", "可以确定的是，权重和原始历史只在首步读取。跨步骤复用这一点与03相同，所以不能拿“权重驻留”单独解释两者的速度差。" ] }, { "title": "同硬件比较揭示的剩余问题", "paragraphs": [ "05生成需要27,497周期，03为24,949；05生成外存写出286,976字节，03为148,480字节。更大的输出与中间搬运量、更多短块，都是检查数据交接的线索。不能把这些统计直接累加成一个因果解释。", "下一步若要判定原因，应追踪哪些写出是接口必需、哪些是临时交接，再查看消费者是否因此等待。当前只有最终程序，未保存做这种对照的过程。" ] } ], "lesson": "对看似相同的方案，不重复列共同技巧；寻找共同技巧之后还剩下的差别，例如中间结果是否落地、累加被切成多少轮、任务何时获得输入。", "process": [], "sources": [] }, "label": "方案05", "trace_file_count": 0, "trace_hashes": [], "area": 23.966565304320003, "programs": { "M1_P1": { "cycles": 260349, "peak_power_w": 19.930454756607116, "hbm_read_bytes": 32701952, "hbm_write_bytes": 5591040, "instruction_count": 102798, "sha256": "2e7da035b47689b6df6bdd57a0acba078c143084b5f08e2c17c871aae61a71f5", "ops": { "WG.BEGIN": 16, "LD": 21444, "VEC": 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"两层生成权重和原始历史仅在首步读取，后续七步继续复用。这把反复搬运变成一次性准备，却保留了启动加载、每步新输入、局部计算与结果交接的成本。", "生成外存读取约2.26 MB、写出约0.31 MB。对这种已大量复用的程序，下一轮调查不能只问能否再缓存权重，还应问剩余传输是谁在等、是否可以更早到达。" ] }, { "title": "与07的差异是两类任务的取舍", "paragraphs": [ "06与07硬件完全一致。06整段输入301,644周期，比07的335,948更快；生成39,484周期，却比07的38,337略慢。06整段读取也较少，约25.63 MB对28.24 MB。", "共同评分取决于两类任务耗时的乘积，因此这次整段收益足以抵消生成退步。这个对照证明软件组合存在取舍；没有单因素实验，不能说增加或减少某一组任务就导致了它。" ] } ], "lesson": "分开保存两类任务的变化，再用真实目标函数决定是否采用。只挑一个更好看的分项，会遗漏另一项付出的代价。", "process": [], "sources": [] }, "label": "方案06", "trace_file_count": 0, "trace_hashes": [], "area": 23.646565304320003, "programs": { "M1_P1": { "cycles": 301644, "peak_power_w": 19.546054411408008, "hbm_read_bytes": 25627648, "hbm_write_bytes": 5343232, "instruction_count": 97942, "sha256": "024cec37fb8891e7b2459a8f85faa94f93c2fb28ca5b14a1ae45e4c0693a322a", "ops": { "WG.BEGIN": 17, "BARRIER": 309, "LD": 20446, "REDUCE": 7800, "VEC": 31026, "SFU": 4380, "ST": 5050, "WAIT": 7872, "MMA.ACC": 21024, "STEP.COMMIT": 1, "WG.END": 17 }, "groups": 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MB，高于06的2.26 MB，写出也更多。这说明即使搬运量上升，安排得更合适仍可能更快。具体原因还需查看数据何时到达和计算重叠。", "两者都仅在首步加载权重和原始历史，也都只有一个搬运引擎。解释差异时应关注程序分工和时间安排，避免拿相同的硬件或相同的驻留策略重复归因。" ] }, { "title": "为什么这份对照值得保留", "paragraphs": [ "整段任务反过来慢了34,304周期，超过生成阶段节省的相对比例。最终总分较低，但这并不抹掉生成子程序的比较价值：它可能包含可移植的调度安排。", "若尝试组合06的整段程序与07的生成程序，需要重新核对兼容配置和完整评估。这只是由同硬件对照引出的后续实验，本页没有替作者运行或声称组合一定更好。" ] } ], "lesson": "保留分项优势，不只保留总分最高的一个文件。后续组合可以有新机会，但要通过实际整合验证。", "process": [], "sources": [] }, "label": "方案07", "trace_file_count": 0, "trace_hashes": [], "area": 23.646565304320003, "programs": { "M1_P1": { "cycles": 335948, "peak_power_w": 19.542478308594525, "hbm_read_bytes": 28238848, "hbm_write_bytes": 5834752, "instruction_count": 118794, "sha256": "037a247e174f9632bc8a07cda360d6e4a077bc3632519f451352238f0a7f1260", "ops": { "WG.BEGIN": 32, "LD": 24070, "VEC": 55392, "REDUCE": 7800, "SFU": 7164, "ST": 5614, "BARRIER": 473, "MMA.ACC": 17760, "STEP.COMMIT": 1, "WAIT": 456, "WG.END": 32 }, "groups": 32, "group_lifecycles": 32, "top_shapes": [ [ "(32, 16, 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"各片产生局部最大分数、指数和、加权值向量。合并时依据各片最大值重新缩放，再合成总分母与结果。不能直接平均六份局部输出，因为每片归一化用的分母不同。拆分减少单任务工作，却增加六份摘要的交换和汇总。" ] }, { "title": "前馈也沿求和方向分给不同任务", "paragraphs": [ "第二次前馈矩阵变换的输入宽度为512，方案把它沿求和方向切开。已核对的一个任务负责176项，按64、64、48项分轮计算，再与另外任务的部分和相加。较短的最终输出因此也能让更多计算单元参与。", "最终生成41,549周期、整段352,305周期。这个结果说明较轻单元配细分工可以形成有效方案；它没有证明更多单元或更多历史分片单独产生了多少收益。" ] } ], "lesson": "当输出太短、沿输出切不出足够任务时，可以检查历史维度或求和维度是否可拆。先写清合并公式，再计算通信与同步代价。", "process": [], "sources": [ "旧版 final-solutions.html 中方案五；与本快照方案08为同一份程序。" ] }, "label": "方案08", "trace_file_count": 0, "trace_hashes": [], "area": 23.64464795648, "programs": { "M1_P1": { "cycles": 352305, "peak_power_w": 16.957613829934928, "hbm_read_bytes": 27695360, "hbm_write_bytes": 7188480, "instruction_count": 71999, "sha256": "7c96cfd20a9cd880cb5e6659098a19aab5f49a9fb6b172fa16cd31904a540aa7", "ops": { "WG.BEGIN": 24, "VEC": 20124, "LD": 14594, "WAIT": 8300, "REDUCE": 1656, "SFU": 1044, "ST": 2472, "MMA.ACC": 23760, "STEP.COMMIT": 1, "WG.END": 24 }, "groups": 24, "group_lifecycles": 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"生成程序跨步骤保留权重与历史，把前馈相邻操作接续执行，历史注意力分成8份并合并。已经不再重复读取的权重仍要在首次装进来，多个计算单元同时发请求会争抢外存、网络和寄存器写口。", "agent进一步按实际搬运字节量分桶，用事件控制发射间隔。按计算单元数量分批可能一批很重、一批很轻，字节量能更直接描述本轮放出的请求规模。K/V也改为直接写最终位置，省掉临时区往返。" ] }, { "title": "最后的关键是定位，而非全面降速", "paragraphs": [ "功率限制检查任意连续1000周期的平均值。一个阶段变快或一个峰被压下去后，决定资格的窗口可能转移到别处。最终整段程序只多付出117周期，就让最高窗口进入预算；它不是凭空找到一个万能等待常数。", "最终整段337,476周期、生成47,049周期。额外输入检查也保留在旧复盘中。下面的过程是依据保存记录重建的判断变化，按问题整理，不表示所有实验严格串行发生。" ] } ], "lesson": "遇到改动无效，先检查是否干预了真正限制；遇到改动成功，再检查限制是否转移。把一条失败当作下一次测量的线索。", "process": [ "先检查数据能否跨步骤保留：总权重加临时区、历史状态，既算总量，也算每个任务的局部容量。模型只保证放置合法，完整程序仍需评估。", "观察到少数任务承担历史注意力，尝试拆分历史并正确合并；整段输入则从供数与重复读取转向二维复用。", "局部变快后多次超功率。一次放慢前部计算让程序变慢，却没有改变最高峰，调查于是转向后续K/V导出，发现并删掉临时区往返。", "进一步复用后再次接近功率上限，agent重建窗口时间线，先调整前馈写回；最高峰随后转移到Q/K/V写回，又只调整新峰附近的节奏。", "最后以小幅等待换取完整合格版本，再固定最终产物核对评分和额外输入。旧复盘保留成功与失败对照。" ], "sources": [ "astra-path.html：容量、代理失配、二维复用、首次装载、最高功率窗口转移。" ] }, "label": "方案09", "trace_file_count": 2, "trace_hashes": [ "ec789a8c865b382d5cc6d03bc8f32a7d4f1db4b4778681cc15c054d473959d46", 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"reading": { "rank": 10, "question": "局部内核几乎减半，为什么完整程序还需要回退和重新安排加载？", "sections": [ { "title": "让任务持有自己负责的权重和结果", "paragraphs": [ "随包知识记录把第二次前馈变换W2沿求和方向拆开：每个任务保留自己负责的权重，计算对应部分和，再合成完整输出。“owner常驻”在这里指数据由明确任务持有，避免每次计算再从外部搬来。", "整段程序最常见的指令为16行×16列×128项，生成则按不同输出宽度和求和长度混用分块。两层生成权重仅首步读取，读取总量约1.97 MB。流量较低，仍不足以单独说明关键路径短。" ] }, { "title": "局部对照与整案发生了什么", "paragraphs": [ "知识记录中的一组W2冷启动局部实验，从60,223周期降到30,689周期，外存读取从6.85 MB降到2.10 MB。这是该局部核的对照，不能乘上层数当作完整模型收益。", "合入完整程序后，一版整段输入达到394,748周期，但峰值21.2578 W，超过20 W门槛。随后保留有用的数据所有权安排，改变其它投影的保留方式、分段加载与生成中的计算搬运重叠，最终整段为403,197周期，功率19.1436 W。" ] }, { "title": "资源忙碌比例给线索，不直接给原因", "paragraphs": [ "最终整段报告的片内网络忙碌比例约48.01%，最忙计算单元的寄存器读比例约36.97%，矩阵阵列约20.70%。这些数字促使agent继续查供数与等待，但没有证明全部时间都卡在外存。", "不同任务可同时工作，累计等待也可能超过总时长；不能把这些比例相加拼成总耗时。该记录明确保留了这种证据边界，是值得继承的分析习惯。" ] } ], "lesson": "把局部实验当成方向筛选，尽早合入完整执行；更快但超限的版本可以提供线索，却不能替换已合格版本。", "process": [ "先在W2局部核验证沿求和方向拆分和任务保留权重是否减少流量。局部收益支持继续整合。", "完整评估暴露功率超限，推翻了“局部加速可直接采用”的判断。", "保留有效结构，重新组织加载与重叠，接受整段输入略慢以恢复合法性；最后同时验证生成程序与共同硬件。" ], 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KiB共享暂存，矩阵阵列为4×8、求和并行4项。最终整段源码中有3,072行直接涉及共享暂存的读取或写出，生成源码中为0行。这里是源码行数，不等于循环展开后的传输次数。", "因此不能把较快的生成直接归功于大共享暂存。生成权重与原始历史仅首步读取，后续更接近由持久任务在寄存器里保存并继续使用。芯片配置与具体程序使用路径必须分开看。" ] }, { "title": "过程记录里的大变化来自生成重写", "paragraphs": [ "当时的阶段总结先报告：同一旧硬件下，重写生成程序将约138k周期降到约58k；之后再调整硬件与加载，进入33,323周期的阶段。这个摘要支持“先有软件变化，再评估资源配置”的顺序，不能把整段收益全部归给换阵列。", "重写包括权重与历史保留、把历史分片交给多个任务，以及把归一化与投影安排在可重叠的路径上。偏置、激活和残差等收尾操作接入相邻计算，减少独立阶段的同步。" ] }, { "title": "这条过程没有止于“多给资源”", "paragraphs": [ "阶段记录中，把全局网络由256加宽到512字节/周期，两类任务都变快，但整段功率达到20.069 W，超限。随后改变加载分批，也出现生成更快、整段更慢且仍超限的组合。这些是历史候选，不能当作最终程序的指标。", "本快照最终生成仍为33,323周期，整段已经降到585,222周期；早期摘要的整段814,670周期不应混作最终结果。目前仅对所列过程片段做了核对，没有把之后每一步整段优化全部还原。" ] } ], "lesson": "有剩余面积或功率，不意味着简单加硬件就能受益。先弄清软件是否产生了可服务的工作，再检查新增资源是否触发另一类任务的最坏窗口。", "process": [ "早期记录把逐步生成列为主要调查对象，先重写跨步骤数据保留与任务安排。", "在软件变化后评估关闭缓存、增加搬运引擎、调整矩阵阵列形状等组合；阶段摘要显示生成进一步缩短。", "扩大网络试验虽更快却超功率，agent转查具体加载阶段并尝试分批。这些失败没有被写成合格成绩。", "最终提交的两份程序以本页服务器报告为准。过程片段来自当时说明，未做整条历史的重新评分。" ], "sources": [ 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"阶段记录将整段最后位置的投影专门处理，以及注意力中的标量广播，列为后期改动。标量广播是把同一个数供给一批元素使用，减少重复准备；它仍应符合指令和输出语义。", "某次编译器变换连带改变了生成程序的寄存器分配，agent因此重新测量生成，即使数学含义相同也没有直接继承旧的性能结果。最终以438,472和53,830周期的服务器报告为准。" ] } ], "lesson": "保留“更快但不可用”的完整证据，让它推动下一次约束修复。一次去缓存失败只限定当时组合，不应变成所有无缓存方案都不可行的先验。", "process": [ "先在已有合格版本上优化整段末位置计算和注意力数据供给；编译输出变化后重新检查受影响程序。", "测得去缓存版本更快，但两项功率都超限，拒绝把它作为当前最佳合格版本。", "由超限结果转向供数节奏与带宽组合，保留单题筛选与完整验收的区别；当前快照选择的是已完成验证的保留缓存版本。" ], "sources": [ "project/STATUS.md 的 Completed experiment conclusions：R182、R183、R191、R192与后续候选状态。" ] }, "label": "方案12", "trace_file_count": 6, "trace_hashes": [ "62aeb75d26ca3ac54cb1831724a35259b36dcc38555c80b99414da104957c6f9", "dabc81c256fca72fe03bf07d2020bee982fb7e86a71bcc4f038c6d83808c6254", "b0974b7f0405b59081a238046925063dbc6b9a8617e6ec531055f783c057107a", "7e3e5a5177b693dd70d48c34a34f775bd81bf7809d236948fc338cf271430fb5", "9cb74e977454e22c832adc3117a0cce6343ca61d61171aa9fbd6ef0179b5bda6", "9916974de8e1474359a2ebf4904b4f863cd2db1a91bf052d5fc15d8013adc3f5" ], "area": 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"一个早期失败的硬件方向，为什么在软件结构改变后值得重试？", "sections": [ { "title": "先消掉反复搬运，再改相邻操作的交接", "paragraphs": [ "生成阶段每步处理新输入，但权重与已存在的历史会反复使用。笔记记录的路线先让历史键K和值V保留在寄存器，再让负责权重的任务连续完成相邻操作，减少中间数据离开原任务后的搬运。", "历史注意力随后分成四片：每片计算自己的局部最大值、指数和及加权向量，最后用一致的缩放合并。这里的“4片”是具体候选选择，分得更多还会增加合并成本。" ] }, { "title": "一个较干净的局部归因", "paragraphs": [ "笔记记载，在硬件和整段程序保持不变的条件下，采用四片历史版本使生成从62,515降到53,736周期。这个对照比跨参赛方案的分差更接近回答一次改动的作用，但本页没有重新运行该历史实验。", "最终程序逐步读取新输入，加载与矩阵指令都存在，权重和原始历史只在第一步加载。它与方案14的固定常量输出有明确区别；两者在本快照都标为本地报告，不代表具有同一种问题。" ] }, { "title": "为什么重新开放之前失败的配置", "paragraphs": [ "记录中，较早的8通道组合没有成为可采用的结果。完成数据保留和融合后，agent重新测试相关硬件组合，得到通过完整本地评估的候选。这说明此前失败依赖当时的软件与资源组合。", "最终缓存为1 MiB、全局网络256字节/周期。笔记的40,076.35分首先来自实验评估路径；本快照仅认定为本地成绩，不用它冒充服务器隐藏输入验证。" ] } ], "lesson": "给失败附上条件：哪个程序、什么配置、因为什么失败。数据流已经改变，就允许重开旧方向；同时保留本地实验与服务器验证的层次。", "process": [ "先保留合格起点，再尝试按输出列分工；这一步单独没有改善，未被当作成功。", "历史保留、任务内相邻操作接续分别带来改善，再把整段复用与调度改动整合。", "软件结构变化后重试此前不成功的硬件方向，完成共同硬件下的评估。", "最后用同硬件、同整段程序的四片历史对照验证生成收益，再核对额外输入、生成产物与报告对应关系。" ], "sources": [ 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"cache_mib": 0, "dma_depth": 1, "dma_engines": 1, "hbm_channels": 4, "multicast": false, "noc_bytes_per_cycle": 256, "reduction_units": 0, "rf_ports": "4R2W", "sfu_lanes": 8, "shared_banks": 4, "shared_kib": 128, "shared_ports": "1R1W", "sm_count": 8, "sm_noc_bytes_per_cycle": 64, "tc_array": "4x8", "tc_count": 2, "tc_k_parallel": 1, "vector_lanes": 16, "version": "challenge-hardware-v0.4" }, "hashes": { "zip": "2f2a9f519b457965424e759a1d1a9631f0b0fd6af3201e8caf7034ee7ca4c878", "report": "a196f1fed2b9c1814dea87f7615250869a9df1ea664eba4b716a6b5e65ddfb41", "hardware": "a37b1544841dc63c237b93c30b791c1eac36cf1d22e05455d7ed65e0c043f717", "canonical_hardware": "419300a7886e7f109a2daa5d97cb2447d324b1ef038f95ef6be8d485bab76f74" }, "title": "报告中的短生成耗时，不可当作通用优化", "mechanism": "最终D1有5,120条向量指令和136条写出指令，没有LD读取，也没有矩阵乘法。它把立即数写入输出；P1仍有15,583,748周期。", "reason": "没有读取输入或权重的程序无法随新的输入改变结果，因此1,996周期不能解释为普适的Transformer生成性能。", "boundary": "榜单目前仅本地成绩。保留原榜位置与数字，但从已验证耗时图、提升拆分和可学习性能结论中排除；本次未重跑隐藏输入。", "family": "需复核", "process": "无过程记录；静态程序检查足以指出上述范围。", "reading": { "rank": 14, "question": "极短的生成耗时，究竟代表计算被优化了，还是预定答案被直接写了出来？", "sections": [ { "title": "正常生成程序需要完成什么", "paragraphs": [ "连续生成8步时，每一步要使用新输入、模型权重和历史状态进行计算，写出新的隐藏向量以及每层的键K和值V，再提交本步结果。输入变化时，相应输出也应能变化。", "这份生成程序没有读取指令，也没有矩阵乘法。它有5,120条向量指令，主要用“固定浮点数加0”的方式把常量放到寄存器，再用136条存储指令写到输出位置。" ] }, { "title": "为何1996周期不能和其它生成程序同读", "paragraphs": [ "程序将这些常量写出后连续执行8次步骤提交。报告外存读取为0，写出20,480字节，正好对应5,120个32位浮点数。它测到的主要是装入常量和写出结果的工作，没有覆盖一般输入下的模型计算。", "因此换新输入时，这段程序仍写同一批值，无法作为通用生成实现。硬件只有8个计算单元、整段耗时15,583,748周期，这些配置与指标也不能替代对生成语义的检查。" ] }, { "title": "结论到哪里为止", "paragraphs": [ "静态程序足以确认生成输出不依赖输入；它不能说明常量最初从哪里得到、作者的动机是什么，也不能证明服务器隐藏验证被绕过。快照成绩来自本地报告。", "本页保留它在冻结榜单中的位置，便于说明数据来源；排除它的已验证性能比较和分差拆分。其它方案的工程经验不能从这个固定输出耗时推导。" ] } ], "lesson": "面对异常大的加速，先核对完成的工作是否一致，再讨论优化机制。结果的输入依赖、必需输出与完整步骤，是比一个漂亮周期数更基础的证据。", "process": [], "sources": [] }, "label": "方案14", "trace_file_count": 0, "trace_hashes": [], "area": 5.59764068864, "programs": { "M1_P1": { "cycles": 15583748, 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"question": "计算单元与共享暂存都不少，为什么仍需从实际流量和任务映射查起？", "sections": [ { "title": "资源数量不等于有效并行", "paragraphs": [ "方案15有20个计算单元，每个配96 KiB共享暂存；矩阵阵列沿求和方向并行1项。整段有40个工作组生命周期，生成有16个。工作组如何映射、哪些阶段同时运行，无法只由这些数量确定。", "更多单元扩大了可用资源，但如果依赖限制可运行任务，或数据集中经过少数通路，规格优势就未必转成速度。需要把任务和数据的实际分配对应起来。" ] }, { "title": "两份程序都实际使用共享暂存", "paragraphs": [ "整段源码直接涉及共享暂存的读写有1,440行，生成有6,480行。共享暂存可以保存可复用数据，也要付出进入和离开它的搬运成本；这些行数仅说明存在路径，不能当作精确传输量。", "生成权重和原始历史仅在首步读取，之后仍会发生内部存储之间的搬运。常见生成指令为1行×8列×32项和1行×16列×32项；小输出块有较低单次容量需求，也可能带来更多指令与交接。" ] }, { "title": "从可见的流量提出下一步问题", "paragraphs": [ "整段外存读取约63.34 MB，高于本页多数已验证方案；最终整段569,942周期、生成61,638周期。这使重复读取、布局放大和中间转存成为值得检查的方向。", "但大流量不是因果证明：应按数据对象检查哪些读取不可避免、哪些由当前任务分工重复产生，再看减少它们能否提前完成时间。该提交没有搜索记录，不能说作者已经做过这套诊断或确定某一级存储就是瓶颈。" ] } ], "lesson": "硬件表只是起点。沿数据经过的每一级存储追踪读取次数、拥有者和等待者，才能判断资源是在帮助复用还是增加交接。", "process": [], "sources": [] }, "label": "方案15", "trace_file_count": 0, "trace_hashes": [], "area": 23.628502758400003, "programs": { "M1_P1": { "cycles": 569942, "peak_power_w": 19.85411954644595, 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80005.4299018277, "p1": 237647, "d1": 26532, "summary_sha256": "695e1f356bafa54f95e9bbc6c497d5f145cd64cc289d464e1da19edaff064536", "plan_sha256": "456cde21848c177e59d35013e62b33f1bd8e6aed9868330dc3d167b469f5af63", "source": "project/copt80_runs/search-state-loose-affine-loads-plan.json", "hardware_sha256": "22cb8a8e5a1515a7e89eb53386a5c16b21c328308f37ed6a6dff164027d0de98", "evidence_kind": "saved complete-grade summary and hardware plan; original historical report omitted from ZIP" }, { "label": "合并已验证的写回与生成改动", "explanation": "整段输入写回按8组一批发出，和上一项生成加载方式组合；固定硬件，最终两题为237,098 / 26,532周期。", "score": 80098.00256063155, "p1": 237098, "d1": 26532, "summary_sha256": "3dc77d213dba0ddc4d8e7652e3243c1cc1399d18736ef6cafb06bbec37c9e34d", "plan_sha256": "12341ec99164c0eb4c2f7dc16ca0e006044568fe4dead4d1f3caf0cdd41e7ac3", "source": "project/copt80_runs/search-state-cohort-output-current-plan.json", "hardware_sha256": "22cb8a8e5a1515a7e89eb53386a5c16b21c328308f37ed6a6dff164027d0de98", "evidence_kind": 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