evolvinglmms-lab Purpose / 用途
Use this skill when analyzing, debugging, or tuning LengthPoolSortDataset — the cross-rank length synchronization mechanism used in this repository's training pipeline.
在分析、调试或调优本仓库训练 pipeline 中的跨 rank 长度同步机制 LengthPoolSortDataset 时，使用这个 skill。
This skill is specifically for:
aiak_training_llm/data/multimodal/length_sort_dataset.py- Understanding why training speed improves when
length_sort_pool_size > 0 - Tuning
pool_sizefor optimal multi-GPU efficiency - Diagnosing rank synchronization bottlenecks
这个 skill 专门用于：
aiak_training_llm/data/multimodal/length_sort_dataset.py- 理解为什么
length_sort_pool_size > 0时训练速度提升 - 调优
pool_size以获得最佳多卡效率 - 排查 rank 间同步瓶颈
Core mechanism / 核心机制
Three-step pipeline / 三步流水线
上游 dataset → 累积 pool_size 个 sample → 按序列长度排序 → 用确定性 seed shuffle → 逐个 yield
for batch_idx, sample in enumerate(self.dataset):
pool.append(sample)
if len(pool) >= self.pool_size:
pool.sort(key=self.key_fn) # 1. 按长度排序
shuffle_seed = 42 + batch_idx # 2. 确定性 seed
random.Random(shuffle_seed).shuffle(pool) # 3. 同 seed shuffle
for s in pool:
yield s
pool.clear()
Pipeline position / 在 pipeline 中的位置
CrudeWebdataset → ShuffleBuffer → cook_crude_sample → encode_sample
→ LengthPoolSortDataset → BatchDataset → EpochizeDataset → LogSampleDataset
Inserted after encode_sample (where total_len / tokens are available) and before BatchDataset.
插在 encode_sample 之后（此时已有 total_len / tokens）、BatchDataset 之前。
Activated by: --length-sort-pool-size N (where N > 0).
通过 --length-sort-pool-size N（N > 0）激活。
Why it accelerates training / 为什么能加速训练
The problem / 问题
In multi-GPU data-parallel training, all ranks must synchronize at each step (gradient all-reduce). If different ranks process samples of very different lengths, fast ranks idle waiting for slow ranks.
多卡数据并行训练中，所有 rank 每步都要同步（梯度 all-reduce）。如果不同 rank 处理的 sample 长度差异很大，快的 rank 空等慢的 rank。
无 pool sort:
Rank 0 step 100: 长度 200 → 0.5s ┐
Rank 1 step 100: 长度 5000 → 3.0s ├→ 所有 rank 等 3.0s
Rank 2 step 100: 长度 300 → 0.6s ┘
The solution / 解决方案
Sort + same-seed shuffle ensures all ranks yield samples of approximately the same length at the same time.
排序 + 同 seed shuffle 保证所有 rank 在同一时刻输出近似相同长度的 sample。
有 pool sort:
Rank 0 step 100: 长度 ~200 → 0.5s ┐
Rank 1 step 100: 长度 ~200 → 0.5s ├→ 所有 rank 等 0.5s
Rank 2 step 100: 长度 ~200 → 0.5s ┘
Why it works — step by step / 为什么有效——逐步分析
Step 1: Sort aligns the i-th position across ranks / 排序对齐各 rank 第 i 个位置
Each rank's data comes from different shards of the same dataset, so length distributions are approximately identical. After sorting, the i-th position in each rank's pool corresponds to the i-th quantile of its length distribution. Similar distributions → similar quantiles → similar lengths.
各 rank 的数据来自同一数据集的不同 shard，长度分布近似相同。排序后，各 rank pool 中第 i 个位置对应各自长度分布的第 i 个分位数。分布相似 → 分位数相似 → 长度相似。
Rank 0 排序后: [100, 102, 105, 108, ..., 4998, 5000]
Rank 1 排序后: [101, 103, 106, 109, ..., 4999, 5001]
Rank 2 排序后: [ 99, 104, 107, 110, ..., 4997, 5002]
Step 2: Same seed preserves alignment after shuffle / 同 seed 保持 shuffle 后的对齐
shuffle_seed = 42 + batch_idx is identical across all ranks → same permutation applied to all pools. Since the i-th position had similar lengths, the shuffled output at the same position still has similar lengths.
shuffle_seed = 42 + batch_idx 对所有 rank 相同 → 相同排列应用于所有 pool。由于第 i 个位置的长度相似，shuffle 后同一位置的输出长度仍然相似。
permutation = [3, 0, 4, 1, 2]:
Rank 0 输出: [108, 100, 5000, 102, 105]
Rank 1 输出: [109, 101, 5001, 103, 106]
Rank 2 输出: [110, 99, 5002, 104, 107]
→ 同一位置长度近似一致
Why both sort AND shuffle are needed / 为什么排序和 shuffle 缺一不可
| 方案 | 效果 |
|---|---|
| 只 sort 不 shuffle | 所有 rank 先跑短 sample 后跑长 sample。前期 step 极快，后期 step 极慢。训练动态不稳定 |
| sort + shuffle | 各 step 长度随机但 rank 间一致。step 耗时平稳，rank 间同步开销小 |
| 不 sort 不 shuffle | 各 rank 同一 step 长度随机且不一致，快的等慢的 |
Sort solves cross-rank synchronization. Shuffle solves temporal uniformity. Both are required.
排序解决跨 rank 同步问题，shuffle 解决时序均匀性问题。二者缺一不可。
pool_size tuning / pool_size 调优
| pool_size | 跨 rank 长度同步精度 | 内存开销 | 首个 pool 输出延迟 |
|---|---|---|---|
| 小（~100） | 较差，各 rank 分位数估计不准 | 低 | 低 |
| 中（~1000-10000） | 好，推荐范围 | 中 | 中 |
| 大（~全量） | 完美同步 | 高 | 高 |
| 极限 = dataset 大小 | 等价全局排序 | 不实际 | 不实际 |
Larger pool_size → more accurate quantile estimation across ranks → better synchronization → less idle time.
pool_size 越大 → 各 rank 分位数估计越准 → 同步越好 → 空等越少。
Rule of thumb: pool_size should be significantly larger than batch_size, ideally 10x-100x.
经验法则：pool_size 应远大于 batch_size，理想情况下 10x-100x。
Multi-worker behavior (num_workers > 1) / 多 worker 行为
When num_workers > 1, each worker runs an independent LengthPoolSortDataset instance with its own pool.
当 num_workers > 1 时，每个 worker 运行独立的 LengthPoolSortDataset 实例，各自维护独立 pool。
Correctness: No issue. Each worker sorts its own shard subset independently.
Synchronization effect: Diluted. DataLoader interleaves outputs from multiple workers, partially disrupting the within-pool length ordering.
Recommendation:
num_workers=1gives the strongest synchronization effect. Ifnum_workers > 1is needed for I/O throughput, increasepool_sizeproportionally.正确性：无问题。每个 worker 独立排序自己的 shard 子集。
同步效果：被稀释。DataLoader 交替取多个 worker 的输出，部分打乱 pool 内的长度顺序。
建议：
num_workers=1同步效果最强。如果需要num_workers > 1提升 I/O 吞吐，可按比例增大pool_size。
Checkpoint resume caveat / checkpoint 恢复注意事项
save_state / restore_state delegate directly to the upstream dataset. The pool's internal state (accumulated but not yet yielded samples) is NOT saved.
save_state / restore_state 直接委托给上游 dataset。pool 内部状态（已累积但未 yield 的 sample）不会被保存。
On resume, up to pool_size - 1 samples may be re-ordered differently or skipped. This is generally negligible for large datasets.
恢复时，最多 pool_size - 1 个 sample 可能会被不同地排序或跳过。对于大数据集来说通常可以忽略。
key_fn / 排序键
key_fn=lambda s: getattr(s, "total_len", len(getattr(s, "tokens")))
Prefers total_len attribute (set by encode_sample), falls back to len(tokens).
优先使用 total_len 属性（由 encode_sample 设置），回退到 len(tokens)。
What to check during debugging / 调试时要检查什么
Is
--length-sort-pool-sizeactually set and > 0?What is the actual length distribution of the dataset? (High variance → more benefit from pool sort)
Are all ranks receiving data from the same underlying dataset with similar length distributions?
Is
num_workers> 1 diluting the synchronization effect?After enabling, did step time variance across steps decrease?
Is memory usage acceptable for the chosen
pool_size?--length-sort-pool-size是否真正设置且 > 0？数据集的实际长度分布如何？（方差越大 → pool sort 收益越大）
所有 rank 是否从同一底层数据集接收长度分布相似的数据？
num_workers> 1 是否稀释了同步效果？启用后，各 step 的 step time 方差是否降低了？
所选
pool_size的内存占用是否可接受？
Expected outputs when using this skill / 使用本 skill 时的期望输出
When asked to analyze a training speed issue related to length sorting, return:
当被要求分析与长度排序相关的训练速度问题时，应返回：
Whether
LengthPoolSortDatasetis active and with whatpool_sizeThe dataset's length distribution characteristics (high/low variance)
The relationship between
pool_size,batch_size, andnum_workersWhether cross-rank synchronization is the bottleneck
Recommended
pool_sizeadjustment if neededLengthPoolSortDataset是否激活，pool_size是多少数据集的长度分布特征（高/低方差）
pool_size、batch_size、num_workers之间的关系跨 rank 同步是否是瓶颈
如需要，推荐的
pool_size调整