Scaling LawScaling Laws

假设你的团队有一笔预算：可以再采 1 万条遥操作示教，可以把策略网络放大 10 倍，也可以买更多 GPU 把现有模型训练得更久。三件事都花钱，花的还是不同的钱：示教花的是人的时间，网络放大花的是训练和机器人上的推理算力，训练更久花的是集群时间。你想在花钱之前知道哪一项最划算。

语言和视觉模型过去几年的一个重要发现是：测试损失随数据量 \(D\)、参数量 \(N\) 和训练算力 \(C\) 的变化，在好几个数量级的范围里都非常规律，近似是幂律 \(L\propto N^{-\alpha}\)。这种规律叫 scaling law。有了它，就能用一串小规模实验拟合出直线，再外推到大规模，预测大模型的损失、决定参数和数据的配比。GPT-3、Chinchilla 这样的大模型就是这样规划出来的。

对机器人，这个问题更尖锐：机器人数据极其昂贵（每一条示教都要人花几十秒到几分钟去遥操作，22.3 节），模型还要在机器人自己的计算机上以控制频率实时运行。16.2 节讨论的是“给定数据，怎样泛化得好”；本节问的是下一个问题：多给多少数据、多大的模型，能换来多少进步，以及这种预测什么时候会失灵。

Suppose your team has a budget: you could collect 10,000 more teleoperated demonstrations, make the policy network 10 times larger, or buy more GPUs and train the current model longer. All three cost money, and different kinds of money: demonstrations cost human time, a bigger network costs training compute and inference compute on the robot, longer training costs cluster time. You want to know which is the best deal before you spend.

One of the key findings for language and vision models in recent years is that the test loss depends on the amount of data \(D\), the number of parameters \(N\) and the training compute \(C\) in a remarkably regular way over many orders of magnitude, approximately as a power law \(L\propto N^{-\alpha}\). Such a regularity is called a scaling law. With one, you can fit a straight line to a series of small experiments, extrapolate it to large scale, predict a large model's loss and decide the ratio of parameters to data. Large models such as GPT-3 and Chinchilla were planned this way.

For robots the question is sharper: robot data are extremely expensive (each demonstration takes a person tens of seconds to minutes of teleoperation, Lesson 22.3), and the model has to run in real time at control rate on the robot's own computer. Lesson 16.2 asked "given the data, how do we generalize well"; this lesson asks the next question: how much progress do more data and a bigger model buy, and when does that prediction break down.

先看幂律长什么样。如果 \(L=A\,n^{-\alpha}\)，两边取对数得 \(\log L=\log A-\alpha\log n\)：在双对数坐标上是一条直线，斜率是 \(\hl{slope}{-\alpha}\)。它的含义是“每多 10 倍数据，损失乘以 \(10^{-\alpha}\)”，不管你现在在哪个位置。\(\alpha=1\) 时 10 倍数据换来 10 倍更低的损失；\(\alpha=0.1\) 时只换来 1.26 倍，想把损失降一半要多 1000 倍的数据。指数 \(\alpha\) 就是“规模的回报率”。

为什么会是幂律？一个能完全看清楚的例子：机器人要从样本里学一个光滑函数（比如某种状态下的抓取质量），而它用最简单的办法：最近邻，查询时直接返回离它最近的那个训练样本的值。误差有多大，取决于最近的样本有多远。如果 \(n\) 个样本均匀撒在一个 \(d\) 维的单位立方体里，每个样本“管辖”的体积约为 \(1/n\)，对应的边长（也就是到最近样本的典型距离）\(h\propto n^{-1/d}\)。光滑函数的误差正比于距离，平方误差正比于 \(h^2\propto n^{-2/d}\)：一条幂律，指数由输入空间的维数决定。

下图在 \(d=1,2,4,8\) 维里做这个实验（目标函数是 \(d\) 个正弦之和，样本和测试点都均匀随机）。拖动 \(n\)，看左边的几何和右边的双对数曲线。

First, what a power law looks like. If \(L=A\,n^{-\alpha}\), taking logs gives \(\log L=\log A-\alpha\log n\): a straight line on log-log axes, with slope \(\hl{slope}{-\alpha}\). It means "every 10 times more data multiplies the loss by \(10^{-\alpha}\)", wherever you are now. With \(\alpha=1\), 10 times the data buys a 10 times lower loss; with \(\alpha=0.1\) it buys only a factor of 1.26, and halving the loss takes 1000 times more data. The exponent \(\alpha\) is the "return on scale".

Why a power law? An example you can see completely: a robot learns a smooth function from samples (say the grasp quality as a function of the state), and it uses the simplest method there is: nearest neighbour, which answers a query with the value of the closest training sample. The error depends on how far away that closest sample is. If \(n\) samples are spread uniformly in a \(d\)-dimensional unit cube, each sample "owns" a volume of about \(1/n\), whose side length (the typical distance to the nearest sample) is \(h\propto n^{-1/d}\). For a smooth function the error is proportional to the distance, so the squared error is proportional to \(h^2\propto n^{-2/d}\): a power law, with an exponent set by the dimension of the input space.

The figure below runs this experiment in \(d=1,2,4,8\) dimensions (the target is a sum of \(d\) sines; samples and test points are uniform at random). Drag \(n\) and watch the geometry on the left and the log-log curves on the right.

拟合斜率（\(n\) 从 64 到 4096）：\(d=1\) 为 −2.02，\(d=2\) 为 −1.03，\(d=4\) 为 −0.53，\(d=8\) 为 −0.24，几乎正好是 \(-2/d\)。换成“要把误差降到十分之一需要多少倍数据”：\(d=1\) 约 3 倍，\(d=2\) 10 倍，\(d=4\) 100 倍，\(d=8\) 一万倍。这就是维数灾难的定量形式。

这个玩具模型抓住了真实 scaling law 的两个要点。第一，指数取决于数据的内在维数（数据实际占据的流形的维数），而不是原始输入的维数：一张 224×224 的图像有 15 万个像素，但同一个场景的图像只在一个低得多的维度里变化。Sharma 与 Kaplan（2022）论证了神经网络的指数近似为 \(4/d\)（网络是分段线性的近似器，误差 \(\propto h^2\)，比最近邻的 \(\propto h\) 好一阶）。第二，指数通常很小：语言模型的实测指数在 0.05 到 0.1 左右（第 7 节），对应很高的内在维数，所以每一点进步都需要成倍的规模。机器人的情况类似：物体、位姿、光照、背景、本体……决定动作的因素很多，所以机器人数据的“填满”格外困难（第 8 节）。

Fitted slopes (for \(n\) from 64 to 4096): −2.02 for \(d=1\), −1.03 for \(d=2\), −0.53 for \(d=4\), −0.24 for \(d=8\), almost exactly \(-2/d\). In terms of "how many times more data to cut the error tenfold": about 3 times for \(d=1\), 10 for \(d=2\), 100 for \(d=4\), ten thousand for \(d=8\). This is the curse of dimensionality in quantitative form.

The toy model captures two essentials of real scaling laws. First, the exponent depends on the intrinsic dimension of the data (the dimension of the manifold the data actually occupy), not on the raw input dimension: a 224×224 image has 150 thousand pixels, but images of one scene vary along far fewer dimensions. Sharma and Kaplan (2022) argued that for neural networks the exponent is about \(4/d\) (a network is a piecewise-linear approximator with error \(\propto h^2\), one order better than nearest neighbour's \(\propto h\)). Second, the exponents are usually small: the measured exponents for language models are around 0.05 to 0.1 (Section 7), which corresponds to a high intrinsic dimension, so every bit of progress takes many times the scale. Robots are in a similar position: many factors decide the action (objects, poses, lighting, background, embodiment…), which is why "filling" the space with robot data is especially hard (Section 8).

最近邻是非参数方法，它的“模型大小”就是数据本身。神经网络有固定数量的参数 \(N\)，训练要花算力。Kaplan 等人（2020）的经典做法是：训练一族大小不同的模型，每个都用新鲜数据（每个样本只用一次）训练，把每个模型的测试损失画成训练算力的函数。算力按标准近似 \(C\approx6ND\) 计：每个样本前向约 \(2N\) 次浮点运算（每个参数一次乘一次加），反向约 \(4N\) 次，\(D\) 是见过的样本数。

下图在浏览器里同时训练 6 个单隐藏层 GELU 网络，宽度 2 到 64（20 到 578 个参数），学习一台两连杆机械臂的逆动力学：输入 6 个数 \((\mathbf q,\dot{\mathbf q},\ddot{\mathbf q})\)，输出 2 个关节力矩 \(\boldsymbol\tau=M(\mathbf q)\ddot{\mathbf q}+\mathbf b(\mathbf q,\dot{\mathbf q})\)（8.3 节的模型：每根连杆 1 kg、1 m，质心在中点，重力 9.81 m/s²；力矩除以 12 归一化）。每一步用 16 个新采的样本，Adam，学习率 0.01 按余弦衰减到零，共 6000 步。

Nearest neighbour is non-parametric: its "model size" is the data itself. A neural network has a fixed number of parameters \(N\), and training it costs compute. The classic recipe of Kaplan et al. (2020): train a family of models of different sizes, each on fresh data (every sample used once), and plot each model's test loss as a function of its training compute. Compute is counted with the standard approximation \(C\approx6ND\): about \(2N\) floating-point operations per sample forward (one multiply and one add per parameter) and about \(4N\) backward, where \(D\) is the number of samples seen.

The figure below trains 6 one-hidden-layer GELU networks at once in your browser, widths 2 to 64 (20 to 578 parameters), on the inverse dynamics of a two-link arm: 6 inputs \((\mathbf q,\dot{\mathbf q},\ddot{\mathbf q})\), 2 outputs, the joint torques \(\boldsymbol\tau=M(\mathbf q)\ddot{\mathbf q}+\mathbf b(\mathbf q,\dot{\mathbf q})\) (the model of Lesson 8.3: links of 1 kg and 1 m with the centre of mass at the midpoint, gravity 9.81 m/s²; torques divided by 12 to normalize). Each step uses 16 freshly drawn samples, Adam, learning rate 0.01 with cosine decay to zero, 6000 steps in all.

读图的三个要点。(1) 每条曲线都有平台：开始时所有网络都卡在约 0.5（这大约是目标的方差：只学到了平均值），然后迅速降到 0.09 左右的一个“肩”（占方差大部分的重力项学会了，惯性项和科氏项还没有），再继续下降，最后在一个由容量决定的水平上变平。小模型下降得早，但很快就“装不下”了；大模型每一步更贵，起步慢，但能降得更低。(2) 下包络是“给定算力能达到的最低损失”，它依次经过越来越大的模型，近似是一条直线：在默认种子下从 \(2.5\times10^6\) FLOPs 的 0.094 降到 \(3.2\times10^8\) FLOPs 的 0.0028，斜率约 −0.77（换种子在 −0.65 到 −0.80 之间）。这就是“算力的 scaling law” \(L^\ast(C)\propto C^{-\gamma}\)。(3) 把每个模型训练到最后的损失对 \(N\) 作图，6 个点（0.089、0.053、0.023、0.011、0.0050、0.0027）也近似在一条斜率约 −1.07 的直线上。

任务 2 揭示了不可约损失：测试目标带噪声时，再大的模型也只能逼近 \(\sigma^2\)，曲线在双对数坐标上向上弯，直线拟合的斜率被低估。正确的形式是 \(L=\hl{floor}{E}+A\,N^{-\alpha}\)，其中 \(E\) 是数据本身的熵（噪声、真正的随机性），只有 \(L-E\) 才服从幂律。任务 3 是 Kaplan 与 Chinchilla 结论不同的原因之一（第 5 节）：超参数，尤其是学习率衰减的长度，必须和每次训练的长度匹配；配错了，大模型显得比实际差，你就会低估规模的回报。

Three things to read off. (1) Every curve has plateaus: at first all networks sit at about 0.5 (roughly the variance of the targets: they have learned only the mean), then drop quickly to a shoulder near 0.09 (the gravity terms, which carry most of the variance, are learned; the inertial and Coriolis terms are not yet), keep falling, and finally flatten at a level set by their capacity. Small models drop early but soon "run out of room"; big models are more expensive per step and start slower, but go lower. (2) The lower envelope is "the lowest loss reachable with a given compute"; it passes through larger and larger models and is close to a straight line: with the default seed it goes from 0.094 at \(2.5\times10^6\) FLOPs to 0.0028 at \(3.2\times10^8\) FLOPs, a slope of about −0.77 (between −0.65 and −0.80 for other seeds). This is the "scaling law in compute" \(L^\ast(C)\propto C^{-\gamma}\). (3) Plotting each model's loss at the end of training against \(N\), the 6 points (0.089, 0.053, 0.023, 0.011, 0.0050, 0.0027) also lie close to a straight line, with slope about −1.07.

Task 2 reveals the irreducible loss: when the test targets are noisy, even an infinitely large model can only approach \(\sigma^2\), so the curve bends upward on log-log axes and a straight-line fit underestimates the slope. The right form is \(L=\hl{floor}{E}+A\,N^{-\alpha}\), where \(E\) is the entropy of the data itself (noise, genuine randomness), and only \(L-E\) follows the power law. Task 3 is one reason the conclusions of Kaplan and of Chinchilla differ (Section 5): hyperparameters, above all the length of the learning-rate decay, must match the length of each training run; get that wrong and big models look worse than they are, so you underestimate the return on scale.

scaling law 的价值在于外推：用便宜的小实验预测昂贵的大实验。做法是1.13 节的最小二乘，在对数空间里做：纯幂律对 \(\log L\) 和 \(\log N\) 拟合一条直线；带地板的形式 \(E+AN^{-\alpha}\) 有三个参数，这里对 \(E\) 做网格搜索、对每个 \(E\) 拟合 \(\log(L-E)\) 的直线，选对数残差平方和最小的那个。在对数空间拟合，是因为不同随机种子之间的波动通常近似是乘性的（下图假设为 1%）。

下图的数据点来自一个隐藏的“真实规律”，加上 1% 的乘性噪声，每半个数量级一个模型，从 \(10^6\) 到 \(10^{12}\) 个参数。真实规律默认是 Chinchilla 拟合公式中的参数项 \(1.69+406.4\,N^{-0.34}\)（数据无限时的损失）。你只能“看到”拟合范围里的点；任务是预测 \(10^{12}\) 处的损失。

The value of a scaling law lies in extrapolation: predicting an expensive large experiment from cheap small ones. The method is the least squares of Lesson 1.13, done in log space: the pure power law fits a straight line to \(\log L\) against \(\log N\); the form with a floor, \(E+AN^{-\alpha}\), has three parameters, and here we grid-search \(E\), fit a straight line to \(\log(L-E)\) for each \(E\), and keep the one with the smallest sum of squared log residuals. We fit in log space because the run-to-run variation between random seeds is usually roughly multiplicative (the figure below assumes 1%).

The data points in the figure below come from a hidden "true law" plus 1% multiplicative noise, one model every half order of magnitude from \(10^6\) to \(10^{12}\) parameters. By default the true law is the parameter term of Chinchilla's fitted formula, \(1.69+406.4\,N^{-0.34}\) (the loss with unlimited data). You only "see" the points inside the fit range; the job is to predict the loss at \(10^{12}\).

具体数字（默认噪声种子）。用 \(10^6\)–\(10^8\) 的 5 个点：带地板的拟合在 \(10^{12}\) 处预测 1.70（自助法范围 1.54–1.85），真实值 1.72，很好；同样的点用直线拟合，预测 0.49，因为直线以为损失可以一直降到零。把范围扩大到 \(10^9\)，直线的预测变成 0.68，它许诺的进步是真实进步的 5 倍：更多的点并不能挽救错误的函数形式。另一面，如果只拟合一个数量级（比如 \(10^{6.5}\)–\(10^{7.5}\)），带地板的拟合的自助法范围从 0.43 一直到 1.61：三个参数在这么短的直线上根本分不开，地板 \(E\) 尤其定不下来。

“规律改变”是最危险的情形：小规模的点和默认情形完全一样，任何拟合方法都看不出来，但真实曲线在 \(10^9\) 之后变平了，用 \(10^6\)–\(10^9\) 拟合的地板模型预测 1.63（范围 1.57–1.68），而真实值是 1.87，许诺的进步是实际的 3 倍。现实中这类“断点”确实出现过：数据开始重复、换了一种评测、模型进入了新的工作区间（Caballero 等，2023 把这种形状叫作 broken neural scaling laws）。外推的距离越远，越要怀疑。

Concrete numbers (default noise seed). With the 5 points from \(10^6\) to \(10^8\): the fit with a floor predicts 1.70 at \(10^{12}\) (bootstrap range 1.54–1.85) against a true 1.72, which is good; a straight line through the same points predicts 0.49, because a straight line believes the loss can fall all the way to zero. Extend the range to \(10^9\) and the straight line predicts 0.68, promising 5 times the real progress: more points cannot rescue the wrong functional form. On the other side, fit only one order of magnitude (say \(10^{6.5}\)–\(10^{7.5}\)) and the bootstrap range of the floor fit runs from 0.43 to 1.61: three parameters cannot be separated on such a short stretch, least of all the floor \(E\).

The "regime change" is the most dangerous case: the small-scale points are identical to the default case, so no fitting method can see it coming, but the true curve flattens beyond \(10^9\). The floor model fitted on \(10^6\)–\(10^9\) predicts 1.63 (range 1.57–1.68), the truth is 1.87, and the promised progress is 3 times the real one. Such "breaks" do happen in practice: data starts to repeat, the evaluation changes, the model enters a new regime (Caballero et al., 2023 call these shapes broken neural scaling laws). The further you extrapolate, the more suspicious you should be.

算力 \(\hl{iso}{C\approx6ND}\) 是参数量和数据量的乘积。预算固定时，模型大一点就只能少看一些数据，反之亦然。最好的配比是多少？Kaplan 等人（2020）的结论是模型应该比数据长得快得多：\(N_{\rm opt}\propto C^{0.73}\)。两年后 Hoffmann 等人（2022，DeepMind）用 400 多个模型、每个规模都把学习率衰减调到与训练长度匹配，得到了不同的结论：参数和数据应该同比例增长，\(N_{\rm opt}\propto C^{a}\)、\(D_{\rm opt}\propto C^{b}\)，\(a\approx b\approx0.5\)，相当于每个参数约 20 个 token。他们按这个配比训练了 700 亿参数、1.4 万亿 token 的 Chinchilla，算力与 2800 亿参数、3000 亿 token 的 Gopher 相近，却在大多数评测上更好。

他们的第三种方法直接拟合 \(L(N,D)=E+A/N^\alpha+B/D^\beta\)。原文给出的系数后来被 Besiroglu 等人（2024）复核时发现与原文的另外两种方法不一致；下图用的是这次复核重新拟合的系数（\(E=1.82\)，\(A=482\)，\(B=2085\)，\(\alpha=0.348\)，\(\beta=0.366\)），它给出的最优配比约为每参数 20 个 token，与 Chinchilla 一致。注意：这些数字属于特定的数据集、分词器和模型族（17.5 节），不能直接搬到机器人上，但形状和推理可以。

Compute \(\hl{iso}{C\approx6ND}\) is the product of parameters and data. With a fixed budget, a bigger model must see less data, and the other way round. What is the best split? Kaplan et al. (2020) concluded that the model should grow much faster than the data: \(N_{\rm opt}\propto C^{0.73}\). Two years later Hoffmann et al. (2022, DeepMind) trained over 400 models, tuning the learning-rate decay at every scale to match the training length, and reached a different conclusion: parameters and data should grow in equal proportion, \(N_{\rm opt}\propto C^{a}\), \(D_{\rm opt}\propto C^{b}\) with \(a\approx b\approx0.5\), which comes to about 20 tokens per parameter. Following that recipe they trained Chinchilla, 70 billion parameters on 1.4 trillion tokens, with about the same compute as Gopher (280 billion parameters, 300 billion tokens), and it did better on most evaluations.

Their third method fits \(L(N,D)=E+A/N^\alpha+B/D^\beta\) directly. When Besiroglu et al. (2024) re-examined it, the published coefficients turned out to be inconsistent with the paper's other two methods; the figure below uses the coefficients of that re-fit (\(E=1.82\), \(A=482\), \(B=2085\), \(\alpha=0.348\), \(\beta=0.366\)), which give an optimal split of about 20 tokens per parameter, consistent with Chinchilla. Mind that these numbers belong to a particular dataset, tokenizer and model family (Lesson 17.5) and do not carry over to robots as they are; the shape and the reasoning do.

几个读数。按复核后的公式，最优点的 token/参数比随算力缓慢变化：\(10^{18}\) 时约 26，\(10^{22}\) 时约 20，\(10^{24}\) 时约 18。等算力线上的谷很平：在 Gopher 的算力下，参数量偏离最优值 2 倍（0.3 个数量级），损失只差 0.25%；但 Gopher 的配比偏了约 4 倍，按这个公式，一个最优配比只需要它一半的算力就能达到同样的损失。GPT-3 式的配比（1750 亿参数、3000 亿 token）效率约 61%。

反过来，“过度训练”一个小模型也有它的道理。Meta 的 Llama 3 8B 用了约 15 万亿 token 训练，远超每参数 20 个 token；按图中的公式，同样的损失用最优配比只要约 18% 的算力。为什么要多花 5 倍多的训练算力？因为推理成本由 \(N\) 决定，而模型要被调用无数次。机器人更是如此：策略必须在机器人自己的计算机上以 10–50 Hz 运行，\(N\) 有硬上限。任务 2 显示了上限的后果：上限下的损失有自己的地板 \(E+A/N_{\max}^\alpha\)，再多的数据也越不过去；在到达那个地板之前，把预算花在数据上是对的。

A few readings. Under the re-fitted formula the optimal tokens-per-parameter ratio drifts slowly with compute: about 26 at \(10^{18}\), about 20 at \(10^{22}\), about 18 at \(10^{24}\). The valley along an iso-compute line is flat: at Gopher's compute, being off the optimal parameter count by a factor of 2 (0.3 orders of magnitude) costs only 0.25% in loss; but Gopher's allocation is off by about a factor of 4, and by this formula an optimal allocation would reach the same loss with half the compute. A GPT-3-like allocation (175 billion parameters, 300 billion tokens) has an efficiency of about 61%.

Conversely, "over-training" a small model has its reasons. Meta's Llama 3 8B was trained on about 15 trillion tokens, far beyond 20 tokens per parameter; by the formula in the figure an optimal allocation would reach the same loss with about 18% of the compute. Why spend over 5 times the training compute? Because inference cost is set by \(N\), and the model will be called countless times. Robots even more so: the policy must run at 10–50 Hz on the robot's own computer, so \(N\) has a hard cap. Task 2 shows the consequence of the cap: under it the loss has its own floor \(E+A/N_{\max}^\alpha\) that no amount of data can cross; until you reach that floor, spending the budget on data is right.

损失平滑下降，可很多能力看起来是突然出现的：模型小于某个规模时某项任务的成功率接近零，越过某个规模后迅速上升。Wei 等人（2022）把这类现象称为涌现能力，并收集了许多例子。Schaeffer、Miranda 与 Koyejo（2023）提出了另一种解释：很多“涌现”来自指标而不是模型。全对才算对的指标（精确匹配、整个任务成功）把平滑的逐步进步变成了陡峭的曲线；换成连续的指标，跳变往往消失。

机器人上的版本：一个长时程任务（比如收拾桌子）由 \(k\) 个子步骤组成，必须每一步都成功。设每一步的对数损失随规模平滑地按幂律下降 \(\hl{ell}{\ell(N)}=(N/10^7)^{-0.5}\)，每一步的成功率 \(\hl{pstep}{p(N)}=e^{-\ell(N)}\)，整个任务的成功率是 \(\hl{pk}{p^k}=e^{-k\ell(N)}\)。下图比较三种指标。

The loss falls smoothly, yet many abilities seem to appear suddenly: below some scale a task's success rate is near zero, beyond it the rate climbs fast. Wei et al. (2022) called such phenomena emergent abilities and collected many examples. Schaeffer, Miranda and Koyejo (2023) proposed another explanation: much "emergence" comes from the metric, not the model. All-or-nothing metrics (exact match, whole-task success) turn smooth step-by-step progress into a steep curve; switch to a continuous metric and the jump often disappears.

The robot version: a long-horizon task (say clearing a table) consists of \(k\) sub-steps, and every one must succeed. Let the per-step log-loss fall smoothly as a power law in scale, \(\hl{ell}{\ell(N)}=(N/10^7)^{-0.5}\), so each step succeeds with probability \(\hl{pstep}{p(N)}=e^{-\ell(N)}\) and the whole task with \(\hl{pk}{p^k}=e^{-k\ell(N)}\). The figure compares three metrics.

数字。\(k=20\) 时，任务成功率在 \(N=10^{8.9}\) 达到 10%、在 \(10^{11.6}\) 达到 90%；在这两点，每一步的成功率分别是 0.891 和 0.995：每一步只是从“还不错”变成“几乎完美”，整个任务却从“几乎从不成功”变成“几乎总成功”。“7 个模型 × 20 次试验”的测量值（\(k=20\)，默认种子）是 0、0、0、0.10、0.35、0.85、0.95：前三个规模都是 0/20，看上去能力是在 \(10^9\) 附近“出现”的。真正平滑的是对数损失 \(k\ell(N)\)，它在双对数坐标上是一条直线，从来没有跳变。

一个反直觉的细节：在对数横轴上，\(p^k\) 从 10% 升到 90% 跨越的宽度与 \(k\) 无关（这里总是 2.68 个数量级，第 7 节推导），\(k\) 只把整条曲线往右推。所以“越长的任务越陡”并不对，真正的变化是：长任务的上升被推迟到逐步成功率已经很高的区间，而在那之前它一直贴着零。这场争论并没有完全结束：也有研究认为某些能力即使用连续指标也表现出急剧变化。实际的教训是：报告连续指标（对数损失、部分得分），把试验次数和置信区间写清楚，不要用两三个规模的成功率去宣称能力“出现”了。机器人评测常常每个条件只做 10–50 次试验：20 次试验、真实成功率 50% 时，95% 置信区间约为 ±22 个百分点。

Numbers. For \(k=20\) the task success reaches 10% at \(N=10^{8.9}\) and 90% at \(10^{11.6}\); at those two points the per-step success is 0.891 and 0.995: each step only goes from "fairly good" to "almost perfect", while the whole task goes from "almost never" to "almost always". The "7 models × 20 trials" measurements (\(k=20\), default seed) are 0, 0, 0, 0.10, 0.35, 0.85, 0.95: the first three scales all score 0/20, so the ability seems to "appear" near \(10^9\). What is really smooth is the log-loss \(k\ell(N)\), a straight line on log-log axes that never jumps.

A counter-intuitive detail: on a logarithmic horizontal axis, the width over which \(p^k\) climbs from 10% to 90% does not depend on \(k\) (here it is always 2.68 orders of magnitude; Section 7 derives it); \(k\) only shifts the whole curve to the right. So "longer tasks are steeper" is not right; what really changes is that the rise of a long task is delayed to where per-step success is already high, and until then it hugs zero. The debate is not entirely settled: some work argues that certain abilities change sharply even under continuous metrics. The practical lesson: report continuous metrics (log-loss, partial credit), state the number of trials and the confidence intervals, and do not claim that an ability "emerged" from success rates at two or three scales. Robot evaluations often run only 10–50 trials per condition: with 20 trials and a true success rate of 50%, the 95% confidence interval is about ±22 percentage points.

纯幂律 \(L(x)=A\,x^{-\alpha}\)（\(x\) 是 \(N\)、\(D\) 或 \(C\)）满足 \(L(\lambda x)=\lambda^{-\alpha}L(x)\)：放大 \(\lambda\) 倍，损失乘以 \(\lambda^{-\alpha}\)，与起点无关，这种尺度不变性正是它在双对数坐标上是直线的原因。带地板的形式 \(L=\hl{floor}{E}+A\,x^{-\alpha}\) 的局部斜率是

$$\frac{d\log L}{d\log x}=-\alpha\,\frac{L-E}{L},$$

远离地板时（\(L\gg E\)）接近 \(-\alpha\)，接近地板时趋于零：这就是 FIG 2 加噪声后和 FIG 3 里曲线向上弯的原因。在一段有限的范围里用直线去拟合，得到的是这段范围里的平均局部斜率，它系统性地小于 \(\alpha\)，而且外推时会一直保持这个斜率，所以会低估地板、高估进步。

A pure power law \(L(x)=A\,x^{-\alpha}\) (with \(x\) being \(N\), \(D\) or \(C\)) satisfies \(L(\lambda x)=\lambda^{-\alpha}L(x)\): scale up by \(\lambda\) and the loss is multiplied by \(\lambda^{-\alpha}\), wherever you start; this scale invariance is exactly why it is a straight line on log-log axes. The form with a floor, \(L=\hl{floor}{E}+A\,x^{-\alpha}\), has the local slope

$$\frac{d\log L}{d\log x}=-\alpha\,\frac{L-E}{L},$$

close to \(-\alpha\) far above the floor (\(L\gg E\)) and tending to zero near it: that is why the curves bend upward in FIG 2 with noise and in FIG 3. A straight line fitted over a finite range measures the average local slope over that range, which is systematically smaller than \(\alpha\), and extrapolation keeps that slope forever, so it underestimates the floor and overestimates the progress.

设测量值 \(L_i=L(x_i)\,e^{\varepsilon_i}\)，\(\varepsilon_i\sim\mathcal N(0,s^2)\)（乘性噪声）。纯幂律的最大似然估计就是 \(y_i=\log L_i\) 对 \(u_i=\log x_i\) 的普通最小二乘（1.13 节）：

$$\hat\alpha=-\frac{\sum_i(u_i-\bar u)(y_i-\bar y)}{\sum_i(u_i-\bar u)^2},\qquad \operatorname{Var}(\hat\alpha)=\frac{s^2}{\sum_i(u_i-\bar u)^2}.$$

分母是拟合点在 \(\log x\) 上的分散程度：范围窄一半，\(\hat\alpha\) 的标准差大一倍；而外推到距离中心 \(\Delta u\) 处，\(\log L\) 的预测误差约为 \(\Delta u\cdot\mathrm{sd}(\hat\alpha)\)，随外推距离线性增长。带地板时 \(\log L=\log(E+Ae^{-\alpha u})\) 对 \(E\) 非线性；FIG 3 对 \(E\) 做网格搜索，对每个 \(E\) 做上面的线性拟合。当拟合范围很窄、\(L-E\) 在范围内变化不大时，\(E\)、\(A\)、\(\alpha\) 几乎可以互相替代（残差几乎一样），这就是 FIG 3 里一个数量级的拟合定不下来的原因。

Let the measurements be \(L_i=L(x_i)\,e^{\varepsilon_i}\) with \(\varepsilon_i\sim\mathcal N(0,s^2)\) (multiplicative noise). The maximum-likelihood estimate of a pure power law is ordinary least squares of \(y_i=\log L_i\) on \(u_i=\log x_i\) (Lesson 1.13):

$$\hat\alpha=-\frac{\sum_i(u_i-\bar u)(y_i-\bar y)}{\sum_i(u_i-\bar u)^2},\qquad \operatorname{Var}(\hat\alpha)=\frac{s^2}{\sum_i(u_i-\bar u)^2}.$$

The denominator is the spread of the fitted points in \(\log x\): halve the range and the standard deviation of \(\hat\alpha\) doubles; and extrapolating to a distance \(\Delta u\) from the centre, the prediction error of \(\log L\) is about \(\Delta u\cdot\mathrm{sd}(\hat\alpha)\), growing linearly with the extrapolation distance. With a floor, \(\log L=\log(E+Ae^{-\alpha u})\) is nonlinear in \(E\); FIG 3 grid-searches \(E\) and does the linear fit above for each \(E\). When the range is narrow and \(L-E\) changes little across it, \(E\), \(A\) and \(\alpha\) can almost substitute for one another (the residuals are almost the same), which is why a one-decade fit in FIG 3 pins nothing down.

\(n\) 个均匀样本在 \(d\) 维单位立方体里，一个测试点 \(\mathbf x\) 到最近样本的距离 \(r\) 满足 \(P(r>t)=(1-V_dt^d)^n\approx e^{-nV_dt^d}\)（\(V_d\) 是单位球体积），所以 \(r\) 的典型大小是 \((nV_d)^{-1/d}\)，\(\mathbb E[r^2]\propto n^{-2/d}\)。对光滑的 \(f\)，最近邻的误差 \(f(\mathbf x_{\rm nn})-f(\mathbf x)\approx\nabla f(\mathbf x)^\top(\mathbf x_{\rm nn}-\mathbf x)\)，方向随机时其平方的期望是 \(\tfrac1d\|\nabla f\|^2\,\mathbb E[r^2]\)，于是

$$\text{MSE}(n)\propto n^{-2/d}.$$

分段线性的近似（ReLU 网络、线性插值）的误差是 \(O(r^2)\)，平方误差 \(\propto n^{-4/d}\)；用 \(N\) 个参数把输入空间分成 \(N\) 块时，同样的论证给出 \(L\propto N^{-4/d}\)，这是 Sharma 与 Kaplan（2022）把实测指数与数据流形内在维数联系起来的思路。一般地，函数越光滑、内在维数越低，指数越大。

With \(n\) uniform samples in a \(d\)-dimensional unit cube, the distance \(r\) from a test point \(\mathbf x\) to its nearest sample satisfies \(P(r>t)=(1-V_dt^d)^n\approx e^{-nV_dt^d}\) (\(V_d\) is the volume of the unit ball), so \(r\) is typically \((nV_d)^{-1/d}\) and \(\mathbb E[r^2]\propto n^{-2/d}\). For a smooth \(f\) the nearest-neighbour error is \(f(\mathbf x_{\rm nn})-f(\mathbf x)\approx\nabla f(\mathbf x)^\top(\mathbf x_{\rm nn}-\mathbf x)\), whose square has expectation \(\tfrac1d\|\nabla f\|^2\,\mathbb E[r^2]\) for a random direction, so

$$\text{MSE}(n)\propto n^{-2/d}.$$

A piecewise-linear approximation (a ReLU network, linear interpolation) has error \(O(r^2)\), squared error \(\propto n^{-4/d}\); when \(N\) parameters cut the input space into \(N\) pieces the same argument gives \(L\propto N^{-4/d}\), which is how Sharma and Kaplan (2022) linked measured exponents to the intrinsic dimension of the data manifold. In general, the smoother the function and the lower the intrinsic dimension, the larger the exponent.

Kaplan 等人（2020）在 Transformer 语言模型上测得（\(N\) 不计嵌入层参数，损失是每个 token 的交叉熵，单位 nat）：

$$\begin{gathered}L(N)\approx\Big(\frac{N_c}{N}\Big)^{\alpha_N},\ \alpha_N\approx0.076;\qquad L(D)\approx\Big(\frac{D_c}{D}\Big)^{\alpha_D},\ \alpha_D\approx0.095;\\ L(C_{\min})\propto C_{\min}^{-0.050},\end{gathered}$$

前两条分别是数据无限（训练到收敛）和模型足够大（早停）时的情形，第三条是 FIG 2 那样的算力前沿 \(\hl{front}{L^\ast(C)}\)。三个指数都很小：\(\alpha_N=0.076\) 意味着参数多 10 倍，损失只乘以 \(10^{-0.076}\approx0.84\)。他们还提出了联合形式 \(L(N,D)=\big[(N_c/N)^{\alpha_N/\alpha_D}+D_c/D\big]^{\alpha_D}\)，并从对算力高效训练的分析（模型大小与训练步数的联合拟合）得出 \(N_{\rm opt}\propto C^{0.73}\)。这些纯幂律没有地板项，只在测量的范围里成立。

Kaplan et al. (2020) measured on Transformer language models (with \(N\) excluding embedding parameters and the loss being the per-token cross-entropy in nats):

$$\begin{gathered}L(N)\approx\Big(\frac{N_c}{N}\Big)^{\alpha_N},\ \alpha_N\approx0.076;\qquad L(D)\approx\Big(\frac{D_c}{D}\Big)^{\alpha_D},\ \alpha_D\approx0.095;\\ L(C_{\min})\propto C_{\min}^{-0.050},\end{gathered}$$

the first two for unlimited data (trained to convergence) and for large enough models (early-stopped), the third the compute frontier \(\hl{front}{L^\ast(C)}\) as in FIG 2. All three exponents are small: \(\alpha_N=0.076\) means 10 times the parameters multiplies the loss by only \(10^{-0.076}\approx0.84\). They also proposed the joint form \(L(N,D)=\big[(N_c/N)^{\alpha_N/\alpha_D}+D_c/D\big]^{\alpha_D}\), and from their analysis of compute-efficient training (a joint fit in model size and training steps) derived \(N_{\rm opt}\propto C^{0.73}\). These pure power laws have no floor term and hold only over the measured range.

在约束 \(6ND=C\) 下最小化 \(L(N,D)=E+AN^{-\alpha}+BD^{-\beta}\)。代入 \(D=C/(6N)\)：

$$\begin{gathered}L(N)=E+AN^{-\alpha}+B\Big(\frac{6N}{C}\Big)^{\beta},\\ \frac{dL}{dN}=-\frac{\alpha A}{N^{\alpha+1}}+\frac{\beta B}{N}\Big(\frac{6N}{C}\Big)^{\beta}=0\ \Longrightarrow\ \alpha A\,N^{-\alpha}=\beta B\,D^{-\beta}.\end{gathered}$$

最优点上，模型项和数据项按指数加权后相等：再多一点参数和再多一点数据带来的边际收益一样。解出

$$\begin{gathered}\hl{nopt}{N_{\rm opt}(C)}=G\Big(\frac C6\Big)^{\frac{\beta}{\alpha+\beta}},\qquad D_{\rm opt}(C)=G^{-1}\Big(\frac C6\Big)^{\frac{\alpha}{\alpha+\beta}},\\ G=\Big(\frac{\alpha A}{\beta B}\Big)^{\frac1{\alpha+\beta}}.\end{gathered}$$

复核后的系数给出 \(a=\beta/(\alpha+\beta)=0.513\)、\(b=0.487\)：几乎对半分，算力多 100 倍，参数和数据各多约 10 倍（Kaplan 的 0.73 则意味着参数多 29 倍、数据只多 3.5 倍）。\(\alpha\approx\beta\) 是“同比例增长”的根源。把 \(N_{\rm opt},D_{\rm opt}\) 代回去，\(L^\ast(C)-E\propto C^{-\alpha\beta/(\alpha+\beta)}\)，指数约 0.18：算力前沿本身也是带地板的幂律。若推理要求 \(N\le N_{\max}\)，当 \(N_{\rm opt}(C)>N_{\max}\) 时最优解贴在约束上，\(L\to E+AN_{\max}^{-\alpha}\)（FIG 4 的任务 2）。

Minimize \(L(N,D)=E+AN^{-\alpha}+BD^{-\beta}\) subject to \(6ND=C\). Substituting \(D=C/(6N)\):

$$\begin{gathered}L(N)=E+AN^{-\alpha}+B\Big(\frac{6N}{C}\Big)^{\beta},\\ \frac{dL}{dN}=-\frac{\alpha A}{N^{\alpha+1}}+\frac{\beta B}{N}\Big(\frac{6N}{C}\Big)^{\beta}=0\ \Longrightarrow\ \alpha A\,N^{-\alpha}=\beta B\,D^{-\beta}.\end{gathered}$$

At the optimum the model term and the data term, weighted by their exponents, are equal: a little more model and a little more data bring the same marginal gain. Solving,

$$\begin{gathered}\hl{nopt}{N_{\rm opt}(C)}=G\Big(\frac C6\Big)^{\frac{\beta}{\alpha+\beta}},\qquad D_{\rm opt}(C)=G^{-1}\Big(\frac C6\Big)^{\frac{\alpha}{\alpha+\beta}},\\ G=\Big(\frac{\alpha A}{\beta B}\Big)^{\frac1{\alpha+\beta}}.\end{gathered}$$

The re-fitted coefficients give \(a=\beta/(\alpha+\beta)=0.513\) and \(b=0.487\): almost an even split, so 100 times the compute means about 10 times the parameters and 10 times the data (Kaplan's 0.73 would mean 29 times the parameters and only 3.5 times the data). \(\alpha\approx\beta\) is the root of "grow both in proportion". Substituting \(N_{\rm opt},D_{\rm opt}\) back, \(L^\ast(C)-E\propto C^{-\alpha\beta/(\alpha+\beta)}\), an exponent of about 0.18: the compute frontier is itself a power law with a floor. If inference demands \(N\le N_{\max}\), then once \(N_{\rm opt}(C)>N_{\max}\) the optimum sits on the constraint and \(L\to E+AN_{\max}^{-\alpha}\) (task 2 of FIG 4).

若每一步独立、对数损失为 \(\hl{ell}{\ell(N)}=cN^{-\alpha}\)，整个任务成功率为 \(y=e^{-k\ell}\)。解出达到成功率 \(y\) 所需的规模：

$$\log_{10}N_y=\frac1\alpha\log_{10}\frac{kc}{\ln(1/y)}.$$

从 10% 到 90% 的宽度 \(\log_{10}N_{0.9}-\log_{10}N_{0.1}=\frac1\alpha\log_{10}\frac{\ln10}{\ln(1/0.9)}=\frac{1.34}{\alpha}\)，与 \(k\) 无关（FIG 5 中 \(\alpha=0.5\)，宽 2.68 个数量级），\(k\) 只使曲线平移 \(\frac1\alpha\log_{10}k\)。而 \(-\log y=k\ell(N)\) 本身是一条幂律。部分得分 \(\frac1k\sum_{j=1}^k p^j\) 在 \(p\) 不大时就已经离开零，上升得更早、更平缓。现实中各步并不独立、错误也可以纠正，但结论的方向不变：非线性的、阈值式的指标会把平滑的进步变成看似突然的跳变。

If the steps are independent with log-loss \(\hl{ell}{\ell(N)}=cN^{-\alpha}\), the whole task succeeds with \(y=e^{-k\ell}\). The scale needed to reach success rate \(y\) is

$$\log_{10}N_y=\frac1\alpha\log_{10}\frac{kc}{\ln(1/y)}.$$

The width from 10% to 90% is \(\log_{10}N_{0.9}-\log_{10}N_{0.1}=\frac1\alpha\log_{10}\frac{\ln10}{\ln(1/0.9)}=\frac{1.34}{\alpha}\), independent of \(k\) (in FIG 5 \(\alpha=0.5\), a width of 2.68 orders of magnitude); \(k\) only shifts the curve by \(\frac1\alpha\log_{10}k\). And \(-\log y=k\ell(N)\) is itself a power law. Partial credit \(\frac1k\sum_{j=1}^k p^j\) leaves zero while \(p\) is still modest, rising earlier and more gently. Real steps are not independent and mistakes can be corrected, but the direction of the conclusion stands: nonlinear, thresholded metrics turn smooth progress into seemingly sudden jumps.

语言模型的 scaling law 是在“数据几乎无限”的前提下测出来的：Chinchilla 用了 1.4 万亿个 token，后来的模型用到十几万亿。机器人没有这样的数据。到 2024 年，公开的大型真实机器人数据集是这样的量级：RT-1（Brohan 等，2022）在 17 个月里用 13 台机器人采了约 13 万条示教、覆盖 700 多个任务；DROID（Khazatsky 等，2024）由 50 位采集者在 12 个月里于 564 个场景采了 7.6 万条轨迹（350 小时）；Open X-Embodiment（2023）汇集了 21 个机构、22 种机器人的 100 多万条轨迹。粗略估一下：即使每条轨迹有 100 个控制步，100 万条也只是 \(10^8\) 个“动作样本”，比 Chinchilla 的 token 数少四个数量级，而且每一条都要人亲手遥操作（22.3 节）。

数据这么贵，问题就变成：同样的示教预算，怎样采最有用？FIG 1 的结论是“数据要填满策略会遇到的空间”。对机器人，最难填满的维度往往不是物体在桌上的位置，而是环境：不同的厨房、桌面、光照、物体实例。下图是一个模拟实验：策略要从示教里学会抓取偏移量，动作取决于物体位置 \(\mathbf s\in[-1,1]^2\) 和 3 个环境因素 \(\mathbf z\in[-1,1]^3\)（比如桌面高度、光照造成的视觉偏差、物体形状），\(\mathbf a=\mathbf s+\mathbf g(\mathbf z)\)，\(\mathbf g\) 是一个平滑的非线性函数。数据集由 \(E\) 个环境、每个环境 \(M\) 条示教组成；策略是一个基于 150 个随机傅里叶特征的岭回归（一种平滑的全局函数逼近器），在 300 个全新环境中的随机状态上测试，误差小于 0.1 算成功。

Language-model scaling laws were measured under "almost unlimited data": Chinchilla used 1.4 trillion tokens, later models over ten trillion. Robots have nothing like it. As of 2024, large public real-robot datasets are of this order: RT-1 (Brohan et al., 2022) collected about 130 thousand demonstrations over 700+ tasks with 13 robots in 17 months; DROID (Khazatsky et al., 2024) has 76 thousand trajectories (350 hours) collected by 50 people in 564 scenes over 12 months; Open X-Embodiment (2023) pooled over a million trajectories from 22 robot types at 21 institutions. A rough estimate: even at 100 control steps per trajectory, a million trajectories is only \(10^8\) "action samples", four orders of magnitude fewer than Chinchilla's tokens, and every one was teleoperated by hand (Lesson 22.3).

With data this expensive the question becomes: for the same demonstration budget, what is most useful to collect? FIG 1's conclusion was "the data must fill the space the policy will meet". For a robot the hardest dimensions to fill are often not where the object sits on the table but the environment: different kitchens, tabletops, lighting, object instances. The figure below is a simulated experiment: a policy must learn a grasp offset from demonstrations, where the action depends on the object position \(\mathbf s\in[-1,1]^2\) and 3 environment factors \(\mathbf z\in[-1,1]^3\) (say table height, a lighting-induced visual bias, object shape): \(\mathbf a=\mathbf s+\mathbf g(\mathbf z)\) with \(\mathbf g\) a smooth nonlinear function. A dataset consists of \(E\) environments with \(M\) demonstrations each; the policy is ridge regression on 150 random Fourier features (a smooth, global function approximator), tested on random states in 300 brand-new environments, with success meaning an error below 0.1.

每一行在 \(M\approx8\) 之后就基本不变了：一个环境里，物体位置这两个维度很快被覆盖，再多的示教只是重复。每一列却一路上升，而且即使有 64 个环境，成功率也只有约 70%：3 维的环境空间还远没有被填满，这和 FIG 1 里 \(n^{-2/d}\) 的逻辑一样。真实实验得出过同样的结论。RT-1 的消融实验比较了“减少每个任务的示教数”和“减少任务种类”（后者只去掉了 3% 的数据，却去掉了 25% 的任务），发现减少多样性的伤害大得多，作者的结论是数据多样性比数据数量更关键。Lin 等人（2025）用手持夹爪（UMI 式，22.3 节）采了 4 万多条示教，在真实机器人上做了 1.5 万多次测试，发现策略在新环境、新物体上的泛化能力随训练环境数和物体数近似按幂律增长，而每个环境或物体的示教数超过某个阈值后，再增加几乎没有效果。

Each row hardly changes beyond \(M\approx8\): within one environment the two dimensions of object position are covered quickly, and more demonstrations only repeat. Each column keeps rising, and even with 64 environments success is only about 70%: the 3-dimensional environment space is still far from filled, the same logic as \(n^{-2/d}\) in FIG 1. Real experiments have reached the same conclusion. RT-1's ablation compared "fewer demonstrations per task" with "fewer kinds of task" (the latter removed only 3% of the data but 25% of the tasks) and found the loss of diversity far more damaging; the authors concluded that data diversity is more essential than data quantity. Lin et al. (2025) collected over 40 thousand demonstrations with hand-held grippers (UMI-style, Lesson 22.3), ran over 15 thousand tests on real robots, and found that a policy's generalization to new environments and objects grows roughly as a power law in the number of training environments and objects, while beyond some threshold more demonstrations per environment or object have little effect.