exp: 轮次37——K-MF 方案核查：Eq.7 抄法不动点不自洽(1.87×)，一步输出比自身2段采样离老师远10倍

2026-10-20 21:38:40 +08:00
parent 349814db7e
commit a4cac354ea
4 changed files with 457 additions and 1 deletions
@@ -0,0 +1,337 @@
+{
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@@ -0,0 +1,97 @@
+{
+ "n": 512,
+ "steps": {
+  "2": {
+   "vs_teacher_l2": 0.4342995882034302,
+   "vs_32step_l2": 0.30502474308013916,
+   "vs_1step_l2": 4.2227020263671875
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+   "vs_32step_l2": 0.29940515756607056,
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+  },
+  "8": {
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+   "vs_1step_l2": 4.195967197418213
+  },
+  "16": {
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+   "vs_32step_l2": 0.12673738598823547,
+   "vs_1step_l2": 4.085569381713867
+  },
+  "32": {
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+   "vs_32step_l2": 0.0,
+   "vs_1step_l2": 3.970090866088867
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+ },
+ "vs_teacher": {
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+   "l2_p95": 4.9564337730407715,
+   "per_axis_norm": [
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+    0.36649173498153687,
+    0.3425973951816559,
+    0.3447648882865906,
+    0.3368167281150818,
+    0.32962310314178467
+   ]
+  },
+  "aU_32step_rneq_t": {
+   "l2_mean": 0.6008902192115784,
+   "l2_p95": 1.9733507633209229,
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+    0.04775547981262207,
+    0.04531646892428398,
+    0.0381026417016983,
+    0.034751877188682556
+   ]
+  },
+  "aF_flow_map_r_eq_t": {
+   "l2_mean": 0.6651337146759033,
+   "l2_p95": 1.9794291257858276,
+   "per_axis_norm": [
+    0.06039007008075714,
+    0.05440700799226761,
+    0.05297258496284485,
+    0.05065540224313736,
+    0.04315900802612305,
+    0.03988185524940491
+   ]
+  }
+ },
+ "mutual": [
+  {
+   "l2_mean": 3.970090866088867,
+   "pair": [
+    "a1",
+    "aU_32"
+   ]
+  },
+  {
+   "l2_mean": 3.891127109527588,
+   "pair": [
+    "a1",
+    "aF_flow"
+   ]
+  },
+  {
+   "l2_mean": 0.08254596590995789,
+   "pair": [
+    "aU_32",
+    "aF_flow"
+   ]
+  }
+ ],
+ "note": "动作块是 ±1 归一化的 16×6；l2_mean 无单位权重，越小越贴老师"
+}
@@ -770,6 +770,27 @@
 - **改动（默认关，`--backoff`）**：全拒时不再站住，把首选候选按 1/2、1/4、1/8…（档数 = backoff）**并进同一次批量认证**（零额外几何调用），只在原候选连第 1 步都不安全时顶上；缩步候选不参与"前缀更长者优先"的比较，所以正常路径的选择逐位不变。新增 `backoff_mean` 指标记录缩步次数。测试 35→38 项（全拒→慢爬、有安全前缀时插手无效、未开批量认证时报错）。
 - **待测**：256 题 A/B（`--inc_cap 1 --cert_batch 1 --backoff 0/1/2/3`）——GPU 先让给 K-MF E1，这轮排在其后。

+## 轮次 37：K-MF（运动学 MeanFlow）方案核查——先测"一步 vs 多步"，结果比论文那条更要命
+
+- **交来的方案里那几条 file:line 断言，逐条回代码核过（都成立）**：`ml/model/meanflow.py:38-39` 是标准 Eq.4 单段自举（`jvp` 切向量 (v,0,1)，目标 `v − (t−r)·du/dt`）；`ml/train.py:162` 调 `meanflow_loss` 时**没接 ada_p** ⇒ 自适应加权一直是默认 `ada_p=1.0`（配置里改不了，与方案说的一致）；`p_neq` 在第 `p_neq_warm=5000` 步从 0 硬切到 0.5；`p_edge=0.1` 把 10% 样本钉在部署角 (r=0,t=1)；`policy.sample_k(..., steps=1)` 是默认，`ml/model/policy.py:153` ⇒ **历史发布的所有数字都是一步推理**。
+- **方案里过时的两条要先改掉，否则会拿旧靶子做新实验**：「闭环 0% 到达、79% 碰撞」是 DAgger 之前的状态，现在是 256 题到达 57.4%、实际执行碰撞 0（轮次 30/33）；「DAgger 全量是主线，先做」也已完成（`base_dag2_dag1` 就是两轮的产物）。所以 E 系列的判据不能再用"能不能从 0 救起来"。
+- **E1 第一版判据是自证的，作废**：我先拿"网络自己的 r≠t 切片走 32 段"当平均速度真值——那正是 MF 损失直接训练的切片，等于用被测公式的收敛解当真值。改成**与损失无关的参考**：平均速度的定义是瞬时场（r=t 切片，由纯流匹配监督）的流图 `u_true = (z_t − Φ_{t→r}(z_t))/(t−r)`，128 段欧拉；离散化自证：128 段与 512 段的真值差 0.019（相对 **0.18%**）⇒ 参考本身稳。
+- **E1 结果（`docs/experiment_round36_kmf_e1.json`，256 个 val 样本，λ=0.5，‖u_true‖ 均值 10.38）**：目标与真值的距离（越小越好）
+
+  | 目标写法 | (r=0,t=1) 部署角 | 最差 ‖du/dt‖ 处 (t=0.05) |
+  |---|---|---|
+  | MF（现行 Eq.4） | **3.851** | 2.629 |
+  | K-MF 按方案抄的 (1−λ)²/λ² | 7.187（1.87×，更差） | 3.808 |
+  | K-MF 把平方换成线性权重 (1−λ)/λ | 14.661（更差） | 2.106 |
+  | 把整条修正移到中点时刻 `du(z_c,r,c)/dc` | **2.364（0.61×）** | 2.468 |
+  | 网络现在的一步输出 u(ε,0,1) | 3.906 | — |
+
+- **两条硬结论**：
+  1. **方案抄来的 Eq.7 系数在收敛解上不自洽**：λ=0.5 时两项系数和 = (1−λ)²+λ² = 0.5，不动点变成 `v − 0.5(t−r)u'`，不是平均速度；实测也印证——部署角 `‖du(z_t,c,t)/dt‖ = 28.0` 而 `‖du/dt‖ = 4.24`（把 r 槽钉在中点后对 t 的导数大 6.6 倍），组合被这一项主导，所以偏差反而更大。**K-MF 仓库只有 README，论文只能引公式，这条是我们自己的数**：要么抄漏了子区间时长因子，要么 c 的定义不同。判据「K-MF 偏差 ≤ MF 的一半」在这个写法下不成立（1.87×）。
+  2. **自举污染是真的，而且就在部署角**：网络一步输出与自己的流图差 3.906（相对 38%），而 MF 目标的偏差 3.851 几乎一样 ⇒ 训练把目标的偏差原样灌进了线上那一步。`‖du/dt‖` 在两端炸开（t=0.05 为 156.6、t=0.95 为 64.3，中段只有 1.6–4.1），与论文 Fig.1a 的形态一致（他们报 76×，我们这里 0.05↔0.4 是 **66×**）。
+- **E1-b（`docs/experiment_round36_kmf_e1b.json`，512 样本，比的是"离老师多远"，归一化动作块 L2）**：一步 **4.326**；同模型 2 段 **0.434**、4 段 0.429、32 段 0.601；流图（r=t，128 段）0.665。**一步比多步差 10 倍**，而 2 段就基本到位（再多段反而偏离老师——那是模型自身场的偏差，不是采样误差）。
+- **这一条比 K-MF 本身更要命**：57.4% 的到达是在"线上一步输出偏离自己场 10 倍"的前提下拿到的，安全层/反馈/DAgger 都在替这个角点擦屁股。所以先把 `--steps` 打通，量它对闭环到达的影响，再决定要不要为"把一步训好"付一次训练。
+
 ## 分支清理（2026-10-20）

 - `feat/fb-mix`（8e5431c）：反馈混合方案的中间分支，**已验证是 `exp/dagger-deploy` 的祖先**（所有提交都在后者历史里），本地/远端分支删除，内容保留在 exp/dagger-deploy。
@@ -93,6 +93,7 @@ def main():
    ap.add_argument('--inc_cap', type=int, default=0, help='1=反馈叠加后再按每轴 VMAX·DT 限幅')
    ap.add_argument('--gap_margin', type=float, default=0.0, help='安全层要求的间隙余量 mm（>0 时宽相位 cap 自动抬到 margin+0.5）')
    ap.add_argument('--cert_batch', type=int, default=0, help='1=一次 gaps 调用覆盖全部候选（压最坏情况，典型路径略贵）')
+    ap.add_argument('--steps', type=int, default=1, help='MeanFlow 采样段数；1=线上一口径（历史发布数字全是 1）')
    ap.add_argument('--backoff', type=int, default=0, help='全拒时把首选候选按 1/2、1/4…缩步重试的档数（需 --cert_batch 1）')
    ap.add_argument('--max_sec', type=float, default=None)
    ap.add_argument('--seed', type=int, default=0)
@@ -116,7 +117,7 @@ def main():
        'kin_fp': kin.fp, 'act_scale': float(model.act_scale),
    }
    t0 = time.perf_counter()
-    r, pers = run(model, kin, tasks, a.device, K=a.K, safety=bool(a.safety), seed=a.seed, max_sec=a.max_sec,
+    r, pers = run(model, kin, tasks, a.device, K=a.K, safety=bool(a.safety), seed=a.seed, max_sec=a.max_sec, steps=a.steps,
            safety_partial=(a.commit if (a.safety and a.partial) else 0), commit_n=a.commit,
            select=a.select, replan=a.replan, fb_gain=a.fb, fb_scale=a.fb_scale, fb_near=a.fb_near, fb_fallback=a.fb_fallback, fb_ki=a.ki, fb_decay=a.kdecay, fb_imax=a.kimax,
            fb_every=a.fb_every, safety_try=a.safety_try, sel_w=a.sel_w, dls_safety=a.dls_safety, fb_clear=a.fb_clear, fb_cmin=a.fb_cmin, sub_refresh=a.sub_refresh, fb_block_gain=a.fb_block_gain,