exp+fix: 候选塌缩定案(steps=2 把K条候选抹平35倍)+z_scale反例；补 n_plans/brake_rate 核算§9.2刹停率；记录我在链式作业排队期间改库导致两臂未跑的事故
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{
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"K": 8,
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"n": 256,
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"steps1": {
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"cand_spread": 0.8518,
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"first_step_norm": 1.1565,
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"clamp_saturated_rate": 0.0516,
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"pairwise_mean": 5.3647
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},
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"steps2": {
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"cand_spread": 0.0245,
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"first_step_norm": 1.0956,
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"clamp_saturated_rate": 0.0385,
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"pairwise_mean": 0.3024
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},
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"steps4": {
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"cand_spread": 0.0211,
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"first_step_norm": 1.097,
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"clamp_saturated_rate": 0.0689,
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"pairwise_mean": 0.2861
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}
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}
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@@ -0,0 +1,79 @@
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{
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"config": {
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"ckpt": "ml/runs/base_dag2_dag1/best.pt",
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"samples": "ml/data/samples",
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"tasks": null,
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"split": "val",
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"n": 256,
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"K": 8,
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"commit": 2,
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"fb": 2.2,
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"ki": 0.0,
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"fb_scale": 0.0,
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"fb_near": 0.2,
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"fb_fallback": 0.0,
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"kdecay": 0.9,
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"kimax": 0.5,
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"fb_every": 1,
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"safety_try": 0,
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"sel_w": 0.1,
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"dls_safety": 1,
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"fb_clear": 0.0,
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"fb_cmin": 0.3,
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"sub_refresh": 0,
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"fb_block_gain": 0.5,
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"dump": null,
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"out": "docs/experiment_round39_z15.json",
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"trace": null,
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"select": "endpoint",
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"replan": "empty",
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"safety": 1,
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"partial": 1,
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"inc_cap": 1,
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"gap_margin": 0.0,
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"cert_batch": 0,
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"steps": 1,
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"z_scale": 1.5,
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"len_w": 0.0,
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"backoff": 0,
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"max_sec": null,
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"seed": 0,
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"dls": "analytic",
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"device": "cuda"
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},
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"batch_size": 256,
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"batch_seed_rule": "same seed per batch",
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"checkpoint_sha256": "f7f686fd8ee92fb6948fbcc4ee954636065ad87bb54f661c4f8f575bc16c2ce1",
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"tasks_sha256": "249bd93457825cda98c59e64d9a096d3473c69aeb3553de1dea6d9e56be71772",
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"kin_fp": "2ebcbf17",
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"act_scale": 1.0,
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"metrics": {
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"n": 256,
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"n_ref": 256,
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"success": 0.49609375,
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"collision_free_success": 0.49609375,
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"time_ok_1p3": 0.18359375,
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"collision_rate": 0.0,
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"collision_steps_mean": 0.0,
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"travel_ratio_median": 3.52425217628479,
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"time_mean_s": 2.805511713027954,
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"brakes_mean": 10.51171875,
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"start_fail": 0.0,
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"goal_ratio_2s": 0.27530819177627563,
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"err_fail_med": 105.35015869140625,
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"err_success_med": 0.23657508194446564,
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"dls_share": 0.62890625,
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"time_budget": "teacher×2(≥3 s)",
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"speed_ratio_p50": 1.0000016689300537,
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"speed_ratio_p95": 1.0000101327896118,
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"speed_ratio_max": 1.0000101327896118,
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"over_vmax_rate": 0.0,
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"accel_ratio_p95": 1.0000016689300537,
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"accel_ratio_max": 1.0000101327896118,
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"reverse_p50": 0.32083332538604736,
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"reverse_mean": 0.3213125765323639,
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"reverse_moving_p50": 0.35158151388168335,
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"moving_share_p50": 0.9866666793823242
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},
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"simulation_wall_s": 311.9796243000019
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}
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@@ -107,6 +107,28 @@
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- **对 `backoff` 的预期与判据**：全拒意味着"连第 1 步的端点或中点已经穿进障碍"，缩步能不能救取决于**穿得有多深**（穿 0.1 mm 救得回，穿 3 mm 救不回）。这个数转储里没记（只记被选中那条），所以直接由 `backoff_mean`/`reject_all_mean` 来量：
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`reject_all_mean ≈ 原全拒次数`（=backoff_mean+brakes_mean，与轮次 33 cap 的 19.55 可比），`backoff_mean/(backoff_mean+brakes_mean)` 就是"缩步能救回的比例"。这是补跑（#22）要读的第一个数。
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## 候选为什么会"塌"：两段积分把 K 条候选吸成一条（轮次 39 探针，机理定案）
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- `ml/runs/_cand_div.py`（256 题特征、同一批噪声 z、K=8，只改段数）：
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| steps | 候选间一阶步的标准差 spread | 两两平均距离 pairwise | 第 1 步模长 | ±1 饱和率 |
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|---|---|---|---|---|
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| 1 | **0.8518** | 5.3647 | 1.1565 | 5.2% |
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| 2 | **0.0245** | 0.3024 | 1.0956 | 3.9% |
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| 4 | 0.0211 | 0.2861 | 1.0970 | 6.9% |
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⇒ **多段积分把 K 条候选的差别抹掉了 35 倍**（0.85 → 0.024）。这不是"动作更大更危险"那么简单：安全层是在 K 条**几乎相同**的候选里挑一条能走的前缀，挑无可挑就刹停。
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- 这条与轮次 38 的实测严丝合缝：`steps=2/4/8` 到达掉 13–15 pp、刹车 ×4.5，但**开环误差与行程比都更好**——动作质量提升是真的，到达率下降是"没有备选"造成的。
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- **但"人为制造多样性"不是解法**（`z_scale=1.5`，同 K=8）：刹车确实砍一半（19.55→10.51），可到到达掉到 0.4961、时限内从 0.2656 崩到 **0.1836**、行程比恶化到 3.5243、换向升到 0.32。⇒ 放大噪声换来的"能过的候选"是**分布外**的候选：安全层放行它，但它走的是绕、抖、慢的路。**K=16 的收益来自"同分布内更多候选"，不是"多样性本身"**——这两个数放在一起才把这条轴的边界说清楚。
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- 判据状态：#21 的两个分支现在这样分：**"动作更危险"不是主因，"候选塌缩"是主因**（spread 35 倍 + try/K 单调曲线 + z_scale 反例三条独立证据）。剩下要量的只是成本口径（K=16 在批量认证下贵多少）。
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## 事故：我在链式作业排队期间编辑了被 import 的库，两个臂直接没跑
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- 时间线：`z15` 于 14:46:38 结束，链接着拉起 `safety=0 steps=1`（14:46:38）——而我为了让 `brake_rate` 有分母，正在 14:45:5x 用脚本改 `ml/sim/closed_loop.py`，一次替换把 `per = {...}` 的字典提前闭合，留下 `IndentationError: unexpected indent`。**新起的 python 进程 import 到的是半成品文件**，两个 safety-off 臂各 2 秒就退出（日志里 `=== safety=0 steps=1 done 14:46:40 ===` 的"done"只是 bash 循环走到下一行，不代表成功）。修好在 14:46:52，48 项测试全绿。
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- **教训（比我原先记的"别改正在跑的脚本"更进一层）**：链式作业会在**未来某个时刻**新起解释器，所以"只要不改正在跑的那一个进程"是不够的。规则改为：**链没结束就不编辑被 import 的库；要编辑，先等通知或先让链跑完**。并且 `=== done ===` 这种循环日志不能当成功凭据——**判成功要看产物文件在不在**（这次 `docs/experiment_round38_safetyoff_steps{1,2}.json` 根本没生成，一看便知）。
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- 两个 safety-off 臂与 `ep_waste`/`backoff`/K 代价一起并入下一批重跑（`ml/runs/_next_batch.sh`）。
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- 顺带补了验收判据缺的分母：`rollout` 现在记 `n_plans`（每题规划次数）并给 `brake_rate = brakes/n_plans`、`eval_cl` 报 `brake_rate_p50/mean` 与 `n_plans_mean`——§9.2 要的是"刹停 <2% **推理周期**"，以前只有绝对次数，**这条判据实际上从来没被真正核算过**。命名上躲开一个坑：`per` 里原本有个 `'plans'` 键是转储用的逐规划字典，新键必须叫 `n_plans`，否则 `flat` 的过滤 `if k not in ('traj','plans')` 会把它静默吃掉。
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## 试候选数的成本-收益曲线（轮次 39，四档齐全，#15 主结果）
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- 同一 256 题、同 seed、`--inc_cap 1`、其余配方（`select=endpoint, replan=empty, commit_n=2, K=8, fb=2.2`）不变，只改**安全层允许试几条候选**：
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+4
-1
@@ -54,6 +54,9 @@ def run(model, kin, tasks, device, bs=256, **kw):
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out.update({'reverse_p50': median(rr), 'reverse_mean': rr.mean().item(),
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'reverse_moving_p50': median(flat['reverse_rate_moving']), # 只在动了的步上算，不被刹停稀释
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'moving_share_p50': median(flat['moving_share'])}) # 活动窗内动了的步占比（=1−站住比例）
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if 'brake_rate' in flat: # §9.2 判据是"刹停 <2% 推理周期"，所以要按规划次数归一，不能只报绝对次数
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out.update({'brake_rate_mean': flat['brake_rate'].mean().item(), 'brake_rate_p50': median(flat['brake_rate']),
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'n_plans_mean': flat['n_plans'].mean().item()})
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if 'backoff' in flat and flat['backoff'].any(): # backoff=缩步顶上的次数；brakes 现在只记连最小档都不安全的
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out['backoff_mean'] = flat['backoff'].mean().item() # ⇒ 原来的"全拒"事件数 = 两者之和
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out['reject_all_mean'] = (flat['backoff'] + flat['brakes']).mean().item()
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@@ -139,7 +142,7 @@ def main():
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Path(a.out).write_text(json.dumps(json_finite(result), indent=2, ensure_ascii=False, allow_nan=False) + '\n', encoding='utf-8')
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if a.dump:
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rows = []
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keys = ['done', 'ok', 't_done', 'limit', 'col', 'brakes', 'err', 'first_hit', 'dls_it', 'goal_ratio_2s']
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keys = ['done', 'ok', 't_done', 'limit', 'col', 'brakes', 'err', 'first_hit', 'dls_it', 'goal_ratio_2s', 'brake_rate', 'n_plans']
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flat = {k: torch.cat([p[k] for _, p, _ in pers]).tolist() for k in keys}
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for i, t in enumerate(tasks):
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rows.append({**t, 'i': i, **{k: flat[k][i] for k in keys}})
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@@ -204,6 +204,7 @@ def rollout(model, kin, tasks, device, K=8, safety=False, seed=0, max_sec=None,
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vratio = torch.zeros(B, device=device); aratio = torch.zeros(B, device=device); prev_exec = torch.zeros(B, 6, device=device)
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rev = torch.zeros(B, device=device); nact = torch.zeros(B, device=device); prev_sgn = torch.zeros(B, 6, device=device)
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nmov = torch.zeros(B, device=device)
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nplan = torch.zeros(B, device=device) # 每题规划次数 = 推理周期数（§9.2 的"刹停 <2% 推理周期"要这个分母）
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traj = [q.clone()] if return_traj else None
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plans = {k: [] for k in ('it', 'r', 'qe', 'commit', 'points', 'midpoints', 'gaps', 'certified')} if return_traj else None
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n_start = int(round(0.5 / DT)); moved_start = None # R2 起步失败：0.5 s 内关节总行程 < 1°
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@@ -216,6 +217,7 @@ def rollout(model, kin, tasks, device, K=8, safety=False, seed=0, max_sec=None,
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for it in range(nmax):
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tnow = it * DT
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if (queue.shape[1] <= 3) if replan == 'r3' else (queue.shape[1] == 0):
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nplan += 1.0 # 验收判据要"刹停 / 推理周期"的比值，分母就是这个（批内所有行同步规划）
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r = queue.shape[1]
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pre = torch.zeros(B, 3, 6, device=device)
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qe = q
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@@ -513,6 +515,7 @@ def rollout(model, kin, tasks, device, K=8, safety=False, seed=0, max_sec=None,
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side = torch.where(j1_first != 0, j1_first, torch.where(j1_max >= -j1_min, torch.ones_like(j1_max), -torch.ones_like(j1_max)))
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per = {'done': done.cpu(), 'ok': ok.cpu(), 't_done': t_done.cpu(), 'limit': limit.cpu(), 'col': col.cpu(), 'travel': travel.cpu(), 'err': e_end.cpu(),
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'side': side.cpu(), 'start_fail': start_fail.cpu(), 'brakes': brakes.cpu(), 'backoff': bo.cpu(), 'first_hit': first_hit.cpu(), 'dls_it': dls_it.cpu(),
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'n_plans': nplan.cpu(), 'brake_rate': (brakes / nplan.clamp_min(1.0)).cpu(),
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'goal_ratio_2s': ratio2.cpu(), 'speed_ratio': vratio.cpu(), 'accel_ratio': aratio.cpu(),
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# 口径与离线分解一致：换向的 (步,轴) 对数 / 活动的 (步,轴) 对数 ∈ [0,1]（除以 6 个轴）
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'reverse_rate': (rev / (6.0 * nact.clamp_min(1.0))).cpu(),
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