diff --git a/research/live/verify_incr_parity.py b/research/live/verify_incr_parity.py
index 57ab500..94317c9 100644
--- a/research/live/verify_incr_parity.py
+++ b/research/live/verify_incr_parity.py
@@ -120,8 +120,72 @@ def run_one(sym: str, cache: Path, n: int, start_at: int | None) -> dict:
             "full_ms": t_full / k * 1000, "incr_ms": t_incr / k * 1000}
 
 
+def interleave(sym: str, cache: Path, n: int, nw: int) -> dict:
+    """模拟多 worker 交错：nw 份独立缓存轮流接同一个币。
+
+    这是单进程对拍覆盖不到的路径。`ProcessPoolExecutor` 不保证同一个币落到
+    同一个 worker，所以每个 worker 只能隔 nw 根才再见到这个币，一次要补 nw
+    根。补根走的是 `for row in df[ts > last_ts]` 那个循环——逻辑上等价于连续
+    追加 nw 次，但「等价」是推理，没实测过。
+
+    缓存是 worker 进程内的 dict、键含 symbol，所以不存在「worker A 的状态被
+    worker B 读到」或「拿到别的币的状态」。亲和性影响的是内存（每个 worker
+    最终缓存全部币）与补根次数，不影响正确性——本函数就是来证这一点的。
+    """
+    import shadow_signal as ss
+    from verify_lean_parity import SKIP, WINDOW_KEYS, canon, load, signal_bars
+
+    skip = SKIP + WINDOW_KEYS + ("stream_bars",)
+    l_all, h_all = load(sym, "1m", cache), load(sym, "5m", cache)
+    l_ts = l_all["timestamp"].to_numpy("int64")
+    h_ts = h_all["timestamp"].to_numpy("int64")
+    try:
+        sb = signal_bars(sym, cache)
+        sb = sb[(sb > BASE_L + n) & (sb < len(l_all) - 1)]
+        start = int(sb[-1]) - n + 5 if len(sb) else BASE_L + 10
+    except Exception:
+        start = BASE_L + 10
+    ends = [e for e in range(start, start + n) if e < len(l_all) - 1]
+
+    caches: list[dict] = [{} for _ in range(nw)]
+    same = diff = n_hits = 0
+    first = None
+    for j, e in enumerate(ends):
+        hi = int(np.searchsorted(h_ts, l_ts[e], side="right"))
+        df_l = l_all.iloc[e - BASE_L + 1:e + 1]
+        df_h = h_all.iloc[max(0, hi - BASE_H):hi]
+        entry = float(l_all["open"].to_numpy(float)[e + 1])
+
+        rf = ss.compute(df_l.copy(), df_h.copy(), entry, incr=False)
+        # 轮流换缓存 = 轮流换 worker
+        ss._STREAMS = caches[j % nw]
+        ri = ss.compute(df_l.copy(), df_h.copy(), entry, sym=sym, incr=True)
+
+        n_hits += len(rf.get("hits") or [])
+        if canon(rf, skip) == canon(ri, skip):
+            same += 1
+        else:
+            diff += 1
+            if first is None:
+                first = (e, canon(rf, skip), canon(ri, skip))
+
+    grown = [len(c[(sym, "1m")][0].dataframe) for c in caches
+             if (sym, "1m") in c]
+    print(f"  {len(ends)} 根 · {nw} 份缓存轮流 · 一致 {same} · 不一致 {diff}"
+          f" · 命中 {n_hits} 个 · 各缓存末窗 {grown}")
+    if first:
+        e, a, b = first
+        print(f"  ⚠ 首个分歧 idx={e}\n    全量: {a[:300]}\n    增量: {b[:300]}")
+    ss._STREAMS = {}
+    del l_all, h_all
+    return {"sym": sym, "n": len(ends), "same": same, "diff": diff,
+            "hits": n_hits, "full_ms": 0.0, "incr_ms": 0.0}
+
+
 def main() -> None:
     ap = argparse.ArgumentParser()
+    ap.add_argument("--interleave", type=int, default=0,
+                    help="模拟这么多个 worker 轮流接同一个币（一次补多根）")
     ap.add_argument("--syms", default="BTC,ETH,SOL")
     ap.add_argument("--cache", default="research/live/cache")
     ap.add_argument("--n", type=int, default=300)
@@ -132,7 +196,10 @@ def main() -> None:
     for sym in a.syms.split(","):
         print(f"\n{'=' * 70}\n{sym}")
         try:
-            rows.append(run_one(sym, Path(a.cache), a.n, a.start))
+            if a.interleave:
+                rows.append(interleave(sym, Path(a.cache), a.n, a.interleave))
+            else:
+                rows.append(run_one(sym, Path(a.cache), a.n, a.start))
         except Exception as e:
             print(f"  跳过：{e!r}")
     if not rows:
@@ -141,9 +208,10 @@ def main() -> None:
     print(f"\n\n{'=' * 70}\n汇总\n")
     print(f"  对拍 {int(d['n'].sum()):,} 根 · 不一致 {int(d['diff'].sum())} · "
           f"命中 {int(d['hits'].sum())} 个")
-    print(f"  单根 全量 {d['full_ms'].mean():.1f}ms → "
-          f"增量 {d['incr_ms'].mean():.1f}ms "
-          f"（{d['full_ms'].sum() / max(d['incr_ms'].sum(), 1e-9):.2f}x）")
+    if d["incr_ms"].sum() > 0:
+        print(f"  单根 全量 {d['full_ms'].mean():.1f}ms → "
+              f"增量 {d['incr_ms'].mean():.1f}ms "
+              f"（{d['full_ms'].sum() / max(d['incr_ms'].sum(), 1e-9):.2f}x）")
     if int(d["diff"].sum()) == 0:
         print("\n  逐字段一致，增量可以上线。")
     else: