面试题库

14 天用 Agent 搭一条 AI 短剧生产线

D10 审片室：能预览、能改词、能重生成单镜的人机协作后台

• 用户改了中间一步的输入，怎么算出哪些下游需要重做？A user edits an intermediate input. How do you compute which downstream steps must rerun?

  国内高频海外高频进阶#dag#incremental-recompute#cost

  分析过程 · 先想清楚再作答

  1. 这题的区分度在方向和收尾两处，很多人只答出中间那段「沿依赖图传播」，前后都丢了。
  2. 方向：从被改的节点**沿着「谁依赖我」正向传播**，不是往上游找依赖。写反的后果很隐蔽——上游会被一起重跑，结果是对的，钱多花了一倍，测试也发现不了。
  3. 落到实现：把种子节点放进集合，反复扫一遍图，只要某个节点的依赖里有一个已经在集合里就把它也加进来，跑到不动点为止；最后按拓扑序返回，调用方顺着数组跑就不会先跑下游后跑上游。
  4. 收尾这一步最容易漏：**没受影响的节点，产物要从上一版复制过来，不是重新生成**。半径算得再准，少了复制这一步就一分钱没省。
  5. 然后是怎么验证。不要比文件哈希——同样的输入很可能生成逐字节相同的结果，哈希相同证明不了没重跑。要数**接口调用次数**，这才是硬证据，而且在离线与真实两种模式下都成立。
  6. 可预期的追问是「输入没变但你想重跑怎么办」。留一个强制重跑的开关，并且把它和自动判定分开记账，否则你会分不清一次重跑是系统判的还是人手动点的。

  How to reason about it · think before answering

  1. The signal lives at the two ends. Most candidates produce the middle part, propagation over a dependency graph, and drop both the direction and the finish.
  2. Direction: propagate forward along who-depends-on-me from the edited node, not backward to its dependencies. Getting it backward is insidious, because upstream nodes rerun, the output is still correct, the bill doubles, and no test catches it.
  3. Implementation: seed a set, sweep the graph repeatedly adding any node with a dependency already in the set until it stops growing, then return in topological order so the caller can just walk the array.
  4. The finish is what people forget: unaffected nodes must have their artifacts copied from the previous version, not regenerated. A perfect radius saves nothing without that copy.
  5. Then verification. Do not compare file hashes, because identical inputs often produce byte-identical output and a matching hash proves nothing. Count API calls instead; that evidence holds both offline and against a real vendor.
  6. Expected follow-up: what about forcing a rerun when nothing changed. Keep an explicit force flag and account for it separately, or you lose the ability to tell system-decided reruns from human-triggered ones.

  答题要点

  • 从被改的节点沿着「谁依赖我」正向传播，不是反向找依赖。
  • 扫图到不动点，结果按拓扑序返回，保证执行顺序不会颠倒。
  • 没受影响的节点要从上一版复制产物，否则半径算得再准也没省钱。
  • 验证要数接口调用次数，不要比文件哈希——同样的输入可能产出逐字节相同的结果。
  • 另留一个强制重跑开关，并与自动判定分开记账。

  Key points

  • Propagate forward along who-depends-on-me from the edited node, never backward.
  • Sweep to a fixed point and return in topological order so execution never runs downstream first.
  • Copy artifacts for unaffected nodes from the previous version, or the computed radius saves nothing.
  • Verify by counting API calls, not by comparing file hashes, since identical inputs can produce byte-identical output.
  • Keep a separate force-rerun switch and account for it apart from automatic decisions.