Interview Bank

How to reason about it · think before answering

1. This looks like a giveaway and is actually a filter. 'Cache the question and answer' earns a third of the credit; the interviewer is waiting for the layering and the per-layer invalidation rules.
2. Lead with a transferable rule: 'when must this be invalidated' is the same question as 'is that thing part of the key'. Leave something out of the key and changes to it will never invalidate the entry. With that rule the three layers derive themselves.
3. Then go layer by layer. The answer layer maps a question to a final answer; its key needs the question, the permission scope, the index version, and the model plus prompt version. The retrieval layer maps a query to a hit list; its key needs the question, scope, topK, index version and embedding backend, but not the generation model. The embedding layer maps text to a vector; its key is just the text and the backend.
4. Emphasize the counterintuitive part of the embedding layer: it is content-addressed, so the index version must not be in its key. Put it there and a single sync invalidates tens of thousands of vectors, which is exactly the full rebuild you added caching to avoid. This is the one layer that can live a long time, even on disk.
5. Offer a concrete invalidation mechanism: version numbers rather than targeted deletion. Bump an index version whenever a sync actually changes something and old keys simply stop being computed. Targeted deletion would require enumerating which questions a change affected, and that list cannot be produced.
6. Expected follow-up: can you give a real 'should have expired but didn't' case? Yes: an answer cache keyed only on the question. A document's limit changes from 200 MB to 500 MB, the index is updated, and the same question still returns 200 MB. Nothing errors; the log shows a clean cache hit. The same key also serves one department's answer to a user from another.

分析过程 · 先想清楚再作答

1. 这题看起来是送分题，实际是筛人题。答成「把问答结果缓存起来」只拿到三分之一，面试官等着听的是「分几层」和「各自什么时候失效」。
2. 先给一条能迁移到别的题上的判断依据：**「什么时候必须失效」这个问题，等价于「key 里有没有把那样东西算进去」。** key 少放一样，那样东西变了缓存就不会失效。有了这条，三层的答案自己就长出来了。
3. 然后逐层给：答案层缓存问题到最终答案，key 要有问题、权限范围、索引版本、模型与提示词版本；检索层缓存检索式到命中块列表，key 要有问题、权限范围、topK、索引版本、向量后端，但不需要模型；向量层缓存文本到向量，key 只有文本和向量后端。
4. 重点讲向量层的反直觉之处：它是**内容寻址**的，文本没变、模型没变，向量就不会变，所以**不能把索引版本放进它的 key**。放进去的话一次同步就作废几万条向量，正好绕回全量重建——你加缓存想省的那笔钱又花回去了。这一层可以放很久甚至持久化。
5. 给一个具体的失效手法：用**索引版本号**而不是精确删除。同步只要真的改动了索引就把版本号加一，旧 key 再也算不出来，自然没人读得到。精确删除要求你能列出「这次改动影响了哪些问题」，而那是列不出来的。
6. 可预期的追问：能举一个「该失效却没失效」的真实例子吗？答：答案缓存的 key 只放了问题本身，文档里的上限从 200 MB 改成 500 MB、索引已经更新，再问同一个问题仍然返回 200 MB。它不报错，日志上是一次漂亮的缓存命中；同一个 key 还会让另一个部门的用户直接命中别人的答案。

How to reason about it · think before answering

1. The discriminator is the second half: what must not go in. People who only say 'hash the inputs' have usually never been burned by a cache. The two failure modes point in opposite directions: never hitting, and hitting when it should not.
2. State the criterion first: include everything that changes the artifact, exclude everything that changes every run without affecting the artifact. Both lists fall out of that.
3. Include four things: node id, implementation version, this node's own inputs (model id, prompt, duration, resolution), and the fingerprints of all dependencies. The version and the dependency fingerprints are the two people forget — miss the version and new code reads old artifacts; miss the dependencies and an upstream script change never propagates.
4. Exclude: run id, timestamps, random values, absolute paths, and anything carrying a hostname or temp directory. Any of those makes every key new, and you will blame the cache instead of the key.
5. Two implementation details worth volunteering: decide 'is it done' by checking the artifacts on disk, not the state file, because files get deleted by hand; and think about granularity — four shots in one node means one failed shot redoes all four, while finer granularity saves money at the cost of a much larger graph.
6. Expect the follow-up 'does hashing dependency keys over-invalidate'. Yes. An upstream wording change that produces an identical artifact still invalidates downstream. Hashing the dependency's artifact content instead is tighter but requires reading the artifact every time — worth it for small files, not for large videos.

分析过程 · 先想清楚再作答

1. 这题的区分度全在「不该放什么」那一半。只答「把输入哈希一下」的人，通常没在真实项目里被缓存坑过——缓存的两种病方向相反，一种是永远不命中，一种是命中了不该命中的。
2. 先给判据：键里应该出现的，是所有会改变产物的东西；不该出现的，是所有每次都会变但不影响产物的东西。这一条能直接推出下面两张清单。
3. 该放的四样：节点标识、实现版本号、本节点的输入（模型 id、提示词、时长、分辨率）、以及全部依赖的指纹。版本号和依赖指纹是最容易漏的两样——漏了版本号，改完代码读到旧产物；漏了依赖指纹，上游换了剧本你还在用旧的镜头。
4. 不该放的：运行标识、时间戳、随机数、绝对路径、以及任何带机器名或临时目录的东西。放进去等于每次都是新键，你会以为缓存写坏了，其实是键设计错了。
5. 还有两条落地细节值得主动说：判断「做没做完」要看磁盘上产物齐不齐，不能只信状态文件，因为文件可能被手删；以及幂等的粒度要想清楚，一个节点里跑四个镜头，第三镜失败就是四镜全重做，粒度更细更省钱但任务图会大很多。
6. 可预期的追问是「依赖指纹会不会失效得太狠」。答：会。上游只是文案改了、产物其实一样，下游也会跟着重做。更省的做法是对依赖的产物内容做哈希而不是对它的键做哈希，代价是每次都要把产物读一遍——小文件划算，大视频不划算，这是要自己量的一笔账。