Interview Bank

How to reason about it · think before answering

1. The hinge is the second half. Defining the three concepts only proves you read the docs; explaining what hurts without a framework proves you know what it buys you. The general move for this family of questions is: describe your hand-rolled version first, then name what the framework collapsed.
2. Hand-rolled version: a loop, a chain of conditionals picking the next step, and one big object carrying data between steps. By the third branch you hit three walls — when two steps write the same field, is it overwrite or append, and you hand-write that merge in every branch; intermediate state lives in local variables so debugging means print statements; a crash restarts from zero and the model calls you already paid for are wasted.
3. Then map them: a node is an ordinary function that reads the whole state and returns a delta containing only what it changed; edges connect nodes, unconditional ones fix the order and conditional ones decide at runtime; state is a table of fields where each field is its own channel carrying a merge rule.
4. Dwell on the third, which is the most skipped and most valuable point: the merge rule is declared on the field, not written inside the node. Adding a node therefore requires no thought about how to combine with other writers, and parallel writes to one field behave deterministically instead of depending on who returns first.
5. Add two concrete traps to show you have actually run this: mutating state in place inside a node bypasses the merge rule — invisible single-threaded, an intermittent overwrite once things run in parallel; and adding a node without wiring an edge raises no error at all, it simply never executes, which only per-node tracing reveals.
6. Expect: so why not just write it yourself? Because the three primitives are genuinely light — a few dozen lines. What the framework actually sells is checkpointing and recovery, parallel execution, and per-step observability, all of which cost far more to build than the primitives. Mention too that there is no official LangGraph for Java or Swift, so in those languages you do hand-roll exactly these three.

分析过程 · 先想清楚再作答

1. 题眼在后半句。只答三个概念的定义，面试官会认为你读过文档；能说出「不用框架会难受在哪」，才证明你知道框架替你解决了什么。这类题的通用解法是：先讲自己手写的版本，再讲框架把哪几处收敛了。
2. 先给手写版：一个循环，里面一串条件判断决定下一步走哪，中间用一个大对象在各步之间传数据。写到第三个分支就会撞上三件事——两步都往同一个字段写，是覆盖还是追加，你要在每个分支里手写一遍合并逻辑；中间过程全在局部变量里，出错只能靠打印；进程一挂就从头重来，已经花掉的模型调用钱白付。
3. 然后一一对上：节点是一个普通函数，读全量状态、返回只含改动字段的增量；边是节点之间的连接，无条件边写死顺序，条件边在运行时决定去哪；状态是一张字段表，每个字段是一条独立通道，通道上挂着合并规则。
4. 重点讲第三条，因为它是最容易被略过、也最值钱的一条：**合并规则是声明在字段上的，不是写在节点里的**。这意味着新增节点时不需要考虑「我该怎么和别人的写入合并」，字段自己知道；也意味着并行写同一个字段时行为是确定的，而不是取决于谁先返回。
5. 配两个具体的坑，证明你真跑过：一是在节点里原地修改状态（比如直接往数组里 push）会绕过合并规则，单线程时察觉不到，并行时变成偶发覆盖；二是加了节点没连边不会报错，表现只是那个节点永远不执行，只能靠逐节点追踪发现。
6. 可以预期的追问：那你为什么不直接自己写？答：三要素本身很轻，核心逻辑几十行就能手写出来——框架真正值钱的是检查点与恢复、并行执行、以及每一步的可观测，这三样自己写的成本远高于三要素本身。顺带说明 Java 和 Swift 没有官方 LangGraph，真要在这两门语言里做，就是把这三要素手写一遍。

How to reason about it · think before answering

1. The crux is that models have no memory. Answering just concatenate previous episodes into the context invites a fatal follow-up: context grows linearly with episode count, so by episode five you pay repeatedly for four full episodes, and the model may still miss details.
2. Break it down by separating what is invariant across episodes from what is recomputed each time. Invariant: the world, each character's appearance, personality, voice id, and a few hard rules. Recomputed: scenes and shots. Extract the invariant part into its own file and load it verbatim before generating each episode.
3. Add the commonly missed point: the fields in that file are not only lore, they are downstream input parameters. Appearance text goes straight into image prompts, the voice id goes straight into the speech API. Keeping them beside the name means consistency is solved in one file rather than restated in three places.
4. Choose the storage boundary by write frequency: the profile is written once and read many times, while the shot list is rewritten on every run. Mixing lifetimes in one file makes it impossible to rerun one episode without disturbing the others.
5. Conclusion and cost: the profile itself can drift. Change a character's appearance mid-season and previously generated assets no longer match, so version the profile and include that version in the asset cache key — editing the profile then invalidates exactly the affected assets. That is only possible because it lives on its own.
6. Likely follow-up: should you use a vector store? Usually not. Cross-episode canon is small, structured, and must be injected in full; retrieval risks dropping the one line that matters. Retrieval fits large corpora where only a few relevant items are needed.

分析过程 · 先想清楚再作答

1. 这题的题眼是「模型没有记忆」。答成「把前一集的输出拼进上下文」的人会被追问到崩——上下文会随集数线性膨胀，第五集时你在为前四集的全文反复付费，而且模型仍然可能漏读。
2. 怎么拆：先分辨哪些是「跨集不变」的，哪些是「每集重算」的。不变的是世界观、人物外貌、性格、音色与几条硬规则；每集重算的是场景与分镜。把不变的那部分抽成单独的档案文件，每一集生成前原样读进去。
3. 接着说一个容易被忽略的点：档案里的字段不只是设定，还是**下游的输入参数**。外貌描述要原样进图像提示词，音色 id 要原样进语音接口。所以它们必须和名字放在同一份档案里，一致性问题才是在一个文件里解决的，而不是散在三处各写一遍。
4. 存放位置的判据是写入频率：档案一次生成、多次读取，分镜每跑一次就重写。生命周期不同的数据放同一个文件，你就没法只重跑一集而不动其他集。按写入频率切分文件，是这类流水线最省事的一条习惯。
5. 结论与代价：档案本身也会漂——中途改了人物外貌，之前生成的资产就对不上了。所以档案要有版本，且资产的缓存键要包含档案版本，改档案等于让相关资产失效。这条也是把它单独存放才做得到的。
6. 可预期的追问：那要不要上向量库做检索？多数情况下不需要。跨集共享的设定是**有限的、结构化的、必须全量注入的**，检索反而可能漏掉关键一条。检索适合的是「素材库很大且只需要相关几条」的场景。