Interview Bank

From Frontend Engineer to Agent Engineer in 30 Days

D15 A Tour of Multi-Agent Patterns (Router/Supervisor, Planner-Executor, Critic, Swarm, Blackboard) and When Not to Use Them; Getting Started With LangGraph

• In LangGraph, what roles do nodes, edges and state play? If you had no framework, how would you implement it yourself?LangGraph 里节点、边、状态分别扮演什么角色？如果不用框架，你自己会怎么实现？

  Common in ChinaCommon overseasIntermediate#langgraph#orchestration#state-management

  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。

  Key points

  • A node is a plain function: read the full state, return a delta of changed fields only, never mutate in place
  • Edges set execution order: unconditional edges are fixed, conditional edges decide the next hop at runtime — that is what a supervisor uses
  • State is a table of fields, each field a channel carrying a merge rule declared on the field rather than inside nodes
  • Without a framework you hit three walls: hand-written merges in every branch, no visibility into intermediate steps, and full restart after a crash
  • Two real traps: in-place mutation bypasses the merge rule and causes intermittent overwrites under parallelism; an unwired node raises no error, it just never runs
  • What the framework really sells is checkpoint recovery, parallel execution and per-step observability — not the three primitives themselves

  答题要点

  • 节点是普通函数：读全量状态，返回只含改动字段的增量，不在节点里原地改状态
  • 边决定执行顺序：无条件边写死，条件边在运行时决定下一步去哪（Supervisor 就靠它）
  • 状态是一张字段表，每个字段一条通道，通道上挂合并规则——规则声明在字段上而不是写在节点里
  • 不用框架会撞三堵墙：合并逻辑在每个分支手写一遍、中间过程只能靠打印、进程挂了从头重来
  • 两个真实的坑：原地改状态绕过合并规则（并行时偶发覆盖）、加了节点没连边不报错只是永不执行
  • 框架真正值钱的不是这三要素，而是检查点恢复、并行执行和逐步可观测

D17 Planner-Executor-Critic Plus a Shared Workspace: Workspace State, toolBudget, Parallel Fan-Out, a Review Loop

• When several subtasks run in parallel and all write the same shared state, how do you design it so they do not clobber each other?多个子任务并行执行、都要写同一份共享状态时，怎么设计才不会互相覆盖？

  Common in ChinaCommon overseasDeep dive#multi-agent#state-management#concurrency

  How to reason about it · think before answering

  1. This one separates people fast, because most candidates answer locks or immutable data structures — instincts carried over from threads. A graph runtime has no concurrent memory writes at all: updates are collected and merged. Answering in the wrong frame is worse than answering incompletely.
  2. Get the mechanism right first: parallel nodes each return a delta, the runtime groups all deltas from the same step by field, then calls that field's reducer to compute the new value. So the question is not how to lock, it is whether that field's reducer is correct.
  3. Then give a reusable chain: how many writers touch this field in one step, and do they write the same record? One writer — last-write-wins is fine. Several writers on different records — appending to a list is fine. Several writers on the same record — upsert by key. Several writers on different fields of the same record — merge per field. Four cases, four reducers, and the chain transfers to any framework.
  4. Land on the common mistake: implementing update this record as append a new version with the same id. The symptom is not an error — the same id exists twice and which one comes first depends on who finished first, so any lookup by id may return the stale version. Clean logs, occasionally wrong results.
  5. Add the trade-off: you can leave the reducer alone and dedupe by id at every read instead. But there are three or four read sites, and missing one is an intermittent stale read; a reducer is written once and every read is clean afterwards. Solve it once on the field, or N times at the read sites.
  6. Expect: does nondeterministic ordering matter? Ideally the reducer is order-insensitive (commutative); if it is not, you must guarantee one writer per record. Upsert-by-id is the latter — it is last-write-wins and is safe only because each record has exactly one executor per round.

  分析过程 · 先想清楚再作答

  1. 这题的区分度极高，因为大多数人会答成「加锁」或者「用不可变数据结构」——都是从多线程经验迁移过来的答案，但图的执行模型里根本没有并发写内存这回事，写入是被收集起来统一合并的。答错方向比答不全更致命。
  2. 先把机制说对：并行节点各自返回一份增量，框架把同一轮里所有增量按字段收集，再逐字段调用这个字段的合并规则（reducer）算出新值。所以问题不是「怎么加锁」，而是**这个字段的合并规则写得对不对**。
  3. 然后给一条可复用的判断链：先问这个字段同一轮会被几个人写；再问他们写的是不是同一条记录。只有一个写者，默认的后写覆盖就够；多个写者写不同记录，数组追加就够；多个写者写同一条记录的同一份数据，要按主键原地更新；多个写者写同一条记录的不同字段，要做字段级合并。四种情况四种 reducer，这条链能直接迁移到任何框架。
  4. 结论落在最容易踩的那一格：把「更新一条记录」写成「往数组里追加一条同 id 的新版本」。它的症状不是报错，是同一个 id 在状态里有两份、而且哪份在前取决于谁先跑完——下游任何按 id 查的地方都可能拿到过期版本，日志干净、结果偶尔错。
  5. 补一句权衡：也可以不动 reducer，改成每处读状态前先按 id 去重。但读取点有三四处，漏一处就是一个偶发脏读；reducer 只写一次，之后所有读取点自动干净。在字段上解决一次，还是在每个读取点解决 N 次，这是同一个问题的两种成本。
  6. 可以预期的追问：那顺序不确定要不要紧？答：合并规则最好对顺序不敏感（可交换），做不到就必须保证每条记录只有一个写者。本课的按 id 原地更新属于后者——它是最后写入者获胜，靠「一轮里一条记录只有一个执行者」这个前提才安全。

  Key points

  • A graph runtime has no concurrent memory writes: nodes return deltas, the runtime groups them per field and calls that field's reducer — so the answer is a correct reducer, not a lock
  • Decision chain: how many writers per step, and same record or not — overwrite, append, upsert by key, or per-field merge
  • The classic bug is implementing update as append-a-new-version-with-the-same-id: two entries per id, order depends on who finished first, lookups return stale data, and nothing ever errors
  • The alternative is deduping at every read site, but there are several and missing one gives an intermittent stale read; a reducer is written once
  • Prefer an order-insensitive reducer; if it is not, guarantee exactly one writer per record per step

  答题要点

  • 图的执行模型里没有并发写内存：节点各返回增量，框架按字段收集后调用该字段的 reducer 合并，所以问题是 reducer 写得对不对，不是加不加锁
  • 判断链：同一轮几个写者、写的是不是同一条记录——单写者用覆盖、多写者写不同记录用追加、多写者写同一条记录用按主键原地更新、写同一条记录的不同字段要字段级合并
  • 最常见的错是把「更新」写成「追加同 id 的新版本」，症状是同 id 两份、顺序取决于谁先跑完、按 id 查会拿到过期版本，而且全程不报错
  • 另一条路是每处读取前手动去重，但读取点有好几处，漏一处就是偶发脏读；reducer 只写一次就一劳永逸
  • 合并规则最好对顺序不敏感；做不到就必须保证一轮里一条记录只有一个写者