面试题库

30 天从前端工程师到 Agent 工程师

D17 Planner–Executor–Critic + 共享工作区：workspace state、toolBudget、并行 fan-out、review 回路

• 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？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?

  国内高频海外高频深入#multi-agent#state-management#concurrency

  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。

  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.

  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

  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