面试题库

14 天 RAG：从检索到可信回答

D11 高级索引：父子文档、摘要索引、上下文检索，以及树状聚合与图检索的取舍

• 上下文检索要给每个块调一次模型，这笔一次性成本怎么估？有哪些办法能压下来？Contextual retrieval needs one model call per chunk. How do you estimate that one-off cost, and what levers bring it down?

  国内高频海外高频深入#contextual-retrieval#prompt-caching#cost

  分析过程 · 先想清楚再作答

  1. 这题考的是你有没有真的算过账。只会说『用提示词缓存就便宜了』属于听过没做过——面试官会追问缓存到底省在哪一项上。
  2. 先把成本拆开：一次性 = 每块的输入 + 输出 + 全量 embedding；每次查询 = 块头在重排和上下文里各被读一遍。**这两笔要分开记**，因为它们随业务量的增长方式完全不同。
  3. 一次性那笔的主项是『同一篇文档被重复读了多少遍』。一篇切成 n 块就要读 n 遍，这是成本的大头。提示词缓存省的正是这一项：把整篇放在提示词最前面并标记为可缓存，第一块付一次缓存写入，后面 n-1 块只付缓存读取，而读取价通常比输入价低一个数量级。
  4. 顺序不能反：缓存按前缀匹配，整篇必须在前、块内容在后。把变化的块放前面，前缀次次都变，缓存一次都不会命中——这是最常见的翻车点。
  5. 我们的实测：30 篇、134 块，不开缓存输入 103017 token，开缓存后拆成写入 17340 加读取 60137，一次性成本降约 29%。**块切得越碎这个比例越高**，因为重复读的次数更多。
  6. 结论反直觉但很实用：一次性那笔是小钱，摊到 217 次查询就降到每次查询成本的一成以下；真正的长期账是每次查询多出来的那几十个 token（我们量到 +12.3%）。所以压成本的第一优先级不是压建索引，而是让块头别进上下文、别过长、别对全库无差别地生成。
  7. 最后一条是加分项：花这笔钱之前先确认你的评估环境**测得出**收益。我们的离线环境里向量路对召回的独立贡献实测为 0，所以它根本没法回答『块头对向量侧有没有用』——在这种环境里做的 A/B 会给你一个看起来有数字支撑的错误结论。

  How to reason about it · think before answering

  1. This checks whether you have actually done the arithmetic. Saying 'prompt caching makes it cheap' without knowing which line item it touches is a tell.
  2. Split the bill first: one-off = per-chunk input + output + full re-embedding; per-query = the header read twice, once by the reranker and once in the context. Keep them separate, because they scale with completely different things.
  3. The dominant term on the one-off side is how many times the same document is re-read. A doc split into n chunks is read n times. Prompt caching attacks exactly that: put the whole document first and mark it cacheable, pay a cache write once, then cache reads for the remaining n-1, typically an order of magnitude cheaper than input.
  4. Order matters. Caching is prefix-matched, so the document must come first and the chunk after. Put the varying part first and the prefix changes every call — zero cache hits. This is the most common way people get it wrong.
  5. Our measurement: 30 docs, 134 chunks. Without caching, 103017 input tokens; with caching, 17340 written plus 60137 read, cutting the one-off cost by roughly 29%. The finer the chunks, the bigger the saving, because re-reads multiply.
  6. The counter-intuitive part is the useful part: the one-off cost amortizes below 10% of per-query cost after about 217 queries. The lasting bill is the extra tokens every query carries (we measured +12.3%). So the first lever is not cheaper index building — it is keeping the header out of the context, keeping it short, and not generating it for the whole corpus indiscriminately.
  7. A bonus point: before spending any of it, confirm your evaluation setup can actually detect the benefit. In our offline harness the vector route contributed exactly zero unique answer documents, so it cannot answer whether headers help embeddings at all — an A/B run there hands you a wrong conclusion that looks numerically supported.

  答题要点

  • 把账拆成一次性（每块的输入输出 + 全量 embedding）和每次查询（块头在重排与上下文里各读一遍）两笔。
  • 一次性的大头是同一篇被重复读 n 遍；提示词缓存把它压成一次写入加 n-1 次读取。
  • 缓存按前缀匹配，整篇必须放在提示词最前面，块内容在后，顺序反了一次都不会命中。
  • 实测 30 篇 134 块，一次性成本降约 29%，块越碎省得越多。
  • 长期账在每次查询：块头别进上下文、控制长度、只对真正需要的文档生成。

  Key points

  • Split into one-off (per-chunk input/output plus re-embedding) and per-query (header read by both reranker and generator).
  • The one-off is dominated by re-reading each document n times; caching turns that into one write plus n-1 reads.
  • Caching is prefix-matched: the full document must come first, the chunk after, or you get zero hits.
  • Measured on 30 docs / 134 chunks, caching cut the one-off cost by about 29%, and finer chunks save more.
  • The lasting cost is per query: keep headers out of the context window, keep them short, and generate them selectively.
• 什么样的问题必须上图检索？给一个该上的具体例子和一个不该上的例子。What kind of question actually requires graph retrieval? Give one concrete case where it is justified and one where it is not.

  国内高频海外高频深入#graph-rag#multi-hop#cost

  分析过程 · 先想清楚再作答

  1. 这题在考你会不会为了用而用。只要答案里出现『多跳问题就要上图检索』，面试官基本就知道你没落地过——多跳只是必要条件，远不是充分条件。
  2. 判据要落在一个可观察的现象上：**答案的第二篇文档和查询之间，有没有字面或语义上的重合**。有重合，普通的混合检索就能捞到它，多跳是假的；完全没有重合，只能靠一条关系边走过去，这才是图检索的领地。
  3. 该上的例子：问『生产库主备切换必须谁书面审批、这个人叫什么』。一篇写着须平台组组长审批，另一篇写着平台组组长是某人。第二篇跟查询一个词都不重合，我们在五种索引结构下测了一遍，它在 20 条候选池里一次都没出现过——换切法、加块头、父子回填全都无效。
  4. 不该上的例子：问『扩容要走哪个流程、最晚提前几个工作日提单』。同样跨两篇文档，但两篇都跟查询有明显字面重合，混合检索把它们分别排在第 2 名，一次检索就凑齐了。为它建图是拿几倍成本买一个已经解决的问题。
  5. 然后说代价，这一段决定了你像不像做过：建图不止一次抽取调用，实体要消歧、关系要去重、文档更新时受影响的子图要重算，还要多维护一套图存储和一套更新链路。
  6. 可预期的追问是『不上图检索还有什么办法』。答案是把多跳交给 Agentic 检索：让模型先查出中间实体，再拿这个实体发起第二次检索。它的一次性成本几乎为零，代价换成了每次查询的延迟与调用次数——先试这条，试不通再考虑建图。

  How to reason about it · think before answering

  1. This one tests whether you reach for tools you don't need. If the answer is 'multi-hop questions need a graph', the interviewer knows you haven't shipped one — multi-hop is necessary, nowhere near sufficient.
  2. Anchor the criterion on something observable: does the second required document share any lexical or semantic overlap with the query? If it does, ordinary hybrid retrieval will surface it and the hop is illusory. If it shares nothing, only a relation edge gets you there — that is graph territory.
  3. Justified case: 'who must sign off on a production failover, and what is that person's name?' One doc says the platform lead must approve; another says who the platform lead is. The second shares not one term with the query. Across all five index structures we tested, it never once appeared in a 20-item candidate pool — rechunking, headers and parent backfill all failed.
  4. Unjustified case: 'which process covers a capacity change, and how many working days ahead must the ticket be filed?' Also two documents, but both overlap the query lexically; hybrid retrieval ranked them second each, and one pass collected both. Building a graph for this buys a solved problem at several times the cost.
  5. Then state the cost, which is what makes the answer sound operational: graph building is not one extraction call. Entities need disambiguation, relations need dedup, updates force recomputing affected subgraphs, and you now run a graph store and its update pipeline.
  6. Expect 'what else could you do instead'. Hand multi-hop to agentic retrieval: let the model retrieve the intermediate entity first, then issue a second query with it. Near-zero build cost, paid back in latency and call count per query. Try that before you build a graph.

  答题要点

  • 判据不是『是不是多跳』，而是『第二篇文档跟查询有没有字面或语义重合』——没有重合才轮得到图检索。
  • 该上：审批人那类问题，中间实体是唯一的桥，第二篇文档在候选池里一次都不出现。
  • 不该上：两篇都跟查询有重合的多跳题，混合检索一次就能凑齐。
  • 建图的真实成本是实体消歧、关系去重、增量重算和一套额外的图存储，不是一次抽取调用。
  • 先试 Agentic 检索的两次查询，走不通再考虑建图。

  Key points

  • The test is not 'is it multi-hop' but 'does the second document overlap the query at all' — only zero overlap earns a graph.
  • Justified: the approver question, where an intermediate entity is the only bridge and the second doc never enters the candidate pool.
  • Not justified: a multi-hop question whose documents both overlap the query — hybrid retrieval collects them in one pass.
  • Real graph cost is entity disambiguation, relation dedup, incremental subgraph recomputation and a whole extra store — not a single extraction call.
  • Try two-pass agentic retrieval first; build the graph only when that fails.

D12 Agentic RAG：把检索做成工具，让模型自己决定查不查、查几次、要不要推翻重来

• 什么情况下你会拒绝把一个 RAG 系统做成 Agentic 的？拿什么数据说服你的团队？When would you refuse to make a RAG system agentic, and what data would you use to convince your team?

  国内高频海外高频进阶#agentic-rag#cost#engineering-judgement

  分析过程 · 先想清楚再作答

  1. 这题在考工程判断力，也在考你会不会算账。凡是答「Agentic 更先进所以要上」的，直接出局；面试官想听的是你能主动说出它的代价，并且用数字划出适用边界。
  2. 怎么拆：先承认收益来自哪一类问题，再看这类问题在你的流量里占多大比例。Agentic 的收益几乎全部集中在多跳和检索失败重试上，单文档可答的问题一次检索就够了，多查一轮纯属浪费。
  3. 所以判据不是感觉，是评估集：跑一遍，看 multi 那一档占多少题、涨了多少个点，再对照总调用次数涨了多少倍。在一份 20 题的集合上，我们量到的是多跳召回从 75% 涨到 100%，可答题整体只从 93.8% 涨到 100%，代价是平均检索调用从 1 次涨到 1.75 次、外加同样次数的自评调用——为 100% 的问题付钱，只有 5% 的问题拿到好处。
  4. 三类明确不上：延迟敏感（每多一轮就是一次检索加一次模型往返，首字延迟拉长一到两倍）；问题模式固定（九成是单文档可答，收益接近零）；成本吃紧（真实模型不像离线替身那样老实，成本方差比均值更难受，按均值做的容量规划会在长尾上被打穿）。
  5. 给出替代方案才算完整：分流。先用一次便宜的判断看这一问像不像多跳，像才进循环，不像走固定流程。九成走一次检索、一成走循环，账完全不一样。这也说明循环是一种能力，不是默认值。
  6. 可预期的追问是「那你怎么知道哪些问题像多跳」。答案是从评估集和线上日志里找模式（问句里同时问了两个事实、问的是某个角色背后的人），先用规则跑，跑不动再上小模型分类——顺序不要反。

  How to reason about it · think before answering

  1. This tests engineering judgement and whether you can do arithmetic. Anyone who says 'agentic is more advanced so we should ship it' is out. The interviewer wants you to name the cost and draw the boundary with numbers.
  2. Decompose it: identify which question types actually benefit, then check how much of your traffic they represent. Agentic gains concentrate in multi-hop questions and retrieval retries; single-document questions are answered by one lookup and every extra round is waste.
  3. So the criterion is the evaluation set, not intuition. On a 20-item set we measured multi-hop recall going from 75% to 100% while overall answerable recall moved only from 93.8% to 100%, at the cost of average retrieval calls going from 1 to 1.75 plus the same number of assessment calls - you pay for 100% of traffic so that 5% of it improves.
  4. Three clear refusals: latency-sensitive surfaces, where each round adds a retrieval plus a model round trip and roughly doubles time to first token; fixed question patterns, where nine in ten questions are single-document and the gain is near zero; and tight cost budgets, where a real model is less disciplined than an offline stand-in and the variance, not the mean, is what breaks your capacity plan.
  5. Finish with the alternative: route. Use one cheap check to decide whether a question looks multi-hop, and only then enter the loop. Nine tenths take a single retrieval, one tenth loops, and the economics change completely. Looping is a capability, not a default.
  6. Expected follow-up: how do you know which questions look multi-hop? Mine the eval set and production logs for patterns - two facts requested in one sentence, or a question about the person behind a role - start with rules, and reach for a small classifier only when rules stop working.

  答题要点

  • 收益集中在多跳与检索失败重试，单文档可答的问题上收益接近零。
  • 用评估集算账：multi 档涨了多少点，对照总调用次数涨了多少倍。
  • 实测过的一组数字：多跳召回 75% 到 100%，整体 93.8% 到 100%，检索调用 1 次到 1.75 次外加等量自评调用。
  • 三类不上：延迟敏感、问题模式固定、成本吃紧（方差比均值更难受）。
  • 替代方案是分流：便宜的判断先过滤，像多跳才进循环。
  • 循环是一种能力，不是默认值。

  Key points

  • Gains concentrate in multi-hop and retry cases; single-document questions gain almost nothing.
  • Settle it with the evaluation set: multi-hop delta against the multiplier on total calls.
  • One measured set: multi-hop recall 75% to 100%, overall 93.8% to 100%, retrieval calls 1 to 1.75 plus the same number of assessment calls.
  • Refuse when latency-sensitive, when question patterns are fixed, or when cost is tight - variance hurts more than the mean.
  • Route instead: a cheap check up front, and only multi-hop-looking questions enter the loop.
  • Looping is a capability, not a default.