面试题库

14 天 RAG：从检索到可信回答

D1 为什么要检索：幻觉、知识截止与长上下文的代价，以及一个纯关键词的最小 RAG

• 上下文窗口已经做到上百万 token 了，检索这一步会被淘汰吗？Context windows are now in the millions of tokens. Does that make the retrieval step obsolete?

  国内高频海外高频深入#long-context#cost#system-design

  分析过程 · 先想清楚再作答

  1. 这是一道立场题，容易答成非黑即白。判断你有没有做过的地方在于：会不会区分「技术上能不能塞进去」和「工程上该不该每次都塞」，只谈前者的答案一听就是纸上谈兵。
  2. 先承认对方有道理的部分：窗口变大确实吃掉了检索的一部分场景。几十篇文档、更新不频繁、调用量不大的内部工具，直接全塞是最省事的选择，为它建一套检索系统是过度设计。
  3. 再给三条它吃不掉的理由。第一是成本：材料是按次计费的，同一份材料被问一万次就要付一万次，而检索只付取回的那几段；预填充缓存能缓解但不能消除，缓存也有有效期和命中率。
  4. 第二是规模：企业知识库动辄几十万篇，再大的窗口也塞不下，检索是唯一的入口。第三是归因与权限：答案要指回具体某一段，以及不同的人只能看到自己有权访问的材料——这两件事必须在把材料喂给模型之前完成，窗口再大也不解决。
  5. 还要补一条经验事实：材料变多之后，模型在长上下文里定位关键信息的稳定性会下降，出现「读了但没读到」。所以「全塞」并不总是等于「效果更好」，很多时候少而准反而更好。
  6. 可预期的追问：那检索的形态会不会变？会——窗口变大之后，取回的块可以更大、条数可以更多，重排与压缩的压力变小，检索从「精挑几句」变成「粗筛一批」。趋势是检索的粒度变粗，不是检索消失。

  How to reason about it · think before answering

  1. This is a position question and it is easy to answer as a binary. The signal is whether you separate what fits technically from what is worth paying for on every request.
  2. Concede the valid half first: bigger windows genuinely absorb part of the use case. For an internal tool over a few dozen stable documents with low traffic, stuffing everything in is the right call and building a retrieval stack would be over-engineering.
  3. Then give three reasons it does not absorb the rest. Cost is the first: context is billed per request, so the same corpus is paid for on every one of ten thousand queries, whereas retrieval only pays for the passages it returns. Prompt caching softens this but does not remove it.
  4. Scale is the second: enterprise corpora run to hundreds of thousands of documents and no window holds them. Attribution and access control are the third: pointing an answer at a specific passage, and showing each user only what they are permitted to see, both have to happen before the material reaches the model.
  5. Add the empirical point: as the supplied material grows, models become less reliable at locating the one relevant fact inside it. More context is not automatically better; fewer and more precise passages often win.
  6. Expected follow-up: does retrieval change shape? Yes. Larger windows allow bigger chunks and more of them, which relieves pressure on reranking and compression. Retrieval gets coarser, it does not disappear.

  答题要点

  • 先区分「能不能塞进去」和「该不该每次都塞」，前者是技术问题，后者是成本问题。
  • 小规模、低频、少变的语料确实可以直接全塞，为它建检索系统是过度设计。
  • 检索不会被淘汰的三个理由：按次计费的成本、几十万篇塞不下的规模、必须在喂给模型之前完成的归因与权限过滤。
  • 材料越多，模型定位关键信息的稳定性越差，全塞不等于效果更好。
  • 趋势是检索粒度变粗——块更大、条数更多、重排压力变小，而不是检索消失。

  Key points

  • Separate whether it fits from whether it is worth paying for on every request.
  • Small, stable, low-traffic corpora can legitimately be stuffed whole; building retrieval for them is over-engineering.
  • Three reasons retrieval survives: per-request cost, corpora too large for any window, and attribution plus access control that must happen before the model sees the material.
  • More supplied context reduces the reliability of locating a single fact, so stuffing everything is not automatically better.
  • The trend is coarser retrieval — bigger chunks, more of them, less reranking pressure — not the removal of retrieval.

D4 切块策略：固定、递归、按结构、父子与语义五种切法，以及用评估而不是直觉来选

• 语义切分比递归切分贵不少，你怎么向团队证明这笔钱值得花？Semantic chunking costs considerably more than recursive splitting. How would you prove to your team that the money is well spent?

  国内高频海外高频深入#chunking#evaluation#cost

  分析过程 · 先想清楚再作答

  1. 这题表面问技术，实际考的是你会不会做一次带对照组的技术论证。上来就讲语义切分原理的人，答的是另一道题。
  2. 第一步是先承认它可能不值。语义切分的收益来自「文档没有可用的结构」；如果知识库是结构良好的文档，作者的标题层级已经免费替你做完了语义切分，这时候花的钱大概率打水漂。**先说清适用前提，再谈证明，这一步就把大多数候选人区分开了。**
  3. 第二步是把「值不值」翻译成可测的三笔账：指标涨了多少（同一批标准问题、同一个 token 预算下的命中率）、延迟涨了多少（切块是离线的，但更新链路的端到端时间会变）、钱涨了多少（首次全量 embedding 的费用，加上按更新频率折算的重算费用）。只报第一笔的论证不成立。
  4. 第三步是设计对照。递归切分是基线，语义切分是实验组，两组必须用同一份语料、同一批问题、同一个上下文预算、同一个检索器，只改切法这一个变量。改两个变量的实验，结论一文不值。
  5. 第四步是给决策一个门槛，而不是给一个感想。比如：命中率相对基线提升低于三个百分点就不上；提升超过五个百分点且重算成本在月度预算内就上；中间地带先在一类文档上灰度。**门槛要在跑数字之前定好**，否则你会不自觉地去迁就已经跑出来的结果。
  6. 可预期的追问是「有没有更便宜的办法拿到同样的收益」。答有：先试按结构切，它零成本且效果常常接近；结构确实不可用时，再考虑只对高价值的那一部分文档做语义切分，而不是全量上。

  How to reason about it · think before answering

  1. This looks like a technical question but it tests whether you can run a controlled technical argument. Launching into how semantic chunking works answers a different question.
  2. Step one is to concede that it may well not be worth it. The gain comes from documents that have no usable structure; if your knowledge base is well-formed documents, the authors' heading hierarchy already did the semantic split for free and the money is likely wasted.
  3. Step two is translating 'worth it' into three measurable numbers: how much the metric moved (hit rate on the same golden set under the same token budget), how much latency moved (chunking is offline, but the end-to-end update path changes), and how much it costs (the initial full embedding pass plus recomputation amortised over update frequency).
  4. Step three is the control. Recursive splitting is the baseline, semantic chunking the treatment, and they must share the corpus, the questions, the context budget and the retriever. Change one variable only; a two-variable experiment proves nothing.
  5. Step four is a decision threshold rather than an impression. For example: below three points of hit-rate gain, no; above five points with recomputation inside the monthly budget, yes; in between, roll it out on one document class first. Fix the threshold before you run the numbers, or you will quietly bend it to fit them.
  6. Expect the follow-up: is there a cheaper way to the same gain. Yes — try structural splitting first, since it is free and often nearly as good, and if the structure really is unusable, apply semantic chunking only to the high-value subset rather than the whole corpus.

  答题要点

  • 先讲适用前提：语义切分的收益来自文档没有可用结构，结构良好的文档上它大概率不值。
  • 把「值不值」翻译成三笔账：命中率涨多少、延迟涨多少、钱涨多少，只报第一笔不算论证。
  • 做对照实验：同语料、同问题集、同上下文预算、同检索器，只改切法一个变量。
  • 决策门槛必须在跑数字之前定好，避免事后迁就结果。
  • 先试零成本的按结构切；确需语义切分时也优先只覆盖高价值文档，而不是全量上。

  Key points

  • Start with the precondition: the gain comes from documents without usable structure, so on well-formed documents it usually is not worth it.
  • Translate 'worth it' into three numbers — hit rate, latency, and cost. Reporting only the first is not an argument.
  • Run a controlled comparison: same corpus, same golden set, same context budget, same retriever, with the splitting strategy as the only variable.
  • Fix the decision threshold before running the numbers so you cannot bend it to fit the result afterwards.
  • Try free structural splitting first, and if semantic chunking is genuinely needed, apply it to the high-value subset rather than the entire corpus.

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.