面试题库

14 天 RAG：从检索到可信回答

D2 embedding 与向量检索：相似度、维度与模型选型，把文本存进 pgvector

• 余弦相似度和内积什么时候等价？如果向量没有归一化，用内积排序会出什么问题？When are cosine similarity and inner product equivalent? What goes wrong if you rank by inner product on vectors that are not normalised?

  国内高频海外高频基础#embeddings#similarity#normalisation

  分析过程 · 先想清楚再作答

  1. 这题是送分题，但区分度藏在后半句。只答「归一化之后两者等价」的人很多，面试官真正想听的是「没归一化会怎么坏」，因为那是线上真的会发生的事。
  2. 先把定义摆出来：余弦相似度等于内积除以两个向量模长的乘积。模长都是 1 时除数就是 1，所以余弦相似度就是内积——这一句话就是等价的全部理由，不需要额外的假设。
  3. 再说没归一化的后果：内积里混着「方向有多一致」和「向量有多长」两层信息。文本越长，模型输出的向量模长往往越大，于是排序会系统性地偏向长文档——这跟 BM25 里 b 参数要压的是同一个毛病，只是换了个地方冒出来。
  4. 点出这类 bug 的性质：它不报错。程序照常跑、结果照常出，只是名次悄悄偏了，你要跑一轮离线评估才可能发现。所以工程上的做法是在 embedding 的出口统一归一化一次，而不是靠每个调用点自觉。
  5. 补一句欧氏距离：向量都归一化之后，欧氏距离的平方等于 2 减去 2 倍内积，也就是余弦距离的单调函数，三种距离排出来的名次完全一致。这一句能说明你理解的是关系而不是三条并列的规则。
  6. 可预期的追问：那 pgvector 里该用哪个运算符？答案是既然已经归一化，`<=>`（余弦距离）和 `<#>`（负内积）名次一样，选 `<=>` 的理由是可读性和「就算哪天有人漏了归一化也不至于错」。

  How to reason about it · think before answering

  1. This starts as a giveaway, but the second half is where candidates separate. Many can say 'they are equivalent after normalisation'; few can describe what breaks without it.
  2. State the definition: cosine similarity is the inner product divided by the product of the two magnitudes. When both magnitudes are 1, the divisor is 1 and cosine reduces to the inner product. That is the whole argument.
  3. Then the failure mode: an un-normalised inner product mixes 'how aligned' with 'how long'. Longer texts tend to produce larger-magnitude vectors, so ranking drifts systematically toward long documents, the same bias BM25's b parameter exists to counter.
  4. Stress that this bug is silent. Nothing throws, results still look plausible, and only an offline evaluation reveals the drift. Hence the engineering rule: normalise once at the embedding boundary, never at each call site.
  5. Add Euclidean distance for completeness: on normalised vectors, squared L2 equals 2 minus twice the inner product, a monotone function of cosine distance, so all three metrics produce the same ranking.
  6. Expected follow-up: which pgvector operator should you use? Since the vectors are normalised, `<=>` and `<#>` rank identically; prefer `<=>` for readability and because it stays correct if someone later forgets to normalise.

  答题要点

  • 余弦相似度 = 内积 / 两个模长之积，模长为 1 时除数为 1，两者等价。
  • 没归一化时内积混入模长信息，长文档的向量模长普遍更大，排序会系统性偏向长文档。
  • 这类错误不报错，只能靠离线评估发现，所以要在 embed 出口统一归一化。
  • 归一化之后欧氏距离与余弦距离互为单调函数，三种运算符名次一致。
  • pgvector 里对应 `<->`（L2）、`<#>`（负内积）、`<=>`（余弦距离）三个运算符。

  Key points

  • Cosine equals inner product divided by both magnitudes; with unit magnitudes the divisor is 1, so they coincide.
  • Without normalisation the inner product carries magnitude, and longer documents usually have larger magnitudes, biasing the ranking.
  • The failure is silent, so normalise once at the embedding boundary and verify with offline evaluation.
  • On normalised vectors L2 and cosine are monotonically related, so all operators rank the same.
  • In pgvector the operators are `<->` for L2, `<#>` for negative inner product and `<=>` for cosine distance.
• 把 embedding 维度从 1536 降到 512，你会损失什么？什么场景下这个损失可以接受？What do you lose when you cut embedding dimensions from 1536 to 512, and when is that loss acceptable?

  国内高频海外高频进阶#embeddings#dimensions#cost

  分析过程 · 先想清楚再作答

  1. 这题考的是你会不会算账。只说「维度越低越省、精度越低」的答案没有区分度，面试官在等一个具体的成本模型和一个决策顺序。
  2. 先把三笔账列出来：存储与内存（向量数量乘维度乘每维字节数，近似最近邻索引要把它放进内存，所以基本等于机器预算）、检索延迟（每次比较就是一轮乘加，维度大致线性影响耗时）、检索质量（收益递减，低维段每加一档提升明显，高维段加倍只换来很小的改善）。
  3. 再说清降维为什么可行：主流模型用套娃式表示训练，重要信息压在靠前的维度上，所以直接截短再归一化仍然可用，这不是另训了一个小模型。截短必然有损失，损失多少只能在自己的数据上跑评估才知道。
  4. 给出决策顺序：先按存储与内存预算倒推一个维度上限，再从上限往下试两三档，看指标掉多少，掉得能接受就用低的。反过来「先选最高维再想办法省钱」基本都会返工。
  5. 点出可接受的典型场景：库很大而单条价值不高（比如日志、工单）、召回之后还有重排兜底（重排能把粗排的损失补回来一部分）、或者对延迟极敏感的在线场景。反过来法务、医疗这类一条都不能漏的场景就要谨慎。
  6. 可预期的追问：能不能不同文档用不同维度？不能——同一个索引里所有向量必须同维，改维度等于全库重建，这跟换模型是同一类迁移成本。

  How to reason about it · think before answering

  1. This is a cost-modelling question. 'Lower dimensions are cheaper but less accurate' earns nothing; the interviewer wants a cost model and a decision order.
  2. Lay out three costs: storage and memory (vector count times dimensions times bytes per dimension, which an ANN index must hold in RAM), query latency (roughly linear in dimensions), and retrieval quality, whose returns diminish sharply at the high end.
  3. Explain why truncation works at all: models trained with Matryoshka representations pack the most important information into the leading dimensions, so truncating and re-normalising keeps the vector usable. It is still lossy, and how lossy is an empirical question on your own data.
  4. Give the decision order: derive a dimension ceiling from your memory budget, then step down two or three notches and measure the metric drop. Choosing the largest model first and optimising cost later usually means redoing the work.
  5. Name the acceptable cases: large corpora of low individual value, pipelines where a reranker recovers some of the loss, and latency-critical online paths. Be conservative where a single miss is expensive, such as legal or clinical retrieval.
  6. Expected follow-up: can different documents use different dimensions? No. Every vector in an index must share one dimension, so changing it means rebuilding the whole index, the same migration cost as changing models.

  答题要点

  • 三笔账：存储与索引内存、检索延迟、检索质量，前两笔随维度近似线性，第三笔收益递减。
  • 套娃式表示让截短再归一化仍然可用，但一定有损失，损失多少要在自己的数据上评估。
  • 决策顺序是先按内存预算定上限，再往下试档位看指标掉多少。
  • 库大、单条价值低、后面还有重排兜底、对延迟敏感的场景，降维划算。
  • 同一索引里维度必须一致，改维度等于全库重建。

  Key points

  • Three costs: storage and index memory, query latency, and retrieval quality; the first two scale with dimensions, the third has diminishing returns.
  • Matryoshka representations make truncation viable, but it is lossy and the loss must be measured on your own data.
  • Decide by deriving a ceiling from the memory budget, then stepping down and measuring.
  • Truncation pays off for large corpora, low-value items, latency-sensitive paths, and pipelines with a reranker.
  • All vectors in one index share a dimension, so changing it forces a full rebuild.
• 为什么有些 embedding 模型要求查询和文档加不同的前缀？不加会怎样，你怎么在上线前发现这个问题？Why do some embedding models require different prefixes for queries and documents? What happens if you skip them, and how would you catch it before shipping?

  国内高频海外高频进阶#embeddings#model-selection#evaluation

  分析过程 · 先想清楚再作答

  1. 这题的题眼是「静默失效」。会背「e5 要加 query 和 passage 前缀」只能拿基础分，能说清它为什么不报错、以及怎么在上线前抓住它，才是做过的人。
  2. 先讲原因：这一族模型是拿成对数据训练的，一侧是短问句、一侧是长段落，两者的分布本来就不一样。前缀是训练时给模型的角色标记，告诉它这一段该按查询编码还是按文档编码。推理时不给，模型就落在了训练分布之外。
  3. 再讲后果的性质：不加前缀模型照样输出向量、照样能算距离、名次照样有先后，只是整体质量下滑。**没有任何报错**——这跟忘了归一化是同一类问题：错误不会自己浮出来。
  4. 怎么发现：唯一可靠的办法是一小份标注问题集，用同一批文档跑两遍（加前缀与不加前缀），比命中率。这就是第 8 天要做的评估闸门，它的价值恰恰在于抓这类静默错误。上线前跑一遍，比读十遍文档管用。
  5. 补一个更容易踩的变体：**建库时加了前缀、查询时忘了加**，或者两边加成同一个前缀。这种情况下所有向量都在同一个坐标系里，看起来更「正常」，但查询与文档的对齐关系是错的，掉分同样查不出来。所以前缀应该封装在 embed 的调用约定里，而不是散在各处手拼。
  6. 可预期的追问：OpenAI 的模型要不要加前缀？不需要——它不属于这一族。所以这不是一条普遍规则，而是**每换一个模型都要重新读模型卡片确认**的事。

  How to reason about it · think before answering

  1. The core of this question is silent failure. Reciting 'e5 needs query: and passage: prefixes' is the baseline; explaining why nothing errors out and how you would catch it is what shows experience.
  2. The reason: these models are trained on pairs, short questions on one side and longer passages on the other, two genuinely different distributions. The prefix is a role marker learned during training. Omit it at inference and you are off-distribution.
  3. The consequence: the model still returns vectors, distances still compute, results still have an order, quality just degrades. Nothing throws, exactly like forgetting to normalise.
  4. How to catch it: run a small labelled question set against the same corpus twice, with and without prefixes, and compare hit rate. That is the evaluation gate built on day 8, and catching silent regressions is precisely what it is for.
  5. Mention the sneakier variant: prefixing at index time but not at query time, or using the same prefix on both sides. Everything sits in one coordinate space and looks healthier, yet the query-document alignment is wrong and the loss is just as invisible. Encapsulate prefixes in the embedding call convention rather than hand-writing them everywhere.
  6. Expected follow-up: do OpenAI models need prefixes? No, they are not in that family, so this is not a universal rule but a per-model detail you re-check on the model card every time you switch.

  答题要点

  • 这类模型用问句与段落的成对数据训练，前缀是区分两种角色的标记，缺了就落在训练分布之外。
  • 不加前缀不会报错，只会整体掉分，属于静默失效。
  • 唯一可靠的发现方式是拿一份标注问题集跑 A/B 对比命中率。
  • 更隐蔽的错法是两边前缀不一致或用了同一个前缀，看起来更正常但对齐是错的。
  • 前缀应封装在 embed 的调用约定里；换模型必须重读模型卡片，它不是普遍规则。

  Key points

  • These models are trained on question-passage pairs; the prefix marks which role a text plays, and omitting it puts you off-distribution.
  • Skipping prefixes never errors, it only degrades quality, so the failure is silent.
  • The reliable detection is an A/B run over a small labelled question set, comparing hit rate.
  • A subtler bug is mismatched or identical prefixes on both sides, which looks healthier but misaligns queries and documents.
  • Keep prefixes inside the embedding call convention, and re-read the model card whenever you switch models.
• 向量检索能完全取代关键词检索吗？举一个向量必然失手的查询，并说说你会怎么补。Can vector search fully replace keyword search? Give a query where vectors are bound to fail, and say how you would fix it.

  国内高频海外高频进阶#hybrid-search#embeddings#retrieval-failure

  分析过程 · 先想清楚再作答

  1. 这题是典型的「立场题」，答「能」或「不能」都不重要，重要的是你能不能举出一个具体到能复现的反例。举不出例子，前面说得再漂亮也会被判成没做过。
  2. 先给失手的类型，一次给全：错误码与状态码（429、E1032）、版本号与型号（v2.3.1、X20 Pro）、人名与工号、订单号与文档编号、以及否定表达。前四类的共同点是**这些词的价值在于字面唯一，而向量只保留语义邻近**，模型会把 429 和「限流」「超时」这些话题相近的东西编到一起，反而把真正写着 429 的那篇挤下去。
  3. 拿一个能复现的例子说：问「限流超了返回 429 吗」，BM25 稳稳命中写着 429 的接口文档，向量却可能把话题相近但没提 429 的产品手册排在前面。这个现象在本课第 2 天的实验里就能亲眼看到。
  4. 否定表达要单独强调：「支持导出 PDF」和「不支持导出 PDF」在向量空间里几乎重合，因为它们谈的是同一件事。指望向量区分肯定与否定一定翻车，这一层要靠生成侧读原文来判断。
  5. 怎么补：两路并行跑再融合，关键词一路用 BM25、向量一路用最近邻，用倒数排名融合把两个名次合成一个。这就是混合检索，本课第 9 天展开。要点是**两套的错法不一样**，所以合起来才有增益——如果两套错在同一批查询上，融合是白做的。
  6. 可预期的追问：那关键词一路能不能扔掉、改成让模型改写查询？可以缓解一部分（第 10 天的查询改写），但改写救不了字面唯一的标识符——你没法把 429 改写成别的说法。

  How to reason about it · think before answering

  1. This is a stance question where the stance matters less than the counter-example. Without a concrete, reproducible failing query, the rest of the answer reads as theory.
  2. Enumerate the failure classes up front: error and status codes, version numbers and SKUs, names and employee IDs, order or document identifiers, and negation. The first four share one property: their value lies in exact literal identity, which embeddings deliberately blur into semantic neighbourhoods.
  3. Give a reproducible example: ask whether rate limiting returns 429. BM25 lands on the API document that literally contains 429, while vector search may rank a topically similar product manual that never mentions the code.
  4. Call out negation separately: 'supports PDF export' and 'does not support PDF export' sit almost on top of each other because they discuss the same thing. Vectors cannot carry that distinction; the generation step reading the source has to.
  5. The fix: run both retrievers and fuse the rankings, BM25 on the lexical side and nearest neighbour on the vector side, combined with reciprocal rank fusion. That is hybrid search, covered on day 9. Fusion helps precisely because the two systems fail on different queries.
  6. Expected follow-up: could you drop the keyword path and rewrite queries instead? Rewriting helps with vocabulary mismatch, but it cannot rescue exact identifiers, since there is no paraphrase of 429.

  答题要点

  • 不能取代：错误码、版本号、人名、单号这类词的价值在于字面唯一，向量只保留语义邻近。
  • 具体反例：问「限流超了返回 429 吗」，BM25 命中写着 429 的文档，向量把话题相近却没提 429 的文档排前面。
  • 否定表达是另一类失手：肯定句与否定句在向量空间里几乎重合。
  • 补法是混合检索：两路并行再用倒数排名融合合并名次。
  • 融合有增益的前提是两套的错法不同；查询改写能缓解词汇不匹配，但救不了字面唯一的标识符。

  Key points

  • No: codes, version numbers, names and IDs matter as exact literals, which embeddings blur into neighbourhoods.
  • Concrete example: asking whether rate limiting returns 429, where BM25 hits the document containing 429 and vectors surface a topically similar one that never mentions it.
  • Negation is a second failure class, since affirmative and negative statements sit almost on top of each other.
  • The remedy is hybrid retrieval: run both paths and merge with reciprocal rank fusion.
  • Fusion pays off because the two paths fail differently; query rewriting helps vocabulary mismatch but not exact identifiers.

D7 第一周综合：把六天的零件装成一个可一键启动的检索问答服务并复盘

• 你会怎么划分一个检索增强生成系统的模块边界？其中哪一层最应该做成可替换的，为什么？How would you draw the module boundaries of a RAG system, and which layer most needs to be swappable? Why?

  国内高频海外高频基础#architecture#modularity#embeddings

  分析过程 · 先想清楚再作答

  1. 这题考的是你有没有真的维护过这类系统。只按「解析、切块、检索、生成」复述一遍流程图，面试官会判定你只搭过 demo——流程图人人都会画，切口画在哪才是经验。
  2. 给一条可复用的判据再往下推：切口应该落在「将来最可能被整个换掉」的地方，而不是按代码量或者功能名称均分。
  3. 用它过一遍：embedding 一年会换好几次，换一次库里所有向量作废、必须全量重算，所以它必须是接口；存储可能从 PostgreSQL 换成专用向量库，而且摄取和查询都要通过它，所以它是两条链路的唯一交界；切块策略在调优期天天改，所以它必须是配置项而不是硬编码。
  4. 结论：最该做成可替换的是 embedding 那一层，理由不是「设计模式」，而是「换模型这件事真的会发生，且发生时代价极高」。
  5. 顺手点出抽象的代价：每多一层间接就多一次跳转和一份心智负担，所以判据是「那件事会不会真的发生」，不会发生的别抽象。
  6. 可预期的追问：那生成模型要不要也抽象？答案是要，但优先级低——换生成模型不需要重算任何存量数据，回滚也便宜，所以它是配置项而不是一层接口。

  How to reason about it · think before answering

  1. This question separates people who have maintained such a system from people who have only built a demo. Reciting the pipeline diagram is not an answer; where you cut it is.
  2. Offer a reusable criterion first: cut where a layer is most likely to be replaced wholesale, not by lines of code or by tidy functional names.
  3. Apply it. Embedding models change several times a year, and each change invalidates every stored vector, so that layer must be an interface. Storage may move from PostgreSQL to a dedicated vector database, and both ingestion and query talk through it, so it is the single shared boundary. Chunking changes daily during tuning, so it belongs in config, not in code.
  4. Conclusion: the embedding layer is the one that must be swappable, because the swap is both likely and expensive, not because interfaces are good style.
  5. Name the cost of abstraction too: every indirection is one more hop while debugging, so the test is whether the change will actually happen.
  6. Expected follow-up: should the generation model be abstracted as well? Yes, but at lower priority, because swapping it does not force recomputation of stored data and rollback is cheap. It is a config value, not a layer.

  答题要点

  • 先给判据：切口落在最可能被整体替换的那一层，不按代码量或功能名称均分。
  • embedding 是最该抽象的一层：换模型意味着存量向量全部作废、必须全量重算，代价高且真的会发生。
  • 存储层是摄取与查询唯一的交界，接口要先定下来再谈两边实现。
  • 切块与检索路数做成配置项，因为它们在调优期改动最频繁，改一次不该动代码。
  • 抽象有成本，判据是那件事会不会真的发生；不会发生的抽象就是过度设计。

  Key points

  • Lead with the criterion: cut where a layer is most likely to be replaced wholesale.
  • The embedding layer is the one to abstract: swapping models invalidates every stored vector and forces a full recompute.
  • Storage is the single boundary shared by ingestion and query, so define its interface before either implementation.
  • Chunking and retrieval routes belong in configuration because they change most often during tuning.
  • Abstraction costs indirection, so only abstract changes that will actually happen.