Interview Bank

From Frontend Engineer to Agent Engineer in 30 Days

D12 Long-Term Memory: pgvector, Embeddings, Chunking, the memory_search Tool

• How does the chunking strategy affect retrieval quality, and how do you pick a chunk size?chunking 的切分策略会怎么影响检索效果？切多大合适？

  Common in ChinaCommon overseasIntermediate#chunking#rag#retrieval-quality

  How to reason about it · think before answering

  1. The hinge is 'how does it affect'. Naming a number alone invites a why, so describe both failure modes first and let the number follow.
  2. Too small: a chunk loses its context. 'He wants size 42' retrieves fine but resolves to nothing — pronouns dangle and the model is more likely to fabricate.
  3. Too large is the counter-intuitive half and the real discriminator: a chunk spanning three topics gets a vector that averages them, so it looks only vaguely like any query and recall drops. Bigger chunks carry more information yet are harder to retrieve.
  4. Give an operational default: target 400 characters with 80 characters of overlap, ending on natural boundaries such as sentence stops or newlines. Explain the overlap — when a key sentence lands on a cut, each side holds half of it, and the overlap guarantees at least one chunk holds it whole.
  5. Add the costs: 80 over 400 is 20% storage amplification plus an extra vector per duplicated span, and near-duplicate chunks can both surface and waste result slots, so deduplicate by content before returning.
  6. Expect: how do you validate a chunking strategy? Build a query set with labelled expected hits and measure recall and top-k hit rate, then re-run after changing parameters — chunking is measurable, not a matter of taste. Second follow-up: should raw dialogue be chunked as-is? No — have the model distil it into standalone statements first, or filler turns flatten the vectors.

  分析过程 · 先想清楚再作答

  1. 题眼在「怎么影响」。只回答一个数字（比如「切 500 字」）会被追着问为什么，所以要先把两个方向的失效模式讲出来，数字才有落点。
  2. 切太碎的失效模式：单张卡片脱离上下文。「他说要 42 码」检索命中了也没用，代词失去指代，模型拿到一句悬空的话反而更容易编。
  3. 切太整的失效模式更反直觉，也是这题真正的区分点：一块横跨三个主题时，它的向量是这几个主题的平均值，结果对哪个 query 都不太像，命中率反而下降。**块越大信息越全，却越难被检索到**——能说出这句话基本就过了。
  4. 然后给可操作的口径：目标 400 字符、相邻块重叠 80 字符，并优先在句号、换行这类自然边界收尾。重叠的作用要说清楚——一句关键的话被切口劈开时，两块各拿半句，重叠保证它至少在其中一块里是完整的。
  5. 补上代价，这是工程视角：重叠 80 除以 400 等于 20% 的存储放大，向量也跟着多一份；内容高度重叠的两块可能一起被检索出来，白占返回名额，所以要按内容去重。
  6. 可以预期的追问：怎么验证切分策略好不好？答案是准备一批 query 与标注好的期望命中，量召回率和 top-k 命中率，改切分参数后重跑对比——切分是可以被度量的，不该靠感觉调。第二个追问是「对话数据要不要原样切」，答不要：先让模型抽成陈述句再切，否则大量寒暄句会把向量拉平。

  Key points

  • Too small: chunks lose context, pronouns dangle, and a hit is useless
  • Too large: one chunk spans several topics, its vector averages them, and recall drops for every query
  • Working default: target 400 characters with 80 characters of overlap, cutting on sentence or newline boundaries
  • Overlap keeps a split sentence whole in at least one chunk, at roughly 20% storage amplification plus possible duplicate hits
  • Distil dialogue into standalone statements before chunking, and validate with a labelled query set measuring recall

  答题要点

  • 切太碎：单块脱离上下文，代词失去指代，命中了也用不上
  • 切太整：一块横跨多个主题，向量被平均，对任何 query 都不够像，命中率反而下降
  • 可操作口径：目标 400 字符、重叠 80 字符，优先在句号或换行这类自然边界收尾
  • 重叠的作用是保证被切口劈开的句子至少在一块里完整；代价是约 20% 的存储放大和可能的重复命中
  • 别直接切对话原文，先抽成陈述句；切分效果要用标注好的 query 集测召回率，而不是凭感觉

D24 RAG, Level Up: Hybrid Search, Reranking, Citations, Recall Evaluation

• How do you merge two retrieval rankings, and why not just take a weighted sum of the scores?两路检索结果怎么合并？为什么不能直接加权求和？

  Common in ChinaCommon overseasIntermediate#rag#rrf#ranking

  How to reason about it · think before answering

  1. The second half is the real question. Anyone can say 'RRF'; explaining why weighted sums fail is what separates people who have looked at the score distributions.
  2. Decompose it: are the two scores even the same unit? Cosine similarity is bounded in 0 to 1 and tightly clustered — candidates often differ by 0.02. BM25 is unbounded and a few rare-term hits reach 12. Adding them lets the larger-magnitude channel decide everything; the weight only tunes how much it dominates.
  3. Worse, it is unstable. Weights tuned on one corpus drift on the next, so you re-tune forever.
  4. Conclusion: fuse ranks, not scores. RRF maps each rank to 1/(k + rank) and sums, with k = 60. Ranks are unitless and need no calibration. k flattens the head of the list so that 'top-ranked in both channels' beats 'first in one channel' — consensus over single-source confidence.
  5. A hand-checkable example helps: rankings [a,b,c] and [c,d,a] give a = 1/61 + 1/63 ≈ 0.0323, while a raw score sum promotes c on the strength of its BM25 12.
  6. Expected follow-up: what about ties? You must break them explicitly, e.g. by id. Otherwise ordering depends on hash-map iteration order and differs across languages and runs, which makes your evaluation numbers irreproducible. Mentioning this signals you actually ran it more than once.

  分析过程 · 先想清楚再作答

  1. 题眼在后半句。前半句答「RRF」谁都会，后半句「为什么不能加权求和」才是筛人的地方——它考的是你有没有真的看过两路分数的分布。
  2. 怎么拆：先问自己两个分数是不是同一个量纲。余弦相似度有界（0 到 1）且分布密集，同一批候选常常只差 0.02；BM25 无上界，命中几个稀有词就能到 12 分。**不同量纲的数相加，等于让量纲大的那一路单方面决定结果**，权重只是在调「它说了算的程度」。
  3. 更麻烦的是它不稳定：权重在这批语料上调好了，换一批语料分布就变了，得重调。这是一个永远还不完的技术债。
  4. 结论：改用名次。RRF 把每一路的名次折算成 `1/(k + rank)` 再相加，k 取 60。名次是无量纲的，不需要任何标定。k 的作用是压平头部差距，让「两路都进前列」压过「一路排第一」——共识优先于单点自信。
  5. 一个能当场手算的例子很加分：两路排名 [a,b,c] 与 [c,d,a]，a 得 1/61 + 1/63 ≈ 0.0323；而分数直接相加的版本会把 BM25 里 12 分的 c 顶到第一。
  6. 可预期的追问：同分了怎么办？必须显式定序（比如按 id），否则结果取决于哈希表遍历顺序，同一份输入在不同语言、不同运行里给出不同排序——评估集量出来的数字也就不可复现了。这一条答出来会非常加分，因为它说明你真的跑过多次。

  Key points

  • Use RRF: map each channel's rank to 1/(k + rank) and sum, with k = 60.
  • Weighted sums fail because the scores are different units — bounded, tightly clustered cosine versus unbounded BM25, so BM25 decides the outcome.
  • Weights also do not transfer: tuned on one corpus, they drift on the next.
  • Ranks are unitless and need no calibration; k flattens the head so cross-channel consensus outweighs single-channel confidence.
  • Break ties explicitly (by id) or ordering depends on hash iteration order and your evaluation numbers stop being reproducible.

  答题要点

  • 用 RRF：每一路的名次折算成 1/(k + rank) 再相加，k 取 60。
  • 不能加权求和是因为两个分数量纲不同——余弦有界密集、BM25 无上界，相加等于让 BM25 单方面决定结果。
  • 而且权重不可迁移：这批语料调好，换一批就得重调，是还不完的债。
  • 名次是无量纲的，不需要标定；k 压平头部差距，让两路共识压过单路自信。
  • 同分必须显式定序（按 id），否则结果依赖哈希表遍历顺序，评估数字不可复现。
• How is reranking usually implemented, what problem does it solve, and what does it cost?重排（rerank）一般怎么实现？它解决了初步检索的什么问题，代价是什么？

  Common in ChinaCommon overseasIntermediate#rag#rerank#latency

  How to reason about it · think before answering

  1. The lazy answer is 'sort again, more accurately'. What the interviewer wants is why the first pass cannot rank well, and why reranking cannot run over the whole corpus.
  2. Decompose: the first pass ranks by retrieval signals — cosine distance or term statistics — which are designed to scan millions of items fast, and coarseness is the price. Reranking changes the algorithm: query and candidate go into one model together (a cross-encoder), which is far more accurate but costs one forward pass per candidate. Hence it must sit behind a wide recall stage.
  3. Distinguish two implementations. For teaching or prototypes, batch-score with an LLM (0-10 for 40 candidates in one call). Production uses a trained cross-encoder reranker. Name the cost: an extra 100-300 ms hop plus an inference box — it is not a per-token API, it consumes capacity.
  4. Framing it as a funnel is clearest: recall sets the ceiling, reranking decides whether what is under the ceiling reaches the top five. Measured: adding the keyword channel lifts recall@20 from 83% to 95%; adding reranking moves recall@20 only to 98%, but recall@5 jumps from 80% to 91% and MRR from 0.732 to 0.908.
  5. Expected follow-up 1: does reranking improve recall? No. It introduces no new candidates, so recall@20 is the wrong metric to judge it by.
  6. Expected follow-up 2: why not ship LLM scoring to production? Unpredictable latency, per-token cost, scores that drift with prompt wording, and no clean path to offline distillation.

  分析过程 · 先想清楚再作答

  1. 这题最容易答成「再排一次序，更准」。面试官想听的是**为什么第一轮不能直接排准**，以及**为什么重排不能对全库做**。
  2. 怎么拆：第一轮的排序依据是「检索信号」——余弦距离或词频统计，它们是为了能在百万条里快速筛选而设计的，代价就是粗。重排换了一种算法：把 query 和候选**拼在一起**送进同一个模型算相关度（cross-encoder），精度高得多，但复杂度是每条候选一次前向，没法对全库做。所以它必须跟在一个宽召回后面。
  3. 结论要区分两种实现：教学 / 原型可以用 LLM 批量打分（一次调用给 40 条打 0 到 10 分），生产用专门训练的 cross-encoder 重排模型。**代价说清楚：多一次 100 到 300 毫秒的调用，外加一台推理机器**——它不是按 token 计费的 API，是要占资源的。
  4. 把它放进漏斗里说最清楚：召回决定天花板，重排决定天花板上的东西能不能排到前五。实测的样子是——加了关键词那一路，recall@20 从 83% 涨到 95%（天花板抬高）；再加重排，recall@20 只到 98%，但 recall@5 从 80% 跳到 91%、MRR 从 0.732 到 0.908。
  5. 可预期的追问一：重排能不能提高召回？不能。它不引入新候选，只重排已有的那批——所以看 recall@20 判断重排效果是错的指标。
  6. 可预期的追问二：为什么不用 LLM 打分上生产？延迟不可控、成本按 token 走、分数会随提示词措辞漂移，而且没法做批量离线蒸馏。

  Key points

  • The first pass ranks by retrieval signals so it can scan a large index fast; coarseness is the trade.
  • Reranking feeds query and candidate through one model together (cross-encoder): much sharper, but one forward pass per candidate, so only tens of items.
  • Batch LLM scoring works for teaching; production uses a dedicated reranker, costing an extra 100-300 ms hop plus an inference box.
  • Reranking does not raise recall — it raises recall@5 and MRR (measured 80% to 91%, 0.732 to 0.908) while recall@20 barely moves from 95% to 98%.
  • So judge a reranker by small-k metrics, never by recall@20.

  答题要点

  • 第一轮按检索信号粗排（余弦、词频），为的是能在大库里快速筛，代价是粗。
  • 重排把 query 和候选拼在一起过同一个模型（cross-encoder），精度高但每条一次前向，只能对几十条做。
  • 教学版可用 LLM 批量打 0 到 10 分；生产用专用重排模型，代价是多一次 100 到 300 毫秒的调用加一台推理机器。
  • 重排不提高召回，它提高的是 recall@5 与 MRR——实测 80% → 91%、0.732 → 0.908，而 recall@20 只从 95% 到 98%。
  • 所以判断重排效果要看前 k 小的指标，不要看 recall@20。

RAG in 14 Days: From Retrieval to Trustworthy Answers

D8 Evaluation First: Building a Golden Set, Computing Recall and Ranking Metrics, Using a Model as Judge for Faithfulness

• You need to build an evaluation set from scratch for a RAG system over a company knowledge base. How would you do it, and how many questions are enough?让你从零给一个公司知识库的 RAG 系统建评估集，你会怎么做？多少题才算够用？

  Common in ChinaCommon overseasIntermediate#evaluation#golden-set#rag

  How to reason about it · think before answering

  1. The discriminator here is the direction you generate questions in, and whether you can justify a size rather than name one.
  2. Go corpus-first: read each document and write the questions it can answer. The answer document is fixed at authoring time, so labeling is nearly free. Question-first gives you items whose answers nobody can locate.
  3. Give the schema: question, answer document ids, and a type. At minimum three types - single-document, multi-hop, and unanswerable. Multi-hop counts as a hit only when every answer document makes it into the context; unanswerable items are scored on abstention, not recall.
  4. Justify the size: 20 items separate 'broken' from 'usable' and are enough for a smoke gate; 100 to 200 are needed before a two-point delta means anything. Then grow the set - every production failure becomes a new item.
  5. Mention cost and decay: roughly two hours for 20 items, and answer labels must be rechecked whenever the corpus changes, or the set rots and you misread the drop as a system regression.
  6. Expected follow-up: how do you avoid overfitting to the eval set? Keep a held-out slice that never informs tuning, and refresh it from real production questions.

  分析过程 · 先想清楚再作答

  1. 这题的区分度在「出题方向」和「规模的理由」两处。开口就说「找几百个用户真实问题」的，多半没真做过——真实问题的答案在哪篇文档里，没人标得出来。
  2. 先给方向：从语料反向出题，打开每一篇读它能回答什么，出题的那一刻答案文档就已经确定了，标注成本几乎为零。反方向（先想问题再找答案）会得到一堆自己都不知道答案的题。
  3. 再给结构：每题记问题、答案文档列表、类型三个字段；类型至少分单文档、多跳、无答案三类，并说明多跳必须全部答案文档命中才算命中，无答案不参与召回率而是考拒答。
  4. 规模的理由要给出来，不能只报一个数字：20 题能把「完全不能用」和「基本能用」分开，够做冒烟；100 到 200 题才有资格判断「涨了两个点」是真的还是噪声。上线之后每次线上出问题就把那个问题补进集合——评估集是长出来的。
  5. 补一句成本与保鲜：出题是人力活，20 题两小时是正常量级；语料更新后要复核答案文档还在不在，否则集合会悄悄腐烂，指标下跌你会误以为是系统坏了。
  6. 可预期的追问是「怎么防止评估集被过拟合」。答案是留一份不参与调优的保留集，并且定期从线上真实问题里补充新题，只用来验收不用来调参。

  Key points

  • Author corpus-first so the answer document is known at authoring time.
  • Label every item with a type: single-document, multi-hop, unanswerable.
  • Multi-hop requires all answer documents; unanswerable items score abstention, not recall.
  • 20 items for a smoke gate, 100 to 200 to trust small deltas, and keep growing it from production failures.
  • Hold out a slice that never informs tuning to avoid overfitting the set.

  答题要点

  • 从语料反向出题，出题时答案文档就已确定，标注成本最低。
  • 每题标类型：单文档、多跳、无答案，三类缺一不可。
  • 多跳要求全部答案文档命中；无答案不算召回率，考的是拒答。
  • 20 题够冒烟，100 到 200 题才能判断小幅变化；线上故障持续补题。
  • 留一份不参与调优的保留集，防止对评估集过拟合。