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.