Question bank — Reranking and hybrid search

False. A reranker only reorders the candidate set it is given; if the document never entered the top-100, no rescoring can conjure it — recall must be fixed upstream. See Evaluation - Recall@k and MRR.

False (mostly). Hybrid's job is recall — widening the net so the right chunk is somewhere in the candidates. Precision-at-top is the reranker's job. H before R: recall then precision.

False. BM25 is unbounded (can be 8, 40, ...) while cosine lives in $[-1,1]$, so BM25 arbitrarily dominates any sum. Use rank-based fusion (RRF) or normalise first.

False. A bi-encoder embeds query and doc separately (cheap, precomputable), a cross-encoder feeds them through the transformer together so every query token attends to every doc token (accurate, not precomputable).

False. With $k=60$: #1-only gives $\frac{1}{61}\approx 0.0164$, while #2-in-both gives $\frac{2}{62}\approx 0.0323$. Cross-verified evidence wins — that is the whole point of fusion.

False. LLMs suffer lost-in-the-middle: a relevant chunk buried among 50 dilutes attention, raises cost/latency, and increases hallucination. Reranking to a tight top-5 is the fix.

True. $\frac{\mathbf{q}\cdot \mathbf{d}}{\Vert \mathbf{q}\Vert \Vert \mathbf{d}\Vert }$ divides out both norms, so it measures direction/angle only — useful because embedding magnitude often reflects token count, not meaning.

False — that's its main virtue. RRF uses only ranks (positions), never raw scores, so BM25's unbounded scale and cosine's $[-1,1]$ scale never have to be reconciled.

A cross-encoder produces no standalone query vector — it only scores a (query, doc) pair jointly. Nearest-neighbour search needs a bi-encoder embedding. Cross-encoders are rerankers, not retrievers.

BM25 is lexical: it matches surface tokens, and "car" ≠ "automobile" as strings. Catching synonyms is the dense/semantic retriever's job — the exact reason we need hybrid.

Cross-encoder scores depend on the query, so they can't be precomputed per document — a new query means a fresh forward pass. That's why it runs only on the ~100 pre-retrieved candidates, not the corpus.

Backwards. $b$ controls normalisation strength: $b=0$ ignores document length entirely, $b=1$ applies full normalisation. The usual $b=0.75$ is a middle ground.

With $k=0$, rank 1 contributes $1$ and rank 2 contributes $0.5$ — a huge, winner-take-all gap. The RRF constant $k\approx 60$ deliberately softens the gap so consensus across lists can compete with a single #1.

Codes and rare tokens have weak, unstable embeddings; semantic search may return "nearby" but literally wrong chunks. Exact-token matching (BM25) is what reliably surfaces ERR_4092.

A reranker cannot find anything new; it can only reorder existing candidates. It improves precision-at-top, never recall. Recall comes from retrieval (hybrid).

Because relevance has diminishing returns: the 50th "cat" adds almost nothing over the 5th. As term frequency $f\to \infty$ the score saturates toward $(k_1+1)\cdot \text{IDF}$, capping any single term's influence.

Rare terms (e.g. "photosynthesis") are discriminating — they separate relevant from irrelevant docs — while common terms ("the") appear everywhere and carry no signal. With $N$ = total docs and $n$ = docs containing the term, IDF $=\ln \frac{N-n+0.5}{n+0.5}+1$ grows as $n$ shrinks.

Bi-encoder doc embeddings are computed offline once and stored; at query time you only embed the query and do fast approximate nearest-neighbour lookup. A cross-encoder must run a fresh joint forward pass per candidate per query.

Each list contributes an additive $\frac{1}{k+\text{rank}}$ term. A doc present in both lists collects two contributions, so consistent-but-middling evidence can outscore a single top hit — fusion prefers cross-verified relevance.

The cross-encoder costs one forward pass per document, which is infeasible over a corpus. Cheap retrieval (BM25 + bi-encoder) narrows to ~100; the expensive scorer then pays its cost only on those 100.

Merging is by union of candidates plus rank fusion; a retriever that misses simply contributes nothing useful, while the other still surfaces the right chunk. You gain the strengths of both without either vetoing the other.

Near zero. With $n\approx N$ (term in all docs), $\ln \frac{N-n+0.5}{n+0.5}\to \ln (\text{small})$, so the term is treated as non-discriminating — exactly right, since it can't distinguish any document.

Just $\frac{1}{63}\approx 0.0159$ — a single contribution. It can still rank respectably, but a doc appearing in both lists at similar ranks will typically edge it out.

The cross-encoder gives no further signal, so ordering falls back to an implementation tie-break (e.g. original candidate order or hybrid RRF score). Genuinely tied relevance means the choice is arbitrary — flag it, don't over-interpret.

The norm $\Vert \mathbf{d}\Vert =0$ makes the cosine formula divide by zero — undefined. Such degenerate chunks must be filtered before indexing; a zero vector has no direction, hence no meaningful angle.

No. With a single candidate there is nothing to reorder and no precision to gain — reranking's value only appears when many candidates compete for the top-$k$ slots.

The dense retriever, since BM25 finds zero matching tokens and scores everything at ~0. This is precisely the "car vs automobile" case that motivates adding semantic search.