Reciprocal Rank Fusion

The RRF Formula

For each document d, across all ranked lists:

  score(d) = Σ  1 / (rank(d, list) + k + 1)
    

Where k = 60 (a constant from the original RRF paper). The constant prevents top-ranked items from completely dominating — a document ranked 1st scores 1/62 ≈ 0.016 while one ranked 10th scores 1/71 ≈ 0.014. The difference is meaningful but not overwhelming.

Full rrfMerge() Implementation

private List<RetrievedChunk> rrfMerge(List<Document> dense,
                                      List<DocumentChunk> keyword, int k) {
    Map<String, Double> scores = new LinkedHashMap<>();
    Map<String, RetrievedChunk> chunkMap = new LinkedHashMap<>();

    // Score dense results by rank position
    for (int i = 0; i < dense.size(); i++) {
        String id = dense.get(i).getId();
        scores.merge(id, 1.0 / (i + 1 + RRF_K), Double::sum);
        chunkMap.computeIfAbsent(id, _ -> fromAiDocument(dense.get(i)));
    }

    // Score FTS results by rank position
    for (int i = 0; i < keyword.size(); i++) {
        String id = keyword.get(i).getQdrantId();
        scores.merge(id, 1.0 / (i + 1 + RRF_K), Double::sum);
        chunkMap.computeIfAbsent(id, _ -> fromDomainChunk(keyword.get(i)));
    }

    // Sort by combined RRF score, cap at topK
    return scores.entrySet().stream()
        .sorted(Map.Entry.<String, Double>comparingByValue().reversed())
        .limit(k)
        .map(e -> withScore(chunkMap.get(e.getKey()), e.getValue()))
        .toList();
}

Implementation Notes

• scores.merge(id, value, Double::sum) — adds the RRF contribution for each list. If a chunk appears in both lists, the scores accumulate.
• chunkMap.computeIfAbsent — stores the first representation encountered. A chunk appearing in both lists is only stored once.
• RRF_K = 60 — the constant from the original paper (Cormack, Clarke, Buettcher 2009).
• The final list is capped at k = topK (typically 10).

Worked Example

Chunk X, despite ranking only 3rd in dense search, wins because it is consistently relevant across both methods — a strong signal of genuine relevance. Chunk Y topped the dense list but was not in FTS at all, suggesting it might be a semantic near-miss.