🎯 场景：聚合 N 条独立采样的推理路径，按等价类投票出 consensus answer + agreement_rate（软置信度）。降低单路径方差，适合算式 / 多步 QA 等错误率高的任务。

Quick Use

Use when: You've sampled N candidate answers to the same question (with temperature) and want to take a majority vote. Fill in: {{question}} = the original question; {{candidate_paths}} = JSON array of N (rationale_summary, final_answer) objects; {{equivalence_hint}} = optional definition of "equivalent" answers (e.g. numerical tolerance). You'll get: The consensus answer, vote counts per equivalence class, an agreement_rate, and a trustworthy flag. Output is JSON.

Purpose

Aggregate N independently-sampled reasoning paths for the same question into a single consensus answer plus agreement metrics — the self-consistency technique. This card does not sample the paths itself; it operates on N pre-sampled (rationale_summary, final_answer) pairs (typically generated with temperature > 0 by cot/structured-reasoning-with-rationale-summary or similar). Used when accuracy on hard reasoning tasks matters more than latency: arithmetic, multi-step QA, code translation. Output is structured so the agreement rate doubles as a confidence signal.

Prompt

You are aggregating N candidate reasoning paths for the same question
into a consensus answer using majority vote. Decide which final answers
are equivalent, count the votes, and return the consensus.

Original question:
{{question}}

Candidate paths (each has its own rationale_summary and final_answer):
{{candidate_paths}}

Equivalence hint (may be empty):
{{equivalence_hint}}

Steps:
1. Group the candidate paths' final_answer values into equivalence
   classes. Two answers are in the same class if and only if they
   would be considered "the same answer" by the equivalence_hint
   (or by exact-match if no hint is given). Examples:
   - "42" and "forty-two" are equivalent under numerical equivalence.
   - "Paris" and "Paris, France" are equivalent under "city-level
     match".
2. The consensus answer is the answer of the largest equivalence
   class. If two classes tie, pick the answer whose member paths have
   the most coherent rationale_summary as a tiebreaker.
3. Compute agreement_rate = size of consensus class / N.
4. List the minority answers and how many paths produced each.

Return ONLY this JSON object:
{
  "consensus_answer": "<the canonical form of the majority answer>",
  "vote_counts": [
    {"answer": "<canonical form>", "count": <integer>}
    /* one entry per equivalence class, sorted by count desc */
  ],
  "agreement_rate": <float between 0.0 and 1.0>,
  "trustworthy": true | false,
  "minority_summary": "<one short phrase: are minority answers obviously wrong, or plausible alternatives, or both?>",
  "decision_basis": "<one sentence, <=30 words, no internal CoT>"
}

Set "trustworthy": true only if agreement_rate >= 0.6 AND the consensus
class contains at least 2 paths. Otherwise trustworthy=false; the
consumer should treat the consensus as a weak signal.

Example

Input:

question: "If a train leaves at 9:00am at 80 km/h and a second train leaves at 10:30am at 100 km/h on the same track, when does the second train catch up?"
candidate_paths: [
  {"rationale_summary": "Lead = 80*1.5 = 120 km. Closing speed = 20 km/h. Time = 120/20 = 6h. Catch-up at 4:30pm.", "final_answer": "4:30pm"},
  {"rationale_summary": "Computed lead 120 km, closing 20 km/h, 6 hours after 10:30am gives 4:30pm.", "final_answer": "4:30pm"},
  {"rationale_summary": "Used relative-speed framing; 6h after 10:30am.", "final_answer": "16:30"},
  {"rationale_summary": "Off-by-one: thought lead was 80 km, got 4 hours later = 2:30pm.", "final_answer": "2:30pm"},
  {"rationale_summary": "Same approach as path 1; 4:30pm.", "final_answer": "4:30pm"}
]
equivalence_hint: "Treat 24-hour and 12-hour clock formats as equivalent."

Expected output:

{
  "consensus_answer": "4:30pm",
  "vote_counts": [
    {"answer": "4:30pm", "count": 4},
    {"answer": "2:30pm", "count": 1}
  ],
  "agreement_rate": 0.8,
  "trustworthy": true,
  "minority_summary": "One path made an arithmetic error on the lead distance; the dissent is clearly wrong, not a plausible alternative.",
  "decision_basis": "Four of five paths agree on 4:30pm under 24h-equivalence; the dissent is an obvious computation error."
}

Failure Modes

• Equivalence over-merging — model decides "4:30pm" and "5:30pm" are equivalent under a vague hint, inflating agreement. Detect by spot-checking equivalence classes that combine numerically distinct values; tighten equivalence_hint.
• Equivalence under-merging — model treats "42" and "forty-two" as different answers because the hint is missing. Always pass an explicit equivalence_hint for non-string outputs.
• Tie-handling bias — when two classes tie, model defaults to the first-listed path; ordering of candidate_paths then influences the result. Mitigation: shuffle candidate_paths before passing in; or implement a deterministic tiebreaker upstream.
• Format normalization drift — consensus_answer is in a different format than any candidate (e.g. all candidates said "4:30pm" but consensus says "16:30 PM"). Mitigation: rule says pick the canonical form FROM the consensus class, not invent a new format.
• Trustworthy inflation — for N=3 with split 2-1, model marks trustworthy=true even though 2 paths is a weak basis. The rule "consensus class contains at least 2 paths AND agreement >= 0.6" helps; on small N (<5) treat trustworthy as a noisy hint.
• Non-categorical answer collapse — for free-text answers (e.g. long-form explanations), majority vote is meaningless. This card is for answers that have a small number of canonical forms; free-text generation needs a different aggregator (e.g. LLM-judge on pairwise of paths).

Tuning Notes

• N 选择：典型 N=5–10。N<3 不显著降低方差；N>20 边际收益小。算式 类问题 N=5 通常足够，open-domain 多步推理 N=10 更稳。
• 采样温度：上游采样建议 0.5–0.8，给路径多样性。本卡 aggregator 本身用 0.0，确保聚合可重现。
• 模型差异：aggregator 不需要强模型——它做的是 string equivalence 和 vote counting，中档模型够用。采样阶段才需要强模型，否则 N 个 path 可能集体一致地犯同一种错。
• 与 cot/structured-reasoning-with-rationale-summary 的关系： 那张卡产出单个 (rationale_summary, final_answer)；本卡聚合多个。 典型工作流：上游把 structured-reasoning 卡跑 N 遍 → 把 N 个 outputs 作为 candidate_paths 喂本卡。
• 与 cot/least-to-most-decomposition 的关系：least-to-most 把单个 问题拆成易子问题，是单条路径的优化；self-consistency 是多条路径 的方差缩减。两者正交，可叠加：每条 self-consistency path 内部用 least-to-most 推理。
• agreement_rate 的解读：可作为答案的"软置信度"，但不要直接当 概率使用（self-consistency 的 calibration 在不同任务上偏差大）。 在校准集上回归一遍再使用。
• trustworthy=false 的样本：不要丢弃；它们是模型的 disagreement zone，是后续 active learning 或人工 review 的高价值候选。

Changelog

• 0.1.0 — Initial card.