feat: flow 第 2 部分，第 r 轮开轮到选批（任务书 14 第二块）

- next-step 按"第 r 轮"列出提案、第一次预测、挑刺、回应与统一、第二次预测的任务（一人对一个对象），判重、冻结协议、选批三步，走到 implementation 停下（任务书 15） - steps/dispatch.ts：同阶段一人任务先后跑、不同人同时跑；提案、挑刺按 proposalId、critiqueId 分份 - steps/round.ts：判重两两问 Jev、连通成组、并进留池候选；冻结协议在回应阶段关闭后一个事务写全；选批照 rules 的 BatchInput 拼，种子用 decisionSeed 版本 1（选格预留版本 2） - materials 导出 Board（构造改收记录数组，给纯函数用）；讨论原文里已作废的入口编号就地标"（已作废）" - 删 SOURCES.md 里参考项目实测事实那一段（用户 2026-10-03 同意） - 测试：下一步做什么 21 条、判重连组 4 条、冻结与选批 2 条、作废标注 1 条；npm test 437 通过全新上下文审阅一轮（只看三个程序步骤与恢复）：没有会污染账本的问题。 Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
2026-10-03 13:59:54 -04:00
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+import assert from "node:assert/strict";
+import { test } from "node:test";
+import { groupSameChange, type SamePair } from "../src/flow/steps/round.ts";
+
+// 判重合并怎么连组（任务书 14 第 2.4 节）：判为相同的连成一组，组里有留池候选就并进它，
+// 新候选按组里编号最小的原提案编号。预期值按规则手写。
+
+const pair = (a: string, b: string, yes: boolean): SamePair => ({
+  a,
+  b,
+  judgment: { question: "same_change", questionHash: "h", threshold: 0.5, probability: yes ? 0.9 : 0.1, yes, requestId: "req" },
+});
+
+test("A 与 B 相同、B 与 C 相同：A、B、C 连成一个候选，即使 A 与 C 判为不同；新候选按组里最小的原提案编号", () => {
+  const ps = ["r1-researcher-1-1", "r1-researcher-1-2", "r1-researcher-2-1", "r1-researcher-3-1"];
+  const pairs = [
+    pair("r1-researcher-1-1", "r1-researcher-1-2", false),
+    pair("r1-researcher-1-1", "r1-researcher-2-1", false),
+    pair("r1-researcher-1-1", "r1-researcher-3-1", false),
+    pair("r1-researcher-1-2", "r1-researcher-2-1", false),
+    pair("r1-researcher-1-2", "r1-researcher-3-1", true),
+    pair("r1-researcher-2-1", "r1-researcher-3-1", true),
+  ];
+  assert.deepEqual(groupSameChange(1, ps, [], pairs), {
+    fresh: [
+      { candidate: "r1-c1", proposals: ["r1-researcher-1-1"] },
+      { candidate: "r1-c2", proposals: ["r1-researcher-1-2", "r1-researcher-2-1", "r1-researcher-3-1"] },
+    ],
+    carried: [],
+  });
+});
+
+test("编号按数字大小比：researcher-10 排在 researcher-2 后面", () => {
+  const ps = ["r1-researcher-10-1", "r1-researcher-2-1"];
+  assert.deepEqual(groupSameChange(1, ps, [], [pair(ps[0]!, ps[1]!, false)]).fresh, [
+    { candidate: "r1-c1", proposals: ["r1-researcher-2-1"] },
+    { candidate: "r1-c2", proposals: ["r1-researcher-10-1"] },
+  ]);
+});
+
+test("和留池候选相同的提案并进它，连带和这条提案相同的别的提案；没并进提案的留池候选列空数组", () => {
+  const ps = ["r2-researcher-1-1", "r2-researcher-2-1", "r2-researcher-3-1"];
+  const pairs = [
+    pair("r2-researcher-1-1", "r2-researcher-2-1", true),
+    pair("r2-researcher-1-1", "r2-researcher-3-1", false),
+    pair("r2-researcher-2-1", "r2-researcher-3-1", false),
+    pair("r2-researcher-1-1", "r1-c1", false),
+    pair("r2-researcher-1-1", "r1-c3", false),
+    pair("r2-researcher-2-1", "r1-c1", false),
+    pair("r2-researcher-2-1", "r1-c3", true),
+    pair("r2-researcher-3-1", "r1-c1", false),
+    pair("r2-researcher-3-1", "r1-c3", false),
+  ];
+  assert.deepEqual(groupSameChange(2, ps, ["r1-c3", "r1-c1"], pairs), {
+    fresh: [{ candidate: "r2-c1", proposals: ["r2-researcher-3-1"] }],
+    carried: [
+      { candidate: "r1-c1", proposals: [] },
+      { candidate: "r1-c3", proposals: ["r2-researcher-1-1", "r2-researcher-2-1"] },
+    ],
+  });
+});
+
+test("一组连到两个留池候选：按内部错误停下", () => {
+  const pairs = [pair("r2-researcher-1-1", "r1-c1", true), pair("r2-researcher-1-1", "r1-c2", true)];
+  assert.throws(() => groupSameChange(2, ["r2-researcher-1-1"], ["r1-c1", "r1-c2"], pairs), /留池候选 r1-c1、r1-c2 连成了一组/);
+});