// 多维钩子（第二擅长 + 性格 → 副天赋并集 → computeRank）的单测。
//
// 守四件事：
//   ① chip → subs 只落在【现有 19 项表达型】副天赋里（排除治学 hobby，防后人映到不存在的项）。
//   ② mapChipsToSubs 纯函数：并集、去重、未知 id 优雅忽略、query 两种形态（"a,b" / 数组）都摊平。
//   ③ 回落：不传 talents/traits ⇒ 合成 subs 与纯职业【逐字相同】，computeRank 过全朝【逐格相同】。
//      —— 这一条就是「不破 /zhiye/:id、611 基线不动」的机器证明。
//   ④ 多维【真改落点】：产品经理 + 沉得住(→沉毅) 让【宋】从「白身」变成走通的科举正途；
//      任务示例 产品经理 + 会写 + 直 至少改变一朝 —— 位阶是 computeRank 真算的，比纯职业更细。
//
// 跑：node --test test/zhiye-compose.test.js
import test from 'node:test';
import assert from 'node:assert/strict';
import { tuiyanProfile, mapChipsToSubs, listChips, CHIPS, VALID_SUBS } from '../src/zhiye.js';
import { computeAxes, computeRank, listDynasties } from '../src/dynasties.js';
import { SUB_REGISTRY } from '../src/talents.js';

// 与 routes/tianxia.js /zhiye/:id 里【逐字同源】的合成逻辑：职业 subs ∪ chip subs。
function composeSubs(zhiyeId, talents, traits) {
  const base = tuiyanProfile(zhiyeId);
  const { subs: extra, chips } = mapChipsToSubs(talents, traits);
  const subs = base.subs.slice();
  for (const s of extra) if (!subs.includes(s)) subs.push(s);
  return { subs, chips };
}

// 一个 subs 过全朝的位阶指纹（真跑 computeRank，不用快照）。
function fingerprint(subs) {
  const profile = { axes: computeAxes(subs) };
  const out = {};
  for (const d of listDynasties()) {
    try {
      const r = computeRank(profile, d.id);
      out[d.id] = r.rank ? `${r.rank.title}|t${r.rank.tier}|${r.channel ? r.channel.id : '-'}` : null;
    } catch { out[d.id] = 'ERR'; }
  }
  return out;
}

// ── ① 白名单：每个 chip 的每个 sub 都在现有表达型副天赋里 ──
test('★ chip → subs 只落在现有 19 项表达型副天赋里（不发明新 sub）', () => {
  const valid = new Set(VALID_SUBS);
  const expressive = SUB_REGISTRY.filter((item) => item.subtype !== 'hobby').map((item) => item.id).sort();
  assert.equal(valid.size, 19, '白名单必须正好 19 项表达型副天赋');
  assert.deepEqual([...valid].sort(), expressive, '白名单必须与 talents.js 的非 hobby 项逐一一致');
  for (const c of CHIPS) {
    assert.ok(c.kind === 'tuiyan', `${c.id} 必须标 kind:'tuiyan'（是推演不是史料）`);
    assert.ok(typeof c.basis === 'string' && c.basis.length > 4, `${c.id} 必须有可反驳的 basis`);
    assert.ok(c.subs.length > 0, `${c.id} 至少映一项`);
    for (const s of c.subs) assert.ok(valid.has(s), `${c.id} 映到了不存在的副天赋: ${s}`);
  }
});

// ── ② mapChipsToSubs 纯函数行为 ──
test('★ mapChipsToSubs：并集 + 去重 + 未知 id 忽略', () => {
  // 会写(story,vivid) + 直(blunt)
  const a = mapChipsToSubs('huixie', 'zhi');
  assert.deepEqual(a.subs, ['story', 'vivid', 'blunt']);
  assert.deepEqual(a.chips.map((c) => c.id), ['huixie', 'zhi']);

  // 直(blunt) + 刚(blunt) ⇒ blunt 去重成一项；两个 chip 都回显
  const b = mapChipsToSubs('', 'zhi,gang');
  assert.deepEqual(b.subs, ['blunt']);
  assert.deepEqual(b.chips.map((c) => c.id), ['zhi', 'gang']);

  // 未知 id 优雅忽略，不抛错
  const c = mapChipsToSubs('nope,huisuan', 'bogus');
  assert.deepEqual(c.subs, ['datadriven', 'rigorous']);
  assert.deepEqual(c.chips.map((c2) => c2.id), ['huisuan']);
});

test('★ mapChipsToSubs：query 两种形态（"a,b" 字符串 / 数组）都摊平', () => {
  const asString = mapChipsToSubs('huixie,chendezhu', 'zhi');
  const asArray = mapChipsToSubs(['huixie', 'chendezhu'], ['zhi']);
  assert.deepEqual(asString.subs, asArray.subs);
  // 重复 chip id 也去重（数组里传两遍）
  const dup = mapChipsToSubs(['huixie', 'huixie'], undefined);
  assert.deepEqual(dup.chips.map((c) => c.id), ['huixie']);
});

// ── ③ 回落：不传 chip ⇒ 与纯职业逐字相同（不破 /zhiye/:id） ──
test('★★ 回落：不传 talents/traits ⇒ subs 与位阶都与纯职业【逐字相同】', () => {
  for (const zid of ['chanpin', 'chengxuyuan', 'yisheng', 'jizhe']) {
    const base = tuiyanProfile(zid).subs;
    for (const args of [[undefined, undefined], ['', ''], [[], []], ['', undefined]]) {
      const { subs, chips } = composeSubs(zid, args[0], args[1]);
      assert.deepEqual(subs, base, `${zid} 回落 subs 必须逐字等于纯职业`);
      assert.equal(chips.length, 0, `${zid} 回落不应有 chip`);
    }
    // 位阶指纹也必须逐格相同 —— 这才是「611 基线不动」的证明
    assert.deepEqual(fingerprint(composeSubs(zid).subs), fingerprint(base));
  }
});

// ── ④ 多维真改落点 ──
test('★★ 产品经理 + 沉得住(→沉毅) 让【宋】从白身变成走通的科举正途', () => {
  const pure = fingerprint(tuiyanProfile('chanpin').subs);
  const { subs, chips } = composeSubs('chanpin', 'chendezhu', '');
  const multi = fingerprint(subs);
  assert.deepEqual(chips.map((c) => c.name), ['沉得住']);
  // 纯产品经理在宋是白身；加一笔「沉得住」后宋这一格【变了】且不再是白身
  assert.ok(/白身/.test(pure.song), '前提：纯产品经理在宋应是白身');
  assert.notEqual(multi.song, pure.song, '加沉得住后宋这一格必须变');
  assert.ok(!/白身/.test(multi.song), '加沉得住后宋不再是白身（走通了科举正途）');
});

test('★★ 任务示例：产品经理 + 会写 + 直 —— 位阶真算、比纯职业更细（至少改一朝）', () => {
  const base = tuiyanProfile('chanpin').subs;
  const pure = fingerprint(base);
  const { subs, chips } = composeSubs('chanpin', 'huixie', 'zhi');
  const multi = fingerprint(subs);
  assert.deepEqual(chips.map((c) => c.name), ['会写', '直']);
  // 合成六轴：会写→wen、直→wu 都长起来了（真算，不是编）
  const ax = computeAxes(subs);
  assert.ok(ax.wen > computeAxes(base).wen, '会写应抬高文采轴');
  assert.ok(ax.wu > computeAxes(base).wu, '直应抬高果决轴');
  // 至少一朝的落点与纯职业不同
  const changed = listDynasties().filter((d) => multi[d.id] !== pure[d.id]).map((d) => d.id);
  assert.ok(changed.length >= 1, `多维应至少改变一朝，实测改变: ${changed.join(',')}`);
});

// ── listChips 分组（供前端渲染） ──
test('★ listChips 按 talent/trait 分组，每笔带 basis/kind', () => {
  const { talents, traits, caveat } = listChips();
  assert.ok(talents.length >= 6 && traits.length >= 6);
  assert.ok(typeof caveat === 'string' && caveat.length > 10);
  for (const c of [...talents, ...traits]) {
    assert.ok(c.id && c.name && Array.isArray(c.subs) && c.subs.length > 0);
    assert.equal(c.kind, 'tuiyan');
    assert.ok(typeof c.basis === 'string' && c.basis.length > 4);
  }
});