麻将胡了

2026-09-22 15:39:53 +08:00
parent 8f8d2e042c
commit 470d1471e5
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+"""麻将胡了 · 引擎单测（不依赖 pytest：`python tests/test_engine.py` 直接跑）。
+
+重点覆盖：§2.1 七个正反例、百搭补链/不能替代胡、金色就地转百搭、
+一个百搭被两条链同时借用、阶梯取档、FS 旋数上限与"发不满就不发"、500x 封顶。
+"""
+from __future__ import annotations
+
+import random
+import sys
+from pathlib import Path
+
+sys.path.insert(0, str(Path(__file__).resolve().parent.parent))
+
+from mahjong import config as C                                    # noqa: E402
+from mahjong.engine import (                                       # noqa: E402
+    assert_wild_only_middle, count_scatters, new_grid, play_round,
+    settle_rotation, ways_wins,
+)
+
+
+def make_grid(rows, golden=()):
+    """rows: 4 个长度 5 的短码字符串（上→下）；golden: 0 基 (行,列) 集合。"""
+    grid = [c for row in rows for c in row]
+    assert len(grid) == C.GRID_SIZE, f"盘面必须是 20 格，收到 {len(grid)}"
+    return grid, {r * C.GRID_COLS + c for r, c in golden}
+
+
+def cols_of(counts, fill="S", sym="B"):
+    """按"每轴 sym 的个数"造盘面（其余填 fill，保证不形成别的链）。"""
+    rows = []
+    for r in range(C.GRID_ROWS):
+        row = []
+        for c in range(C.GRID_COLS):
+            # 从底部往上摆 sym
+            row.append(sym if r >= C.GRID_ROWS - counts[c] else fill)
+        rows.append("".join(row))
+    return make_grid(rows)
+
+
+def mul(x):
+    return round(x, 4)
+
+
+# ---------------------------------------------------------------- §2.1 前缀链
+CASES = [
+    ([2, 3, 1, 0, 0], 3, 6),      # ✅ 3 连
+    ([2, 3, 1, 2, 0], 4, 12),     # ✅ 4 连（第 5 轴才断）
+    ([2, 0, 3, 4, 4], None, None),  # ❌ 第 2 轴断
+    ([0, 4, 4, 4, 4], None, None),  # ❌ 第 1 轴必需
+    ([1, 1, 0, 4, 4], None, None),  # ❌ 不能跳轴
+    ([1, 2, 3, 4, 4], 5, 96),     # ✅ 满链
+    ([1, 1, 1, 1, 1], 5, 1),      # ✅ 最低 ways
+]
+
+
+def test_prefix_chain():
+    for counts, run, ways in CASES:
+        grid, _ = cols_of(counts)
+        wins = [w for w in ways_wins(grid) if w.symbol == "B"]
+        if run is None:
+            assert not wins, f"{counts} 不该成奖，却得到 {wins}"
+        else:
+            assert len(wins) == 1, f"{counts} 应恰好一笔，得到 {wins}"
+            assert wins[0].run == run and wins[0].ways == ways, f"{counts} → {wins[0]}"
+    print("[ok] §2.1 前缀链 7 个正反例")
+
+
+def test_scatter_breaks_chain_and_no_substitute():
+    grid, _ = make_grid(["ASASS", "SSSSS", "SSSSS", "SSSSS"])
+    assert ways_wins(grid) == (), "發,胡,發 不能成链（胡不进链、也挡链）"
+    grid, _ = make_grid(["AWASS", "SSSSS", "SSSSS", "SSSSS"])
+    wins = [w for w in ways_wins(grid) if w.symbol == "A"]
+    assert len(wins) == 1 and wins[0].run == 3 and wins[0].ways == 1, "百搭要能补链中间的断点"
+    print("[ok] 胡不成链 / 百搭补链")
+
+
+def test_paytable_grid():
+    for code, rule in C.SYMBOL_RULES.items():
+        for run, pay in rule.payouts.items():
+            assert abs(pay / 0.05 - round(pay / 0.05)) < 1e-9, f"{code} {run} 连 {pay} 不是 0.05 的整数倍"
+    print("[ok] 赔付表全部落在 0.05 网格上")
+
+
+# ---------------------------------------------------------------- 金色 → 百搭
+def test_golden_converts_in_place():
+    """§15 第 1 波：两张金色「中」各留一个百搭，且是就地改写的。"""
+    grid, golden = make_grid(
+        ["BCBES", "BBBDF", "DCCGE", "CEHFG"], golden=[(1, 1), (1, 2)])
+    rng = random.Random(7)
+    rot = settle_rotation(grid, golden, 1, False, rng)
+    assert rot is not None
+    paid = {w.symbol: mul(w.pay_base) for w in rot.wins}
+    assert paid == {"B": 2.0, "C": 0.8}, f"本波两笔奖应为 2.00 / 0.80，得到 {paid}"
+    assert mul(rot.win_mul) == 2.8, "x1 阶梯下本波 2.80 倍"
+    assert len(rot.converted) == 2, f"两个金色格各留一个百搭，得到 {rot.converted}"
+    assert {s for _, s in rot.converted} == {"B"}, "转化前的自然符号应记进 wild_converted"
+    assert all(grid[i] != C.WILD for i in golden), "百搭永不回填金色标记（每格只转化一次）"
+    assert sum(1 for v in grid if v == C.WILD) == 2, "两个百搭仍在盘面上（跟着重力落）"
+    assert_wild_only_middle(grid)
+    print("[ok] 金色格就地转百搭（不空洞/不补牌，2 个金色格 → 2 个百搭）")
+
+
+def test_one_wild_borrowed_by_three_chains():
+    """§15 第 4 波：同一个百搭同时撑起「筒」5 连、「索红绿」5 连、「白板」3 连。
+
+    注：设计文档原来只算了前两笔（漏了白板链），本测试按规则把它补上；
+    文档已同步勘误（本波 ¥1200、本旋总账 ¥2380）。
+    """
+    grid, golden = make_grid(["GAWGC", "HCAAS", "CFBGF", "FGHFG"])
+    rng = random.Random(11)
+    rot = settle_rotation(grid, golden, 4, False, rng)
+    assert rot is not None
+    got = {w.symbol: (w.run, w.ways, mul(w.pay_base)) for w in rot.wins}
+    assert got == {"G": (5, 2, 1.0), "F": (5, 1, 1.0), "C": (3, 1, 0.4)}, f"三笔并赔，得到 {got}"
+    shared = set(rot.wins[0].loc)
+    for w in rot.wins[1:]:
+        shared &= set(w.loc)
+    assert shared == {3}, f"三条链必须共用第 3 格那个百搭，实际 {shared}"
+    assert mul(rot.base_mul) == 2.4 and mul(rot.win_mul) == 12.0, "本波 2.40 倍 × x5 = 12.00 倍"
+    assert sum(1 for v in grid if v == C.WILD) == 0, "参与成链的百搭要随本波被消除"
+    print("[ok] 一个百搭被三条链同时借用，并随波被消除")
+
+
+def test_involved_golden_only():
+    """只有"参与成奖"的金色格才转化；没被涉及的金色格原地不动。"""
+    # B 链在轴1~轴3；轴4 的 F 带金色但没参与任何链 → 不转化
+    grid, golden = make_grid(["BCBES", "BBBDF", "DCCGE", "CEHFG"], golden=[(3, 3)])
+    rng = random.Random(9)
+    rot = settle_rotation(grid, golden, 1, False, rng)
+    assert rot is not None and rot.converted == (), f"未被涉及的金色格不该转化：{rot.converted}"
+    assert 3 * C.GRID_COLS + 3 in golden, "未涉及的金色标记要保留"
+    print("[ok] 只有参与成奖的金色格才就地转百搭")
+
+
+def test_golden_travels_with_symbol():
+    """金色是镀在符号上的：金色格没被消除但它下方的牌被消 → 它下落，金标记跟着走。"""
+    grid, golden = make_grid(["BDBSS", "BBBSS", "BBBSS", "SSSSS"], golden=[(0, 1)])
+    rot = settle_rotation(grid, golden, 1, False, random.Random(1))
+    assert rot is not None, "B 应为 3 连（col0=3 / col1=2 / col2=3）"
+    # D 从行1 落到行3（col1 有 2 格空洞），金标记必须跟着
+    assert grid[2 * C.GRID_COLS + 1] == "D", "金色 D 应该落到行3"
+    assert 2 * C.GRID_COLS + 1 in golden, "金标记要跟着符号落到新位置"
+    assert 0 * C.GRID_COLS + 1 not in golden, "不能留在原格（否则会传染给新补的牌）"
+    print("[ok] 金色标记跟着符号下落")
+
+
+def test_no_wild_without_conversion():
+    """原生盘面（卷轴+补位）永不落百搭；金色只在第 2/3/4 轴。"""
+    for seed in range(300):
+        g, gold = new_grid(random.Random(seed))
+        assert C.WILD not in g, "原生盘面永远不落百搭"
+        assert all(c % C.GRID_COLS in C.GOLDEN_COLS for c in gold), "金色只在第 2/3/4 轴"
+    print("[ok] 原生不落百搭 / 金色只在第 2/3/4 轴")
+
+
+# ---------------------------------------------------------------- 阶梯 / 封顶 / FS
+def test_ladder():
+    base = [C.ladder_mul(n, False) for n in range(1, 7)]
+    fs = [C.ladder_mul(n, True) for n in range(1, 7)]
+    assert base == [1, 2, 3, 5, 5, 5], base
+    assert fs == [2, 4, 6, 10, 10, 10], fs
+    print("[ok] 阶梯：基础 1/2/3/5，FS 2/4/6/10，第 4 波起封顶")
+
+
+def test_example_2_9_1_wave1():
+    """§2.9.1 第 1 波：中 3 连 + 白板 3 连 = 0.90 倍（x1）。"""
+    grid, golden = make_grid(["BCBEH", "DBAGF", "EDCAG", "CEHFD"])
+    rot = settle_rotation(grid, golden, 1, False, random.Random(5))
+    assert rot is not None and mul(rot.base_mul) == 0.9 and mul(rot.win_mul) == 0.9
+    print("[ok] §2.9.1 第 1 波 = 0.90 倍")
+
+
+def test_fs_award_and_cap():
+    assert [C.fs_award(n) for n in (3, 4, 5)] == [12, 14, 16]
+    rng = random.Random(2026)
+    reachable = set()
+    for _ in range(40000):
+        r = play_round(rng, box_id=5)
+        assert r.actual_mul <= C.MAX_WIN_CAP_MUL, "单局封顶 500x"
+        assert mul(r.actual_mul) == mul(min(r.raw_mul, C.MAX_WIN_CAP_MUL))
+        if r.free_spins:
+            assert r.free_spins <= C.MAX_FREE_SPINS, f"FS 上限 36，得到 {r.free_spins}"
+            assert r.free_spins in (12, 14, 16, 24, 26, 28, 30, 32, 36), r.free_spins
+            reachable.add(r.free_spins)
+        if r.capped:
+            assert r.actual_mul == C.MAX_WIN_CAP_MUL
+    assert reachable, "4 万局都没触发过一次 FS，权重/参数明显不对"
+    print(f"[ok] 封顶与 FS 上限（4 万局内出现的赠送旋数：{sorted(reachable)}）")
+
+
+if __name__ == "__main__":
+    test_prefix_chain()
+    test_scatter_breaks_chain_and_no_substitute()
+    test_paytable_grid()
+    test_golden_converts_in_place()
+    test_one_wild_borrowed_by_three_chains()
+    test_involved_golden_only()
+    test_golden_travels_with_symbol()
+    test_no_wild_without_conversion()
+    test_ladder()
+    test_example_2_9_1_wave1()
+    test_fs_award_and_cap()
+    print("\n全部通过 ✅")