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Video-Tree-TRM5/tests/unit/test_gate_prefix.py
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iomgaa 21c360bc87 feat: gate prefix-ordered consumption core (algo #6)
CE-Gate 语义修订获批:块序贯 → 阶梯序前缀逐对序贯。新增 GateSpec/_UnitSlot/
_GateRun 数据结构与 _advance_prefix 纯逻辑(乱序到达下统计严格按预声明阶梯序
消费,INFRA 剔除后重判防 continue 悬置,过线即冻结)。旧块路径共存,Task 6 删。
2026-07-16 23:44:32 -04:00

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4.9 KiB
Python

"""连续并发 gate 的前缀消费纯逻辑测试。"""
from __future__ import annotations
from app.harness.validate import GateSpec, _advance_prefix, _GateRun
from core.evolution import GateParams
from core.types import GeneratedQuestion, QuestionUnit
def _mk_question(qid: str, task_type: str = "Action Reasoning") -> GeneratedQuestion:
"""构造最小可用的 single 题(字段以 core.types 真实定义为准,缺省值从简)。"""
return GeneratedQuestion(
question_id=qid,
video_id="v1",
task_type=task_type,
question=f"q-{qid}",
options=("A. x", "B. y"),
answer="A",
source_nodes=(),
difficulty="easy",
)
def _mk_unit(qid: str, task_type: str = "Action Reasoning") -> QuestionUnit:
"""由单条题目构造 single 单元(unit_id 回填为 question_id)。"""
return QuestionUnit.from_single(_mk_question(qid, task_type))
def _mk_run(n_units: int) -> _GateRun:
"""构造含 n_units 个 single 单元的初始 gate 运行时状态。"""
spec = GateSpec(
task_type="Action Reasoning",
target_file="action-reasoning.md",
candidate_content="cand",
base_skill_content="base",
units=[_mk_unit(f"q{i}") for i in range(n_units)],
gate_run_prefix="r_e1_s0_gate_action-reasoning",
)
return _GateRun.from_spec(spec)
_PARAMS = GateParams(
e_confirm=20.0,
e_provisional=3.0,
w_net_min=2,
delta_min=0.02,
lambda_dir=-0.642,
e_rollback=10.0,
)
def test_prefix_blocks_on_unresolved_head() -> None:
"""阶梯头部单元未配齐时,即使尾部全部配齐也一个都不消费。"""
run = _mk_run(4)
for i in (1, 2, 3): # 尾部三个先到
run.slots[i].base = False
run.slots[i].cand_per_q = {f"q{i}": True}
_advance_prefix(run, _PARAMS)
assert run.n_used == 0 and run.w == 0 and run.verdict is None
def test_prefix_consumes_in_ladder_order_after_head_arrives() -> None:
"""头部补齐后一次性顺序消费到最长已配齐前缀。"""
run = _mk_run(4)
for i in (0, 1, 2):
run.slots[i].base = False
run.slots[i].cand_per_q = {f"q{i}": True}
_advance_prefix(run, _PARAMS)
assert run.n_used == 3 and run.w == 3 and run.l == 0
assert [r["ladder_rank"] for r in run.evidence_rows] == [0, 1, 2]
def test_freeze_on_terminal_verdict_stops_consumption() -> None:
"""过线即冻结,后续已配齐单元不再消费。
数值:W 连胜 L=0 时 E=(2^(W+1)-1)/(W+1),W=6→18.14<20,W=7→31.875≥20,
故 7 连胜恰好 confirmed 过线(Codex 复核)。
"""
run = _mk_run(12)
for i in range(12):
run.slots[i].base = False
run.slots[i].cand_per_q = {f"q{i}": True}
_advance_prefix(run, _PARAMS)
assert run.frozen and run.verdict is not None
assert run.verdict.decision == "accept_confirmed"
assert run.n_used == 7 # 第 7 个净胜恰好过线,早停不吃满
def test_tail_infra_reaches_terminal_not_continue() -> None:
"""尾部全 INFRA:剔除后须重判(n_remaining 归 0 → 题尽第四出口),
verdict 不得停留在 continue(Codex plan 审 C1 回归锁)。"""
run = _mk_run(4)
run.slots[0].base = False
run.slots[0].cand_per_q = {"q0": True}
run.slots[1].base = True
run.slots[1].cand_per_q = {"q1": True}
for i in (2, 3):
run.slots[i].base_infra = True
run.slots[i].cand_per_q = {f"q{i}": True}
_advance_prefix(run, _PARAMS)
assert run.verdict is not None and run.verdict.decision != "continue"
assert run.frozen
def test_infra_unit_skipped_not_counted() -> None:
"""INFRA 单元(任一臂)剔除:不入 (W,L)、计入 n_excluded、前缀继续推进。
注意 futility 出口在小 n_remaining 下很敏感,用 6 单元(首个 INFRA、其余 5 个
W 翻转)保证消费全程不提前触发 futility:题尽走 accept_provisional 终态。
"""
run = _mk_run(6)
run.slots[0].base_infra = True
run.slots[0].cand_per_q = {"q0": True}
for i in range(1, 6):
run.slots[i].base = False
run.slots[i].cand_per_q = {f"q{i}": True}
_advance_prefix(run, _PARAMS)
assert run.n_excluded == 1 and run.n_used == 5 and run.w == 5 and run.l == 0
assert run.verdict is not None and run.verdict.decision == "accept_provisional"
def test_ties_hit_futility_early_and_freeze() -> None:
"""全打平(无翻转对)时 futility 出口尽早触发并冻结——早停语义(数值:W=L=0
时乐观 E = E(n_remaining, 0),n 小易 <e_provisional=3)。"""
run = _mk_run(3)
for i in range(3):
run.slots[i].base = True
run.slots[i].cand_per_q = {f"q{i}": True}
_advance_prefix(run, _PARAMS)
assert run.frozen
# 精确锁定 futility 出口:首个消费后 W=L=0,n_remaining=2,
# 乐观 E=E(2,0)=(2^3-1)/3=2.33<3 → 立即 reject_futility(Codex 复核)
assert run.verdict is not None and run.verdict.decision == "reject_futility"
assert run.n_used == 1 and run.w == 0 and run.l == 0