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。
This commit is contained in:
2026-07-16 23:44:32 -04:00
parent b02db74237
commit 21c360bc87
2 changed files with 269 additions and 0 deletions
+135
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@@ -793,3 +793,138 @@ async def validate_skill_local(
shutil.rmtree(cand_dir)
except OSError as e:
logger.warning("候选临时目录清理失败 {}: {}", cand_dir, e)
# ---------------------------------------------------------------------------
# 连续并发 gate:数据结构 + 前缀消费(algo #6 语义修订:块序贯 → 阶梯序前缀逐对序贯)
# ---------------------------------------------------------------------------
@dataclass(frozen=True)
class GateSpec:
"""单题型 gate 验证规格(runner 装配阶段产物,调度器输入)。
字段:
task_type: 题型。
target_file: 解析后生效 skill 文件名(候选物化写此文件)。
candidate_content: 候选 skill 全文。
base_skill_content: 基线侧生效 skill 全文(skill_hash 作缓存键)。
units: 阶梯序单元列表(已排除案例单元、截断 gate_n_max)。
gate_run_prefix: run_id 前缀,必须含 "_gate_"(防泄露过滤依赖)。
"""
task_type: str
target_file: str
candidate_content: str
base_skill_content: str
units: list[QuestionUnit]
gate_run_prefix: str
@dataclass
class _UnitSlot:
"""单个阶梯单元的双臂到达状态。
base 为单元级对错(AR pair 已折叠);cand_per_q 为逐题对错(折叠交给消费时,
以复用 unit_correctness_view 并保留逐题溯源)。INFRA 标志与结果互斥。
"""
unit: QuestionUnit
base: bool | None = None
cand_per_q: dict[str, bool] | None = None
base_infra: bool = False
cand_infra: bool = False
def resolved(self) -> bool:
"""双臂均已出结果(含 INFRA 判定)。"""
base_done = self.base is not None or self.base_infra
cand_done = self.cand_per_q is not None or self.cand_infra
return base_done and cand_done
def excluded(self) -> bool:
"""任一臂 INFRA 即整单元剔除(不入配对)。"""
return self.base_infra or self.cand_infra
@dataclass
class _GateRun:
"""单题型 gate 的运行时状态(计数器 + 前缀指针 + 证据)。"""
spec: GateSpec
slots: list[_UnitSlot]
s_hash: str
prefix_ptr: int = 0
w: int = 0
l: int = 0 # noqa: E741
n_used: int = 0
n_excluded: int = 0
errors: int = 0
infra_denom: int = 0
frozen: bool = False
verdict: GateVerdict | None = None
base_obs: dict[str, bool] = field(default_factory=dict)
cand_obs: dict[str, bool] = field(default_factory=dict)
candidate_per_q: dict[str, bool] = field(default_factory=dict)
evidence_rows: list[dict] = field(default_factory=list)
@classmethod
def from_spec(cls, spec: GateSpec) -> _GateRun:
"""由规格构造初始状态(slots 与阶梯序一一对应)。"""
return cls(
spec=spec,
slots=[_UnitSlot(unit=u) for u in spec.units],
s_hash=skill_hash(spec.base_skill_content),
)
def _advance_prefix(run: _GateRun, params: GateParams) -> None:
"""沿阶梯序消费"已配齐前缀",逐单元更新 (W,L) 并判定,过线即冻结。
统计合法性关键(设计 v3 §1 / Codex C1):严禁按到达序消费——base 臂缓存命中
瞬间返回、cand 臂必新鲜跑,两臂延迟不对称,若 cand 延迟与对错相关,早到翻转
对系统性偏向 W 型 → e-值虚高假接受。前缀消费把判定顺序钉回预声明阶梯序,
anytime-valid 无条件成立;INFRA 单元视为"已解决(剔除)"不阻塞前缀。
"""
while not run.frozen and run.prefix_ptr < len(run.slots):
slot = run.slots[run.prefix_ptr]
if not slot.resolved():
return
rank = run.prefix_ptr
run.prefix_ptr += 1
if slot.excluded():
run.n_excluded += 1
# 剔除使 n_remaining 缩小,必须重判(Codex plan 审 C1):否则尾部全 INFRA
# 时 verdict 停留在 "continue",绕过题尽第四出口且 _finalize_outcome
# 查 stop_reason 映射 KeyError。n_used==0(纯前导 INFRA)时无证据可判,跳过。
if run.n_used > 0:
n_remaining = (len(run.slots) - run.n_excluded) - run.n_used
run.verdict = gate_decision(run.w, run.l, run.n_used, n_remaining, params=params)
if run.verdict.decision != "continue":
run.frozen = True
continue
uid = slot.unit.unit_id
assert slot.base is not None and slot.cand_per_q is not None
c_units = unit_correctness_view([slot.unit], slot.cand_per_q)
pair_result = pair_block({uid: slot.base}, c_units, [uid])
run.candidate_per_q.update(slot.cand_per_q)
for u, (b, c) in pair_result.observed.items():
run.base_obs[u] = b
run.cand_obs[u] = c
run.w += pair_result.w
run.l += pair_result.l
run.n_used += 1
n_remaining = (len(run.slots) - run.n_excluded) - run.n_used
run.verdict = gate_decision(run.w, run.l, run.n_used, n_remaining, params=params)
run.evidence_rows.append(
{
"question_id": uid,
"task_type": run.spec.task_type,
"ladder_rank": rank,
"baseline_correct": slot.base,
"candidate_correct": c_units[uid],
"e_value": run.verdict.e_value,
"stop_reason": "",
}
)
if run.verdict.decision != "continue":
run.frozen = True
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@@ -0,0 +1,134 @@
"""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