坏信封重投不得重置澄清轮次，额度按连续次数算
无头验证抓到的回归。上一版把坏信封的免费额度做成整条谱系一个总额，超出后 落回质量返工分支——而那一支会把 round 归零。 实测复现的链：坏信封 → 第1轮 → 第2轮 → 坏信封 → 目标。第二次截断超出总额， round 被清零，planning_coordinator 于是按 current_round + 1 要求 「第1轮·关键决定」，子 Agent 按真实轮号写的 header 被拒，父 run 卡进 needs-reconciliation： PLAN_INVALID_CLARIFICATION: plan question header 必须精确等于 第1轮·关键决定 两处都改： - 坏信封跳**永远不重置** round。一次输出被截断既没有推进也没有否定任何已 确认的决定，凭什么把用户已经答过的两轮作废。 - 额度按**连续**次数算，不是整条谱系一个总额。分散在链上的多次截断各自独立； 连续多次才说明子 Agent 真的不会写这个契约，那时才计返工深度。 跳类型文案跟着改：新增 static_delegate_next_hop_is_free_envelope_retry， 只在本跳确实免费时才说「不消耗返工深度」，否则退回质量返工文案——否则文案 在超额那一跳上是假的。 Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
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@@ -938,10 +938,7 @@ pub(crate) fn observe_agent_runtime_agent_delegate_at_locked(
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// 读不到 delivery 时退回 Repair 文案：宁可保守，也不凭空宣布这是免费重投。
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let envelope_retry = repair_of_delegation_id.as_deref().is_some_and(|original| {
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list_static_delegate_deliveries_at(root).is_ok_and(|deliveries| {
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deliveries
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.iter()
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.find(|delivery| delivery.delegation_id == original)
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.is_some_and(static_delegate_original_envelope_was_unparsable)
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static_delegate_next_hop_is_free_envelope_retry(&deliveries, original)
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})
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});
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let hop_note = match (
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@@ -1334,22 +1334,31 @@ pub(crate) fn static_delegate_lineage_counters(
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chain.reverse();
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let mut depth = 0u32;
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let mut round = 0u32;
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let mut envelope_retries = 0u32;
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let mut consecutive_envelope_retries = 0u32;
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for parent in &chain[..chain.len().saturating_sub(1)] {
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if static_delegate_original_is_awaiting_clarification(*parent) {
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round += 1;
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consecutive_envelope_retries = 0;
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} else if static_delegate_original_is_user_revision_requested(*parent) {
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// 用户明确触发的修订不是 runaway-agent 返工；保留两个运行时派生计数。
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} else if static_delegate_original_envelope_was_unparsable(*parent)
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&& envelope_retries < STATIC_DELEGATE_ENVELOPE_RETRY_LIMIT
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{
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// 坏信封重投既不吃返工深度，也不重置澄清轮次：它没有推进也没有否定
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// 任何已确认的决定。给的额度是有限的——超出后落回下面的返工分支，
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// 由 `repair_depth >= 1` 这道既有的门兜住，不会无限重投。
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envelope_retries += 1;
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consecutive_envelope_retries = 0;
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} else if static_delegate_original_envelope_was_unparsable(*parent) {
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// 坏信封重投**永远不重置澄清轮次**：一次输出被截断既没有推进也没有否定
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// 任何已确认的决定，把 round 归零会让 planning_coordinator 回头要求
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// 「第1轮·关键决定」，子 Agent 按真实轮号写的 header 随即被拒，整条
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// 父 run 卡进 needs-reconciliation（实测复现过）。
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//
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// 额度按**连续次数**算，不是整条谱系一个总额：分散在链上的多次截断
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// 各自独立，连续多次才说明子 Agent 真的不会写这个契约，那时才按质量
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// 返工计费，由既有的 `repair_depth >= 1` 兜住。
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consecutive_envelope_retries += 1;
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if consecutive_envelope_retries > STATIC_DELEGATE_ENVELOPE_RETRY_LIMIT {
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depth += 1;
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}
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} else {
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depth += 1;
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round = 0;
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consecutive_envelope_retries = 0;
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}
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}
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(depth, round)
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@@ -1372,6 +1381,26 @@ pub(crate) fn static_delegate_lineage_contains_unknown_contract_status(
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}))
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}
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/// 以 `original_delegation_id` 为父的下一跳，是否还落在坏信封免费重投额度内。
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///
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/// 额度按连续次数算，所以只需要数**紧挨着链尾**的那串连续坏信封节点。跳类型
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/// 文案用它决定说「不消耗返工深度」还是退回质量返工文案——说反了就是撒谎。
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pub(crate) fn static_delegate_next_hop_is_free_envelope_retry(
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deliveries: &[StaticDelegateDeliveryRecord],
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original_delegation_id: &str,
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) -> bool {
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let Some(chain) = static_delegate_lineage_nodes(deliveries, original_delegation_id) else {
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return false;
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};
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// chain 是 [original .. 根]，正好按「从链尾往回数」的顺序排列。
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let consecutive = chain
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.iter()
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.take_while(|node| static_delegate_original_envelope_was_unparsable(node))
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.count();
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consecutive > 0
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&& u32::try_from(consecutive).unwrap_or(u32::MAX) <= STATIC_DELEGATE_ENVELOPE_RETRY_LIMIT
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}
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fn static_delegate_lineage_nodes<'a>(
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deliveries: &'a [StaticDelegateDeliveryRecord],
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delegation_id: &str,
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@@ -3428,6 +3457,113 @@ mod tests {
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assert!(blocked.contains("深度最多为 1"), "{blocked}");
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}
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/// 坏信封重投永远不重置澄清轮次，额度也按**连续**次数算而不是整条谱系一个总额。
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///
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/// 生产实测复现过：链是 坏信封 → 第1轮 → 第2轮 → 坏信封 → 目标，第二次截断把
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/// round 归零，planning_coordinator 回头要求「第1轮·关键决定」，子 Agent 按真实
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/// 轮号写的 header 被拒，父 run 卡进 needs-reconciliation。
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#[test]
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fn static_delegate_envelope_retry_never_resets_the_clarification_round() {
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let root = std::env::temp_dir().join(format!(
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"genarrative-static-envelope-round-{}-{}",
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std::process::id(),
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std::time::SystemTime::now()
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.duration_since(std::time::UNIX_EPOCH)
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.expect("system time after unix epoch")
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.as_nanos()
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));
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init_local_game_project_at(&root, "m1c0-envelope-round", "坏信封不重置轮次测试")
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.expect("project init");
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let parent_run_id = "m1c0-envelope-round-parent-run";
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let write = |delegation_id: &str,
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repair_of: Option<&str>,
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status: StaticDelegateContractStatus,
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unparsable: bool| {
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let needs_user_input = status == StaticDelegateContractStatus::NeedsUserInput;
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let mut delivery = claimed_static_delegate_for_lineage_test(
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parent_run_id,
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delegation_id,
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repair_of,
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status,
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);
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let result = delivery
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.structured_result
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.as_mut()
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.expect("structured result");
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result.user_input_envelope_unparsable = unparsable;
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if needs_user_input {
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// needs-user-input 的落盘校验要求问题非空且指纹匹配。
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let envelope = format!(
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"{STATIC_DELEGATE_USER_INPUT_PREFIX}{}",
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serde_json::json!({
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"questions": [{
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"id": "core_loop",
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"header": "关键决定",
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"question": "本轮要定什么？",
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"options": [
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{"label": "A", "description": "方案 A"},
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{"label": "B", "description": "方案 B"}
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]
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}]
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})
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);
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let (questions, sha256) = parse_static_delegate_user_input_request(Some(&envelope))
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.expect("fixture clarification envelope");
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result.user_input_questions = questions.expect("fixture questions");
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result.user_input_questions_sha256 = sha256;
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}
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write_static_delegate_delivery_at(&root, &delivery).expect("write delivery");
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delivery
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};
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let d0 = write(
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"m1c0-envelope-round-d0",
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None,
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StaticDelegateContractStatus::NeedsRepair,
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true,
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);
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let d1 = write(
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"m1c0-envelope-round-d1",
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Some(&d0.delegation_id),
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StaticDelegateContractStatus::NeedsUserInput,
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false,
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);
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let d2 = write(
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"m1c0-envelope-round-d2",
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Some(&d1.delegation_id),
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StaticDelegateContractStatus::NeedsUserInput,
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false,
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);
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let d3 = write(
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"m1c0-envelope-round-d3",
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Some(&d2.delegation_id),
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StaticDelegateContractStatus::NeedsRepair,
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true,
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);
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let d4 = write(
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"m1c0-envelope-round-d4",
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Some(&d3.delegation_id),
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StaticDelegateContractStatus::NeedsUserInput,
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false,
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);
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let deliveries = list_static_delegate_deliveries_at(&root).expect("list deliveries");
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assert_eq!(
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static_delegate_lineage_counters(&deliveries, &d4.delegation_id),
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(0, 2),
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"两次截断被 2 轮澄清隔开，各自都在免费额度内，轮次必须保持 2"
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);
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assert!(
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static_delegate_next_hop_is_free_envelope_retry(&deliveries, &d3.delegation_id),
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"链尾只有一个连续坏信封，下一跳仍是免费重投"
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);
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assert!(
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!static_delegate_next_hop_is_free_envelope_retry(&deliveries, &d2.delegation_id),
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"父节点不是坏信封时不得声称是免费重投"
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);
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}
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#[test]
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fn static_delegate_user_revision_preserves_counters_and_bypasses_depth_gate() {
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let root = std::env::temp_dir().join(format!(
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