{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261012-advance-20261017-r06-tree-0043", "pilot_number": "R0015-08493", "number": 8493, "family_id": "qf-density17-tree-unique-path-gate", "title": { "zh-Hant": "第 8493 題：樹與端點門檻後的唯一道路", "en": "Question 8493: Original-Tree and Endpoint Guards before a Unique Path" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261012-advance-20261017-r06-tree-0043", "batch_id": "batch-20261012-advance-20261017-r06-tree-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-tree" ], "title_zh": "樹與端點門檻後的唯一道路", "stem_zh": "簡單無向圖，Fin頂點從0起。V=Fin(6)=[0, 1, 2, 3, 4, 5]，無向邊列項=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]；各邊兩端皆合法且相異。start=5、end=2。答 (n, 正規邊集, 原圖樹資格, 兩端點資格, 道路或不成樹或不合頂點, 邊數或不求值)。原圖非樹優先答不成樹；合法自身道路僅列[start]。", "answer": { "kind": "expression", "value": "(6, [[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]], 是, [是, 是], [5, 2], 1)", "explanation_zh": "離散結構核對=(6, [[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]], 是, [是, 是], [5, 2], 1)；先驗整張原圖與端點，再用不重複頂點的DFS尋路。；有限精確核對不是一般形式證明；空圖非樹單點度0非葉；邊數不單獨判樹；刪除保留原標籤；加邊先核原樹與不存在邊。" }, "difficulty": { "band": "undergraduate_core", "level": 3, "skills": [ "樹與端點門檻後的唯一道路" ] }, "variant": { "family_id": "qf-density17-tree-unique-path-gate", "seed": 202610176043, "parameters": { "n": 6, "edges": [ [ 0, 1 ], [ 0, 4 ], [ 0, 5 ], [ 1, 3 ], [ 2, 5 ] ], "start": 5, "end": 2, "domain": "Fin_tree_qualification", "operation": "tree-unique-path-gate", "semantic_template": "qf-density17-tree-unique-path-gate/{\"edges\":[[0,1],[0,4],[0,5],[1,3],[2,5]],\"end\":2,\"n\":6,\"start\":5}" } }, "verification": { "status": "candidate", "checks": [ "producer_exact_integer_rational_arithmetic", "producer_finite_enumeration", "producer_source_snapshot_check", "producer_substantive_template_deduplication", "producer_json_schema" ], "warning_zh": "僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。" }, "provenance": { "generator_role": "question_producer", "generator_model": "model_unknown", "generated_at": "2026-10-12", "contract_version": "ocme-question-factory-v0.1" } }, "original_english_edition": { "schema_version": "ocme-question-localization-v0.1", "question_id": "q-20261012-advance-20261017-r06-tree-0043", "batch_id": "batch-20261012-advance-20261017-r06-tree-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "46caae0c9e7a9ac12fa3fbb0ea12d33bb805a41f09bab977ea328385740c4f2a", "family_id": "qf-density17-tree-unique-path-gate", "seed": 202610176043, "semantic_template": "qf-density17-tree-unique-path-gate/{\"edges\":[[0,1],[0,4],[0,5],[1,3],[2,5]],\"end\":2,\"n\":6,\"start\":5}" }, "fields": { "title": "Original-Tree and Endpoint Guards before a Unique Path", "stem": "Use a complete simple undirected graph on Fin(6)={0,…,5} with raw edges=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]. Endpoints are typed and loops are prohibited. Reversed or repeated edges count once; normalize endpoints and edge order. Do not infer adjacency from a drawing. A tree is nonempty, connected and acyclic. Edge count alone does not decide this. The empty graph is not a tree; a singleton is a tree, but its degree-0 vertex is not a leaf. Preserve original vertex labels after deletion. start=5, end=2. Give (n, normalized edges, tree?, two endpoint type flags, unique path or not-a-tree/outside-vertex-set, path edge count or not-evaluated). A failed original-tree guard takes precedence over the path lookup; equal valid endpoints give a one-vertex path.", "answer": { "kind": "expression", "value": "(6, [[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]], Yes, [Yes, Yes], [5, 2], 1)", "explanation": "Establish the original tree and both endpoint types before finding the path; its edge count is one less than its vertex count. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (6, [[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]], Yes, [Yes, Yes], [5, 2], 1)." }, "skills": [ "Original-Tree and Endpoint Guards before a Unique Path" ], "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review." } }, "givens": [ { "id": "G001", "text": { "zh-Hant": "簡單無向圖，Fin頂點從0起。V=Fin(6)=[0, 1, 2, 3, 4, 5]，無向邊列項=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]；各邊兩端皆合法且相異。start=5、end=2。答 (n, 正規邊集, 原圖樹資格, 兩端點資格, 道路或不成樹或不合頂點, 邊數或不求值)。原圖非樹優先答不成樹；合法自身道路僅列[start]。", "en": "Use a complete simple undirected graph on Fin(6)={0,…,5} with raw edges=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]. Endpoints are typed and loops are prohibited. Reversed or repeated edges count once; normalize endpoints and edge order. Do not infer adjacency from a drawing. A tree is nonempty, connected and acyclic. Edge count alone does not decide this. The empty graph is not a tree; a singleton is a tree, but its degree-0 vertex is not a leaf. Preserve original vertex labels after deletion. start=5, end=2. Give (n, normalized edges, tree?, two endpoint type flags, unique path or not-a-tree/outside-vertex-set, path edge count or not-evaluated). A failed original-tree guard takes precedence over the path lookup; equal valid endpoints give a one-vertex path." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "n=6;raw edges=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]", "after": "canonical edges=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]", "relation": "implies", "rule_id": "simple_graph_edge_normalization", "conditions": [], "explanation": { "zh-Hant": "整張簡單無向圖先查端點合型且無自環；邊兩端排序、反向與重複邊只算一次，不以畫法猜相鄰。", "en": "Validate every simple undirected edge for typed distinct endpoints.Sort endpoints and count reversed or repeated edges once;do not infer adjacency from a drawing." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S002", "uses": [ "G001", "S001" ], "before": "phase=original;vertex=0;all edges=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]", "after": "neighbors=[1, 4, 5];degree=3", "relation": "implies", "rule_id": "complete_graph_neighbor_inventory", "conditions": [], "explanation": { "zh-Hant": "掃描所有正規邊，只有以本頂點為端點的邊才給一個不同鄰居；原頂點標籤保留。", "en": "Scan all canonical edges.An incident edge contributes its other endpoint as a distinct neighbor.Keep original vertex labels." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "phase=original;vertex=1;all edges=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]", "after": "neighbors=[0, 3];degree=2", "relation": "implies", "rule_id": "complete_graph_neighbor_inventory", "conditions": [], "explanation": { "zh-Hant": "掃描所有正規邊，只有以本頂點為端點的邊才給一個不同鄰居；原頂點標籤保留。", "en": "Scan all canonical edges.An incident edge contributes its other endpoint as a distinct neighbor.Keep original vertex labels." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S004", "uses": [ "G001", "S003" ], "before": "phase=original;vertex=2;all edges=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]", "after": "neighbors=[5];degree=1", "relation": "implies", "rule_id": "complete_graph_neighbor_inventory", "conditions": [], "explanation": { "zh-Hant": "掃描所有正規邊，只有以本頂點為端點的邊才給一個不同鄰居；原頂點標籤保留。", "en": "Scan all canonical edges.An incident edge contributes its other endpoint as a distinct neighbor.Keep original vertex labels." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S005", "uses": [ "G001", "S004" ], "before": "phase=original;vertex=3;all edges=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]", "after": "neighbors=[1];degree=1", "relation": "implies", "rule_id": "complete_graph_neighbor_inventory", "conditions": [], "explanation": { "zh-Hant": "掃描所有正規邊，只有以本頂點為端點的邊才給一個不同鄰居；原頂點標籤保留。", "en": "Scan all canonical edges.An incident edge contributes its other endpoint as a distinct neighbor.Keep original vertex labels." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S006", "uses": [ "G001", "S005" ], "before": "phase=original;vertex=4;all edges=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]", "after": "neighbors=[0];degree=1", "relation": "implies", "rule_id": "complete_graph_neighbor_inventory", "conditions": [], "explanation": { "zh-Hant": "掃描所有正規邊，只有以本頂點為端點的邊才給一個不同鄰居；原頂點標籤保留。", "en": "Scan all canonical edges.An incident edge contributes its other endpoint as a distinct neighbor.Keep original vertex labels." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S007", "uses": [ "G001", "S006" ], "before": "phase=original;vertex=5;all edges=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]", "after": "neighbors=[0, 2];degree=2", "relation": "implies", "rule_id": "complete_graph_neighbor_inventory", "conditions": [], "explanation": { "zh-Hant": "掃描所有正規邊，只有以本頂點為端點的邊才給一個不同鄰居；原頂點標籤保留。", "en": "Scan all canonical edges.An incident edge contributes its other endpoint as a distinct neighbor.Keep original vertex labels." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S008", "uses": [ "G001", "S007" ], "before": "phase=original;root=0;depth=0;frontier=[0]", "after": "next newly reached=[1, 4, 5];component seen=[0, 1, 4, 5]", "relation": "implies", "rule_id": "tree_complete_reachability_layer", "conditions": [], "explanation": { "zh-Hant": "只沿正規邊擴展到尚未見的原標籤；前緣清空才完成這個分量，然後再從其他未見頂點開始。", "en": "Expand along canonical edges only to unseen original labels.A component is complete only when its frontier empties;then start at another unseen vertex." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S009", "uses": [ "G001", "S008" ], "before": "phase=original;root=0;depth=1;frontier=[1, 4, 5]", "after": "next newly reached=[3, 2];component seen=[0, 1, 2, 3, 4, 5]", "relation": "implies", "rule_id": "tree_complete_reachability_layer", "conditions": [], "explanation": { "zh-Hant": "只沿正規邊擴展到尚未見的原標籤；前緣清空才完成這個分量，然後再從其他未見頂點開始。", "en": "Expand along canonical edges only to unseen original labels.A component is complete only when its frontier empties;then start at another unseen vertex." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S010", "uses": [ "G001", "S009" ], "before": "phase=original;root=0;depth=2;frontier=[3, 2]", "after": "next newly reached=[];component seen=[0, 1, 2, 3, 4, 5]", "relation": "implies", "rule_id": "tree_complete_reachability_layer", "conditions": [], "explanation": { "zh-Hant": "只沿正規邊擴展到尚未見的原標籤；前緣清空才完成這個分量，然後再從其他未見頂點開始。", "en": "Expand along canonical edges only to unseen original labels.A component is complete only when its frontier empties;then start at another unseen vertex." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S011", "uses": [ "G001", "S010" ], "before": "phase=original;spanning forest=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]];all edges=[[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]]", "after": "nonforest edges=[];cyclic=否", "relation": "implies", "rule_id": "spanning_forest_cycle_witness", "conditions": [], "explanation": { "zh-Hant": "逐層首次到達邊形成各分量的樹；其餘邊連接已有樹道路的兩端並產生環。這先建立分量，不是只靠邊數判原圖為樹。", "en": "First-discovery edges form a tree in each reached component.Any remaining edge closes an existing tree path into a cycle.Components were established first;this is not an edge-count-only tree test." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S012", "uses": [ "G001", "S011" ], "before": "phase=original;vertices=[0, 1, 2, 3, 4, 5];components=[[0, 1, 2, 3, 4, 5]];cyclic=否", "after": "pairwise reachable=是;nonempty=是;tree=是", "relation": "implies", "rule_id": "nonempty_connected_acyclic_tree_gate", "conditions": [], "explanation": { "zh-Hant": "樹要求非空、單一分量且無環。空頂點集的兩兩可達條件空泛為真，卻不滿足非空；單點無邊是樹，度0不是葉。", "en": "A tree must be nonempty,have one component and be acyclic.Pairwise reachability is vacuously true on an empty vertex set,but nonemptiness fails.A singleton with no edges is a tree;its degree0 vertex is not a leaf." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S013", "uses": [ "G001", "S012" ], "before": "original tree=是;endpoints=[5, 2];typed=[是, 是]", "after": "path lookup qualified=是", "relation": "implies", "rule_id": "tree_path_whole_graph_and_endpoint_guard", "conditions": [], "explanation": { "zh-Hant": "原圖非樹優先不成樹；若原圖為樹才依端點資格決定能否讀路徑，不能拿局部路徑存在代替全圖是樹。", "en": "A failed original-tree guard takes precedence.Only for an original tree do endpoint types permit path lookup.A local path's existence does not establish whole-graph treehood." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S014", "uses": [ "G001", "S013" ], "before": "phase=query;prefix=[5];next=2", "after": "edge=[2, 5] present;new prefix=[5, 2]", "relation": "implies", "rule_id": "qualified_tree_path_edge", "conditions": [], "explanation": { "zh-Hant": "每一步沿一條原邊到未重複頂點；樹中兩點的簡單道路唯一，端點相同則不走任何邊。", "en": "Each step follows an original edge to a nonrepeated vertex.The simple path between tree vertices is unique;equal endpoints need no edge." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S015", "uses": [ "G001", "S014" ], "before": "phase=query;start=5;end=2", "after": "unique path=[5, 2]", "relation": "implies", "rule_id": "qualified_tree_path_result", "conditions": [], "explanation": { "zh-Hant": "此道路的唯一性依賴前面原樹資格，不對一般含環圖宣稱唯一道路。", "en": "Path uniqueness relies on the earlier original-tree qualification;it is not asserted for a general cyclic graph." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S016", "uses": [ "G001", "S015" ], "before": "qualified path edge count := sub(2, 1)", "after": "qualified path edge count = 1", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "按本步指定順序精確代入原數據，分母非零才允許除法。", "en": "Substitute the source data exactly in the stated operand order;division requires a nonzero denominator." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 qualified path edge count，不以小數近似替代。", "en": "Recompute qualified path edge count with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sub", "inputs": [ "2", "1" ], "output": "1", "label": "qualified path edge count" } } ], "math_model": { "result": [ 6, [ [ 0, 1 ], [ 0, 4 ], [ 0, 5 ], [ 1, 3 ], [ 2, 5 ] ], true, [ true, true ], [ 5, 2 ], "1" ] }, "conclusion": { "canonical_value": "(6, [[0, 1], [0, 4], [0, 5], [1, 3], [2, 5]], 是, [是, 是], [5, 2], 1)", "explanation": { "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。", "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged." } }, "definition_note": { "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。", "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate." }, "source": { "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db", "question_binding": { "batch_id": "batch-20261012-advance-20261017-r06-tree-foundations", "batch_path": "public/data/questions/batches/batch-20261012-advance-20261017-r06-tree-foundations/questions.jsonl", "batch_sha256": "57dbe1d0a91e1f9b6da54012ecd1ebff389301cee21695dc83d8168fd614d96a", "original_line": 43, "original_line_sha256": "68a79fd4c4304e0b8fcf6356b9d050330301fe3bb7c143f747c078a8ee154b56", "question_compact_sha256": "46caae0c9e7a9ac12fa3fbb0ea12d33bb805a41f09bab977ea328385740c4f2a", "english_row_compact_sha256": "9b69033b203a0154cb3b581fbc0078423d1e58fb422c04e56fe14944eda7053c", "english_manifest_path": "public/data/i18n/editions/batch-20261012-advance-20261017-r06-tree-foundations.en.json", "english_rows_path": "public/data/i18n/questions/batch-20261012-advance-20261017-r06-tree-foundations.en.jsonl" }, "mko_bindings": { "mko-tree": { "sha256": "48c7310136a8fd9749e9a9af278b0ade15fdecb8f4cc5657f4f0895eef89aebf", "english_sha256": "9b6b7c0d53aa18952a811e36d538baf58041cc6982489dd77fe0f5e4535379f6", "original_path": "public/data/mko/tree.json", "english_path": "public/data/i18n/editions/mko-tree.en.json" } }, "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff", "selected_part_sha256": "ca1078dc467a0f3769865919aadc46c87712d098f1dfbde90f787471e22bcfec", "original_candidate_sha256": "c21053c9b897d24d003770c03f62774a33cd73ea8ec28e5499223f1df73baef6", "original_candidate_line_sha256": "c21053c9b897d24d003770c03f62774a33cd73ea8ec28e5499223f1df73baef6", "mko_binding_revision": { "id": "edition-symbol-bindings-20261101", "path": "content/edition-symbol-bindings-20261101/revision.json", "sha256": "e9b4c69f2c4732074c37abb624c8ab8a70afc3645c12bf9df2268ef650ad7b6c", "previous_english_sha256": { "mko-tree": "85d883b1c7551ca05ff4319fb1234a359116ddd647ba5ff4094409cbc9578788" } } }, "historical_source_state": { "review_state": { "derivation": "candidate_authored_and_locally_checked", "source_question_state": "candidate", "independent_AI_or_human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN" }, "scope_note": { "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。", "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged." }, "label": "Frozen producer metadata; current per-question review is separate below." }, "revision": { "id": "review-20261026-005-p16", "review_date": "2026-10-26", "owner_authorized_reader_publication": true, "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.", "review_receipts": { "semantic_review_sha256": "6ecad9d909199221a07e923f87425e80b3bdec3eed46fe5910d04fca694cae18", "question_first_witness_sha256": "e8f00458977a09493b5f9f05ab9e6feec72debaf4c7d1c609d1fc95900a35a1b" }, "status": "second_context_reviewed", "mathematical_semantics": true, "bilingual_semantics": true, "steps_reviewed": 16, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "8d3d79b5ef51e64a69c90248c589dcdebc09475f46a687d7fc496b6740f17774", "arithmetic_steps": 1 }, "uses_answer_as_premise": false, "human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN", "Canonical_truth_promotion": false, "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged." } }