{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261014-advance-20261109-r06-mobius-function-0054", "pilot_number": "R0042-21729", "number": 21729, "family_id": "qf-mod40-mobius-divisor-sum", "title": { "zh-Hant": "第 21729 題：Möbius函數", "en": "Question 21729: All Positive Divisors Supply the Exact Mobius Sum Identity" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261014-advance-20261109-r06-mobius-function-0054", "batch_id": "batch-20261014-advance-20261109-r06-mobius-function-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-mobius-function" ], "title_zh": "Möbius函數", "stem_zh": "positive integer domain與完整UPF multiplicities。操作=mobius-divisor-sum；資料={\"n\":4}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "answer": { "kind": "expression", "value": "(4, [[1, 1], [2, -1], [4, 0]], 0, 0, 是)", "explanation_zh": "Wave6第三層核對=(4, [[1, 1], [2, -1], [4, 0]], 0, 0, 是)；依完整明示資料獨立重算；mu(1)=1，square prime factor gives0，squarefree sign由distinct prime count parity；factor order不影響；zero negative或incomplete trial result不可分類。" }, "difficulty": { "band": "undergraduate_core", "level": 1, "skills": [ "Möbius函數" ] }, "variant": { "family_id": "qf-mod40-mobius-divisor-sum", "seed": 202611096054, "parameters": { "n": 4, "domain": "Exact_positive_integer_Mobius_function", "operation": "mobius-divisor-sum", "semantic_template": "qf-mod40-mobius-divisor-sum/{\"n\":4}" } }, "verification": { "status": "candidate", "checks": [ "producer_exact_finite_or_symbolic_model", "producer_source_snapshot_check", "producer_substantive_template_deduplication", "producer_json_schema" ], "warning_zh": "候選須主端重算；不構成人工審定、single-element stabilizer、finite-prefix BW、single-cover compactness、zero-mass division、noncommutative binomial、incomplete Möbius factorization、bounded-search nonmembership、Lean或Evidence。" }, "provenance": { "generator_role": "question_producer", "generator_model": "model_unknown", "generated_at": "2026-10-14", "contract_version": "ocme-question-factory-v0.1" } }, "original_english_edition": { "schema_version": "ocme-question-localization-v0.1", "question_id": "q-20261014-advance-20261109-r06-mobius-function-0054", "batch_id": "batch-20261014-advance-20261109-r06-mobius-function-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "235f00462b9cdfd6feb858f227c46073b98703199c7a60aa0fb322844205ed61", "family_id": "qf-mod40-mobius-divisor-sum", "seed": 202611096054, "semantic_template": "qf-mod40-mobius-divisor-sum/{\"n\":4}" }, "fields": { "title": "All Positive Divisors Supply the Exact Mobius Sum Identity", "stem": "Use positive integer n=4. Give (n, divisor_mu_rows, sum, target, equal).", "answer": { "kind": "expression", "value": "(4, [[1, 1], [2, -1], [4, 0]], 0, 0, Yes)", "explanation": "Enumerate every positive divisor d increasingly and its complete mu(d). Their sum is1 when n=1 and0 otherwise. This exact instance check does not justify classifying an incompletely factored input. The exact result, retaining the requested order, is (4, [[1, 1], [2, -1], [4, 0]], 0, 0, Yes)." }, "skills": [ "All Positive Divisors Supply the Exact Mobius Sum Identity" ], "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": "positive integer domain與完整UPF multiplicities。操作=mobius-divisor-sum；資料={\"n\":4}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "en": "Use positive integer n=4. Give (n, divisor_mu_rows, sum, target, equal)." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "source operation=mobius-divisor-sum;original snapshot and mathematics retained", "after": "explicit ordered source output fields=[n, divisor_mu_rows, sum, target, equal]", "relation": "implies", "rule_id": "source_ordered_interface_explicit_without_javascript", "conditions": [], "explanation": { "zh-Hant": "原題歷史介面標記保留，本頁直接列來源有序欄位，不需JavaScript。tag不是額外證據，固定No保留其原聲稱範圍。", "en": "Keep the original historical interface marker while listing ordered source fields without JavaScript.Tags are not additional evidence;fixed No markers retain their source claim scope." }, "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": "ORIGINAL n=4;complete positive candidate range1..n", "after": "ALL actual positive divisors INCREASING=[1, 2, 4]", "relation": "implies", "rule_id": "mobius_ALL_positive_divisors_not_sample", "conditions": [], "explanation": { "zh-Hant": "逐完整正範圍保整除者，不漏1或n；後續每列mu亦須完整分解。", "en": "Keep divisors over the complete positive range,including1 and n;every subsequent mu row needs a complete factorization." }, "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": "mobius divisor1 of original n=4:actual positive input for mu=1", "after": "complete prime multiplicities required for this input;mu(1) uses empty product,not a partial factor search", "relation": "implies", "rule_id": "mobius_complete_UPF_input_role", "conditions": [], "explanation": { "zh-Hant": "先明示正在算原n還是其正因數的mu；每次需要完整prime重數，不能用找到的幾個因數猜squarefree。", "en": "Identify whether the input is the original n or one of its positive divisors.Every mu evaluation requires complete prime multiplicities,not a partial factor search." }, "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": "mobius divisor1 of original n=4:input=1;COMPLETE prime multiset=[];distinct primes=[]", "after": "mu=1;empty product case=是;repeated prime=否", "relation": "implies", "rule_id": "mobius_value_one_repeated_prime_or_squarefree_parity", "conditions": [], "explanation": { "zh-Hant": "1的空prime多重集給mu1=1；任何prime重複表示square divisor而給0；否則依不同prime個數奇偶給正負。排序不丟重數。", "en": "The empty prime multiset for1 gives mu1=1.A repeated prime gives a square divisor and value0;otherwise distinct-prime parity determines the sign.Sorting must not discard multiplicities." }, "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": "mobius divisor2 of original n=4:actual positive input for mu=2", "after": "complete prime multiplicities required for this input;mu(1) uses empty product,not a partial factor search", "relation": "implies", "rule_id": "mobius_complete_UPF_input_role", "conditions": [], "explanation": { "zh-Hant": "先明示正在算原n還是其正因數的mu；每次需要完整prime重數，不能用找到的幾個因數猜squarefree。", "en": "Identify whether the input is the original n or one of its positive divisors.Every mu evaluation requires complete prime multiplicities,not a partial factor search." }, "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": "ORIGINAL positive integer absolute magnitude=2>1", "after": "initialize remaining=n,prime multiset empty;invariant product(extracted)*remaining=n;every smaller trial divisor has been exhausted", "relation": "implies", "rule_id": "upf_factorization_remaining_product_invariant", "conditions": [], "explanation": { "zh-Hant": "每次只除真正整除的d，並以已窮盡更小因數維持d為prime的理由，不先假定原n為prime。", "en": "Divide only by an actual divisor and exhaust smaller candidates,which justifies primality of each extracted divisor without assuming n is prime." }, "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": "upf canonical complete product factor0 := mul(1, 2)", "after": "upf canonical complete product factor0 = 2", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原題角色與已核中間量精確代值，保留原順序、正負號及量詞條件。", "en": "Substitute the original roles and established intermediates exactly,retaining order,signs and quantified conditions." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 upf canonical complete product factor0，不以小數近似替代。", "en": "Recompute upf canonical complete product factor0 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "1", "2" ], "output": "2", "label": "upf canonical complete product factor0" } }, { "id": "S008", "uses": [ "G001", "S007" ], "before": "exhausted trial divisors below2;terminal remaining=2;d*d>remaining;ORIGINAL n=2", "after": "complete increasing prime multiset=[2];full product=2;a composite terminal remainder would have an already tested smaller divisor", "relation": "implies", "rule_id": "upf_final_prime_remainder_complete_existence_instance", "conditions": [], "explanation": { "zh-Hant": "若remaining>1仍合數，其最小prime因數會在已驗範圍，與除盡不變式矛盾，所以補入的末項為prime。這證本例完整分解；一般存在也可對n作強歸納，合數拆成較小正因數後遞歸。", "en": "If a terminal remainder greater than1 were composite,its least prime factor would already have been exhausted,a contradiction.The final entry is therefore prime.This certifies this instance;general existence follows by strong induction,recursively splitting a composite into smaller positive factors." }, "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": "mobius divisor2 of original n=4:input=2;COMPLETE prime multiset=[2];distinct primes=[2]", "after": "mu=-1;empty product case=否;repeated prime=否", "relation": "implies", "rule_id": "mobius_value_one_repeated_prime_or_squarefree_parity", "conditions": [], "explanation": { "zh-Hant": "1的空prime多重集給mu1=1；任何prime重複表示square divisor而給0；否則依不同prime個數奇偶給正負。排序不丟重數。", "en": "The empty prime multiset for1 gives mu1=1.A repeated prime gives a square divisor and value0;otherwise distinct-prime parity determines the sign.Sorting must not discard multiplicities." }, "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": "mobius divisor4 of original n=4:actual positive input for mu=4", "after": "complete prime multiplicities required for this input;mu(1) uses empty product,not a partial factor search", "relation": "implies", "rule_id": "mobius_complete_UPF_input_role", "conditions": [], "explanation": { "zh-Hant": "先明示正在算原n還是其正因數的mu；每次需要完整prime重數，不能用找到的幾個因數猜squarefree。", "en": "Identify whether the input is the original n or one of its positive divisors.Every mu evaluation requires complete prime multiplicities,not a partial factor search." }, "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": "ORIGINAL positive integer absolute magnitude=4>1", "after": "initialize remaining=n,prime multiset empty;invariant product(extracted)*remaining=n;every smaller trial divisor has been exhausted", "relation": "implies", "rule_id": "upf_factorization_remaining_product_invariant", "conditions": [], "explanation": { "zh-Hant": "每次只除真正整除的d，並以已窮盡更小因數維持d為prime的理由，不先假定原n為prime。", "en": "Divide only by an actual divisor and exhaust smaller candidates,which justifies primality of each extracted divisor without assuming n is prime." }, "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": "upf extract divisor2 occurrence0 := div(4, 2)", "after": "upf extract divisor2 occurrence0 = 2", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原題角色與已核中間量精確代值，保留原順序、正負號及量詞條件。", "en": "Substitute the original roles and established intermediates exactly,retaining order,signs and quantified conditions." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 upf extract divisor2 occurrence0，不以小數近似替代。", "en": "Recompute upf extract divisor2 occurrence0 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "div", "inputs": [ "4", "2" ], "output": "2", "label": "upf extract divisor2 occurrence0" } }, { "id": "S013", "uses": [ "G001", "S012" ], "before": "upf extract divisor2 occurrence1 := div(2, 2)", "after": "upf extract divisor2 occurrence1 = 1", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原題角色與已核中間量精確代值，保留原順序、正負號及量詞條件。", "en": "Substitute the original roles and established intermediates exactly,retaining order,signs and quantified conditions." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 upf extract divisor2 occurrence1，不以小數近似替代。", "en": "Recompute upf extract divisor2 occurrence1 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "div", "inputs": [ "2", "2" ], "output": "1", "label": "upf extract divisor2 occurrence1" } }, { "id": "S014", "uses": [ "G001", "S013" ], "before": "next divisor=2;previous remaining=4;all smaller divisors exhausted", "after": "extracted multiplicity=2;next remaining=1;remaining modulo2=1;extracted multiset=[2, 2]", "relation": "implies", "rule_id": "upf_complete_trial_division_remaining_invariant", "conditions": [], "explanation": { "zh-Hant": "除盡同一prime的所有重數才前進；不把n的原平方根與remaining的停止責任混淆，也不漏最後的prime remainder。", "en": "Exhaust every multiplicity before advancing.The stopping test concerns the current remainder,and the final prime remainder must not be omitted." }, "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": "upf canonical complete product factor0 := mul(1, 2)", "after": "upf canonical complete product factor0 = 2", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原題角色與已核中間量精確代值，保留原順序、正負號及量詞條件。", "en": "Substitute the original roles and established intermediates exactly,retaining order,signs and quantified conditions." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 upf canonical complete product factor0，不以小數近似替代。", "en": "Recompute upf canonical complete product factor0 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "1", "2" ], "output": "2", "label": "upf canonical complete product factor0" } }, { "id": "S016", "uses": [ "G001", "S015" ], "before": "upf canonical complete product factor1 := mul(2, 2)", "after": "upf canonical complete product factor1 = 4", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原題角色與已核中間量精確代值，保留原順序、正負號及量詞條件。", "en": "Substitute the original roles and established intermediates exactly,retaining order,signs and quantified conditions." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 upf canonical complete product factor1，不以小數近似替代。", "en": "Recompute upf canonical complete product factor1 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "2", "2" ], "output": "4", "label": "upf canonical complete product factor1" } }, { "id": "S017", "uses": [ "G001", "S016" ], "before": "exhausted trial divisors below3;terminal remaining=1;d*d>remaining;ORIGINAL n=4", "after": "complete increasing prime multiset=[2, 2];full product=4;a composite terminal remainder would have an already tested smaller divisor", "relation": "implies", "rule_id": "upf_final_prime_remainder_complete_existence_instance", "conditions": [], "explanation": { "zh-Hant": "若remaining>1仍合數，其最小prime因數會在已驗範圍，與除盡不變式矛盾，所以補入的末項為prime。這證本例完整分解；一般存在也可對n作強歸納，合數拆成較小正因數後遞歸。", "en": "If a terminal remainder greater than1 were composite,its least prime factor would already have been exhausted,a contradiction.The final entry is therefore prime.This certifies this instance;general existence follows by strong induction,recursively splitting a composite into smaller positive factors." }, "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": "S018", "uses": [ "G001", "S017" ], "before": "mobius divisor4 of original n=4:input=4;COMPLETE prime multiset=[2, 2];distinct primes=[2]", "after": "mu=0;empty product case=否;repeated prime=是", "relation": "implies", "rule_id": "mobius_value_one_repeated_prime_or_squarefree_parity", "conditions": [], "explanation": { "zh-Hant": "1的空prime多重集給mu1=1；任何prime重複表示square divisor而給0；否則依不同prime個數奇偶給正負。排序不丟重數。", "en": "The empty prime multiset for1 gives mu1=1.A repeated prime gives a square divisor and value0;otherwise distinct-prime parity determines the sign.Sorting must not discard multiplicities." }, "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": "S019", "uses": [ "G001", "S018" ], "before": "Mobius complete divisor mu sum := sum(1, -1, 0)", "after": "Mobius complete divisor mu sum = 0", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "保留原題角色、順序與條件，逐項精確代值，不作浮點近似。", "en": "Substitute the original roles,order and conditions term by term with exact arithmetic,not floating-point approximation." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Mobius complete divisor mu sum，不以小數近似替代。", "en": "Recompute Mobius complete divisor mu sum with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sum", "inputs": [ "1", "-1", "0" ], "output": "0", "label": "Mobius complete divisor mu sum" } }, { "id": "S020", "uses": [ "G001", "S019" ], "before": "ORIGINAL n=4;ALL increasing [divisor,mu] rows=[[1, 1], [2, -1], [4, 0]]", "after": "exact sum=0;target=0;squarefree divisors correspond to subsets of DISTINCT primes,so sum=(-1)^subset_size summed over all subsets=(1-1)^r;for n1,r0 gives1", "relation": "implies", "rule_id": "mobius_divisor_sum_general_subset_cancellation", "conditions": [], "explanation": { "zh-Hant": "非squarefree因數貢獻0，其餘恰為不同prime子集合。r>0可配對是否含指定prime使正負消去；n1沒有prime而唯一空子集合貢獻1。這是一般理由，不由一次數值相等外推。", "en": "Nonsquarefree divisors contribute0;all others correspond to subsets of distinct primes.For r>0,pair subsets by one fixed prime to cancel signs.For n1,the sole empty subset contributes1.This supplies the general argument rather than extrapolating one equality." }, "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 } ], "math_model": { "result": [ 4, [ [ 1, 1 ], [ 2, -1 ], [ 4, 0 ] ], 0, 0, true ] }, "conclusion": { "canonical_value": "(4, [[1, 1], [2, -1], [4, 0]], 0, 0, 是)", "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-20261014-advance-20261109-r06-mobius-function-foundations", "batch_path": "public/data/questions/batches/batch-20261014-advance-20261109-r06-mobius-function-foundations/questions.jsonl", "batch_sha256": "7ed895d51f7c25fef70946caed5452254a4a2e02ad05f2e1c4b765f1ccdba585", "original_line": 54, "original_line_sha256": "5a5bb760ce94824c80111a7121426db4880f6432aba2674fff58238072b01700", "question_compact_sha256": "235f00462b9cdfd6feb858f227c46073b98703199c7a60aa0fb322844205ed61", "english_row_compact_sha256": "f988d5d5cae3deb8e45d8763fad49c968e7b345c712fe88b2dba670e4cd251c5", "english_manifest_path": "public/data/i18n/editions/batch-20261014-advance-20261109-r06-mobius-function-foundations.en.json", "english_rows_path": "public/data/i18n/questions/batch-20261014-advance-20261109-r06-mobius-function-foundations.en.jsonl" }, "mko_bindings": { "mko-mobius-function": { "sha256": "8991dece49cf8a59e05519f80e840deb92afafba5fef912edc1a03ad30438f84", "english_sha256": "c602666c47bdf330fa90517c2baa6b3bb4811deb05ff4c991c801c6ba0793623", "original_path": "public/data/mko/mobius-function.json", "english_path": "public/data/i18n/editions/mko-mobius-function.en.json" } }, "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff", "selected_part_sha256": "a84fbe07a4e46e8e47a8ad331f35fd3f81b821aef6cb27e7852de854aaa96b9a", "original_candidate_sha256": "353cd4085a32dda224dcd125e652992bf1f231a35846a99c619ae78461c49e7e", "original_candidate_line_sha256": "353cd4085a32dda224dcd125e652992bf1f231a35846a99c619ae78461c49e7e" }, "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-20261026-009-p43", "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": "dd44555041dfbb6fe9936aa00b9a5f8c5982c2784ef61dd0cbd5d1a303a49c1a", "question_first_witness_sha256": "9a6011263f6ac53b9a1fd57addf2f3e2f6088b54802c539f332d689480dc2d17" }, "status": "second_context_reviewed", "mathematical_semantics": true, "bilingual_semantics": true, "steps_reviewed": 20, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "612b09d8deb44b0dc0a91192e6a8cc9884d7c532c2947578a9020aa0cefe8b0d", "arithmetic_steps": 6 }, "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." } }