{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261013-advance-20261021-r03-orbit-fixed-periodic-0012", "pilot_number": "R0004-02512", "number": 2512, "family_id": "qf-app21-orbit-fixed-audit", "title": { "zh-Hant": "第 2512 題：有限軌道與週期", "en": "Question 2512: One State’s Fixed-Point Condition" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261013-advance-20261021-r03-orbit-fixed-periodic-0012", "batch_id": "batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-orbit-fixed-periodic" ], "title_zh": "有限軌道與週期", "stem_zh": "完整有限狀態自映射。操作=orbit-fixed-audit；資料={\"modulus\":10,\"a\":4,\"b\":13,\"x\":1}。請依 [modulus,a,b,x,T_x,fixed] 欄序回答。", "answer": { "kind": "expression", "value": "(10, 4, 13, 1, 7, 否)", "explanation_zh": "應用擴展核對=(10, 4, 13, 1, 7, 否)；依完整明示資料重算；完整明示的有限／精確切片不是一般定理或人員判定；固定點與最小正週期分開；只分類指定有限模型。" }, "difficulty": { "band": "undergraduate_core", "level": 2, "skills": [ "有限軌道與週期" ] }, "variant": { "family_id": "qf-app21-orbit-fixed-audit", "seed": 202610213012, "parameters": { "modulus": 10, "a": 4, "b": 13, "x": 1, "domain": "Finite_orbit_periodicity", "operation": "orbit-fixed-audit", "semantic_template": "qf-app21-orbit-fixed-audit/{\"a\":4,\"b\":13,\"modulus\":10,\"x\":1}" } }, "verification": { "status": "candidate", "checks": [ "producer_exact_finite_model", "producer_source_snapshot_check", "producer_substantive_template_deduplication", "producer_json_schema" ], "warning_zh": "僅為產生端候選及自檢；主端須另驗，不構成人工審定、一般定理、真實學生診斷、經驗預測、Lean執行或形式Evidence。" }, "provenance": { "generator_role": "question_producer", "generator_model": "model_unknown", "generated_at": "2026-10-13", "contract_version": "ocme-question-factory-v0.1" } }, "original_english_edition": { "schema_version": "ocme-question-localization-v0.1", "question_id": "q-20261013-advance-20261021-r03-orbit-fixed-periodic-0012", "batch_id": "batch-20261013-advance-20261021-r03-orbit-fixed-periodic-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "47c5ee1317ec690301cef34990afcc49355f04b842696c671dd39f47c9d8d7ae", "family_id": "qf-app21-orbit-fixed-audit", "seed": 202610213012, "semantic_template": "qf-app21-orbit-fixed-audit/{\"a\":4,\"b\":13,\"modulus\":10,\"x\":1}" }, "fields": { "title": "One State’s Fixed-Point Condition", "stem": "Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=4, b=13. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,T(x),T(x)=x?).", "answer": { "kind": "expression", "value": "(10, 4, 13, 1, 7, No)", "explanation": "Evaluate one update exactly; a fixed point has least positive period 1 in this finite self-map. This explicit finite/exact slice is neither a general theorem nor a judgment about a person. The exact result, retaining the requested order, is (10, 4, 13, 1, 7, No)." }, "skills": [ "One State’s Fixed-Point Condition" ], "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": "完整有限狀態自映射。操作=orbit-fixed-audit；資料={\"modulus\":10,\"a\":4,\"b\":13,\"x\":1}。請依 [modulus,a,b,x,T_x,fixed] 欄序回答。", "en": "Use the complete finite self-map on Fin(10)={0,…,9}, T(x)=(a*x+b) mod 10, a=4, b=13. Take least nonnegative residues even for negative a*x+b. Fixed points, return times and least positive periods are different notions; the conclusion concerns this finite model only. x=1. Give (modulus,a,b,x,T(x),T(x)=x?)." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "n=10; a=4; b=13", "after": "S={0,...,9}; T(z)=(a·z+b) mod n", "relation": "implies", "rule_id": "finite_self_map", "conditions": [], "explanation": { "zh-Hant": "餘數取0到n−1，故每個狀態都有唯一後繼且仍在同一有限集合；負數輸入也不取負餘數。", "en": "Remainders lie in0,...,n−1,so every state has one successor in the same finite set;negative dividends do not produce negative canonical remainders." }, "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": "fixed audit product := mul(4, 1)", "after": "fixed audit product = 4", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "先將目前狀態乘以原始係數a，保留其正負號。", "en": "Multiply the current state by the original coefficient a,retaining its sign." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 fixed audit product，不以小數近似替代。", "en": "Recompute fixed audit product with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "4", "1" ], "output": "4", "label": "fixed audit product" } }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "fixed audit affine value := sum(4, 13)", "after": "fixed audit affine value = 17", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "再加上原始常數b，尚未把此整數當成狀態。", "en": "Then add the original constant b;this integer is not yet the canonical state." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 fixed audit affine value，不以小數近似替代。", "en": "Recompute fixed audit affine value with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sum", "inputs": [ "4", "13" ], "output": "17", "label": "fixed audit affine value" } }, { "id": "S004", "uses": [ "G001", "S003" ], "before": "Euclidean division of 17 by 10", "after": "17=1·10+7", "relation": "implies", "rule_id": "orbit_successor", "conditions": [], "explanation": { "zh-Hant": "用整除商與非負餘數驗算轉移；n>0且0≤餘數0 and0≤remainder