{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261013-advance-20261019-r06-asymptotic-resource-bound-0057",
  "pilot_number": "R0017-09332",
  "number": 9332,
  "family_id": "qf-ext19-asymptotic-linear-witness",
  "title": {
    "zh-Hant": "第 9332 題：漸近資源界",
    "en": "Question 9332: An Explicit All-Input Linear Resource Witness"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261013-advance-20261019-r06-asymptotic-resource-bound-0057",
    "batch_id": "batch-20261013-advance-20261019-r06-asymptotic-resource-bound-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-asymptotic-resource-bound"
    ],
    "title_zh": "漸近資源界",
    "stem_zh": "明示非負函數族。操作=asymptotic-linear-witness；資料={\"a\":6,\"b\":13}。請依 [a,b,C,N,valid,witness_inequality] 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(6, 13, 19, 0, 是, 6n+13≤19(n+1) 對所有 n≥0)",
      "explanation_zh": "第二批擴展核對=(6, 13, 19, 0, 是, 6n+13≤19(n+1) 對所有 n≥0)；依完整明示資料重算；有限／合成資料核對不是一般形式證明；全尾段量化；有限benchmark不證Big-O或Theta；成本模型必須固定。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 1,
      "skills": [
        "漸近資源界"
      ]
    },
    "variant": {
      "family_id": "qf-ext19-asymptotic-linear-witness",
      "seed": 202610196057,
      "parameters": {
        "a": 6,
        "b": 13,
        "domain": "Nat_nonnegative_asymptotics",
        "operation": "asymptotic-linear-witness",
        "semantic_template": "qf-ext19-asymptotic-linear-witness/{\"a\":6,\"b\":13}"
      }
    },
    "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、一般程式證明、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-20261019-r06-asymptotic-resource-bound-0057",
    "batch_id": "batch-20261013-advance-20261019-r06-asymptotic-resource-bound-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "c88c4e54c7c156993e951f087939f5d00dcef1a6ecc5a2c26673b845ebb88d2c",
      "family_id": "qf-ext19-asymptotic-linear-witness",
      "seed": 202610196057,
      "semantic_template": "qf-ext19-asymptotic-linear-witness/{\"a\":6,\"b\":13}"
    },
    "fields": {
      "title": "An Explicit All-Input Linear Resource Witness",
      "stem": "Use natural n≥0, nonnegative integers a=6, b=13, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0).",
      "answer": {
        "kind": "expression",
        "value": "(6, 13, 19, 0, Yes, 6n+13≤19(n+1) for all n≥0)",
        "explanation": "The specified nonnegative-coefficient bound is an all-input algebraic witness, not a conclusion inferred from a few measured points. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (6, 13, 19, 0, Yes, 6n+13≤19(n+1) for all n≥0)."
      },
      "skills": [
        "An Explicit All-Input Linear Resource Witness"
      ],
      "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": "明示非負函數族。操作=asymptotic-linear-witness；資料={\"a\":6,\"b\":13}。請依 [a,b,C,N,valid,witness_inequality] 欄序回答。",
        "en": "Use natural n≥0, nonnegative integers a=6, b=13, f(n)=a*n+b and g(n)=n+1 in a fixed cost model. Choose C=max(1,a+b), N=0. Give (a, b, C, N, witness valid?, the exact inequality a*n+b≤C*(n+1) for all n≥0)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "fixed cost model;explicit nonnegative functions on natural n",
      "after": "O,Omega and Theta quantify positive constants over a full tail",
      "relation": "implies",
      "rule_id": "full_tail_not_finite_benchmark_scope",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先固定成本模型與函數族。O要求存在C>0與N，使所有n≥N都滿足上界；Ω為對應正下界，Θ兩者同時。有限量測只描述有限窗口。",
        "en": "Fix the cost model and function family first.O requires C>0 and N with an upper bound for every n≥N;Omega is the corresponding positive lower bound,and Theta requires both.Finite measurements describe only a finite window."
      },
      "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": "a+b for prescribed constant := sum(6, 13)",
      "after": "a+b for prescribed constant = 19",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依本步原始操作數及指定次序精確計算，不用近似值或替換候選。",
        "en": "Calculate exactly in the specified source-operand order without approximation or substituting a different candidate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 a+b for prescribed constant，不以小數近似替代。",
          "en": "Recompute a+b for prescribed constant with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "6",
          "13"
        ],
        "output": "19",
        "label": "a+b for prescribed constant"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "C-a := sub(19, 6)",
      "after": "C-a = 13",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依本步原始操作數及指定次序精確計算，不用近似值或替換候選。",
        "en": "Calculate exactly in the specified source-operand order without approximation or substituting a different candidate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 C-a，不以小數近似替代。",
          "en": "Recompute C-a with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sub",
        "inputs": [
          "19",
          "6"
        ],
        "output": "13",
        "label": "C-a"
      }
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "C-b := sub(19, 13)",
      "after": "C-b = 6",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依本步原始操作數及指定次序精確計算，不用近似值或替換候選。",
        "en": "Calculate exactly in the specified source-operand order without approximation or substituting a different candidate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 C-b，不以小數近似替代。",
          "en": "Recompute C-b with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sub",
        "inputs": [
          "19",
          "13"
        ],
        "output": "6",
        "label": "C-b"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "a=6;b=13;C=max(1,a+b)=19;N=0",
      "after": "C(n+1)-(a*n+b)=(13)*n+(6)>=0 for every natural n",
      "relation": "implies",
      "rule_id": "linear_all_input_nonnegative_residual",
      "conditions": [],
      "explanation": {
        "zh-Hant": "C≥a且C≥b，所以差式兩個係數都非負，n≥0時整式非負。這涵蓋所有自然n，不是挑幾點；a=b=0時仍選正C=1。",
        "en": "C≥a and C≥b make both residual coefficients nonnegative for every natural n.This covers all n rather than selected points;when a=b=0,C=1 remains positive."
      },
      "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": [
      6,
      13,
      19,
      0,
      true,
      "6n+13≤19(n+1) 對所有 n≥0"
    ]
  },
  "conclusion": {
    "canonical_value": "(6, 13, 19, 0, 是, 6n+13≤19(n+1) 對所有 n≥0)",
    "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."
  },
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      "original_line_sha256": "ca2c9ff121d623309fe515a2a3827c26b4f05b4a269ada7a4fe6c8396bcc0367",
      "question_compact_sha256": "c88c4e54c7c156993e951f087939f5d00dcef1a6ecc5a2c26673b845ebb88d2c",
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      "english_manifest_path": "public/data/i18n/editions/batch-20261013-advance-20261019-r06-asymptotic-resource-bound-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261013-advance-20261019-r06-asymptotic-resource-bound-foundations.en.jsonl"
    },
    "mko_bindings": {
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        "original_path": "public/data/mko/asymptotic-resource-bound.json",
        "english_path": "public/data/i18n/editions/mko-asymptotic-resource-bound.en.json"
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    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "20e59f30c0b6218e6d9614c16170d770e56a2f1e4ae5ce39d0a8a3db206afa61",
    "original_candidate_sha256": "a47bc14399f756a9713b54ea481115a1cc94001007ca031fc278c3a7d6519fd4",
    "original_candidate_line_sha256": "a47bc14399f756a9713b54ea481115a1cc94001007ca031fc278c3a7d6519fd4",
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      "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-20261026-006-p18",
    "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": {
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    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 5,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
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    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
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}