第 25827 題：線性規劃對偶

原題與已知條件

固定max/≤/nonnegative primal與min/≥/nonnegative dual convention。操作=lpdual-strong-boundary；資料={"mode":"dual_infeasible_equal_objective","A":[[1]],"b":[1],"c":[3],"x":[[1,3]],"y":[1]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 固定max/≤/nonnegative primal與min/≥/nonnegative dual convention。操作=lpdual-strong-boundary；資料={"mode":"dual_infeasible_equal_objective","A":[[1]],"b":[1],"c":[3],"x":[[1,3]],"y":[1]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

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。

• mko-linear-program-duality

逐步推導

source roles = [lpdual-strong-boundary]

derived = [[mode, x, y, Ax, transpose_A_y, primal_feasible, dual_feasible, primal_objective, middle_value, dual_objective, first_inequality, second_inequality, zero_gap, unsupported_optimality_or_general_strong_duality_claim, boundary]]

source roles = [[[1]], [1], [3]]

derived = [1, 1, [[1]], primal:max_cTx_Ax_le_b_x_ge_0, dual:min_bTy_ATy_ge_c_y_ge_0, 是]

LPdual ORIGINAL Ax row0 product0 := mul(1, 1/3)

LPdual ORIGINAL Ax row0 product0 = 1/3

LPdual ORIGINAL Ax row0 sum := sum(1/3)

LPdual ORIGINAL Ax row0 sum = 1/3

LPdual ORIGINAL ATy row0 product0 := mul(1, 1)

LPdual ORIGINAL ATy row0 product0 = 1

LPdual ORIGINAL ATy row0 sum := sum(1)

LPdual ORIGINAL ATy row0 sum = 1

LPdual ORIGINAL primal objective product0 := mul(3, 1/3)

LPdual ORIGINAL primal objective product0 = 1

LPdual ORIGINAL primal objective sum := sum(1)

LPdual ORIGINAL primal objective sum = 1

LPdual ORIGINAL middle yTAx product0 := mul(1, 1/3)

LPdual ORIGINAL middle yTAx product0 = 1/3

LPdual ORIGINAL middle yTAx sum := sum(1/3)

LPdual ORIGINAL middle yTAx sum = 1/3

LPdual ORIGINAL dual objective product0 := mul(1, 1)

LPdual ORIGINAL dual objective product0 = 1

LPdual ORIGINAL dual objective sum := sum(1)

LPdual ORIGINAL dual objective sum = 1

source roles = [[1/3], [1], [1/3], [1], [1], [3]]

derived = [是, 否, 1, 1/3, 1, [[0, 0, 1, 1, 1/3, 1/3]], 1/3, [[0, 1/3, -2, -2/3]], -2/3, [[0, 1, 2/3, 2/3]], 2/3, 否, 是, 是]

source roles = [dual_infeasible_equal_objective, 是, 否, 1, 1, 是]

derived = [否]

逐步反向核對

• S014 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S013 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S012 · premise_and_result_audit — 以精確有理數重算 LPdual ORIGINAL dual objective sum，不以小數近似替代。
• S011 · premise_and_result_audit — 以精確有理數重算 LPdual ORIGINAL dual objective product0，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 LPdual ORIGINAL middle yTAx sum，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 LPdual ORIGINAL middle yTAx product0，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 LPdual ORIGINAL primal objective sum，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 LPdual ORIGINAL primal objective product0，不以小數近似替代。
• S006 · premise_and_result_audit — 以精確有理數重算 LPdual ORIGINAL ATy row0 sum，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 LPdual ORIGINAL ATy row0 product0，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 LPdual ORIGINAL Ax row0 sum，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 LPdual ORIGINAL Ax row0 product0，不以小數近似替代。
• S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

(dual_infeasible_equal_objective, [1/3], [1], [1/3], [1], 是, 否, 1, 1/3, 1, 否, 是, 是, 否, objective_equality_without_both_feasibilities_is_not_optimality_and_finite_grids_do_not_prove_general_strong_duality_or_handle_infeasible_unbounded_cases)

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。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。