第 24354 題：有限流網路

原題與已知條件

finite loop-free directed network、逐edge capacity與逐internal-vertex conservation。操作=flow-network-qualification；資料={"vertices":["s3","a3","t3","u3"],"edges":[["s3","a3"],["a3","s3"],["a3","t3"],["t3","a3"]],"source":"s3","sink":"t3","capacities":[20,20,20,20],"flows":[16,8,12,4]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — finite loop-free directed network、逐edge capacity與逐internal-vertex conservation。操作=flow-network-qualification；資料={"vertices":["s3","a3","t3","u3"],"edges":[["s3","a3"],["a3","s3"],["a3","t3"],["t3","a3"]],"source":"s3","sink":"t3","capacities":[20,20,20,20],"flows":[16,8,12,4]}。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-finite-flow-network

逐步推導

source roles = [flow-network-qualification]

derived = [[vertices, source, sink, edge_capacity_flow_rows, antiparallel_pair_count, network_qualified, convention]]

source roles = [[s3, a3, t3, u3], [[s3, a3], [a3, s3], [a3, t3], [t3, a3]], [20, 20, 20, 20], [16, 8, 12, 4], s3, t3]

derived = [[a3, s3, t3, u3], [[[a3, s3], 20, 8], [[a3, t3], 20, 12], [[s3, a3], 20, 16], [[t3, a3], 20, 4]], 2, 是]

source roles = [0, [a3, s3], 8, 20]

derived = [是, 是, 是]

source roles = [1, [a3, t3], 12, 20]

derived = [是, 是, 是]

source roles = [2, [s3, a3], 16, 20]

derived = [是, 是, 是]

source roles = [3, [t3, a3], 4, 20]

derived = [是, 是, 是]

Flow ALL incoming at vertex a3 := sum(16, 4)

Flow ALL incoming at vertex a3 = 20

Flow ALL outgoing at vertex a3 := sum(8, 12)

Flow ALL outgoing at vertex a3 = 20

source roles = [a3, [2, 3], [0, 1]]

derived = [20, 20, 是, 是]

Flow ALL incoming at vertex s3 := sum(8)

Flow ALL incoming at vertex s3 = 8

Flow ALL outgoing at vertex s3 := sum(16)

Flow ALL outgoing at vertex s3 = 16

source roles = [s3, [0], [2]]

derived = [8, 16, 否, 否]

Flow ALL incoming at vertex t3 := sum(12)

Flow ALL incoming at vertex t3 = 12

Flow ALL outgoing at vertex t3 := sum(4)

Flow ALL outgoing at vertex t3 = 4

source roles = [t3, [1], [3]]

derived = [12, 4, 否, 否]

Flow ALL incoming at vertex u3 := sum()

Flow ALL incoming at vertex u3 = 0

Flow ALL outgoing at vertex u3 := sum()

Flow ALL outgoing at vertex u3 = 0

source roles = [u3, [], []]

derived = [0, 0, 是, 是]

Flow source NET outflow not gross outgoing := sub(16, 8)

Flow source NET outflow not gross outgoing = 8

Flow sink NET inflow not gross incoming := sub(12, 4)

Flow sink NET inflow not gross incoming = 8

source roles = [s3, t3, [[[a3, s3], 8, 20, 是], [[a3, t3], 12, 20, 是], [[s3, a3], 16, 20, 是], [[t3, a3], 4, 20, 是]], [[a3, 20, 20, 是], [u3, 0, 0, 是]], 8, 8]

derived = [是, 是, 是, 是, 否]

逐步反向核對

• S021 · 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。
• S020 · premise_and_result_audit — 以精確有理數重算 Flow sink NET inflow not gross incoming，不以小數近似替代。
• S019 · premise_and_result_audit — 以精確有理數重算 Flow source NET outflow not gross outgoing，不以小數近似替代。
• S018 · 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。
• S017 · premise_and_result_audit — 以精確有理數重算 Flow ALL outgoing at vertex u3，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 Flow ALL incoming at vertex u3，不以小數近似替代。
• S015 · 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。
• S014 · premise_and_result_audit — 以精確有理數重算 Flow ALL outgoing at vertex t3，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 Flow ALL incoming at vertex t3，不以小數近似替代。
• S012 · 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。
• S011 · premise_and_result_audit — 以精確有理數重算 Flow ALL outgoing at vertex s3，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 Flow ALL incoming at vertex s3，不以小數近似替代。
• S009 · 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。
• S008 · premise_and_result_audit — 以精確有理數重算 Flow ALL outgoing at vertex a3，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 Flow ALL incoming at vertex a3，不以小數近似替代。
• S006 · 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。
• S005 · 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。
• S004 · 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。
• S003 · 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。
• 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

([a3, s3, t3, u3], s3, t3, [[[a3, s3], 20, 8], [[a3, t3], 20, 12], [[s3, a3], 20, 16], [[t3, a3], 20, 4]], 2, 是, finite_loop_free_simple_digraph_with_explicit_antiparallel_convention_zero_capacities_and_disconnected_vertices_allowed)

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。