第 26634 題：道路連通空間

原題與已知條件

unit-interval path的continuity、兩端點與全range containment責任。操作=path-concatenation；資料={"left":-11,"right":12,"start":-8,"middle":10,"end":8}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — unit-interval path的continuity、兩端點與全range containment責任。操作=path-concatenation；資料={"left":-11,"right":12,"start":-8,"middle":10,"end":8}。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-path-connected-space

逐步推導

source roles = [path-concatenation]

derived = [[start, middle, end, first_branch, second_branch, join, continuous_path]]

Path first ORIGINAL middle minus start := sub(10, -8)

Path first ORIGINAL middle minus start = 18

Path second ORIGINAL end minus middle := sub(8, 10)

Path second ORIGINAL end minus middle = -2

Path first branch JOIN at half := sum(-8, 18)

Path first branch JOIN at half = 10

Path second branch JOIN at half := sum(10, 0)

Path second branch JOIN at half = 10

source roles = [-11, 12, -8, 10, 8, 18, -2]

derived = [first weight=2t in[0,1], second weight=2t-1 in[0,1], 10, 10, 是]

逐步反向核對

• S006 · premise_and_result_audit — 反查原題的完整資料、型別、欄序與本步條件；有限核對不升格為無限定理的形式證明。
• S005 · premise_and_result_audit — 以精確有理數重算 Path second branch JOIN at half，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 Path first branch JOIN at half，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 Path second ORIGINAL end minus middle，不以小數近似替代。
• S002 · premise_and_result_audit — 以精確有理數重算 Path first ORIGINAL middle minus start，不以小數近似替代。
• S001 · premise_and_result_audit — 反查原題的完整資料、型別、欄序與本步條件；有限核對不升格為無限定理的形式證明。

結論

(-8, 10, 8, 0<=t<=1/2:-8+2t(18), 1/2<=t<=1:10+(2t-1)(-2), both_branches_meet_at_middle, 是)

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。