第 22453 題：下推自動機

原題與已知條件

transition=[from,input-or-null,required-top,to,replacement-bottom-to-top]的finite nondeterministic PDA。操作=pda-qualification；資料={"states":["push2","pop2","accept2"],"input_alphabet":["a2","b2"],"stack_alphabet":["Z2","A2"],"start_state":"push2","bottom_symbol":"Z2","final_states":["accept2"],"transitions":[["push2","a2","Z2","push2",["Z2","A2"]],["push2","a2","A2","push2",["A2","A2"]],["push2","b2","A2","pop2",[]],["pop2","b2","A2","pop2",[]],["push2",null,"Z2","accept2",["Z2"]],["pop2",null,"Z2","accept2",["Z2"]]]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — transition=[from,input-or-null,required-top,to,replacement-bottom-to-top]的finite nondeterministic PDA。操作=pda-qualification；資料={"states":["push2","pop2","accept2"],"input_alphabet":["a2","b2"],"stack_alphabet":["Z2","A2"],"start_state":"push2","bottom_symbol":"Z2","final_states":["accept2"],"transitions":[["push2","a2","Z2","push2",["Z2","A2"]],["push2","a2","A2","push2",["A2","A2"]],["push2","b2","A2","pop2",[]],["pop2","b2","A2","pop2",[]],["push2",null,"Z2","accept2",["Z2"]],["pop2",null,"Z2","accept2",["Z2"]]]}。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-pushdown-automaton

逐步推導

source operation=pda-qualification;original snapshot and mathematics retained

explicit ordered source output fields=[states, input_alphabet, stack_alphabet, start_state, bottom_symbol, final_states, transition_count, qualified, convention]

ORIGINAL transition index=0;[from,input_or_null,required_top,to,replacement_bottom_to_top]=[push2, a2, Z2, push2, [Z2, A2]];bottom=Z2

states/input/stack symbols legal;null input consumes0;top=bottom requires replacement first bottom,otherwise replacement contains NO bottom;ORIGINAL transition position retained

ORIGINAL transition index=1;[from,input_or_null,required_top,to,replacement_bottom_to_top]=[push2, a2, A2, push2, [A2, A2]];bottom=Z2

states/input/stack symbols legal;null input consumes0;top=bottom requires replacement first bottom,otherwise replacement contains NO bottom;ORIGINAL transition position retained

ORIGINAL transition index=2;[from,input_or_null,required_top,to,replacement_bottom_to_top]=[push2, b2, A2, pop2, []];bottom=Z2

states/input/stack symbols legal;null input consumes0;top=bottom requires replacement first bottom,otherwise replacement contains NO bottom;ORIGINAL transition position retained

ORIGINAL transition index=3;[from,input_or_null,required_top,to,replacement_bottom_to_top]=[pop2, b2, A2, pop2, []];bottom=Z2

states/input/stack symbols legal;null input consumes0;top=bottom requires replacement first bottom,otherwise replacement contains NO bottom;ORIGINAL transition position retained

ORIGINAL transition index=4;[from,input_or_null,required_top,to,replacement_bottom_to_top]=[push2, null, Z2, accept2, [Z2]];bottom=Z2

states/input/stack symbols legal;null input consumes0;top=bottom requires replacement first bottom,otherwise replacement contains NO bottom;ORIGINAL transition position retained

ORIGINAL transition index=5;[from,input_or_null,required_top,to,replacement_bottom_to_top]=[pop2, null, Z2, accept2, [Z2]];bottom=Z2

states/input/stack symbols legal;null input consumes0;top=bottom requires replacement first bottom,otherwise replacement contains NO bottom;ORIGINAL transition position retained

PDA complete ORIGINAL transition count := sum(1, 1, 1, 1, 1, 1)

PDA complete ORIGINAL transition count = 6

ORIGINAL states=[push2, pop2, accept2],input alphabet=[a2, b2],stack alphabet=[Z2, A2];start=push2,bottom=Z2,final states=[accept2]

finite unique pairwise-disjoint alphabets qualified;transition count=6;acceptance convention FULL input in final state,NOT empty stack;qualification alone proves NO run,determinism or CFG equivalence

逐步反向核對

• 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 PDA complete ORIGINAL transition count，不以小數近似替代。
• S007 · 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。
• 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

([push2, pop2, accept2], [a2, b2], [Z2, A2], push2, Z2, [accept2], 6, 是, accept_only_after_full_input_in_final_state_bottom_marker_preserved)

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。