# 正規語言基礎 · 脈衝 R07

本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。

本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。

batch-20261014-advance-20261102-r07-regular-language-foundations · 2026-10-14 · 100 題



## 1. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":2,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1], [[0, 1], [1, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1], [[0, 1], [1, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0001/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 2. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":3,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2], [[0, 1], [1, 2], [2, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2], [[0, 1], [1, 2], [2, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0002/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 3. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":4,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3], [[0, 1], [1, 2], [2, 3], [3, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3], [[0, 1], [1, 2], [2, 3], [3, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0003/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 4. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":5,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0004/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 5. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":6,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0005/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 6. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":7,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0006/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 7. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":8,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0007/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 8. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":9,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0008/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 9. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":10,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0009/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 10. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":11,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0010/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 11. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":12,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0011/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 12. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":13,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0012/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 13. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":14,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0013/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 14. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":15,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0014/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 15. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":16,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0015/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 16. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":17,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0016/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 17. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":18,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0017/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 18. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":19,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 18], [18, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 18], [18, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0018/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 19. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":20,"residue":0,"target_symbol":1,"alphabet_count":2}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [19, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [19, 0]], 0, [0], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0019/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 20. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":2,"residue":1,"target_symbol":1,"alphabet_count":3}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1], [[0, 1, 0], [1, 0, 1]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1], [[0, 1, 0], [1, 0, 1]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0020/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 21. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":3,"residue":1,"target_symbol":1,"alphabet_count":3}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2], [[0, 1, 0], [1, 2, 1], [2, 0, 2]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2], [[0, 1, 0], [1, 2, 1], [2, 0, 2]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0021/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 22. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":4,"residue":1,"target_symbol":1,"alphabet_count":3}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 0, 3]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 0, 3]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0022/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 23. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":5,"residue":1,"target_symbol":1,"alphabet_count":3}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 0, 4]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 0, 4]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0023/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 24. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":6,"residue":1,"target_symbol":1,"alphabet_count":3}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 5, 4], [5, 0, 5]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 5, 4], [5, 0, 5]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0024/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 25. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-mod-count-certificate；資料={"modulus":7,"residue":1,"target_symbol":1,"alphabet_count":3}。請依 [states,transition,start,accepting,regular,universal_invariant] 欄序回答。

查看答案與解釋([0, 1, 2, 3, 4, 5, 6], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 5, 4], [5, 6, 5], [6, 0, 6]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

Wave5第二層核對=([0, 1, 2, 3, 4, 5, 6], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 5, 4], [5, 6, 5], [6, 0, 6]], 0, [1], 是, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0025/)

正規語言 · qf-mod33-regular-mod-count-certificate · candidate





## 26. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":2,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[0,0,0,0,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], 0, 是)

Wave5第二層核對=([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], 0, 是)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0026/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 27. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":3,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[1,0,0,0,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 0, 0, 0, 0], [0, 1, 1, 1, 1, 1], 1, 否)

Wave5第二層核對=([1, 0, 0, 0, 0], [0, 1, 1, 1, 1, 1], 1, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0027/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 28. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":4,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[0,1,0,0,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 1, 0, 0, 0], [0, 0, 1, 1, 1, 1], 1, 否)

Wave5第二層核對=([0, 1, 0, 0, 0], [0, 0, 1, 1, 1, 1], 1, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0028/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 29. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":5,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[1,1,0,0,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 1, 0, 0, 0], [0, 1, 2, 2, 2, 2], 2, 否)

Wave5第二層核對=([1, 1, 0, 0, 0], [0, 1, 2, 2, 2, 2], 2, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0029/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 30. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":6,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[0,0,1,0,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 0, 1, 0, 0], [0, 0, 0, 1, 1, 1], 1, 否)

Wave5第二層核對=([0, 0, 1, 0, 0], [0, 0, 0, 1, 1, 1], 1, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0030/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 31. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":7,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[1,0,1,0,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 0, 1, 0, 0], [0, 1, 1, 2, 2, 2], 2, 否)

Wave5第二層核對=([1, 0, 1, 0, 0], [0, 1, 1, 2, 2, 2], 2, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0031/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 32. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":8,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[0,1,1,0,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 1, 1, 0, 0], [0, 0, 1, 2, 2, 2], 2, 否)

Wave5第二層核對=([0, 1, 1, 0, 0], [0, 0, 1, 2, 2, 2], 2, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0032/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 33. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":9,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[1,1,1,0,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 1, 1, 0, 0], [0, 1, 2, 3, 3, 3], 3, 否)

Wave5第二層核對=([1, 1, 1, 0, 0], [0, 1, 2, 3, 3, 3], 3, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0033/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 34. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":10,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[0,0,0,1,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 1], 1, 否)

Wave5第二層核對=([0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 1], 1, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0034/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 35. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":11,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[1,0,0,1,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 0, 0, 1, 0], [0, 1, 1, 1, 2, 2], 2, 否)

Wave5第二層核對=([1, 0, 0, 1, 0], [0, 1, 1, 1, 2, 2], 2, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0035/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 36. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":12,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[0,1,0,1,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 1, 0, 1, 0], [0, 0, 1, 1, 2, 2], 2, 否)

Wave5第二層核對=([0, 1, 0, 1, 0], [0, 0, 1, 1, 2, 2], 2, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0036/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 37. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":13,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[1,1,0,1,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 1, 0, 1, 0], [0, 1, 2, 2, 3, 3], 3, 否)

Wave5第二層核對=([1, 1, 0, 1, 0], [0, 1, 2, 2, 3, 3], 3, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0037/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 38. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":14,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[0,0,1,1,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 0, 1, 1, 0], [0, 0, 0, 1, 2, 2], 2, 否)

Wave5第二層核對=([0, 0, 1, 1, 0], [0, 0, 0, 1, 2, 2], 2, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0038/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 39. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":15,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[1,0,1,1,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 0, 1, 1, 0], [0, 1, 1, 2, 3, 3], 3, 否)

Wave5第二層核對=([1, 0, 1, 1, 0], [0, 1, 1, 2, 3, 3], 3, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0039/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 40. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":16,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[0,1,1,1,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 1, 1, 1, 0], [0, 0, 1, 2, 3, 3], 3, 否)

Wave5第二層核對=([0, 1, 1, 1, 0], [0, 0, 1, 2, 3, 3], 3, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0040/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 41. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":17,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[1,1,1,1,0]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 1, 1, 1, 0], [0, 1, 2, 3, 4, 4], 4, 否)

Wave5第二層核對=([1, 1, 1, 1, 0], [0, 1, 2, 3, 4, 4], 4, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0041/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 42. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":18,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[0,0,0,0,1]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], 1, 否)

Wave5第二層核對=([0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], 1, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0042/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 43. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":19,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[1,0,0,0,1]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 0, 0, 0, 1], [0, 1, 1, 1, 1, 2], 2, 否)

Wave5第二層核對=([1, 0, 0, 0, 1], [0, 1, 1, 1, 1, 2], 2, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0043/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 44. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":20,"residue":0,"target_symbol":1,"alphabet_count":2,"word":[0,1,0,0,1]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 1, 0, 0, 1], [0, 0, 1, 1, 1, 2], 2, 否)

Wave5第二層核對=([0, 1, 0, 0, 1], [0, 0, 1, 1, 1, 2], 2, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0044/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 45. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":2,"residue":1,"target_symbol":1,"alphabet_count":3,"word":[1,1,0,0,1]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 1, 0, 0, 1], [0, 1, 0, 0, 0, 1], 1, 是)

Wave5第二層核對=([1, 1, 0, 0, 1], [0, 1, 0, 0, 0, 1], 1, 是)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0045/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 46. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":3,"residue":1,"target_symbol":1,"alphabet_count":3,"word":[0,0,1,0,1]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 0, 1, 0, 1], [0, 0, 0, 1, 1, 2], 2, 否)

Wave5第二層核對=([0, 0, 1, 0, 1], [0, 0, 0, 1, 1, 2], 2, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0046/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 47. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":4,"residue":1,"target_symbol":1,"alphabet_count":3,"word":[1,0,1,0,1]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 0, 1, 0, 1], [0, 1, 1, 2, 2, 3], 3, 否)

Wave5第二層核對=([1, 0, 1, 0, 1], [0, 1, 1, 2, 2, 3], 3, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0047/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 48. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":5,"residue":1,"target_symbol":1,"alphabet_count":3,"word":[0,1,1,0,1]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 1, 1, 0, 1], [0, 0, 1, 2, 2, 3], 3, 否)

Wave5第二層核對=([0, 1, 1, 0, 1], [0, 0, 1, 2, 2, 3], 3, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0048/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 49. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":6,"residue":1,"target_symbol":1,"alphabet_count":3,"word":[1,1,1,0,1]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([1, 1, 1, 0, 1], [0, 1, 2, 3, 3, 4], 4, 否)

Wave5第二層核對=([1, 1, 1, 0, 1], [0, 1, 2, 3, 3, 4], 4, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0049/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 50. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-word-membership；資料={"modulus":7,"residue":1,"target_symbol":1,"alphabet_count":3,"word":[0,0,0,1,1]}。請依 [word,state_trace,final_state,accepted] 欄序回答。

查看答案與解釋([0, 0, 0, 1, 1], [0, 0, 0, 0, 1, 2], 2, 否)

Wave5第二層核對=([0, 0, 0, 1, 1], [0, 0, 0, 0, 1, 2], 2, 否)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0050/)

正規語言 · qf-mod33-regular-word-membership · candidate





## 51. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,0,0,0,0]],"probe":[1,0,0,0,0,0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 0, 0, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 0, 0, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0051/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 52. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,0,0,0,0]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 0], [1, 1, 0, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 0], [1, 1, 0, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0052/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 53. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,1,0,0,0]],"probe":[1,0,1,0,0,0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 0], [1, 0, 1, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 0, 0, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 0], [1, 0, 1, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 0, 0, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0053/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 54. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,1,0,0,0]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0054/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 55. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,0,1,0,0]],"probe":[1,0,0,1,0,0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 0], [1, 0, 0, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 1, 0, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 0], [1, 0, 0, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 1, 0, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0055/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 56. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,0,1,0,0]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 0], [1, 1, 0, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 0], [1, 1, 0, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0056/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 57. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,1,1,0,0]],"probe":[1,0,1,1,0,0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 0], [1, 0, 1, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 1, 0, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 0], [1, 0, 1, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 1, 0, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0057/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 58. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,1,1,0,0]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 0], [1, 1, 1, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 0], [1, 1, 1, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0058/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 59. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,0,0,1,0]],"probe":[1,0,0,0,1,0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 1], [1, 0, 0, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 0, 1, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 1], [1, 0, 0, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 0, 1, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0059/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 60. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,0,0,1,0]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 1], [1, 1, 0, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 1], [1, 1, 0, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0060/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 61. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,1,0,1,0]],"probe":[1,0,1,0,1,0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 1], [1, 0, 1, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 0, 1, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 1], [1, 0, 1, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 0, 1, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0061/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 62. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,1,0,1,0]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 1], [1, 1, 1, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 1], [1, 1, 1, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0062/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 63. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,0,1,1,0]],"probe":[1,0,0,1,1,0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 1], [1, 0, 0, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 1, 1, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 1], [1, 0, 0, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 1, 1, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0063/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 64. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,0,1,1,0]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 1], [1, 1, 0, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 1], [1, 1, 0, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0064/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 65. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,1,1,1,0]],"probe":[1,0,1,1,1,0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 1], [1, 0, 1, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 1, 1, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 1], [1, 0, 1, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 1, 1, 0], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0065/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 66. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,1,1,1,0]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0066/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 67. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,0,0,0,1]],"probe":[1,0,0,0,0,1]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 0, 0, 1], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 0, 0, 1], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0067/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 68. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,0,0,0,1]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 0], [1, 1, 0, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 0], [1, 1, 0, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0068/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 69. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,1,0,0,1]],"probe":[1,0,1,0,0,1]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 0], [1, 0, 1, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 1, 0, 0, 1], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 0], [1, 0, 1, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 1, 0, 0, 1], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0069/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 70. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,1,0,0,1]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0070/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 71. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,0,1,0,1]],"probe":[1,0,0,1,0,1]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 0], [1, 0, 0, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 1, 0, 1], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 0], [1, 0, 0, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 1, 0, 1], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0071/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 72. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,0,1,0,1]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0072/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 73. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,1,1,0,1]],"probe":[1,0,1,1,0,1]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 0], [1, 0, 1, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 1, 1, 0, 1], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 0], [1, 0, 1, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 1, 1, 0, 1], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0073/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 74. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,1,1,1,0,1]],"probe":[0]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 0], [1, 1, 1, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 0], [1, 1, 1, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0074/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 75. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-language-construction；資料={"alphabet_count":2,"words":[[0],[1,0,0,0,1,1]],"probe":[1,0,0,0,1,1]}。請依 [prefix_states,dead_state,accepting,transition,probe,trace,accepted,certificate] 欄序回答。

查看答案與解釋([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 1], [1, 0, 0, 0, 1, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [8, 6], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 0, 1, 1], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)

Wave5第二層核對=([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 1], [1, 0, 0, 0, 1, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [8, 6], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 0, 1, 1], [0, 2, 3, 4, 5, 6, 7], 是, explicit_DFA_recognizes_exact_finite_language)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0075/)

正規語言 · qf-mod33-regular-finite-language-construction · candidate





## 76. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":2,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[0,0,0,0,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[0, 0, 0, 0, 0], 是]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[0, 0, 0, 0, 0], 是]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0076/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 77. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":3,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[1,0,0,0,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[1, 0, 0, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[1, 0, 0, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0077/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 78. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":4,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[0,1,0,0,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[0, 1, 0, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[0, 1, 0, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0078/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 79. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":5,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[1,1,0,0,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[1, 1, 0, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[1, 1, 0, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0079/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 80. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":6,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[0,0,1,0,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[0, 0, 1, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[0, 0, 1, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0080/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 81. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":7,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[1,0,1,0,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[1, 0, 1, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[1, 0, 1, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0081/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 82. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":8,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[0,1,1,0,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[0, 1, 1, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[0, 1, 1, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0082/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 83. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":9,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[1,1,1,0,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[1, 1, 1, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[1, 1, 1, 0, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0083/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 84. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":10,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[0,0,0,1,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[0, 0, 0, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[0, 0, 0, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0084/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 85. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":11,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[1,0,0,1,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[1, 0, 0, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[1, 0, 0, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0085/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 86. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":12,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[0,1,0,1,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[0, 1, 0, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[0, 1, 0, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0086/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 87. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":13,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[1,1,0,1,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[1, 1, 0, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[1, 1, 0, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0087/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 88. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":14,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[0,0,1,1,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[0, 0, 1, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[0, 0, 1, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0088/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 89. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":15,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[1,0,1,1,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[1, 0, 1, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[1, 0, 1, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0089/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 90. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":16,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[0,1,1,1,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[0, 1, 1, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[0, 1, 1, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0090/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 91. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":17,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[1,1,1,1,0]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[1, 1, 1, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[1, 1, 1, 1, 0], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0091/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 92. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":18,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[0,0,0,0,1]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[0, 0, 0, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[0, 0, 0, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0092/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 93. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":19,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[1,0,0,0,1]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[1, 0, 0, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[1, 0, 0, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0093/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 94. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":20,"residue":0,"target_symbol":1,"alphabet_count":2,"test_words":[[],[0,1,0,0,1]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 是], [[0, 1, 0, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 是], [[0, 1, 0, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0094/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 95. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":2,"residue":1,"target_symbol":1,"alphabet_count":3,"test_words":[[],[1,1,0,0,1]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 否], [[1, 1, 0, 0, 1], 是]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 否], [[1, 1, 0, 0, 1], 是]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0095/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 96. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":3,"residue":1,"target_symbol":1,"alphabet_count":3,"test_words":[[],[0,0,1,0,1]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 否], [[0, 0, 1, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 否], [[0, 0, 1, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0096/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 97. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":4,"residue":1,"target_symbol":1,"alphabet_count":3,"test_words":[[],[1,0,1,0,1]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 否], [[1, 0, 1, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 否], [[1, 0, 1, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0097/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 98. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":5,"residue":1,"target_symbol":1,"alphabet_count":3,"test_words":[[],[0,1,1,0,1]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 否], [[0, 1, 1, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 否], [[0, 1, 1, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0098/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 99. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":6,"residue":1,"target_symbol":1,"alphabet_count":3,"test_words":[[],[1,1,1,0,1]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 否], [[1, 1, 1, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 否], [[1, 1, 1, 0, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0099/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate





## 100. 正規語言

完整DFA總轉移與任意word invariant或明示有限語言構造。操作=regular-finite-test-boundary；資料={"modulus":7,"residue":1,"target_symbol":1,"alphabet_count":3,"test_words":[[],[0,0,0,1,1]]}。請依 [tested_words,sufficient,boundary] 欄序回答。

查看答案與解釋([[[], 否], [[0, 0, 0, 1, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)

Wave5第二層核對=([[[], 否], [[0, 0, 0, 1, 1], 否]], 否, finite_test_suite_not_regular_language_or_equivalence_certificate)；依完整明示資料重算；mod-count證書須涵蓋任意word length而不只是測試集；不引入regex、NFA、epsilon、minimization或pumping lemma；有限測試不足。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r07-regular-language-0100/)

正規語言 · qf-mod33-regular-finite-test-boundary · candidate



[原始題目 JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261102-r07-regular-language-foundations/questions.jsonl) · [驗收報告](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261102-r07-regular-language-foundations/validation-report.json)