{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0001","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25595ca971fbae33663ff2e077dbc4d46b19722ac3b8d95b5dbebf5c8e014b85","family_id":"qf-mod32-dfa-qualification","seed":202611017001,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[0,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0002","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c6ba5b85f583955dbc487c74730f4343dbbfccb93ff1bf11baba95a20240e7c","family_id":"qf-mod32-dfa-qualification","seed":202611017002,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[0,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0003","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c4a3cfed3ddaa792237a9410a55061caadab2075df308e1052aceeb3954a095","family_id":"qf-mod32-dfa-qualification","seed":202611017003,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[0,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0004","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8a760a2060cf3213010a2738910a1de66c9d6fe7b63dfd04560a95b396e6389","family_id":"qf-mod32-dfa-qualification","seed":202611017004,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[0,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0005","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4bd6ab063e303ea1b92ce26789d488c0be5dbc53e2413299f755e999979d6f2f","family_id":"qf-mod32-dfa-qualification","seed":202611017005,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[0,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0006","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e7a96bd0ae15ecfb01b615183cc5c5451d53cb52154f3ef3c954106532f7345","family_id":"qf-mod32-dfa-qualification","seed":202611017006,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[0,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0007","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4e3c649981ea88e06e620c883e047335f7c14319491f8c11fffe8b1464038509","family_id":"qf-mod32-dfa-qualification","seed":202611017007,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[0,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0008","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb467c34cc9d59a86c35dca6fb1b5df5999d286f069f1ef70d39d7f2aaf6e355","family_id":"qf-mod32-dfa-qualification","seed":202611017008,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,2],[0,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 2], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0009","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd6710071873810616ba634b45ca2e0ca0ecde2ed8b4fe2e16da88d05ee2409a","family_id":"qf-mod32-dfa-qualification","seed":202611017009,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,2],[0,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 2], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0010","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9ecfc8e4bf84fc84fca3a61dab1913b64c35a2f2c106b3160fa1ff1f4d4469d","family_id":"qf-mod32-dfa-qualification","seed":202611017010,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[1,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0011","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f06ba2d713705770971727c468e79b9b9433de9841124aafdc250a53239e3119","family_id":"qf-mod32-dfa-qualification","seed":202611017011,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[1,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0012","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26f5640d4b80d211c781d3fb18c6f15afb48a25655f786dcd95da55b34576ae1","family_id":"qf-mod32-dfa-qualification","seed":202611017012,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[1,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0013","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"598cc6d456d20e0b888a1936b26dcd8c5b818814ddbf01d0ded2e7c838ff0e6c","family_id":"qf-mod32-dfa-qualification","seed":202611017013,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[1,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0014","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"363d4ef13b808a6a106d07dbe780da9415e512a2c297c918fbf32b1041a8c6b2","family_id":"qf-mod32-dfa-qualification","seed":202611017014,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[1,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0015","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ea194c9b3295d6bcc4f27236bea096b68081858ad73b79c29c561cf9b4ae7b8","family_id":"qf-mod32-dfa-qualification","seed":202611017015,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[1,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0016","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"79b3353be46a1c143bdaf09b085f06574f546e87457202f7d8a9874dbf7e22f6","family_id":"qf-mod32-dfa-qualification","seed":202611017016,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[1,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0017","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"190e876cfc9b151ba05004d77012c94ef25ed9aea97b292d74586d3617675be6","family_id":"qf-mod32-dfa-qualification","seed":202611017017,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,2],[1,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 2], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0018","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7581eff3d4a2b7dff50d807a1b5ca41ed5c23ccd0d503680afe8997cbab436cf","family_id":"qf-mod32-dfa-qualification","seed":202611017018,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,2],[1,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 2], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0019","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d96083f6d07d4ade425bd8b53d0de4237d2da7aac5df66f36c23e426769d5fe","family_id":"qf-mod32-dfa-qualification","seed":202611017019,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[2,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0020","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e8bbe4ecc8ef9bd192c2d4cdd21530afd6dc1269343afde54999ed41b7eaea0","family_id":"qf-mod32-dfa-qualification","seed":202611017020,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[2,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [2, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0021","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"50d8fd0a3b8602a3c69151073f0aa980203304dc0f28e8b77b3a707efc137374","family_id":"qf-mod32-dfa-qualification","seed":202611017021,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[2,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [2, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0022","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2f37e8e1051810397265d226f5ac4956553bf98bdeb8ab1a41d654127ebc05e","family_id":"qf-mod32-dfa-qualification","seed":202611017022,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[2,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0023","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5473138d724cd933609ec5b4808092763cab7f35daf5854f4c20cef4186f876b","family_id":"qf-mod32-dfa-qualification","seed":202611017023,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[2,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [2, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0024","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"218cf8f41ed5384d421a90b94ec5af677cb7586c7c869c768552f972f9fb66ea","family_id":"qf-mod32-dfa-qualification","seed":202611017024,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[2,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [2, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0025","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"32d0a04935b263cd957db1b2788e3438aa157813bdef888160e5f69b560e45d0","family_id":"qf-mod32-dfa-qualification","seed":202611017025,"semantic_template":"qf-mod32-dfa-qualification/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[2,0],[0,0]]}"},"fields":{"title":"A DFA Needs a Complete Total Deterministic Transition Table","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Give (state_count,alphabet_count,start,accepting,transition_cell_count,qualified?,complete_total_deterministic_table).","answer":{"kind":"expression","value":"(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)","explanation":"Check every state-symbol cell, the unique start and the accepting subset. A partial or branching table is rejected, not silently completed or interpreted as an NFA. The exact result, retaining the requested order, is (3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)."},"skills":["A DFA Needs a Complete Total Deterministic Transition Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0026","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bcf49aa7c9d0153f07dd3ee8d8a68e205730038d26283bbe9375d4d4b6304887","family_id":"qf-mod32-dfa-word-trace","seed":202611017026,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[0,0],[0,0]],\"word\":[0,0,0,0,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 0, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0027","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f24af4546180302187c3577e9d79524bcb03dfe6554390515e4e9a09038b545e","family_id":"qf-mod32-dfa-word-trace","seed":202611017027,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[0,0],[0,0]],\"word\":[1,0,0,0,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 0, 0, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 0, 0, 0, 0], [1, 0, 1, 0, 1, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 0, 0, 0, 0], [1, 0, 1, 0, 1, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0028","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"951c4c8e79c85659ed0aa4e2041fd37a46a355d993235951f9782b99854d3cb5","family_id":"qf-mod32-dfa-word-trace","seed":202611017028,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[0,0],[0,0]],\"word\":[0,1,0,0,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 0, 0, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 1, 0, 0, 0], [2, 0, 0, 2, 0, 2], 2)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 1, 0, 0, 0], [2, 0, 0, 2, 0, 2], 2)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0029","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"333eb1332966521e3d6c3490e0c44622c8737e78e3669b0e2fed71c7d017f02b","family_id":"qf-mod32-dfa-word-trace","seed":202611017029,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[0,0],[0,0]],\"word\":[1,1,0,0,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 0, 0, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0030","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ee92a52a6db3a323ffee4e772f977dfc3e60e9fb2fe7f0e05e0c5861b97400f","family_id":"qf-mod32-dfa-word-trace","seed":202611017030,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[0,0],[0,0]],\"word\":[0,0,1,0,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 1, 0, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 0, 1, 0, 0], [1, 0, 1, 0, 1, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 0, 1, 0, 0], [1, 0, 1, 0, 1, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0031","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5a9f7419cfb2fe450ae47ff0bf6d27e6170480cc0f2a62a58111b5536a3c0a3","family_id":"qf-mod32-dfa-word-trace","seed":202611017031,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[0,0],[0,0]],\"word\":[1,0,1,0,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 1, 0, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 0, 1, 0, 0], [2, 0, 2, 0, 2, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 0, 1, 0, 0], [2, 0, 2, 0, 2, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0032","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b72af35c169accb6e5b67b458c37509c2d0e899c2adc057776fef956ac6516c0","family_id":"qf-mod32-dfa-word-trace","seed":202611017032,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[0,0],[0,0]],\"word\":[0,1,1,0,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 1, 0, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 1, 1, 0, 0], [0, 0, 2, 0, 0, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 1, 1, 0, 0], [0, 0, 2, 0, 0, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0033","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3723fbc2987405e7005807fb487318d8a8f63f3250ecffabf741c8d65aef8307","family_id":"qf-mod32-dfa-word-trace","seed":202611017033,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,2],[0,0],[0,0]],\"word\":[1,1,1,0,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 2], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 1, 0, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 1, 1, 0, 0], [1, 0, 2, 0, 1, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 1, 1, 0, 0], [1, 0, 2, 0, 1, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0034","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c53965269c8cbf87ae3668978369276bfb019ac8abdd36f89d08ff03185b5bc3","family_id":"qf-mod32-dfa-word-trace","seed":202611017034,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,2],[0,0],[0,0]],\"word\":[0,0,0,1,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 2], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 1, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 0, 0, 1, 0], [2, 0, 2, 0, 2, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 0, 0, 1, 0], [2, 0, 2, 0, 2, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0035","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ade4c9d9a6c96458967a816a98d25e610dd2cb23db36420af5b39293f22443c","family_id":"qf-mod32-dfa-word-trace","seed":202611017035,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[1,0],[0,0]],\"word\":[1,0,0,1,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 0, 1, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0036","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e908df3dd35c816726b2f2aeb7c7681fe04d6bf7a71d1f25ca5c6316ef6674e","family_id":"qf-mod32-dfa-word-trace","seed":202611017036,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[1,0],[0,0]],\"word\":[0,1,0,1,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 0, 1, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1], 1)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1], 1)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0037","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e9aa35aac9f91c04fe508f258d5edc9e9ae0bf149946eaf54d95381faa79fa7","family_id":"qf-mod32-dfa-word-trace","seed":202611017037,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[1,0],[0,0]],\"word\":[1,1,0,1,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 0, 1, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 1, 0, 1, 0], [2, 0, 0, 2, 0, 2], 2)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 1, 0, 1, 0], [2, 0, 0, 2, 0, 2], 2)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0038","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"721acaa8407915deb72dd121a08e97c31ed780a7fd2b87c3ccc6084b6cde28ea","family_id":"qf-mod32-dfa-word-trace","seed":202611017038,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[1,0],[0,0]],\"word\":[0,0,1,1,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 1, 1, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 0, 1, 1, 0], [0, 0, 0, 1, 0, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 0, 1, 1, 0], [0, 0, 0, 1, 0, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0039","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d769adae519dca5d58a88bc4b31992521b5ddeffa535838e200ee6562a6abe61","family_id":"qf-mod32-dfa-word-trace","seed":202611017039,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[1,0],[0,0]],\"word\":[1,0,1,1,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 1, 1, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1], 1)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1], 1)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0040","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af072796b485fea5db715c4049a8a7cd1251dc8826ec15af61cc98f4fc3fff78","family_id":"qf-mod32-dfa-word-trace","seed":202611017040,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[1,0],[0,0]],\"word\":[0,1,1,1,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 1, 1, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 1, 1, 1, 0], [2, 0, 1, 0, 1, 1], 1)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 1, 1, 1, 0], [2, 0, 1, 0, 1, 1], 1)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0041","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f7c0bc058e332cc305535bff60b807ef39e29ac03b45030e75775b891f22d332","family_id":"qf-mod32-dfa-word-trace","seed":202611017041,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[1,0],[0,0]],\"word\":[1,1,1,1,0]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 1, 1, 0]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 1, 1, 1, 0], [0, 2, 0, 2, 0, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 1, 1, 1, 0], [0, 2, 0, 2, 0, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0042","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"77bc7d0de2f9fa0e7fd0cba2e34f62008777c0dd1993b0eb3c9c3f115183e5fe","family_id":"qf-mod32-dfa-word-trace","seed":202611017042,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,2],[1,0],[0,0]],\"word\":[0,0,0,0,1]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 2], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 0, 1]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0043","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2dd65cade6813af72169704342d9f86b1c9b5133455de5a4e94a81c6a156d2c4","family_id":"qf-mod32-dfa-word-trace","seed":202611017043,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,2],[1,0],[0,0]],\"word\":[1,0,0,0,1]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 2], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 0, 0, 1]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 0, 0, 0, 1], [2, 0, 2, 0, 2, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 0, 0, 0, 1], [2, 0, 2, 0, 2, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0044","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"473a011c40c3d76f549d8f288014eeb6389907c0a8f4dc0e970351aa8775681b","family_id":"qf-mod32-dfa-word-trace","seed":202611017044,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[2,0],[0,0]],\"word\":[0,1,0,0,1]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 0, 0, 1]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 1, 0, 0, 1], [0, 0, 0, 0, 0, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 1, 0, 0, 1], [0, 0, 0, 0, 0, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0045","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f3894243e798f59659dcf31298de16582fbd4ae56f1e534335b7618b469978f","family_id":"qf-mod32-dfa-word-trace","seed":202611017045,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[2,0],[0,0]],\"word\":[1,1,0,0,1]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [2, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 0, 0, 1]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 1, 0, 0, 1], [1, 0, 0, 1, 2, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 1, 0, 0, 1], [1, 0, 0, 1, 2, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0046","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"99f046a2676479cd3b32e4cdede6884135bce791c509b6a632fc7a1e068c2ddb","family_id":"qf-mod32-dfa-word-trace","seed":202611017046,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[2,0],[0,0]],\"word\":[0,0,1,0,1]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [2, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 1, 0, 1]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 0, 1, 0, 1], [2, 0, 2, 0, 2, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 0, 1, 0, 1], [2, 0, 2, 0, 2, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0047","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5013c70fa47f2eda648ca4e3b182ccb4f03e77d674c5e781d2f66d2737c7e41d","family_id":"qf-mod32-dfa-word-trace","seed":202611017047,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[2,0],[0,0]],\"word\":[1,0,1,0,1]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 1, 0, 1]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 0, 1, 0, 1], [0, 1, 2, 0, 0, 1], 1)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 0, 1, 0, 1], [0, 1, 2, 0, 0, 1], 1)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0048","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36985cd0512e7298d9685599d9765eb209047aa40ab85df9eeec121ed646bffd","family_id":"qf-mod32-dfa-word-trace","seed":202611017048,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[2,0],[0,0]],\"word\":[0,1,1,0,1]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [2, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 1, 0, 1]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 1, 1, 0, 1], [1, 2, 0, 1, 2, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 1, 1, 0, 1], [1, 2, 0, 1, 2, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0049","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"654da4515394a07481332176c0f9ed6e656255c8d4eb9e062c5938c14d925a03","family_id":"qf-mod32-dfa-word-trace","seed":202611017049,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[2,0],[0,0]],\"word\":[1,1,1,0,1]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [2, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 1, 0, 1]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([1, 1, 1, 0, 1], [2, 0, 1, 0, 2, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([1, 1, 1, 0, 1], [2, 0, 1, 0, 2, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0050","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c4ea320ca5a00748c60910e8bd842a2cb5f766328d395bccab7711231167eaa","family_id":"qf-mod32-dfa-word-trace","seed":202611017050,"semantic_template":"qf-mod32-dfa-word-trace/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[2,0],[0,0]],\"word\":[0,0,0,1,1]}"},"fields":{"title":"Retain the Initial State and Every Ordered Symbol Transition","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 1, 1]. Give (word,state_trace,final_state), with state_trace length equal to word length plus one.","answer":{"kind":"expression","value":"([0, 0, 0, 1, 1], [0, 0, 0, 0, 2, 0], 0)","explanation":"Start at the specified state and apply one table lookup per symbol. Include the initial state even for an empty word. A trace is not a language-equivalence proof. The exact result, retaining the requested order, is ([0, 0, 0, 1, 1], [0, 0, 0, 0, 2, 0], 0)."},"skills":["Retain the Initial State and Every Ordered Symbol Transition"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0051","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"426400c85f391bc554b5eeb6dfeafad30b633dc4fc207a09db529f34cef9e78e","family_id":"qf-mod32-dfa-acceptance","seed":202611017051,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[0,0],[0,0]],\"word\":[0,0,0,0,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 0, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], 0, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], 0, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0052","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"62f5cec1b1643705088caaeb09f463b9bf2dc9b9a159c888b9d1d29d52597f2c","family_id":"qf-mod32-dfa-acceptance","seed":202611017052,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[0,0],[0,0]],\"word\":[1,0,0,0,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 0, 0, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 0, 0, 0, 0], [1, 0, 1, 0, 1, 0], 0, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 0, 0, 0, 0], [1, 0, 1, 0, 1, 0], 0, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0053","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c45c6323dd4b22349138d08f420c58ebdc9af914e6cb494089bba03de4f01fb2","family_id":"qf-mod32-dfa-acceptance","seed":202611017053,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[0,0],[0,0]],\"word\":[0,1,0,0,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 0, 0, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 1, 0, 0, 0], [2, 0, 0, 2, 0, 2], 2, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 1, 0, 0, 0], [2, 0, 0, 2, 0, 2], 2, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0054","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36698cdd7b6702db5f52e1859deafb8012c4610a5c0d0b4520b6444f9d5f604f","family_id":"qf-mod32-dfa-acceptance","seed":202611017054,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[0,0],[0,0]],\"word\":[1,1,0,0,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 0, 0, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0], 0, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0], 0, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0055","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d23f963d73e70b0bd9213eff341c1e25b31dcee996bb109dd76df120923e318","family_id":"qf-mod32-dfa-acceptance","seed":202611017055,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[0,0],[0,0]],\"word\":[0,0,1,0,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 1, 0, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 0, 1, 0, 0], [1, 0, 1, 0, 1, 0], 0, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 0, 1, 0, 0], [1, 0, 1, 0, 1, 0], 0, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0056","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ebe2e00c37aacad17dd1641c65d31c028b529d8cdac5474c26351db7fc4bfba8","family_id":"qf-mod32-dfa-acceptance","seed":202611017056,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[0,0],[0,0]],\"word\":[1,0,1,0,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 1, 0, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 0, 1, 0, 0], [2, 0, 2, 0, 2, 0], 0, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 0, 1, 0, 0], [2, 0, 2, 0, 2, 0], 0, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0057","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b7698b955c4d6ed026f0139003fff04a562bf39af1f3db1656b9b1a4b84deed","family_id":"qf-mod32-dfa-acceptance","seed":202611017057,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[0,0],[0,0]],\"word\":[0,1,1,0,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 1, 0, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 1, 1, 0, 0], [0, 0, 2, 0, 0, 0], 0, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 1, 1, 0, 0], [0, 0, 2, 0, 0, 0], 0, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0058","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"22de1b6929876b4e55e1d7fb9c80b84099de31c0920e4bd11b77974e9fbca8fa","family_id":"qf-mod32-dfa-acceptance","seed":202611017058,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,2],[0,0],[0,0]],\"word\":[1,1,1,0,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 2], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 1, 0, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 1, 1, 0, 0], [1, 0, 2, 0, 1, 0], 0, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 1, 1, 0, 0], [1, 0, 2, 0, 1, 0], 0, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0059","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9c47fdff41da76bacf06bb3245d12245836288bbb9eb4d2113309f9372da404","family_id":"qf-mod32-dfa-acceptance","seed":202611017059,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,2],[0,0],[0,0]],\"word\":[0,0,0,1,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 2], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 1, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 0, 0, 1, 0], [2, 0, 2, 0, 2, 0], 0, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 0, 0, 1, 0], [2, 0, 2, 0, 2, 0], 0, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0060","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"725962505d52e527a56bdef85fb507ad289ec257979afc0ff413c00fd03f1fe1","family_id":"qf-mod32-dfa-acceptance","seed":202611017060,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[1,0],[0,0]],\"word\":[1,0,0,1,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 0, 1, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0], 0, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0], 0, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0061","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1cc9781d014fe6e8af4c59a4a8c73175234691c9e4f8fc99d9ec3da3aa755cc4","family_id":"qf-mod32-dfa-acceptance","seed":202611017061,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[1,0],[0,0]],\"word\":[0,1,0,1,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 0, 1, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1], 1, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1], 1, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0062","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39dd91cc913135aa0dcfa9750e7946e492778dcde5781d691e4fd1d6d0831a9b","family_id":"qf-mod32-dfa-acceptance","seed":202611017062,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[1,0],[0,0]],\"word\":[1,1,0,1,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 0, 1, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 1, 0, 1, 0], [2, 0, 0, 2, 0, 2], 2, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 1, 0, 1, 0], [2, 0, 0, 2, 0, 2], 2, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0063","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6622f72fd3d2176cfadc184a8f34c4e3735776345fd05f04be701983847ab585","family_id":"qf-mod32-dfa-acceptance","seed":202611017063,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[1,0],[0,0]],\"word\":[0,0,1,1,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 1, 1, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 0, 1, 1, 0], [0, 0, 0, 1, 0, 0], 0, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 0, 1, 1, 0], [0, 0, 0, 1, 0, 0], 0, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0064","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"816a0e00afb0b9176718cf037fb88379e2bdb84f6e7d92b7bafcfbed93b664df","family_id":"qf-mod32-dfa-acceptance","seed":202611017064,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[1,0],[0,0]],\"word\":[1,0,1,1,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 1, 1, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1], 1, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1], 1, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0065","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ae415582e6d5e86e6135fb6635fcd9083c183dc8f92a2be4d4e71e788b920e2","family_id":"qf-mod32-dfa-acceptance","seed":202611017065,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[1,0],[0,0]],\"word\":[0,1,1,1,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 1, 1, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 1, 1, 1, 0], [2, 0, 1, 0, 1, 1], 1, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 1, 1, 1, 0], [2, 0, 1, 0, 1, 1], 1, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0066","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5f14663f2b116530c9d8a8a8eec7b0671863addd33193370dcd0e221db8ca073","family_id":"qf-mod32-dfa-acceptance","seed":202611017066,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[1,0],[0,0]],\"word\":[1,1,1,1,0]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 1, 1, 0]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 1, 1, 1, 0], [0, 2, 0, 2, 0, 0], 0, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 1, 1, 1, 0], [0, 2, 0, 2, 0, 0], 0, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0067","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3363bc305dfe449e97fceeba31f7e3c15ce5ffb9a069fdbd851273d30957fe3","family_id":"qf-mod32-dfa-acceptance","seed":202611017067,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,2],[1,0],[0,0]],\"word\":[0,0,0,0,1]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 2], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 0, 1]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 0], 0, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 0], 0, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0068","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3aed8a97f7517f85dea44757d99170c00a596cac2e6efe842aedf06e8a30fec2","family_id":"qf-mod32-dfa-acceptance","seed":202611017068,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,2],[1,0],[0,0]],\"word\":[1,0,0,0,1]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 2], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 0, 0, 1]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 0, 0, 0, 1], [2, 0, 2, 0, 2, 0], 0, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 0, 0, 0, 1], [2, 0, 2, 0, 2, 0], 0, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0069","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"da4b220140bd4da68667f47a5c414eef2bef7ff5692fdcb6aab6c721bf601ee6","family_id":"qf-mod32-dfa-acceptance","seed":202611017069,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[2,0],[0,0]],\"word\":[0,1,0,0,1]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 0, 0, 1]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 1, 0, 0, 1], [0, 0, 0, 0, 0, 0], 0, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 1, 0, 0, 1], [0, 0, 0, 0, 0, 0], 0, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0070","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c9a3ff2447dc4bcfbd0fe256b373ecfe7ed8d08ccab46237fa3dc4a0c6bf323","family_id":"qf-mod32-dfa-acceptance","seed":202611017070,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[2,0],[0,0]],\"word\":[1,1,0,0,1]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [2, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 0, 0, 1]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 1, 0, 0, 1], [1, 0, 0, 1, 2, 0], 0, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 1, 0, 0, 1], [1, 0, 0, 1, 2, 0], 0, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0071","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09aa55ad223d4a30369a0cabdcfd212a8485cc572da7fa1c95cb67be39c76e92","family_id":"qf-mod32-dfa-acceptance","seed":202611017071,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[2,0],[0,0]],\"word\":[0,0,1,0,1]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [2, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 1, 0, 1]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 0, 1, 0, 1], [2, 0, 2, 0, 2, 0], 0, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 0, 1, 0, 1], [2, 0, 2, 0, 2, 0], 0, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0072","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7dd84aebe84719970d60b34a8042ab9ac15a6227eee77c51b28be3a229a5edf","family_id":"qf-mod32-dfa-acceptance","seed":202611017072,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[2,0],[0,0]],\"word\":[1,0,1,0,1]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 1, 0, 1]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 0, 1, 0, 1], [0, 1, 2, 0, 0, 1], 1, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 0, 1, 0, 1], [0, 1, 2, 0, 0, 1], 1, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0073","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"affdd7772c5656e46b2a161381be91c6e5d458ad0afbff349632d8cada09dbe2","family_id":"qf-mod32-dfa-acceptance","seed":202611017073,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[2,0],[0,0]],\"word\":[0,1,1,0,1]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [2, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 1, 0, 1]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 1, 1, 0, 1], [1, 2, 0, 1, 2, 0], 0, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 1, 1, 0, 1], [1, 2, 0, 1, 2, 0], 0, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0074","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1d0ff69b7861e56ee6eac76c55ba1fc75f6d2f09d7a4b8482c253d76487ec4c5","family_id":"qf-mod32-dfa-acceptance","seed":202611017074,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[2,0],[0,0]],\"word\":[1,1,1,0,1]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [2, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 1, 0, 1]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([1, 1, 1, 0, 1], [2, 0, 1, 0, 2, 0], 0, No)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([1, 1, 1, 0, 1], [2, 0, 1, 0, 2, 0], 0, No)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0075","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e6a6970e35ee878571217ea196cf8fd2b50b185ff740fe00cd9489776c9e18f9","family_id":"qf-mod32-dfa-acceptance","seed":202611017075,"semantic_template":"qf-mod32-dfa-acceptance/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[2,0],[0,0]],\"word\":[0,0,0,1,1]}"},"fields":{"title":"The Empty Word Is Accepted Exactly When the Start State Is Accepting","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 1, 1]. Give (word,state_trace,final_state,accepted?).","answer":{"kind":"expression","value":"([0, 0, 0, 1, 1], [0, 0, 0, 0, 2, 0], 0, Yes)","explanation":"Follow the complete ordered trace and test its final state in the accepting subset. For the empty word, perform no transition and test the start state directly. The exact result, retaining the requested order, is ([0, 0, 0, 1, 1], [0, 0, 0, 0, 2, 0], 0, Yes)."},"skills":["The Empty Word Is Accepted Exactly When the Start State Is Accepting"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0076","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff2094181ba4508e29a8832c79bc2ea7eff9373520032da80e9bf551771c33db","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017076,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[0,0],[0,0]],\"word\":[0,0,0,0,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 0, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0077","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"59e8e44eb42d176e8159daa7f48612824aaab10d44948ff86fb835e4a9a6677b","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017077,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[0,0],[0,0]],\"word\":[1,0,0,0,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 0, 0, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 0, 0, 0, 0], [1, 0, 1, 0, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 0, 0, 0, 0], [1, 0, 1, 0, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0078","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b35b931be108edd54b08cd4e2a8c27952ddd6604fabff9d6b642b8e7ef2606e","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017078,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[0,0],[0,0]],\"word\":[0,1,0,0,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 0, 0, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 1, 0, 0, 0], [2, 0, 0, 2, 0, 2], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 1, 0, 0, 0], [2, 0, 0, 2, 0, 2], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0079","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"635992f16b912b34818f08326a34242d2df3ed37ce80231448f99f193710160a","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017079,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[0,0],[0,0]],\"word\":[1,1,0,0,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 0, 0, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0080","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df6e682bb9484f1b70806fc72097d6cd06fe92f43ec06a99f2ca9a00648898d8","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017080,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[0,0],[0,0]],\"word\":[0,0,1,0,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 1, 0, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 0, 1, 0, 0], [1, 0, 1, 0, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 0, 1, 0, 0], [1, 0, 1, 0, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0081","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f15cf7b02dfa4c7c6d2e450dce3bbab0782d9d8701f3efd6775b17d5a1d97d4","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017081,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[0,0],[0,0]],\"word\":[1,0,1,0,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 1, 0, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 0, 1, 0, 0], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 0, 1, 0, 0], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0082","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f965a9b6e81b576fd30c2f4b5396a8b67b6e7aebb9965603e35740f6be3443e","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017082,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[0,0],[0,0]],\"word\":[0,1,1,0,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [0, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 1, 0, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 1, 1, 0, 0], [0, 0, 2, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 1, 1, 0, 0], [0, 0, 2, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0083","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b36eee1d415a83952329fa1dced0fde47e5f82753dc5b2378912901b8088976","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017083,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,2],[0,0],[0,0]],\"word\":[1,1,1,0,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 2], [0, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 1, 0, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 1, 1, 0, 0], [1, 0, 2, 0, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 1, 1, 0, 0], [1, 0, 2, 0, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0084","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be010c9f39b98082d8249a446ba43750afe54fcef4b9f8db4174e284213df264","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017084,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,2],[0,0],[0,0]],\"word\":[0,0,0,1,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 2], [0, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 1, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 0, 0, 1, 0], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 0, 0, 1, 0], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0085","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"95bb6781d101a170ae340ef7362d80e649e5c26fd464b68a4bd9f3dd17a390e3","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017085,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[1,0],[0,0]],\"word\":[1,0,0,1,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 0, 1, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0086","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8ac9632bc4ef31bbd385e50024e66351c97789b9986f8eb22325abce6a5ae4e","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017086,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[1,0],[0,0]],\"word\":[0,1,0,1,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 0, 1, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0087","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7fcc5a871ee7ed9da25c40b398bd3771feac0ca6561fbb36eeb6b62fe14458e8","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017087,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[1,0],[0,0]],\"word\":[1,1,0,1,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 0, 1, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 1, 0, 1, 0], [2, 0, 0, 2, 0, 2], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 1, 0, 1, 0], [2, 0, 0, 2, 0, 2], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0088","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e561190b2b3da3f9ee7ea27e64cc57d52657df26f9f36a7bddf46ff954208d3","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017088,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[1,0],[0,0]],\"word\":[0,0,1,1,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 1, 1, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 0, 1, 1, 0], [0, 0, 0, 1, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 0, 1, 1, 0], [0, 0, 0, 1, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0089","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"667d2e4ffbfa2a87450f21632c643b350a77f5e08534c26e4b55f73bf9d79ea2","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017089,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[1,0],[0,0]],\"word\":[1,0,1,1,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 1, 1, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0090","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca7902a5bae97ab217be136b7456e22eb3cf370abe7938958b2fe3bcd3fabe39","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017090,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[1,0],[0,0]],\"word\":[0,1,1,1,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 1, 1, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 1, 1, 1, 0], [2, 0, 1, 0, 1, 1], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 1, 1, 1, 0], [2, 0, 1, 0, 1, 1], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0091","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8da806bac6478fd70e9e0871fc9ff69ae181b61543e9809a27a69c0df340a73","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017091,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[1,0],[0,0]],\"word\":[1,1,1,1,0]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [1, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 1, 1, 0]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 1, 1, 1, 0], [0, 2, 0, 2, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 1, 1, 1, 0], [0, 2, 0, 2, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0092","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e4646886731b9db068e94e1b548ae1ed85bdeee6dbd243441faabed578cddb9","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017092,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,2],[1,0],[0,0]],\"word\":[0,0,0,0,1]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 2], [1, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 0, 1]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0093","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c21d40a001050da538743f8c383c59b72056103c4ec73a0d1eaeea468302e5ff","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017093,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,2],[1,0],[0,0]],\"word\":[1,0,0,0,1]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 2], [1, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 0, 0, 1]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 0, 0, 0, 1], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 0, 0, 0, 1], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0094","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed51d118b3e40ad4b0c67db00ba9f5f787c0ae8d2927f752494f3f0ad2212e55","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017094,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,0],[2,0],[0,0]],\"word\":[0,1,0,0,1]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 0], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 0, 0, 1]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 1, 0, 0, 1], [0, 0, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 1, 0, 0, 1], [0, 0, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0095","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3050c58c7ae64d455a63fb697e14f657e9f28a9ccd906b86882d3759e64abd48","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017095,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,0],[2,0],[0,0]],\"word\":[1,1,0,0,1]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 0], [2, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 0, 0, 1]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 1, 0, 0, 1], [1, 0, 0, 1, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 1, 0, 0, 1], [1, 0, 0, 1, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0096","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"23f6cef5bc5c2c648a8ac13f7823c5d7ab3539a9b537ce17df200e8cca8ef6ee","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017096,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,0],[2,0],[0,0]],\"word\":[0,0,1,0,1]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 0], [2, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 1, 0, 1]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 0, 1, 0, 1], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 0, 1, 0, 1], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0097","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f4f6ae37a44c109396755af5f25d09249f29aba60d544b2f2466bba9ba66375","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017097,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,1],[2,0],[0,0]],\"word\":[1,0,1,0,1]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 1], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 0, 1, 0, 1]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 0, 1, 0, 1], [0, 1, 2, 0, 0, 1], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 0, 1, 0, 1], [0, 1, 2, 0, 0, 1], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0098","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc1613dc03a16cc684f5f03b66f70944659a0ab5b58f3afbbdbbd092fd1c3dc0","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017098,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[1],\"alphabet_count\":2,\"start\":1,\"state_count\":3,\"transition\":[[1,1],[2,0],[0,0]],\"word\":[0,1,1,0,1]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[1, 1], [2, 0], [0, 0]], unique start=1, and increasing accepting subset=[1]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 1, 1, 0, 1]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 1, 1, 0, 1], [1, 2, 0, 1, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 1, 1, 0, 1], [1, 2, 0, 1, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0099","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"380238af91351d027226b63dece03cb2297b45059607a83e2151610b61f01c6d","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017099,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[2],\"alphabet_count\":2,\"start\":2,\"state_count\":3,\"transition\":[[2,1],[2,0],[0,0]],\"word\":[1,1,1,0,1]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[2, 1], [2, 0], [0, 0]], unique start=2, and increasing accepting subset=[2]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[1, 1, 1, 0, 1]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([1, 1, 1, 0, 1], [2, 0, 1, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([1, 1, 1, 0, 1], [2, 0, 1, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r07-finite-automaton-0100","batch_id":"batch-20261014-advance-20261101-r07-finite-automaton-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2feb1c8fb69a610e9e24d2bda1993581cca8ca5a228ea137aef6278181c2654f","family_id":"qf-mod32-dfa-single-trace-boundary","seed":202611017100,"semantic_template":"qf-mod32-dfa-single-trace-boundary/{\"accepting\":[0],\"alphabet_count\":2,\"start\":0,\"state_count\":3,\"transition\":[[0,2],[2,0],[0,0]],\"word\":[0,0,0,1,1]}"},"fields":{"title":"One Trace Does Not Establish Equivalence to a Target Language","stem":"Use a complete deterministic finite automaton with states=Fin(3)={0,…,2}, alphabet=Fin(2)={0,…,1}, total row-state/column-symbol transition=[[0, 2], [2, 0], [0, 0]], unique start=0, and increasing accepting subset=[0]. Every transition output lies in this same state set. There are no epsilon transitions or nondeterministic branches. Read a finite word in symbol order; the empty word leaves the automaton at its start state. Word=[0, 0, 0, 1, 1]. Give (word,state_trace,accepted?,proves_language_equivalence?,one_trace_does_not_prove_language_equivalence_or_regularity).","answer":{"kind":"expression","value":"([0, 0, 0, 1, 1], [0, 0, 0, 0, 2, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)","explanation":"Report acceptance of this word only. The final No rejects an inference from one tested trace to an unspecified language or a second automaton. It does not deny that a complete DFA recognizes a regular language by definition; full-language equivalence and constructions need separate obligations. The exact result, retaining the requested order, is ([0, 0, 0, 1, 1], [0, 0, 0, 0, 2, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)."},"skills":["One Trace Does Not Establish Equivalence to a Target Language"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}