# Complete DFA Tables, Ordered Word Traces and Empty-Word Acceptance

This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.

This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261014-advance-20261101-r07-finite-automaton-foundations · 2026-10-14 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.



## 1. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0001/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 2. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0002/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 3. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0003/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 4. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0004/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 5. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0005/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 6. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0006/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 7. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0007/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 8. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0008/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 9. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0009/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 10. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0010/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 11. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0011/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 12. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0012/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 13. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0013/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 14. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0014/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 15. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0015/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 16. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0016/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 17. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0017/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 18. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0018/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 19. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0019/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 20. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0020/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 21. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0021/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 22. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0022/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 23. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 1, [1], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0023/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 24. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 2, [2], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0024/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 25. A DFA Needs a Complete Total Deterministic Transition Table

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).

View answer and explanation(3, 2, 0, [0], 6, Yes, complete_total_deterministic_table)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0025/)

Deterministic finite automaton · qf-mod32-dfa-qualification · candidate





## 26. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0026/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 27. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 0, 0, 0, 0], [1, 0, 1, 0, 1, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0027/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 28. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 1, 0, 0, 0], [2, 0, 0, 2, 0, 2], 2)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0028/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 29. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0029/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 30. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 0, 1, 0, 0], [1, 0, 1, 0, 1, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0030/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 31. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 0, 1, 0, 0], [2, 0, 2, 0, 2, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0031/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 32. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 1, 1, 0, 0], [0, 0, 2, 0, 0, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0032/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 33. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 1, 1, 0, 0], [1, 0, 2, 0, 1, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0033/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 34. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 0, 0, 1, 0], [2, 0, 2, 0, 2, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0034/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 35. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0035/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 36. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1], 1)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0036/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 37. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 1, 0, 1, 0], [2, 0, 0, 2, 0, 2], 2)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0037/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 38. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 0, 1, 1, 0], [0, 0, 0, 1, 0, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0038/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 39. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1], 1)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0039/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 40. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 1, 1, 1, 0], [2, 0, 1, 0, 1, 1], 1)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0040/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 41. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 1, 1, 1, 0], [0, 2, 0, 2, 0, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0041/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 42. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0042/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 43. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 0, 0, 0, 1], [2, 0, 2, 0, 2, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0043/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 44. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 1, 0, 0, 1], [0, 0, 0, 0, 0, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0044/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 45. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 1, 0, 0, 1], [1, 0, 0, 1, 2, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0045/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 46. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 0, 1, 0, 1], [2, 0, 2, 0, 2, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0046/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 47. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 0, 1, 0, 1], [0, 1, 2, 0, 0, 1], 1)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0047/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 48. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 1, 1, 0, 1], [1, 2, 0, 1, 2, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0048/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 49. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([1, 1, 1, 0, 1], [2, 0, 1, 0, 2, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0049/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 50. Retain the Initial State and Every Ordered Symbol Transition

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.

View answer and explanation([0, 0, 0, 1, 1], [0, 0, 0, 0, 2, 0], 0)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0050/)

Deterministic finite automaton · qf-mod32-dfa-word-trace · candidate





## 51. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0051/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 52. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 0, 0, 0, 0], [1, 0, 1, 0, 1, 0], 0, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0052/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 53. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 1, 0, 0, 0], [2, 0, 0, 2, 0, 2], 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0053/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 54. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0], 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0054/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 55. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 0, 1, 0, 0], [1, 0, 1, 0, 1, 0], 0, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0055/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 56. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 0, 1, 0, 0], [2, 0, 2, 0, 2, 0], 0, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0056/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 57. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 1, 1, 0, 0], [0, 0, 2, 0, 0, 0], 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0057/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 58. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 1, 1, 0, 0], [1, 0, 2, 0, 1, 0], 0, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0058/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 59. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 0, 0, 1, 0], [2, 0, 2, 0, 2, 0], 0, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0059/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 60. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0], 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0060/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 61. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1], 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0061/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 62. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 1, 0, 1, 0], [2, 0, 0, 2, 0, 2], 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0062/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 63. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 0, 1, 1, 0], [0, 0, 0, 1, 0, 0], 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0063/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 64. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1], 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0064/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 65. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 1, 1, 1, 0], [2, 0, 1, 0, 1, 1], 1, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0065/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 66. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 1, 1, 1, 0], [0, 2, 0, 2, 0, 0], 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0066/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 67. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 0], 0, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0067/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 68. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 0, 0, 0, 1], [2, 0, 2, 0, 2, 0], 0, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0068/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 69. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 1, 0, 0, 1], [0, 0, 0, 0, 0, 0], 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0069/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 70. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 1, 0, 0, 1], [1, 0, 0, 1, 2, 0], 0, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0070/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 71. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 0, 1, 0, 1], [2, 0, 2, 0, 2, 0], 0, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0071/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 72. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 0, 1, 0, 1], [0, 1, 2, 0, 0, 1], 1, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0072/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 73. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 1, 1, 0, 1], [1, 2, 0, 1, 2, 0], 0, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0073/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 74. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([1, 1, 1, 0, 1], [2, 0, 1, 0, 2, 0], 0, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0074/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 75. The Empty Word Is Accepted Exactly When the Start State Is Accepting

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?).

View answer and explanation([0, 0, 0, 1, 1], [0, 0, 0, 0, 2, 0], 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0075/)

Deterministic finite automaton · qf-mod32-dfa-acceptance · candidate





## 76. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0076/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 77. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 0, 0, 0, 0], [1, 0, 1, 0, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0077/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 78. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 1, 0, 0, 0], [2, 0, 0, 2, 0, 2], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0078/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 79. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0079/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 80. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 0, 1, 0, 0], [1, 0, 1, 0, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0080/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 81. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 0, 1, 0, 0], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0081/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 82. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 1, 1, 0, 0], [0, 0, 2, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0082/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 83. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 1, 1, 0, 0], [1, 0, 2, 0, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0083/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 84. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 0, 0, 1, 0], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0084/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 85. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0085/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 86. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0086/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 87. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 1, 0, 1, 0], [2, 0, 0, 2, 0, 2], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0087/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 88. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 0, 1, 1, 0], [0, 0, 0, 1, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0088/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 89. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0089/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 90. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 1, 1, 1, 0], [2, 0, 1, 0, 1, 1], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0090/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 91. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 1, 1, 1, 0], [0, 2, 0, 2, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0091/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 92. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0092/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 93. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 0, 0, 0, 1], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0093/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 94. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 1, 0, 0, 1], [0, 0, 0, 0, 0, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0094/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 95. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 1, 0, 0, 1], [1, 0, 0, 1, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0095/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 96. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 0, 1, 0, 1], [2, 0, 2, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0096/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 97. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 0, 1, 0, 1], [0, 1, 2, 0, 0, 1], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0097/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 98. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 1, 1, 0, 1], [1, 2, 0, 1, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0098/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 99. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([1, 1, 1, 0, 1], [2, 0, 1, 0, 2, 0], No, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0099/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate





## 100. One Trace Does Not Establish Equivalence to a Target Language

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).

View answer and explanation([0, 0, 0, 1, 1], [0, 0, 0, 0, 2, 0], Yes, No, one_trace_does_not_prove_language_equivalence_or_regularity)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261101-r07-finite-automaton-0100/)

Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261101-r07-finite-automaton-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261101-r07-finite-automaton-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261101-r07-finite-automaton-foundations/validation-report.json)