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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Deterministic finite automaton · qf-mod32-dfa-single-trace-boundary · candidate
Canonical question JSONL · English JSONL · Validation report