Decision Properties & Algorithms for Regular Languages
“Emptiness, finiteness, membership, and equivalence decision procedures for regular sets.”
1.Complete Decidability of Regular Language Properties
§ 2.1.A decision property is an algorithmic question: 'Does a given language satisfy property P?' with a guaranteed YES/NO termination.
Every fundamental property of Regular Languages is DECIDABLE:
1. Membership (w \in L):
- Run string w on the DFA. If it terminates in a final state, YES; otherwise NO. Complexity: O(|w|).
2. Emptiness (L = \emptyset):
- Perform BFS or DFS from the start state q_0. If NO accepting state is reachable, L is empty. Complexity: O(|V| + |E|).
3. Finiteness (Is L finite?):
- Eliminate unreachable states and dead states. If the remaining transition graph contains a directed cycle, L is infinite. If acyclic, L is finite! Complexity: O(|V| + |E|).
4. Equivalence (L_1 = L_2):
- Construct the symmetric difference: L_{\Delta} = (L_1 \cap \overline{L_2}) \cup (\overline{L_1} \cap L_2).
- L_1 = L_2 \iff L_{\Delta} = \emptyset. Test emptiness on L_{\Delta}!
5. Subset / Containment (L_1 \subseteq L_2):
- L_1 \subseteq L_2 \iff L_1 \cap \overline{L_2} = \emptyset. Test emptiness of L_1 \cap \overline{L_2}!
2. GATE Worked Numerical: Equivalence via Symmetric Difference
To test if two DFAs accept the same language, you do NOT have to test infinitely many strings. Minimize both DFAs and check graph isomorphism, or test emptiness of their symmetric difference!
Problem: Let M1 and M2 be two DFAs over {0, 1}. Describe the exact algorithmic procedure to decide whether L(M1) = L(M2).
Step 1: Construct complement of M2: Invert accepting and non-accepting states of M2 to obtain M2_bar.
Step 2: Construct product automaton for L1 cap L2_bar: States are Q1 x Q2, accepting states are F1 x (Q2 \ F2).
Step 3: Construct product automaton for L1_bar cap L2: Accepting states are (Q1 \ F1) x F2.
Step 4: Combine both product automata via union: This yields the symmetric difference L_diff = (L1 \ L2) union (L2 \ L1).
Step 5: Apply Graph Reachability (BFS/DFS) on L_diff starting from (q01, q02):
If any accepting state is reachable: L_diff != empty set ==> L1 != L2 (Output: NO).
If no accepting state is reachable: L_diff = empty set ==> L1 = L2 (Output: YES).
Primary Academic References & Exam Sources
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