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Consider the language L given by the regular expression (a + b) *b(a +b) over the alphabet {a, b}. The smallest number of states needed in a deterministic finite-state automaton (DFA) accepting L is ______.
Note: This questions appeared as Numerical Answer Type.
Note: This questions appeared as Numerical Answer Type.
- a)3
- b)4
- c)5
- d)6
Correct answer is option 'B'. Can you explain this answer?
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Consider the language L given by the regular expression (a + b) *b(a +...
NFA to DFA Conversion:
In the STT below you can see 4 states are there but before confirming the ans lets try to minimize, It cant be minimized further because A goes to Nonfinal Nonfinal & AB goes to Final Final hence can not be merged. Similarly ∗AC goes NF NF & ∗ABC goes to F F hence can not be merged. Hence, 4 is the final answer.
Most Upvoted Answer
Consider the language L given by the regular expression (a + b) *b(a +...
Solution:
The given regular expression is (a b)*b(a b).
We need to find the smallest number of states needed in a DFA accepting L.
We can construct a DFA for L as follows:
1. Start with an initial state q0.
2. For each input symbol a, create a transition from q0 to a new state q1, labeled with a.
3. For each input symbol b, create a transition from q0 to q2, labeled with b.
4. For each input symbol a, create a transition from q1 to q2, labeled with a.
5. For each input symbol b, create a transition from q1 to q3, labeled with b.
6. For each input symbol a, create a transition from q2 to q1, labeled with a.
7. For each input symbol b, create a transition from q2 to q2, labeled with b.
8. For each input symbol a, create a transition from q3 to q2, labeled with a.
9. For each input symbol b, create a transition from q3 to q3, labeled with b.
10. Mark q3 as the final state.
The DFA constructed above has 4 states.
Therefore, the smallest number of states needed in a DFA accepting L is 4.
Hence, option B is the correct answer.
The given regular expression is (a b)*b(a b).
We need to find the smallest number of states needed in a DFA accepting L.
We can construct a DFA for L as follows:
1. Start with an initial state q0.
2. For each input symbol a, create a transition from q0 to a new state q1, labeled with a.
3. For each input symbol b, create a transition from q0 to q2, labeled with b.
4. For each input symbol a, create a transition from q1 to q2, labeled with a.
5. For each input symbol b, create a transition from q1 to q3, labeled with b.
6. For each input symbol a, create a transition from q2 to q1, labeled with a.
7. For each input symbol b, create a transition from q2 to q2, labeled with b.
8. For each input symbol a, create a transition from q3 to q2, labeled with a.
9. For each input symbol b, create a transition from q3 to q3, labeled with b.
10. Mark q3 as the final state.
The DFA constructed above has 4 states.
Therefore, the smallest number of states needed in a DFA accepting L is 4.
Hence, option B is the correct answer.
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Consider the language L given by the regular expression (a + b) *b(a +b) over the alphabet {a, b}. The smallest number of states needed in a deterministic finite-state automaton (DFA) accepting L is ______.Note:This questions appeared as Numerical Answer Type.a)3b)4c)5d)6Correct answer is option 'B'. Can you explain this answer?
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