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Questions and Answers
You can assign the states q, q0, q00, and ______ to these possibilities.
You can assign the states q, q0, q00, and ______ to these possibilities.
q001
The start state is ______, and the only accept state is q001.
The start state is ______, and the only accept state is q001.
q
The star operation is a bit different from the other two because it applies to a ______ language rather than to two different languages.
The star operation is a bit different from the other two because it applies to a ______ language rather than to two different languages.
single
The empty string " is always a member of A*, no matter what ______ is.
The empty string " is always a member of A*, no matter what ______ is.
The union operation simply takes all the strings in both A and B and lumps them together into one ______.
The union operation simply takes all the strings in both A and B and lumps them together into one ______.
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Study Notes
Formal Definition of Computation
- A finite automaton M is defined as a 5-tuple: M = (Q, Σ, δ, q0, F).
- A string w = w1w2...wn consists of symbols from the alphabet Σ.
- M accepts w if there exists a sequence of states r0, r1,..., rn in Q meeting three conditions:
- r0 is the initial state q0.
- Transition δ(r_i, w_{i+1}) leads to r_{i+1} for i = 0,..., n-1.
- The final state rn must be in the set of accepting states F.
- M recognizes language A defined as A = {w | M accepts w}.
Designing Finite Automata
- To design a finite automaton to recognize strings with an odd number of 1s:
- Track whether the count of 1s so far is even or odd instead of remembering the entire string.
- Transition states are defined as:
- State q_even (even number of 1s).
- State q_odd (odd number of 1s).
- On reading a 1, transition from q_even to q_odd, and vice versa. Reading a 0 retains the current state.
- The start state for the automaton is q_even, accepting state is q_odd (as it represents odd occurrences).
Example of Recognizing Substrings
- A finite automaton E2 for recognizing the pattern "001":
- Represents four states based on the substring seen:
- q (no symbols seen).
- q0 (one 0 seen).
- q00 (two 0s seen).
- q001 (substring "001" seen).
- Transition rules:
- From q, stay in q for 1, move to q0 for 0.
- From q0, move to q for 1, go to q00 for 0.
- From q00, move to q001 for 1, stay in q00 for 0.
- In q001, remain in q001 for any input.
- The start state is q, and only the state q001 is accepting.
- Represents four states based on the substring seen:
Regular Operations on Languages
- Union: Combines strings from two languages A and B into one language.
- Concatenation: Attaches strings from A in front of strings from B producing new strings in the resulting language.
- Star Operation: A unary operation applied to a single language A, concatenating any number of strings in A to create new strings in the language. The empty string "" is always included in A*.
- Example with the alphabet {a, b, ..., z}:
- A = {good, bad} and B = {boy, girl} results in:
- Union: A ∪ B = {good, bad, boy, girl}
- Concatenation: A ∩ B = {goodboy, goodgirl, badboy, badgirl}
- Star: A* = {", good, bad, goodgood, badbad,...}
- A = {good, bad} and B = {boy, girl} results in:
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