Software Engineering Deterministic Finite Automaton
Laith Jayyousi I. Regular Expressions 1. In the context of the alphabet {a, b}, the regular expression below captures all strings whose length is not divisible by 3: (a | b)^*(a | b | ab). 2. For the alphabet {p, q}, the following regular expression describes all strings that include an odd quantity of ‘q’s: (p | q)(p | q)^*q. 3. In the context of the binary alphabet {0, 1}, the following regular expression encompasses all strings in which consecutive characters are distinct: (0 | 1)(0 | 1)^*. II. DFA 1. The language defined over the alphabet {a, b}, including strings not multiples of 3, can be represented by a DFA (Deterministic Finite Automaton) as follows: DFA: States: q0 (initial), q1, q2 (accepting) Alphabet: {a, b} Transition Table: q0 q1 a,b > q1 a,b > q2 Laith Jayyousi a,b > q1 q2 The DFA starts processing in state q0 and accepts the input in state q2. With each ‘a’ or ‘b’ input, it cycles between states q1 and q2, ensuring that the input string is never a multiple of 3. 2. In the language over the alphabet {p, q}, which comprises strings with an odd count of ‘q’s, a deterministic finite automaton (DFA) can be defined as follows: DFA Details: States: q0 (initial), q1 (accepting) Alphabet: {p, q} Transition Table: q0 p > q0 q > q1 p > q1 q > q0 q0 q1 q1 Explanation: The DFA initiates in state q0 and recognizes a string if it concludes in state q1. It alternates between states q0 and q1 upon encountering each ‘q’ input, ensuring that the string contains an odd number of ‘q’s. Laith Jayyousi Consider the language over the binary alphabet {0, 1} comprising strings where consecutive characters are distinct. The following deterministic finite automaton (DFA) recognizes this language: DFA Details: States: q0 (initial), q1 Alphabet: {0, 1} Transition Table: q0 0 > q1 1 > q1 0 > q0 1 > q0 q0 q1 q1 The DFA evaluates from state q0 and accepts input if it ends there. If it detects two identical characters, it moves to state q1, and any other input returns to q0. No two consecutive characters in the string are the same with this design. III. RE>NFA>DFA 1. Converting the regular expression to an NFA Laith Jayyousi The provided regular expression is represented as an NFA (Nondeterministic Finite Automaton) through the following steps: 1. Begin with a single start state, denoted as S. 2. For each character in the regular expression, generate a new state and establish a transition from S to the new state, labeled with the respective character. 3. When encountering the Kleene star operator (*), create new states for each accepting state, and for each of these new states, add a transition to itself labeled with the character. 4. Designate the accept states as those states that conclude with the string ‘ab’. This results in the construction of the NFA for the given regular expression.: S / a b / ab 2. Converting the NFA to a DFA Applying the subset construction algorithm turns the NFA into a DFA. Sets of states are created for each subset of the NFA’s accepting states. Create a set of only the starting state. We then go through the NFA’s transitions, adding new states to sets as needed. The NFA subset construction algorithm is outlined here: Laith Jayyousi 1. Start with an initial set containing only the start state, denoted as S. 2. Add the state ab to the set containing S. 3. Proceed to examine the NFA’s transitions: The transition from state S to ‘a’ appends the state ‘a’ to the set containing S. The transition from state ‘a’ to ‘b’ includes the state ‘b’ in the set containing ‘a’. The transition from state ‘b’ to ‘ab’ includes the state ‘ab’ in the set containing ‘b’. This process results in the creation of the following DFA: S / a b The transition table for the DFA is as follows: State | Input | Next State S |a |a S |b |b a |a |a a |b |b b | ab | ab Laith Jayyousi IV. contextfree Grammars 1. L = {an bm | n < m or n = 2m} S → AB A → aA | ε B → bB | ab The variable A generates a sequence of 'a's in this grammar. The variable B generates a sequence of 'b's. The sequence of 'b's can be empty (ε) or a series of 'a's followed by a single 'b'. A and B always produce strings with fewer 'a's than 'bs', proving n < m or n = 2m. 2. The language described by the regular expression ((ab) (c|d))** S → AB A → aA | ε B → bB | cB | dB Two phases make up this grammar. The first step is producing 'ab' pairs with A. Then it creates a string of 'c's, 'd's, or 'ab' pairs starting with B. Empty () or a sequence of 'ab' pairs separated by a 'c', 'd', or another 'ab' pair sequence is valid. The regular expression ((ab)* (c|d))* is true because B's string can have a 'c' or 'd' after 'ab'. V. CFG S → SA → AaB → aaB → aabab The derivation process unfolds through the following steps: 1. Initiate with the starting symbol S. 2. Utilize the production rule S → SA to substitute S with SA. Laith Jayyousi 3. Employ the production rule A → Aa to transform A into Aa. 4. Implement the production rule A → aB to replace A with aB. 5. Utilize the production rule B → aB to transform B into aB. 6. Apply the production rule B → ab to substitute B with ab. These steps ultimately yield the resultant string: aabab.. TABULAR SUMMARY Step Production rule used 1 | S → SA 2 | A → Aa 3 | A → aB 4 | B → aB 5 | B → ab
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