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1.4 Each of the following languages is the intersection of two simpler languages. In each part, construct DFAs for the simpler languages, then combine them using the construction discussed in footnote 3 (page 46) to give the state diagram of a DFA for the language given. In all parts, Σ = {a,b}. a. {w w has at least three a's and at least two b's} Ab. {w w has exactly two a's and at least two b's} c. {w w has an even number of a's and one or two b's}

1.4 Each of the following languages is the intersection of two simpler languages. In each part, construct DFAs for the simpler languages, then combine them using the construction discussed in footnote 3 (page 46) to give the state diagram of a DFA for the language given. In all parts, Σ = {a,b}. a. {w w has at least three a's and at least two b's} Ab. {w w has exactly two a's and at least two b's} c. {w w has an even number of a's and one or two b's}

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Hi I added page 46 because I don't understand what it means but on the question it says u need page 46 but I don't know.

can you please help me with this problem and the parts because I just don't understand and I am so lost. 
can you draw a picture or do something visual so I can understand it better, thank you so much. I would greatly appreciate it.

PROOF Let M₁ recognize A₁, where M₁ = (Q₁, E, 61, 91, F₁), and M₂ recognize A2, where M₂ = (Q2, Σ, 62, 92, F₂). Construct M to recognize A₁ U A2, where M 1. Q = {(r1, r₂)| r₁ € Q₁ and r₂ € Q₂}. This set is the Cartesian product of sets Q₁ and Q2 and is written Q₁ × Q2. It is the set of all pairs of states, the first from Q₁ and the second from Q2. 2. Σ, the alphabet, is the same as in M₁ and M₂. In this theorem and in all subsequent similar theorems, we assume for simplicity that both M₁ and M₂ have the same input alphabet Σ. The theorem remains true if they have different alphabets, Σ₁ and Σ2. We would then modify the proof to let Σ = Σ1 U 22. 3. 6, the transition function, is defined as follows. For each (r1, 2) = Q and each a € Σ, let 4. (Q, Σ, δ, qo, F). 8((r₁, r₂), a) = (8₁ (r₁, a), 82 (r2, a)). Hence d gets a state of M (which actually is a pair of states from M₁ and M₂), together with an input symbol, and returns M's next state. 90 is the pair (91, 92). 5. F is the set of pairs in which either member is an accept state of M₁ or M2. We can write it as F = {(r₁, r₂) | r₁ € F₁ or r₂ = F₂}. This expression is the same as F = (F₁ × Q2) U (Q1 × F₂). (Note that it is not the same as F = F₁ × F₂. What would that give us instead?³) 3 This expression would define M's accept states to be those for which both members of the pair are accept states. In this case, M would accept a string only if both M₁ and M₂ accept it, so the resulting language would be the intersection and not the union. In fact, this result proves that the class of regular languages is closed under intersection.
Transcribed Image Text:PROOF Let M₁ recognize A₁, where M₁ = (Q₁, E, 61, 91, F₁), and M₂ recognize A2, where M₂ = (Q2, Σ, 62, 92, F₂). Construct M to recognize A₁ U A2, where M 1. Q = {(r1, r₂)| r₁ € Q₁ and r₂ € Q₂}. This set is the Cartesian product of sets Q₁ and Q2 and is written Q₁ × Q2. It is the set of all pairs of states, the first from Q₁ and the second from Q2. 2. Σ, the alphabet, is the same as in M₁ and M₂. In this theorem and in all subsequent similar theorems, we assume for simplicity that both M₁ and M₂ have the same input alphabet Σ. The theorem remains true if they have different alphabets, Σ₁ and Σ2. We would then modify the proof to let Σ = Σ1 U 22. 3. 6, the transition function, is defined as follows. For each (r1, 2) = Q and each a € Σ, let 4. (Q, Σ, δ, qo, F). 8((r₁, r₂), a) = (8₁ (r₁, a), 82 (r2, a)). Hence d gets a state of M (which actually is a pair of states from M₁ and M₂), together with an input symbol, and returns M's next state. 90 is the pair (91, 92). 5. F is the set of pairs in which either member is an accept state of M₁ or M2. We can write it as F = {(r₁, r₂) | r₁ € F₁ or r₂ = F₂}. This expression is the same as F = (F₁ × Q2) U (Q1 × F₂). (Note that it is not the same as F = F₁ × F₂. What would that give us instead?³) 3 This expression would define M's accept states to be those for which both members of the pair are accept states. In this case, M would accept a string only if both M₁ and M₂ accept it, so the resulting language would be the intersection and not the union. In fact, this result proves that the class of regular languages is closed under intersection.
1.4 Each of the following languages is the intersection of two simpler languages. In each part, construct DFAs for the simpler languages, then combine them using the construction discussed in footnote 3 (page 46) to give the state diagram of a DFA for the language given. In all parts, Σ = {a,b}. a. {w w has at least three a's and at least two b's} Ab. {w w has exactly two a's and at least two b's} c. {w w has an even number of a's and one or two b's}
Transcribed Image Text:1.4 Each of the following languages is the intersection of two simpler languages. In each part, construct DFAs for the simpler languages, then combine them using the construction discussed in footnote 3 (page 46) to give the state diagram of a DFA for the language given. In all parts, Σ = {a,b}. a. {w w has at least three a's and at least two b's} Ab. {w w has exactly two a's and at least two b's} c. {w w has an even number of a's and one or two b's}
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