Introduction to the Theory of Computation This is an electronic version of the print textbook. Due to electronic rights restrictions, some third party content may be suppressed. Editorial review has...

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Answered Same DaySep 27, 2021

Answer To: Introduction to the Theory of Computation This is an electronic version of the print textbook. Due...

Sandeep Kumar answered on Oct 08 2021
146 Votes
NFA STATE
    DFA STATE
    TYPE
    a
    b
    {0,1,3}
    A
    
    B
    C
    {2,8}
    B
    
    D
    
    {4,5,7,8}
    C
    
    D
    E
    {9,10,12}
    D
    

    F
    G
    {5,6,7,8}
    E
    
    D
    E
    {10,11,12}
    F
    
    F
    G
    {13,14,16}
    G
    accept
    
    H
    {14,15,16}
    H
    accept
    
    H
1.
2. In that part we first add a new start state and a new accept state. The solution is
Ɛ ∪ ((a ∪ b) a*b((b ∪a(a ∪b))a*b)*( Ɛ ∪a))
3. 1. (a) A = B = (0 ∪ 1)*
Perfect shuffle => 10
3.1.(b) A = (01)∗ and B = (10)∗
Perfect shuffle => A = 1∪0, B = 0∪1
3.2. Let DA = (QA, Σ, δA, qA, FA) and DB = (QB, Σ, δB, qB, FB) be two DFAs that
recognize A and B, respectively. Similar to the previous question, we shall prove by
construction. However, the key difference is that D may now switch from running DA
and running DB after each character is read. To allow this flexibility and simplify the
construction, we design an NFA N = (Q, Σ, δ, q, F) that recognizes the shuffle of A and B
instead of directly designing a DFA.
At any time, N needs to keep track of the current states of DA and DB. Then, when a
character is read, N may make a move in DA or DB accordingly. After the whole string is
processed, if both DFAs are in the accept states, the input string is accepted; otherwise,
the input string is rejected. In addition, N...
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