====== 4. Regex to DFA conversion ======

==== 4.1. Nondeterministic Finite Automata ====

4.1.1. Consider the following NFA:

{{:lfa:lab03-nfa2.png|}}        

What are all reachable configurations from (0,abba) ?


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(0, abba) -> (0, bba) -> (0, ba) -> (0, a) -> (0, $\varepsilon$) reject

(0, abba) -> (0, bba) -> (0, ba) -> (0, a) -> (1, $\varepsilon$) reject

(0, abba) -> (1, bba) -> (2, ba) -> (3, a) stuck
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4.1.2. What is the accepted language of the previous NFA? 


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L( (a $\cup$ b)* a (a $\cup$ b) (a $\cup$ b $\cup$ $\varepsilon$) )
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4.1.3. Write an NFA //without// $ \varepsilon$-transitions, which accepts the language $ L = \{abc,abd,aacd\}$ over the alphabet $ \Sigma = \{a,b,c,d\}$.


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{{:lfa:2022:lfa2022_lab2_ex1_4.png?400|}}
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4.1.4. Consider the following NFA:

{{:lfa:lab02-nfa2.png|}}

What are all reachable configurations from $ (0,abbabbb)$?


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{{:lfa:2022:lfa2022_lab4_ex1_4.png?300|}}

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==== 2. Regex to NFA =====

4.2.1. What language does $ (1 \cup \varepsilon)(00^*1)^*0^*$ generate?


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L($ (1 \cup \varepsilon)(00^*1)^*0^*$) = { w $\in$ {0, 1}* | w has the property that there are no sequences of 2 or more consecutive "1" }
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4.2.2. Convert the previous regex to an NFA.


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{{:lfa:2022:lfa2022_lab4_ex2_2_v2.png?600|}}
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==== 3. The subset construction =====

4.3.1. Write the $ \varepsilon$-closure ($ E(q)$) for each state q in the NFA from exercise 4.1.4.


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E(0) = {0, 2}

E(1) = {1, 3}

E(2) = {2}

E(3) = {1, 3}

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4.3.2. Convert the NFA from exercise 4.1.1 to a DFA.


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{{:lfa:2022:lfa2022_lab4_ex3_2.png?400|}}
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4.3.3. Convert the NFA from exercise 4.1.4 to a DFA.


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{{:lfa:2022:lfa2022_lab4_ex3_3.png?400|}}

Barred transitions go to a sink state.
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4.3.4. Convert the NFA from exercise 4.2.2 to a DFA.


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{{:lfa:2022:lfa2022_lab4_ex3_4_v2.png?400|}}

Barred transitions go to a sink state.
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