Introduction to Computability Theory Discussion1: Non-Deterministic Finite Automatons Prof. Amos Israeli 1 Reminder An NFA is a finite automaton in which: 1. 0, 1, many transitions with the same label emanating out of the same state are allowed. 2. transitions are allowed. 2 Exercise (omitted) Construct the following NFA-s: 1. An NFA accepting all words ending with abbab . 2. An NFA accepting all words whose third or second word from the end is a . 3. An NFA accepting all words of the form a i b j c k satisfying a 0, b 0, c 0 a b c mod 2 0 3 Reminder A computation an NFA a tree like collection of paths induced by following all possible transitions and forking when either the current input symbol allows several transitions more than a single transition or when an transition is enabled. 4 Exercise Construct the 1 a computation of this NFA on some short words. 5 b a 3 2 a,b Reminder Proposition DFA-s and NFA-s are equivalent. Meaning: DFA-s and NFA-s recognize the same class of languages, called the Regular Languages. 6 Reminder Every DFA is (a special case of) an NFA, thus La | La is recognized La | La is recognized by some DFA by some NFA In order to prove equality we have to prove: La | La is recognized La | La is recognized by some DFA by some NFA 7 Reminder The proof works as follows: Given some arbitrary NFA N, we construct a DFA N, such that LM LN Yesterday we started to look at the proof. Today we will demonstrate the construction on one particular NFA: 8 Exercise Given NFA N Construct the 1 a b satisfying LM LN . a 3 2 a DFA M 9 a,b Reminder For each set S, the Power Set of S, P( S ) , is the set containing all subsets of S. The set S satisfy: | PS | 2|S| Why: Each subset corresponds to a binary vector of | S | elements. 10 Reminder For each set S, the Power Set of S, P( S ) , is the set containing all subsets of S. The set S satisfy: | PS | 2|S| Why: Each subset corresponds to a binary vector of | S | elements. 11 Reminder The state set of M is the power set of the state set of N . 1 a b a 3 2 a,b Q 1,2,3 PQ , 1 , 2 , 3 , 1,2 , 1,3 , 2,3 , 1,2,3 12
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