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Theorem The class of recognizable languages is closed under Boolean operations.
The construction of the proof of Lemma 3 gives us a DFA that keeps track of whether or not a given string is in either or both of any pair of recognizable languages. We can modify the construction for other Boolean operations simply by selecting the appropriate set of accepting states:
• Union: Let F′
= {(q, p) | q ∈ F1 or p ∈ F2}. Then L(A′ ) = L1 ∪ L2.
• Relative complement: Let F′ = F1 × (Q2 - F2). Then L(A′ ) = L1 -L2.
• Complement: Let L1 = Σ* and use the construction for relative complement.
The computation of an SL 2 automaton A = ( Σ, T) on a string w is the maximal sequence of IDs in which each sequential pair of IDs is related by |- A and which starts with the in
Automaton (NFA) (with ε-transitions) is a 5-tuple: (Q,Σ, δ, q 0 , F i where Q, Σ, q 0 and F are as in a DFA and T ⊆ Q × Q × (Σ ∪ {ε}). We must also modify the de?nitions of th
The fact that the Recognition Problem is decidable gives us another algorithm for deciding Emptiness. The pumping lemma tells us that if every string x ∈ L(A) which has length grea
Given any NFA A, we will construct a regular expression denoting L(A) by means of an expression graph, a generalization of NFA transition graphs in which the edges are labeled with
design an automata for strings having exactly four 1''s
Can you say that B is decidable? If you somehow know that A is decidable, what can you say about B?
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turing machine for prime numbers
As de?ned the powerset construction builds a DFA with many states that can never be reached from Q′ 0 . Since they cannot be reached from Q′ 0 there is no path from Q′ 0 to a sta
For example, the question of whether a given regular language is positive (does not include the empty string) is algorithmically decidable. "Positiveness Problem". Note that
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