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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.
(c) Can you say that B is decidable? (d) If you somehow know that A is decidable, what can you say about B?
c program to convert dfa to re
design an automata for strings having exactly four 1''s
One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included
Paths leading to regions B, C and E are paths which have not yet seen aa. Those leading to region B and E end in a, with those leading to E having seen ba and those leading to B no
write grammer to produce all mathematical expressions in c.
I want a proof for any NP complete problem
One of the first issues to resolve, when exploring any mechanism for defining languages is the question of how to go about constructing instances of the mechanism which define part
How useful is production function in production planning?
design a turing machine that accepts the language which consists of even number of zero''s and even number of one''s?
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