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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.
#can you solve a problem of palindrome using turing machine with explanation and diagrams?
Since the signi?cance of the states represented by the nodes of these transition graphs is arbitrary, we will allow ourselves to use any ?nite set (such as {A,B,C,D,E, F,G,H} or ev
It is not hard to see that ε-transitions do not add to the accepting power of the model. The underlying idea is that whenever an ID (q, σ v) directly computes another (p, v) via
You are required to design a system that controls the speed of a fan's rotation. The speed at which the fan rotates is determined by the ambient temperature, i.e. as the temperatur
De?nition Deterministic Finite State Automaton: For any state set Q and alphabet Σ, both ?nite, a ?nite state automaton (FSA) over Q and Σ is a ?ve-tuple (Q,Σ, T, q 0 , F), w
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s->0A0|1B1|BB A->C B->S|A C->S|null find useless symbol?
Sketch an algorithm for the universal recognition problem for SL 2 . This takes an automaton and a string and returns TRUE if the string is accepted by the automaton, FALSE otherwi
implementation of operator precedence grammer
Let L1 and L2 be CGF. We show that L1 ∩ L2 is CFG too. Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the sec
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