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We got the class LT by taking the class SL and closing it under Boolean operations. We have observed that LT ⊆ Recog, so certainly any Boolean combination of LT languages will also be recognizable. But what about the class of recognizable languages as a whole? Are Boolean combinations of recognizable (not just LT) languages also recognizable. In answering we can use the same methodology we use to show that any language is recognizable: consider what we need to keep track of in scanning a string in order to determine if it belongs to the language or not and then use that information to build our state set.
Suppose, then, that L = L1 ∩ L2, where L1 and L2 are both recognizable. A string w will be in L iff it is in both L1 and L2. Since they are recognizable there exist DFAs A1 and A2 for which L1 = L(A1) and L2 = L(A2). We can tell if the string is in L1 or L2 simply by keeping track of the state of the corresponding automaton. We can tell if it is in both by keeping track of both states simultaneously.
A problem is said to be unsolvable if no algorithm can solve it. The problem is said to be undecidable if it is a decision problem and no algorithm can decide it. It should be note
s->0A0|1B1|BB A->C B->S|A C->S|null find useless symbol?
DEGENERATE OF THE INITIAL SOLUTION
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
We now add an additional degree of non-determinism and allow transitions that can be taken independent of the input-ε-transitions. Here whenever the automaton is in state 1
(c) Can you say that B is decidable? (d) If you somehow know that A is decidable, what can you say about B?
matlab v matlab
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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
design an automata for strings having exactly four 1''s
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