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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.
explain turing machine .
The Myhill-Nerode Theorem provided us with an algorithm for minimizing DFAs. Moreover, the DFA the algorithm produces is unique up to isomorphism: every minimal DFA that recognizes
We can then specify any language in the class of languages by specifying a particular automaton in the class of automata. We do that by specifying values for the parameters of the
what exactly is this and how is it implemented and how to prove its correctness, completeness...
prove following function is turing computable? f(m)={m-2,if m>2, {1,if
Applying the pumping lemma is not fundamentally di?erent than applying (general) su?x substitution closure or the non-counting property. The pumping lemma is a little more complica
The SL 2 languages are speci?ed with a set of 2-factors in Σ 2 (plus some factors in {?}Σ and some factors in Σ{?} distinguishing symbols that may occur at the beginning and en
write short notes on decidable and solvable problem
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
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
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