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We saw earlier that LT is not closed under concatenation. If we think in terms of the LT graphs, recognizing the concatenation of LT languages would seem to require knowing, while scanning a string in L1 . L2, for instance, when to switch from keeping track of factors for L1 to keeping track of factors from L2.
Assuming that the alphabets were not disjoint, there is (evidently, since LT is not closed under concatenation) no way, in general, to know that. For the recognizable languages, on the other hand, we have the convenience of being able to work with non-determinism. We don't actually have to know when to switch from one automaton to the next. Whenever we get to a point in the string that could possibly be the end of the pre?x that is in L1 we can just allow for a non-deterministic choice of whether to continue scanning for A1 (the machine recognizing L1) or to switch to scanning for A2. Since whenever the string is in L1 . L2 there will be some correct place to switch and since acceptance by a NFA requires only that there some accepting computation, the combined automaton will accept every string in L1 . L2. Moreover, the combined automaton will accept a string iff there is some point at which it can be split into a string accepted by A1 followed by one accepted by A2: it accepts all and only the strings in L1 . L2.
short application for MISD
This close relationship between the SL2 languages and the recognizable languages lets us use some of what we know about SL 2 to discover properties of the recognizable languages.
All that distinguishes the de?nition of the class of Regular languages from that of the class of Star-Free languages is that the former is closed under Kleene closure while the lat
Let ? ={0,1} design a Turing machine that accepts L={0^m 1^m 2^m } show using Id that a string from the language is accepted & if not rejected .
LTO was the closure of LT under concatenation and Boolean operations which turned out to be identical to SF, the closure of the ?nite languages under union, concatenation and compl
value chain
For every regular language there is a constant n depending only on L such that, for all strings x ∈ L if |x| ≥ n then there are strings u, v and w such that 1. x = uvw, 2. |u
proof ogdens lemma .with example i am not able to undestand the meaning of distinguished position .
I want a proof for any NP complete problem
automata of atm machine
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