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The upper string r ∈ Q+ is the sequence of states visited by the automaton as it scans the lower string w ∈ Σ*. We will refer to this string over Q as the run of A on w. The automaton A accepts w iff the run of A on w ends in an accepting state. (If A is non-deterministic there will potentially be many runs with the automaton accepting if any one of them ends in an accepting state.) Note that the set of runs of an automaton is an SL2 language, recognized by the SL2 automaton (over Q) one gets by projecting away the third component of the triples of GA. Thus there is some kind of close relationship between the strictly local languages and the recognizable languages.
To get at this we will start by working in the other direction, extending our tiles to hold four symbols. The idea is to include, for each tile (q, p, σ) ∈ GA, a tile extended with σ′ for each σ′ ∈ Σ. (We don't actually need tiles for all such σ′ , only for those that occur on tiles (x, q, σ′) which might precede this one in a tiling, but including all of them will be harmless-the ones that do not occur on such tiles will just be useless.)
proof ogdens lemma .with example i am not able to undestand the meaning of distinguished position .
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
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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
Construct a PDA that accepts { x#y | x, y in {a, b}* such that x ? y and xi = yi for some i, 1 = i = min(|x|, |y|) }. For your PDA to work correctly it will need to be non-determin
The Recognition Problem for a class of languages is the question of whether a given string is a member of a given language. An instance consists of a string and a (?nite) speci?cat
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
s->0A0|1B1|BB A->C B->S|A C->S|null find useless symbol?
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