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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.)
draw pda for l={an,bm,an/m,n>=0} n is in superscript
how to write program Minimum Cost Calculation - Vogel Approximation Method(VAM
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State and Prove the Arden's theorem for Regular Expression
(c) Can you say that B is decidable? (d) If you somehow know that A is decidable, what can you say about B?
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
Claim Under the assumptions above, if there is an algorithm for checking a problem then there is an algorithm for solving the problem. Before going on, you should think a bit about
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Differentiate between DFA and NFA. Convert the following Regular Expression into DFA. (0+1)*(01*+10*)*(0+1)*. Also write a regular grammar for this DFA.
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