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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.)
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
example of multitape turing machine
In Exercise 9 you showed that the recognition problem and universal recognition problem for SL2 are decidable. We can use the structure of Myhill graphs to show that other problems
Computations are deliberate for processing information. Computability theory was discovered in the 1930s, and extended in the 1950s and 1960s. Its basic ideas have become part of
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.
Let L 3 = {a i bc j | i, j ≥ 0}. Give a strictly 2-local automaton that recognizes L 3 . Use the construction of the proof to extend the automaton to one that recognizes L 3 . Gi
design an automata for strings having exactly four 1''s
conversion from nfa to dfa 0 | 1 ___________________ p |{q,s}|{q} *q|{r} |{q,r} r |(s) |{p} *s|null |{p}
As de?ned the powerset construction builds a DFA with many states that can never be reached from Q′ 0 . Since they cannot be reached from Q′ 0 there is no path from Q′ 0 to a sta
write short notes on decidable and solvable problem
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