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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 noted that an unsolvable problem might be partially solvable by an algorithm that makes a complete search for a solution. In such case the solution is eventually found whenever it is defined, but the search might continue forever whenever the solution is undefined. Similarly, an undecidable problem might also be partially decidable by an algorithm that makes an exhaustive search.
Both L 1 and L 2 are SL 2 . (You should verify this by thinking about what the automata look like.) We claim that L 1 ∪ L 2 ∈ SL 2 . To see this, suppose, by way of con
Give DFA''s accepting the following languages over the alphabet {0,1}: i. The set of all strings beginning with a 1 that, when interpreted as a binary integer, is a multiple of 5.
A context free grammar G = (N, Σ, P, S) is in binary form if for all productions A we have |α| ≤ 2. In addition we say that G is in Chomsky Normaml Form (CNF) if it is in bi
Exercise Show, using Suffix Substitution Closure, that L 3 . L 3 ∈ SL 2 . Explain how it can be the case that L 3 . L 3 ∈ SL 2 , while L 3 . L 3 ⊆ L + 3 and L + 3 ∈ SL
While the SL 2 languages include some surprisingly complex languages, the strictly 2-local automata are, nevertheless, quite limited. In a strong sense, they are almost memoryless
program in C++ of Arden''s Theorem
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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Our DFAs are required to have exactly one edge incident from each state for each input symbol so there is a unique next state for every current state and input symbol. Thus, the ne
s-> AACD A-> aAb/e C->aC/a D-> aDa/bDb/e
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