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Intuitively, closure of SL2 under intersection is reasonably easy to see, particularly if one considers the Myhill graphs of the automata. Any path through both graphs will be a path through the intersection of the graphs (by which we mean the graph resulting by taking the intersection of the vertex sets and the intersection of the edge sets).
For the union, on the other hand, the corresponding construction won't work. An automaton built from the union of the two automata will still recognize all of the strings in L1 and all of the strings in L2, but it is likely to also recognize strings made up of adjacent pairs from L1 combined with adjacent pairs from L2 that aren't in either language. And, in fact, we can use Suffx Substitution Closure to show that there are languages that are the union of two SL2 languages that are not, themselves, SL2.
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
explain turing machine .
First model: Computer has a ?xed number of bits of storage. You will model this by limiting your program to a single ?xed-precision unsigned integer variable, e.g., a single one-by
DEGENERATE OF THE INITIAL SOLUTION
State and Prove the Arden's theorem for Regular Expression
program in C++ of Arden''s Theorem
(c) Can you say that B is decidable? (d) If you somehow know that A is decidable, what can you say about B?
Construct a Moore machine to convert a binary string of radix 4.
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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.
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