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One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included in its positive closure (that is, L2 ⊆ L+). The intuitive idea is that if we had a counterexample for closure under concatenation that uses just a single language L, then if there was some pair of strings in L2 that invalidates suffx substitution closure-that yields a string not in L2 when the suffx of one is substituted into the other-then that pair would invalidate suffx substitution closure for L* as well. But this argument doesn't work. The fact that the pair yields a string that is not in L2 does not rule out the possibility of string being in Li for some i = 2.
If one thinks in terms of strictly local generation, it should be clear that a language L is strictly 2-local language i? it includes all and only the strings that start with a symbol from some particular subset of Σ and end with a symbol from another such subset, with only particular pairs of adjacent symbols occurring in between-equivalently, some particular set of forbidden pairs not occurring (see Section 3 of Part 1).
Consider, then L+. Strings in L+ will also start and end with symbols from those subsets of Σ and the adjacent pairs of symbols occurring strictly within the string from a given iteration of L will be only those that are permitted. The only di?erence is that there may be additional adjacent pairs where the strings from successive iterations meet. These we can admit by simply permitting them as well. The question is whether they will allow pairs in the middle of a string from L which should be forbidden. But, since we are only adding pairs in which the left symbol is a permissible ending symbol for a string from L and the right symbol is a permissible starting symbol, everywhere such a pair occurs is a permissible boundary between strings of L. Finally, to extend the construction to get L* all we need to do is add the pair ?? as well.
Kleene called this the Synthesis theorem because his (and your) proof gives an effective procedure for synthesizing an automaton that recognizes the language denoted by any given r
This was one of the ?rst substantial theorems of Formal Language Theory. It's maybe not too surprising to us, as we have already seen a similar equivalence between LTO and SF. But
Consider a water bottle vending machine as a finite–state automaton. This machine is designed to accept coins of Rs. 2 and 5 only. It dispenses a single water bottle as soon as the
De?nition (Instantaneous Description) (for both DFAs and NFAs) An instantaneous description of A = (Q,Σ, δ, q 0 , F) , either a DFA or an NFA, is a pair h q ,w i ∈ Q×Σ*, where
Given any NFA A, we will construct a regular expression denoting L(A) by means of an expression graph, a generalization of NFA transition graphs in which the edges are labeled with
1. Simulate a TM with infinite tape on both ends using a two-track TM with finite storage 2. Prove the following language is non-Turing recognizable using the diagnolization
s-> AACD A-> aAb/e C->aC/a D-> aDa/bDb/e
We'll close our consideration of regular languages by looking at whether (certain) problems about regular languages are algorithmically decidable.
dfa for (00)*(11)*
One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included
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