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To see this, note that if there are any cycles in the Myhill graph of A then L(A) will be infinite, since any such cycle can be repeated arbitrarily many times. Conversely, if the Myhill graph is acyclic, then no path from x to x can be longer than card(Σ) + 2, since otherwise some node would have to occur at least twice in the path.
The question of finiteness of L(A), then, can be reduced to the question of acyclicity of the corresponding Myhill graph. And we established that there is an algorithm for testing acyclicity of graphs in Algorithms and Data Structures. Our algorithm for deciding finiteness of L(A) just interprets A as a graph and calls the algorithm for deciding acyclicity as a subroutine.
Prove xy+yz+ýz=xy+z
A.(A+C)=A
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.
To see this, note that if there are any cycles in the Myhill graph of A then L(A) will be infinite, since any such cycle can be repeated arbitrarily many times. Conversely, if the
We got the class LT by taking the class SL and closing it under Boolean operations. We have observed that LT ⊆ Recog, so certainly any Boolean combination of LT languages will also
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Myhill graphs also generalize to the SLk case. The k-factors, however, cannot simply denote edges. Rather the string σ 1 σ 2 ....... σ k-1 σ k asserts, in essence, that if we hav
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Find the Regular Grammar for the following Regular Expression: a(a+b)*(ab*+ba*)b.
The key thing about the Suffx Substitution Closure property is that it does not make any explicit reference to the automaton that recognizes the language. While the argument tha
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