L = {ab^k | k element of the natural numbers without 0} i.e. 1,2,3,4… ab, abb, abbb, abbbb, ab… are all valid words in the language. I could now add any number of b's to a as a suffix. How many equivalence classes are there here? Please explain. Infinitely many?
Hello, I would be very grateful if someone could help me with part b) of the task: My solution a) 0(1)*(101|01(0)*0)(0*1*)* as a regular expression b) I have marked the non-deterministic places in the NFA Now we need to convert the NFA into a DFA. Normally, you would form the power set, which would be…
The usual procedure is to construct an NFA (nondeterministic automaton) from a language L(G) with a grammar. Then, from the NFA, a DFA (deterministic automaton) is constructed, and from this, the language with its grammar L(G) is derived. L(G) to NFA to DFA to LG? What I am wondering now is, can one also proceed…