Counter-state machines, continued
In a previous post I illustrated a way to count how many strings there are in a formally defined "regular language". One question was, supposing the state machine that recognizes the language has states from which no accepting state is reachable; is its observability matrix still full rank? The answer is no, and here's an example of that, and how to work around it. First, the regular language. Define the alphabet of symbols as "0" and "1", and the grammar rule is this: a finite string is in the language if and only if it contains no "11" substring. Thereafter, I'll use the notions introduced in that previous post. The resulting state machine is given below (initial state omitted), where the "_0" state corresponds to a string ending with "0" and "_1" corresponds to a string ending with "01". "meh" is a rejecting state, meaning the string contains a "11" substring. I'll li