![]() Now reduce the size of the GFA by one stateat each step. Each state has transitions to all other states (including itself), except the super start state, with no incoming transitions, and the super final state, which has no outgoing transitions. Proof: use “generalized finite automata” where a transition can be a regular expression (not just a symbol), and: Only 1 super start state and 1 (separate) super final state. Theorem: Any FA accepts a language denoted by some RE. Theorem: Minimizing the number of states in a pushdown automaton (or TM) is undecidable. ![]() (Is this because non-deterministic FA corresponds to deterministic FA with an exponential number of nodes?) Ĭonjecture: Minimizing the number of states in a nondeterministic FA can not be done in polynomial time. Theorem : the number N of states in a FA can be minimized within time O(N log N). Note here suppose to be an example but I’m lazy to draw it. Theorem: Any regular expression is accepted by some FA.Ī FA for a regular expression can be built by composition. Let’s start with refinement of terminology from part 1.Īlphabet ( Σ - sigma) is a finite, non-empty set of symbols letters. Yes Close Home Privacy Policy RSS Notes on parsing theory, part 3 Jan 28, 2021īased on Theory of Computation (CS3102), Spring 2017, videos are here. For recognizer sometime we use deciders to decide, if machine is in looping then decider will reject according to our description. ![]() Can I store cookies on your device? See Privacy Policy for details. The simple Answer as I thinks is: Decider always halts, accept or reject But Recognizer do not always halt, Machine can accept, reject or loop. ![]()
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