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De?nition Deterministic Finite State Automaton: For any state set Q and alphabet Σ, both ?nite, a ?nite state automaton (FSA) over Q
and
Σ is a ?ve-tuple (Q,Σ, T, q0, F), where:
• T ⊆ Q × Q × Σ,
• q0 ∈ Q is the initial state (also know as the start state) and
• F ⊆ Q is the set of accepting states (also spuriously known as ?nal states).
The FSA is deterministic (a DFA) if for all q ∈ Q and σ ∈ Σ, there is exactly one p ∈ Q such that (q, p, σ) ∈ T.
Each triple in T = hq, p, σi represents an edge from state q to p labeled σ in the transition graph. The state q0 is the initial state of the transition graph (marked by the "edge from nowhere") and the states in F are the states distinguished by being circled. An FSA is deterministic if there is never any choice of what the next state is, given the current state and input symbol and there is never no choice. In terms of the transition graph, this means that every node will have exactly one out-edge for each symbol of the alphabet.
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The fact that SL 2 is closed under intersection but not under union implies that it is not closed under complement since, by DeMorgan's Theorem L 1 ∩ L 2 = We know that
When an FSA is deterministic the set of triples encoding its edges represents a relation that is functional in its ?rst and third components: for every q and σ there is exactly one
A.(A+C)=A
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In Exercise 9 you showed that the recognition problem and universal recognition problem for SL2 are decidable. We can use the structure of Myhill graphs to show that other problems
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The initial ID of the automaton given in Figure 3, running on input ‘aabbba' is (A, aabbba) The ID after the ?rst three transitions of the computation is (F, bba) The p
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