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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 state p such that hq, p, σi ∈ T. This function is called the transition function of the automaton and is usually denoted δ:
For any state q and symbol σ, then, δ(q, σ) is the state reached from q by following a single edge labeled σ. This can be extended to the path function, a function taking a state q and any string w ∈ Σ∗ which returns the statereached from q by following a path labeled w:
Note that ˆ δ is total (has some value for all q and w) and functional (that value is unique) as a consequence of the fact that δ is, which, in turn, is a consequence of the fact that the automaton is deterministic. In terms of the transition graph, this means that for any string w and any node q, there will always be exactly one path labeled w from q (which leads to δ(q,w)) and this is a consequence of the fact that there is always exactly one edge labeled σ from each node q of the graph and every σ ∈ Σ (which leads to δ(q, σ)).
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how to prove he extended transition function is derived from part 2 and 3
Another striking aspect of LTk transition graphs is that they are generally extremely ine?cient. All we really care about is whether a path through the graph leads to an accepting
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Normal forms are important because they give us a 'standard' way of rewriting and allow us to compare two apparently different grammars G1 and G2. The two grammars can be shown to
A.(A+C)=A
We'll close our consideration of regular languages by looking at whether (certain) problems about regular languages are algorithmically decidable.
The Myhill-Nerode Theorem provided us with an algorithm for minimizing DFAs. Moreover, the DFA the algorithm produces is unique up to isomorphism: every minimal DFA that recognizes
Proof (sketch): Suppose L 1 and L 2 are recognizable. Then there are DFAs A 1 = (Q,Σ, T 1 , q 0 , F 1 ) and A 2 = (P,Σ, T 2 , p 0 , F 2 ) such that L 1 = L(A 1 ) and L 2 = L(
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