By Ivanyi A. (ed.)
Ivanyi A. (ed.) Algorithms of informatics, vol.1.. foundations (2007)(ISBN 9638759615)
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Extra resources for Algorithms of informatics
An−1 An−1 =⇒ a1 a2 . . an . This derivation is based on productions S → a1 A1 , A1 → a2 A2 , . . , An−2 → an−1 An−1 , An−1 → an . 2. 9. 14. Then, by the denition of the transitions of NFA A there exists a walk a a an−1 a a 1 2 3 n S −→ A1 −→ A2 −→ · · · −→ An−1 −→ Z, Z ∈ F. Thus, u ∈ L(A). If ε ∈ L(G), there is production S → ε, but in this case the initial state is also a nal one, so ε ∈ L(A). Therefore, L(G) ⊆ L(A). Let now u = a1 a2 . . an ∈ L(A). Then there exists a walk a a an−1 a a 1 2 3 n S −→ A1 −→ A2 −→ · · · −→ An−1 −→ Z, Z ∈ F.
After merging them we get an equivalent minimum state automaton (see Fig. 16). 6. Pumping lemma for regular languages The following theorem, called pumping lemma for historical reasons, may be eciently used to prove that a language is not regular. It is a sucient condition for a regular language. 15 (pumping lemma). For any regular language L there exists a natural number n ≥ 1 (depending only on L), such that any word u of L with length at least n may be written as u = xyz such that (1) |xy| ≤ n, (2) |y| ≥ 1, (3) xy i z ∈ L for all i = 0, 1, 2, .
K We can prove by induction that sets Rij can be described by regular expressik ons. Indeed, if k = −1, then for all i and j languages Rij are nite, so they can be expressed by regular expressions representing exactly these languages. Moreover, if k−1 for all i and j language Rij can be expressed by regular expression, then language k Rij can be expressed also by regular expression, which can be corresponding constk−1 k−1 k−1 k−1 ructed from regular expressions representing languages Rij , Rik , Rkk and Rkj k respectively, using the above formula for Rij .