By Ivanyi A. (ed.)

**Read Online or Download Algorithms of informatics, vol. 1 PDF**

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**Extra info for Algorithms of informatics, vol. 1**

**Example text**

Therefore, no word in L can be decomposed in the form xyz, where |xyz| ≥ n, |xy| ≤ n, |y| ≥ 1, and this is why we can not obtain an infinite number of words in L. 7. Regular expressions In this subsection we introduce for any alphabet Σ the notion of regular expressions over Σ and the corresponding representing languages. A regular expression is a formula, and the corresponding language is a language over Σ. For example, if Σ = {a, b}, then a∗ , b∗ , a∗ +b∗ are regular expressions over Σ which represent respectively languages {a}∗ , {b}∗ , {a}∗ ∪ {b}∗ .

If x = (x1 + x2 ), then L = L1 ∪ L2 , where L1 and L2 are the languages which represent the regular expressions x1 and x2 respectively. By the induction hypothesis languages L1 and L2 are regular, so L is also regular because regular languages are closed on union. Cases x = (x1 x2 ) and x = (x∗1 ) can be proved by similar way. Conversely, we prove that if L is a regular language, then a regular expression x can be associated to it, which represent exactly the language L. If L is regular, then there exists a DFA A = (Q, Σ, E, {q0 }, F ) for which L = L(A).

Aj y = aj+1 aj+2 . . ak z = ak+1 ak+2 . . am . This decomposition immediately yields to |xy| ≤ n and |y| ≥ 1. We will prove that xy i z ∈ L for any i. Because u = xyz ∈ L, there exists an walk x y z q0 −→ qj −→ qk −→ qm , qm ∈ F, and because of qj = qk , this may be written also as x y z q0 −→ qj −→ qj −→ qm , qm ∈ F . y From this walk qj −→ qj can be omitted or can be inserted many times. So, there are the following walks: x z q0 −→ qj −→ qm , qm ∈ F , x y y y z q0 −→ qj −→ qj −→ . . −→ qj −→ qm , qm ∈ F .