By Francine Blanchet-Sadri

ISBN-10: 1420060929

ISBN-13: 9781420060928

The research of combinatorics on phrases is a comparatively new examine zone within the fields of discrete and algorithmic arithmetic. that includes an easy, available type, Algorithmic Combinatorics on Partial phrases offers combinatorial and algorithmic innovations within the rising box of phrases and partial phrases. This publication includes a wealth of routines and difficulties that assists with a number of set of rules tracing, set of rules layout, mathematical proofs, and application implementation. it is usually quite a few labored instance and diagrams, making this a necessary textual content for college kids, researchers, and practitioners trying to comprehend this complicated topic the place many difficulties stay unexplored.

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**Additional resources for Algorithmic Combinatorics on Partial Words**

**Example text**

7 Let x = ab a a b and y = a babba a b. Show that xy ↑ yx and that xy is not (|x|, |y|)-special. Find a word z and integers m, n such that x ⊂ z m and y ⊂ z n . 3. 9 xy is not (|x|, |y|)-special. 10 S Give an example of a partial word that is {3, 6}-special without being (3, 6)-special. 11 S Give partial words u, v, w such that w ⊂ uv, w ⊂ vu and uv = vu. Is w (|u|, |v|)-special? Is it {|u|, |v|}-special? 5? 12 What can be said if u is a full word over the alphabet {0, 1} and satisfies u0 = 0u?

Clearly, rev(xy) = rev(x) = εrev(x) = rev(ε)rev(x) = rev(y)rev(x) 36 Algorithmic Combinatorics on Partial Words Now assume that our result holds for all words y where |y| = n for some nonnegative integer n. According to the process of induction, it remains for us to show that the result holds for words of length n + 1. Let |y| = n and a ∈ A. Then ya is a word of length n + 1. Now, rev(x(ya)) = rev((xy)a) = arev(xy) by definition = arev(y)rev(x) by inductive hypothesis = rev(ya)rev(x) Thus, the result holds for all words ya of length n + 1, and consequently the result is proved for all words x, y in A∗ .

9 Consider the factorization (u, v) = (abb bab, bb) of w = abb babbb. Is abb ba ∈ C(S(u))? Is b ∈ C(P (v))? 2. S A nonempty partial word u is unbordered if no nonempty words x, v, w exist such that u ⊂ xv and u ⊂ wx. Otherwise, it is bordered. If u is a nonempty unbordered partial word, then show that p(u) = |u| and consequently, unbordered partial words are primitive. 12 Different occurrences of the same unbordered factor u in a partial word w never overlap. True or false? 13 S Two partial words u and v are called conjugate if there exist partial words x and y such that u ⊂ xy and v ⊂ yx.

### Algorithmic Combinatorics on Partial Words by Francine Blanchet-Sadri

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