By Francine Blanchet-Sadri
The learn of combinatorics on phrases is a comparatively new study zone within the fields of discrete and algorithmic arithmetic. that includes an easy, obtainable kind, Algorithmic Combinatorics on Partial phrases provides combinatorial and algorithmic options within the rising box of phrases and partial phrases. This publication encompasses a wealth of workouts and difficulties that assists with a number of set of rules tracing, set of rules layout, mathematical proofs, and software implementation. it's also quite a few labored instance and diagrams, making this a necessary textual content for college students, researchers, and practitioners looking to comprehend this complicated topic the place many difficulties stay unexplored.
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Additional info for Algorithmic Combinatorics on Partial Words
8 Show that if for partial words u, v we have that u ⊂ v, then P(v) ⊂ P(u) and P (v) ⊂ P (u). 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.
5 Let x and y be nonempty partial words such that xy has at most one hole. If xy ↑ yx, then there exists a word z such that x ⊂ z m and y ⊂ z n for some integers m, n. However, if xy possesses more than one hole, the situation becomes more subtle. Indeed, it is easy to produce a counterexample when xy contains just one more hole. 6 Let x = bb and y = abb . Then xy = bbabb ↑ abb bb = yx Since gcd(|x|, |y|) = 1, if x and y were contained in powers of a common word z, then |z| would be equal to 1, which is not possible for y.
If i ≥ r, then z(i) ⊂ x(i) and z(i) ⊂ y(i), z(i + k) ⊂ y(i) and z(i + k) ⊂ y(i + k), z(i + 2k) ⊂ y(i + k) and z(i + 2k) ⊂ y(i + 2k), .. z(i + (m − 1)k) ⊂ y(i + (m − 2)k) and z(i + (m − 1)k) ⊂ y(i + (m − 1)k), z(i + mk) ⊂ y(i + (m − 1)k) and z(i + mk) ⊂ x(i − r), z(i − r) ⊂ x(i − r) and z(i − r) ⊂ y(i − r), .. 56 Algorithmic Combinatorics on Partial Words Applying the above repeatedly, we can show that x(i) = y(i). More precisely, in the case where i < r, seqk,l (i) = (i, i + k, . . , i + mk, i + (m + 1)k, i + k − r, .