By Francine Blanchet-Sadri

The learn of combinatorics on phrases is a comparatively new study sector within the fields of discrete and algorithmic arithmetic. that includes an easy, obtainable kind, Algorithmic Combinatorics on Partial phrases offers combinatorial and algorithmic suggestions within the rising box of phrases and partial phrases. This e-book features a wealth of workouts and difficulties that assists with numerous set of rules tracing, set of rules layout, mathematical proofs, and application implementation. it's also a variety of labored instance and diagrams, making this a helpful textual content for college students, researchers, and practitioners trying to comprehend this advanced topic the place many difficulties stay unexplored.

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

**Sample text**

Using a partial function, we can define a function u : {0, 1, 2, 3, 4} → A and then acknowledge that u(2) and u(4) are undefined. We make the following definition. 2 A partial word (or, pword) of length n over A is a partial function u : {0, . . , n − 1} → A. For 0 ≤ i < n, if u(i) is defined, we say that i belongs to the domain of u (denoted by i ∈ D(u)). Otherwise we say that i belongs to the set of holes of u (denoted by i ∈ H(u)). Just as every total function is a partial function, every total word is itself a partial word with an empty set of holes.

Case 1 refers to 0 ≤ i < k, Case 2 to k ≤ i < l, and Case 3 to l ≤ i < l + k (Cases 1 and 3 are symmetric as is seen by putting i = l + j where 0 ≤ j < k). The following diagram pictures the containments xy ⊂ u and yx ⊂ u: 52 Algorithmic Combinatorics on Partial Words xy x(0) . . x(k − 1) y(0) . . y(l − k − 1) y(l − k) . . y(l − 1) yx y(0) . . y(k − 1) y(k) . . y(l − 1) x(0) . . x(k − 1) u u(0) . . u(k − 1) u(k) . . u(l − 1) u(l) . . u(l + k − 1) We prove the result for Case 1 under the assumption that r > 0.

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. Definition of (k, l)-special partial word To extend this theorem to the case when xy has at least two holes, we will need to inspect the structure of the partial word xy more carefully by stepping through a sequence of positions.