By Francine Blanchet-Sadri

ISBN-10: 1420060929

ISBN-13: 9781420060928

The learn of combinatorics on phrases is a comparatively new examine sector within the fields of discrete and algorithmic arithmetic. that includes an easy, obtainable type, Algorithmic Combinatorics on Partial phrases offers combinatorial and algorithmic ideas within the rising box of phrases and partial phrases. This booklet features a wealth of routines and difficulties that assists with various set of rules tracing, set of rules layout, mathematical proofs, and application implementation. it is also various labored instance and diagrams, making this a necessary textual content for college students, researchers, and practitioners trying to comprehend this complicated topic the place many difficulties stay unexplored.

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

**Example text**

If this occurs, we say xy is (k, l)-special, where k = |x| and l = |y|. We adopt the convention that k ≤ l, as we can assume without loss of generality that |x| ≤ |y|. 50 Algorithmic Combinatorics on Partial Words For a partial word of length k + l, we can test for gcd(k, l)-periodicity by checking sequences of letters that are p positions apart, where p = gcd(k, l). Note that, because we are only interested in testing for periodicity, the order of checking these positions is irrelevant. For 0 ≤ i < k + l, we define the sequence of i relative to k, l as seqk,l (i) = (i0 , i1 , i2 , .

Rev(ε) = ε, and 2. rev(xa) = arev(x) where x ∈ A∗ and a ∈ A. In a similar fashion, we provide a recursive description of A∗ , the set of all words over an alphabet A: 1. ε ∈ A∗ 2. If x ∈ A∗ and a ∈ A, then xa ∈ A∗ . It is often very useful to use mathematical induction in order to prove results related to partial words. Below we provide an example of using induction on the length of a partial word to prove a result related to the reversal of the product of two words. 21 Let x, y be words over an alphabet A.

18 Prove that if u is a nonempty partial word, then there exists a S primitive word v and a positive integer n such that u ⊂ v n . (Hint: Use induction on the length of u). Show that uniqueness holds for full words, that is, if u is a nonempty full word, then there exists a unique primitive word v and a unique positive integer n such that u = v n . Does uniqueness hold for partial words? 19 S Let x and y be partial words that are compatible. Show that (xy ∨ yx) ⊂ (x ∨ y)2 . Is the reverse containment true?

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