Combinatorics on Words – A Tutorial

Berstel, Jean; Karhumäki, Juhani

doi:10.1142/9789812562494_0059

Cited by 33 publications

(42 citation statements)

References 113 publications

(161 reference statements)

Supporting

Mentioning

Contrasting

Unclassified

Order By: Relevance

“…Литературу можно посмотреть в [4,5] в контексте более широкой области исследования, называемой Combinatorics on words .…”

Section: а а евдокимовunclassified

An algorithm for recognizing the completeness of a set of words and dynamics of prohibitions

Евдокимов¹

2016

Applied Discrete Mathematics. Supplement

View full text Add to dashboard Cite

Рассмотрим критерий проверки гипотезы H 0 против H 1 , основанный на статисти-ке (1):где α = P S t χ 2 2 ,1−α H 0 − вероятность ошибки первого рода; t χ 2 2 ,α − квантиль уровня α распределения χ-квадрат с двумя степенями свободы.Теорема 1. Пусть в модели вкраплений p a = π a , a ∈ {0, 1}, и среди элементов матрицы переходных вероятностей Π есть хотя бы один, отличный от 1/2. Тогда при выполнении условийкритерий (2) проверки гипотезы H 0 против альтернативы H 1 является состоятельным.Замечание 1. При отсутствии вкраплений (τ ≡ 0) и при наличии вкраплений во всех позициях последовательности X (τ ≡ 1) гипотезы H 0 и H 1 неразличимы, по-скольку в обоих случаях Y является простой однородной цепью Маркова (с глубиной зависимости 1 и 0 соответственно). Критерий будет состоятельным, когда вкраплений не слишком много , что гарантируется условием (3), но в то же время когда чис-ло вкраплений превосходит по порядку квадратный корень из длины наблюдаемого отрезка последовательности X (условие (4)). А. А. ЕвдокимовВводятся инвариантные операции и даётся описание алгоритма распознавания полноты множества слов. Приводится теорема о результатах работы алгоритма и их отношении к свойству полноты исходного множества слов. Формулируется 1 Работа поддержана Новосибирским государственным университетом и грантом РФФИ, проект № 14-01-00507.

show abstract

Section: а а евдокимовunclassified

An algorithm for recognizing the completeness of a set of words and dynamics of prohibitions

Евдокимов¹

2016

Applied Discrete Mathematics. Supplement

View full text Add to dashboard Cite

show abstract

“…This hierarchy comes from combinatorics on words, where these classes are being investigated intensively (cf. [2,8,9,10]). Bi-ideal sequences have been considered, with different names, by several authors in algebra and combinatorics [1,3,7,13,17].…”

Section: Introductionmentioning

confidence: 99%

From Bi-ideals to Periodicity

Buls

Lorencs

2008

RAIRO-Theor. Inf. Appl.

View full text Add to dashboard Cite

show abstract

“…Auxiliary results. We shall need some terminology which is usually used in combinatorics on words (see, e.g., [2], [4], [13]). Any sequence (finite or infinite) of letters of an alphabet A is called a word.…”

mentioning

confidence: 99%

Powers of a rational number modulo 1 cannot lie in a small interval

Dubickas

2009

Acta Arith.

View full text Add to dashboard Cite

Powers of a rational number modulo 1 cannot lie in a small interval by Artūras Dubickas (Vilnius)1. Introduction. Let throughout R, Z and N be the sets of real numbers, integers and positive integers, respectively. We will denote by [x] and {x} the integral part and the fractional part of x ∈ R, respectively. For an interval [s, s + t) ⊂ [0, 1) and two integers p, q, where 1 < q < p, putIn [14] Mahler asked whether the set Z 3/2 (0, 1/2) is empty or not. A hypothetical ξ ∈ Z 3/2 (0, 1/2) is called a Z-number. It seems very likely that Z-numbers do not exist. An important step towards solution of this problem has been made by Flatto, Lagarias and Pollington [12] (see also [11]). It was proved in [12] that for coprime positive integers p > q > 1 and any ξ = 0 the inequalityholds. A generalization of (1) to powers of algebraic numbers is given in [9]. The case of positive integers, namely, p ≥ 2, q = 1 was studied in [7]. Inequality (1) implies that the fractional parts {ξ(p/q) n }, n = 0, 1, 2, . . . , cannot lie in an interval of length strictly smaller than 1/p. Can they all lie in an interval of length 1/p? This small step towards Mahler's problem turns out to be very difficult. It was shown in [12] that the set of s ∈ [0, 1−1/p] for which Z p/q (s, s + 1/p) is empty is everywhere dense in [0, 1 − 1/p]. Naturally, it was conjectured that Z p/q (s, s + 1/p) is empty for each s ∈ [0, 1 − 1/p] (see p. 138 in [12]).This problem is still open, although Bugeaud has made some progress in this direction in [6]. He was able to prove that Z p/q (s, s + 1/p) is empty for almost all s ∈ [0, 1 − 1/p]. Moreover, he showed that the set Z 3/2 (s, s + 1/3) 2000 Mathematics Subject Classification: 11J71, 68R15.

show abstract

Combinatorics on Words – A Tutorial

Cited by 33 publications

References 113 publications

An algorithm for recognizing the completeness of a set of words and dynamics of prohibitions

An algorithm for recognizing the completeness of a set of words and dynamics of prohibitions

From Bi-ideals to Periodicity

Powers of a rational number modulo 1 cannot lie in a small interval

Contact Info

Product

Resources

About