نتایج جستجو برای: recovery empty period
تعداد نتایج: 694024 فیلتر نتایج به سال:
An overlap is a word of the form axaxa, where a is a single letter and x is a (possibly empty) word. In 1980, Earl Fife [8] proved a theorem characterizing the infinite binary overlap-free words as encodings of paths in a finite automaton. Berstel [4] later simplified the exposition, and both Carpi [6] and Cassaigne [7] gave an analogous analysis for the case of finite words. In a previous pape...
Let us observe that there exists a 2-sorted which is strict and non empty and there exists a 2-sorted which is strict and non quasi-empty. Let us observe that there exists a 2-sorted which is strict, empty, and quasi-empty. We consider ContextStr as extensions of 2-sorted as systems 〈 objects, attributes, an information 〉, where the objects and the attributes constitute sets and the information...
If u and v are words, uv denotes their concatenation. For a positive integer n, u = uuu...u (n times). By ǫ we denote the only word of length 0, the empty word. A is the set of all words over A of length n. A = ∪n≥1A n is a free semigroup (an associative set with a binary operation) over A with the group operation being string concatenation. We define A = A ∪ {ǫ}. Definition 1.2. A word u is a ...
A tandem repeat (or square) is a string aa; where a is a non-empty string. We present an OðjSjÞ-time algorithm that operates on the suffix tree TðSÞ for a string S; finding and marking the endpoint in TðSÞ of every tandem repeat that occurs in S: This decorated suffix tree implicitly represents all occurrences of tandem repeats in S; and can be used to efficiently solve many questions concernin...
Proof (complete induction on n): If n = 0, then P (0) asserts that there is f(0) = 1 distinct output starting with an empty string. This is certainly true, since the unique empty string is output, so the base case holds. Induction step: Assume that P (f0; : : : ; n 1g) is true for some arbitrary natural number n. I need to prove that this implies P (n) is true. If n = 0 there is nothing to prov...
Let R be a non empty poset and let A, B be many sorted sets indexed by R. One can check that every many sorted relation between A and B which is os-compatible is also order-sorted. Let R be a non empty poset and let A be a many sorted set indexed by the carrier of R. An order sorted relation of A is an order sorted relation of A, A. Let S be an order sorted signature and let U1 be an order sort...
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