Jobe Smith, Joan · Dipòsit Digital de Documents de la UAB (Universitat Autònoma de Barcelona) · 1990
I give her mornings and bedtime in apple juice more potent to her 70 pounds than a $100 heroin hit to a prickled L.A. hype.Sad and ashamed of her addiction as much as her disease, sometimes she weeps as she sucks through a straw the last drop of morphine from the cup, and sometimes I imagine Deukmejian and the DEA boys breaking down my mother's bedroom door -conquistadores roaring "Eureka!" -coming to prick their spears at u s , a couple pagans all right, red-eyed and doped-up, naked with sin and death.
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Oxford University Press eBooks · 2023
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Manuel Lafond, Wenfeng Lai, Adiesha Liyanage et al. · Information and Computation · 2025
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Md. Mahbubul Hasan, A. S. M. Sohidull Islam, M. Sohel Rahman et al. · Mathematics in Computer Science · 2017
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Binhai Zhu, Manuel Lafond, Wenfeng Lai et al. · SSRN Electronic Journal · 2025
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Norbert Hegyvári · 1998
A is said to be complete if every suÆciently large integer belongs to sumset of A. A 0 is thin comlete subsequence of A if A 0 is complete and A 0 (x) = (1 + o(1)) log 2 x. It is proved that lim n!1 a n+1 =a n = 1 implies the existence of thin complete subsequence. 1. Introduction The well known theorem of Lagrange states that every nonnegative integer n is the sum of four squares. In other words the sequence S = f1; 4; : : : ; n 2 ; : : :g is bases of order four. Generally if A is a bases of order h then A(x) x 1=h , where A(x) is the counting function of A. Wirsing dened the notion of thin bases; A is thin bases of order h if A(x) < c 0 x 1=h ; (c 0 > 0). By a non-constructive method Wirsing proved [1] that S has a subbases S 0 which is almost thin, proving S 0 (x) = O(x 1=4 (log x) 1=4 ). Later J. Spencer [2] gave a short proof for it, using an important tool of probabilistic method, the Janson's inequality. We merely mention it is not even known an ...
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Longin Jan Latecki, Qiang Wang, Suzan Köknar-Tezel et al. · 2007
We consider the problem of elastic matching of sequences of real numbers. Since both a query and a target sequence may be noisy, i.e., contain some outlier elements, it is desirable to exclude the outlier elements from matching in order to obtain a robust matching performance. Moreover, in many applications like shape alignment or stereo correspondence it is also desirable to have a one-to-one and onto correspondence (bijection) between the remaining elements. We propose an algorithm that determines the optimal subsequence bijection (OSB) of a query and target sequence. The OSB is efficiently computed since we map the problem's solution to a cheapest path in a DAG (directed acyclic graph). We obtained excellent results on standard benchmark time series datasets. We compared OSB to Dynamic Time Warping (DTW) with and without warping window. We do not claim that OSB is always superior to DTW. However, our results demonstrate that skipping outlier elements as done by OSB can significantly improve matching results for many real datasets. Moreover, OSB is particularly suitable for partial matching. We applied it to the object recognition problem when only parts of contours are given. We obtained sequences representing shapes by representing object contours as sequences of curvatures.
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Rok Seon Choung (2506954), Eric V. Marietta (169348), Carol T. Van Dyke (2506942) et al. · Figshare · 2016
<p>The 3-mer subsequences with maximum occurrences among sequences with high sensitivity and specificity for among IgG and IgA reactivity were determined, and the best combinations of subsequences were plotted as a matrix table. The nine 3-mers with the highest score were combined in each possible iteration of two 3-mers to form 81 6-mer sequences, with N-terminal sub-sequences listed in the column, and the C-terminal sub-sequences listed in the row. A) IgG 3-mer subsequences, maximum occurrences. B) IgA 3-mer subsequences, maximum occurrences. % implies any number of amino acid(s) present prior to the subsequence portion of the sequence. For example, ‘%QPE’ could be WALLTYQPE while ‘%QPE%’ could be WAQPELLTY.</p>
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Roozbeh Zarei (16839266), Guangyan Huang (10230665), Junfeng Wu (23234) · Figshare · 2023
<p>(a) Score of each subsequence in the sample time series in <a href="http://www.plosone.org/article/info:doi/10.1371/journal.pone.0290092#pone.0290092.g002" target="_blank">Fig 2(a)</a>. The red points indicate the four lowest scores corresponding to four subsequences with the length of <i>ℓ</i> = 110 starting at different positions <i>i</i> in time series. The exclusion-zones are highlighted in gray. (b) Top-4 abnormal subsequences reported and ranked (1<sup><i>st</i></sup> − 4<sup><i>th</i></sup>) by the GraphTS.</p>
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Yung-hsing Peng, Hsing-yen Ann, Communicated L. Boasson · 2006
The longest common subsequence (LCS) problem can be used to measure the relationship between sequences. In general, the inputs of the LCS problem are two sequences. For finding the relationship between one sequence and a set of sequences, we cannot apply the traditional LCS algorithms immediately. In this paper, we define the mosaic LCS (MLCS) problem of finding a mosaic sequence C, composed of repeatable k sequences in source sequence set S, such that the LCS of C and the target sequence T is maximal. Based on the concept of break points in sequence T, we first propose a divide-and-conquer algorithm with O(n2m|S|+n3 log k) time for solving this problem, where n and m are the length of T and the maximal length of sequences in S, respectively. Furthermore, an improved algorithm with O(n(m+k)|S|) time is proposed by applying an efficient preprocessing for the MLCS problem.
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Tsubasa Oizumi, Takeshi Kai, Takuya Mieno et al. · arXiv (Cornell University) · 2022
Park et al. [TCS 2020] observed that the similarity between two (numerical) strings can be captured by the Cartesian trees: The Cartesian tree of a string is a binary tree recursively constructed by picking up the smallest value of the string as the root of the tree. Two strings of equal length are said to Cartesian-tree match if their Cartesian trees are isomorphic. Park et al. [TCS 2020] introduced the following Cartesian tree substring matching (CTMStr) problem: Given a text string $T$ of length $n$ and a pattern string of length $m$, find every consecutive substring $S = T[i..j]$ of a text string $T$ such that $S$ and $P$ Cartesian-tree match. They showed how to solve this problem in $\tilde{O}(n+m)$ time. In this paper, we introduce the Cartesian tree subsequence matching (CTMSeq) problem, that asks to find every minimal substring $S = T[i..j]$ of $T$ such that $S$ contains a subsequence $S'$ which Cartesian-tree matches $P$. We prove that the CTMSeq problem can be solved efficiently, in $O(m n p(n))$ time, where $p(n)$ denotes the update/query time for dynamic predecessor queries. By using a suitable dynamic predecessor data structure, we obtain $O(mn \log \log n)$-time and $O(n \log m)$-space solution for CTMSeq. This contrasts CTMSeq with closely related order-preserving subsequence matching (OPMSeq) which was shown to be NP-hard by Bose et al. [IPL 1998].
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Łukasz Mazurkiewicz, Szymon Żeberski · Topology and its Applications · 2024
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Jason Chen · IGI Global eBooks · 2011
Clustering analysis is a tool used widely in the Data Mining community and beyond (Everitt et al. 2001). In essence, the method allows us to “summarise” the information in a large data set X by creating a very much smaller set C of representative points (called centroids) and a membership map relating each point in X to its representative in C. An obvious but special type of data set that one might want to cluster is a time series data set. Such data has a temporal ordering on its elements, in contrast to non-time series data sets. In this article we explore the area of time series clustering, focusing mainly on a surprising recent result showing that the traditional method for time series clustering is meaningless. We then survey the literature of recent papers and go on to argue how time series clustering can be made meaningful. Request access from your librarian to read this chapter's full text.
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Štěpán Ondřej · Digital Repository (National Repository of Grey Literature) · 2024
The aim of the thesis is to give complete and thorough proofs of some well-known results from the measure theory. Oftentimes, arguments from functional analysis will be used to prove these results. For example, we will use an enhanced version of Schur's theorem to prove the Nikodym theorem and the Vitali-Hahn-Saks theorem. Then we will focus on the weak compactness in L1 and we will present a proof of the Biting Lemma and its corollary Rosenthal's subsequence splitting lemma. 1
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Xiang Li, Qian Shi, Ke Chen et al. · Bioinformatics · 2023
SubseqHash is freely available at https://github.com/Shao-Group/subseqhash.
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Prasad Tetali · Advances in Applied Probability · 1992
An abstract is not available for this content so a preview has been provided. Please use the Get access link above for information on how to access this content.
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Satoshi Matsumoto, Ayumi Shinohara · Kyoto University Research Information Repository (Kyoto University) · 1996
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Thomas Tzvi Scheckter, Fedor Sukochev, Dmitriy Zanin · Studia Mathematica · 2024
Generalised subsequence principles extend almost everywhere convergence results for sequences of independent random variables, satisfying a moment condition, to subsequences of an arbitrary sequence of functions which satisfies the same moment conditi
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Oizumi, Tsubasa, Kai, Takeshi, Mieno, Takuya et al. · DROPS (Schloss Dagstuhl – Leibniz Center for Informatics) · 2022
Park et al. [TCS 2020] observed that the similarity between two (numerical) strings can be captured by the Cartesian trees: The Cartesian tree of a string is a binary tree recursively constructed by picking up the smallest value of the string as the root of the tree. Two strings of equal length are said to Cartesian-tree match if their Cartesian trees are isomorphic. Park et al. [TCS 2020] introduced the following Cartesian tree substring matching (CTMStr) problem: Given a text string T of length n and a pattern string of length m, find every consecutive substring S = T[i..j] of a text string T such that S and P Cartesian-tree match. They showed how to solve this problem in Õ(n+m) time. In this paper, we introduce the Cartesian tree subsequence matching (CTMSeq) problem, that asks to find every minimal substring S = T[i..j] of T such that S contains a subsequence S' which Cartesian-tree matches P. We prove that the CTMSeq problem can be solved efficiently, in O(m n p(n)) time, where p(n) denotes the update/query time for dynamic predecessor queries. By using a suitable dynamic predecessor data structure, we obtain O(mn log log n)-time and O(n log m)-space solution for CTMSeq. This contrasts CTMSeq with closely related order-preserving subsequence matching (OPMSeq) which was shown to be NP-hard by Bose et al. [IPL 1998].
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Alban Mancheron, Hal Id Inria · 2008
HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et a ̀ la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.
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