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The submatrices character count problem: an efficient solution using separable values

机译:子矩阵字符计数问题:使用可分数值的有效解决方案

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The subsequence character count problem has as its input an array S = S[1],..., S[n] of symbols over alphabet E and a natural number m. Its output is: for every i, i = 1,..., n - m + 1, the number of different alphabet symbols occurring in the subsequence S[i], S[i + 1],..., S[i + m - 1]. The subsequence character count problem is a natural problem that has many uses. It can be solved in linear time for finite alphabets and in time O(n log m) for infinite alphabets. When the character count problem is generalized to two dimensions it becomes the submatrix character count problem. Its input is an n x n matrix T over alphabet E and a natural number m. Its output is: for every i, j, i, j = 1,..., n - m + 1, the number of different alphabet symbols occurring in the submatrix T [i + k, j + l], k = 0,...,m - 1; l = 0,...,m - 1. The straightforward one-dimensional solution slides a window along the text adding an element and deleting an element at every step. The problem with two dimensions is that at every move of the window there are m elements added and m deleted. In this paper, we present an alternate one-dimensional solution that generalizes to two dimensions. We achieve a O(n(2)) time solution to the submatrix character count problem over a finite alphabet and a O(n(2) log m) solution over an infinite alphabet. (C) 2003 Elsevier Inc. All rights reserved. [References: 13]
机译:子序列字符计数问题的输入是在字母E和自然数m上的符号数组S = S [1],...,S [n]。它的输出是:对于每个i,i = 1,...,n-m + 1,在子序列S [i],S [i + 1],...,S [中出现的不同字母符号的数量i + m-1]。子序列字符计数问题是具有许多用途的自然问题。对于有限字母,可以在线性时间中求解;对于无限字母,可以在时间O(n log m)中求解。当字符计数问题被推广到二维时,它将成为子矩阵字符计数问题。它的输入是字母E和自然数m上的n x n矩阵T。其输出为:对于每个i,j,i,j = 1,...,n-m + 1,在子矩阵T [i + k,j + l]中出现的不同字母符号的数目,k = 0 ,...,m-1; l = 0,...,m-1。简单的一维解决方案使窗口沿文本滑动,并在每个步骤添加元素和删除元素。二维的问题是,在窗口的每一次移动中,都会添加m个元素,删除m个元素。在本文中,我们提出了可替代的一维解决方案,该解决方案可推广到二维。我们在有限字母上实现了子矩阵字符计数问题的O(n(2))时间解,在无限字母上实现了O(n(2)log m)解决方案。 (C)2003 Elsevier Inc.保留所有权利。 [参考:13]

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