median-of-medians | 易学教程

median of median implementation

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问题 Here is pseudo code for implementation of median by dividing array into 5 groups select(int A[],int first, int last, int i) { n = last - first + 1; /* n is the number elements to select from */ if (i > n) {return ERROR;} /* there is no ith smallest element */ if( n < = 100 ) { /********************* For Small n *********************/ Run selection on A[first..last] taking at most n(n-1)/2 < 50n comparisons; swap A[first+i-1] with A[first] /* put ith smallest in A[first] */ } else /* n > 100 *

Why is the median-of-medians algorithm described as using O(1) auxiliary space?

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问题 Wikipedia lists the median-of-medians algorithm as requiring O(1) auxiliary space. However, in the middle of the algorithm, we make a recursive call on a subarray of size n/5 to find the median of medians. When this recursive call returns, we use the returned median of medians as a pivot to partition the array. Doesn't this algorithm push O(lg n) activation records onto the run-time stack as a part of the recursion? From what I can see, these recursive calls to find successive medians of

Time complexity of median of medians algorithm

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问题 Hello I am taking Introduction to algorithm class this semeseter. However I have some problem in calculating time complexity of median of medians algorithm (here). I'm wondering how to get T(n)<=10cn from T(n)<=T(0.2n)+T(0.7n)+cn.. I think I cannot apply mater theorem to the expression above and wikipedia says I should use induction but I don't know How.. 回答1: It is using induction. Assume for less than or equal n we have T(n) <= 10*c*n (we know the base of induction is true for equal or less

Median of Medians space complexity

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问题 I implemented an nth_number selection algorithm using Medians of Medians. On wikipedia, it states that it's space complexity is O(1) I had to store the medians in a temporary array in order to find the median amongst those medians. How would you be able to do it without using any extra memory? If it does not count as increasing its space complexity, please explain. function nth_number(v, n) { var start = 0; var end = v.length - 1; var targetIndex = n - 1; while(true) { var medians = []; /*

Something I dont understand about median of medians algorithm

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问题 There is something I don't understand about the algorithm of median of medians. One key step about this algorithm is to find an approximate median, and according to Wikipedia, we have the guarantee that this approximate median is greater than 30% of elements of the initial set. To find this approximate median, we compute the median of each group of 5 elements, we gather these medians in a new set, and we recompute the medians until the obtained set have least than 5 elements. In this case, we

Design of an algorithm problems of Clog n[C++ code]

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问题 Two sorted arrays of integers A[1..N] and B[1..N] are provided in ascending order . Q:Design an O(log N)-time algorithm for finding out the median of all 2N integers. N may not power of 2 . To make thing easy, we can assume O(1) algorithm which return m such that: 2^m < N < 2^m+1 My problems: N may not be power of 2 , what does that mean? (understood) I don't know how to change the input and make the length to power of 2 after found min and max elements from both array A and B . 回答1: You can

median of median implementation

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Here is pseudo code for implementation of median by dividing array into 5 groups select(int A[],int first, int last, int i) { n = last - first + 1; /* n is the number elements to select from */ if (i > n) {return ERROR;} /* there is no ith smallest element */ if( n < = 100 ) { /********************* For Small n *********************/ Run selection on A[first..last] taking at most n(n-1)/2 < 50n comparisons; swap A[first+i-1] with A[first] /* put ith smallest in A[first] */ } else /* n > 100 */ { /********** main recursion *************************/ numGroups = n / 5; /* integer division, round

Explanation of the Median of Medians algorithm

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问题 The Median of medians approach is very popular in quicksort type partitioning algorithms to yield a fairly good pivot, such that it partitions the array uniformly. Its logic is given in Wikipedia as: The chosen pivot is both less than and greater than half of the elements in the list of medians, which is around n/10 elements (1/2 * (n/5)) for each half. Each of these elements is a median of 5, making it less than 2 other elements and greater than 2 other elements outside the block. Hence, the

Worst-case O(n) algorithm for doing k-selection

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Apart from the median-of-medians algorithm, is there any other way to do k-selection in worst-case O(n) time? Does implementing median-of-medians make sense; I mean, is the performance advantage good enough for practical purposes ? There is another algorithm for computing kth order statistics based on the soft heap data structure, which is a variant on a standard priority queue that is allowed to "corrupt" some number of the priorities it stores. The algorithm is described in more detail on the Wikipedia article, but the basic idea is to use the soft heap to efficiently (O(n) time) pick a

Worst-case O(n) algorithm for doing k-selection

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问题 Apart from the median-of-medians algorithm, is there any other way to do k-selection in worst-case O(n) time? Does implementing median-of-medians make sense; I mean, is the performance advantage good enough for practical purposes ? 回答1: There is another algorithm for computing kth order statistics based on the soft heap data structure, which is a variant on a standard priority queue that is allowed to "corrupt" some number of the priorities it stores. The algorithm is described in more detail