找到N ^ 2个数字的中位数,其中有N个存储器

你为什么不建立直方图？即属于几个类别的案例(值)的数量.类别应该是变量的连续,非重叠区间.

使用此直方图,您可以首先估计中位数(即中位数在[a,b]之间),并知道有多少值落入此区间(H).如果H <= N,则再次读取数字,在此间隔之外忽略这些数字,并将间隔内的数字移至RAM.找到中位数.

如果H> N,则执行间隔的新分区并重复该过程.它不应该超过2或3次迭代.

请注意,对于每个分区,您只需要存储a,b,Delta和具有落入每个子区间的值数的数组.

编辑.它转出来比我想象的要复杂一点.在估计中位数落入的间隔后的每次迭代中,我们还应该考虑我们在该区间的右侧和左侧留下的"多少"直方图.我也改变了停止条件.无论如何,我做了一个C++实现.

#include <iostream>
#include <algorithm>
#include <time.h>
#include <stdlib.h>

//This is N^2... or just the number of values in your array,
//note that we never modify it except at the end (just for sorting
//and testing purposes).
#define N2 1000000
//Number of elements in the histogram. Must be >2
#define HISTN 1000

double findmedian (double *values, double min, double max);
int getindex (int *hist);
void put (int *hist, double min, double max, double val, double delta);


int main ()
{
    //Set max and min to the max/min values your array variables can hold,
    //calculate it, or maybe we know that they are bounded
    double max=1000.0;
    double min=0.0;
    double delta;
    double values[N2];
    int hist[HISTN];
    int ind;
    double median;
    int iter=0;
    //Initialize with random values   
    srand ((unsigned) (time(0)));
    for (int i=0; i<N2; ++i)
        values[i]=((double)rand()/(double)RAND_MAX);

    double imin=min;
    double imax=max;

    clock_t begin=clock(); 
    while (1) {
        iter++;
        for (int i=0; i<HISTN; ++i)
            hist[i]=0;

        delta=(imax-imin)/HISTN;
        for (int j=0; j<N2; ++j)
            put (hist, imin, imax, values[j], delta);

        ind=getindex (hist);
        imax=imin;
        imin=imin+delta*ind;
        imax=imax+delta*(ind+1);

        if (hist[ind]==1 || imax-imin<=DBL_MIN) {
            median=findmedian (values, imin, imax);
            break;
        }   
    }

    clock_t end=clock();
    std::cout << "Median with our algorithm: " << median << " - " << iter << "iterations of the algorithm" << std::endl; 
    double time=(double)(end-begin)/CLOCKS_PER_SEC;
    std::cout << "Time: " << time << std::endl;  

    //Let's compare our result with the median calculated after sorting the
    //array
    //Should be values[(int)N2/2] if N2 is odd
    begin=clock();
    std::sort (values, values+N2);
    std::cout << "Median after sorting: " << values[(int)N2/2-1] << std::endl;
    end=clock();
    time=(double)(end-begin)/CLOCKS_PER_SEC;
    std::cout << "Time: " << time << std::endl;  

    return 0;
}

double findmedian (double *values, double min, double max) {
    for (int i=0; i<N2; ++i) 
        if (values[i]>=min && values[i]<=max)
            return values[i];

    return 0;
}

int getindex (int *hist)
{
    static int pd=0;
    int left=0;
    int right=0; 
    int i;

    for (int k=0; k<HISTN; k++)
        right+=hist[k];

    for (i=0; i<HISTN; i++) {
        right-=hist[i];
        if (i>0)
            left+=hist[i-1];
        if (hist[i]>0) {
            if (pd+right-left<=hist[i]) {
                pd=pd+right-left;
                break;
            }
        }

    }

    return i;
}

void put (int *hist, double min, double max, double val, double delta)
{
    int pos;
    if (val<min || val>max)
        return;

    pos=(val-min)/delta;
    hist[pos]++;
    return;
}

为了与算法的结果进行比较,我还包括了中位数(排序)的简单计算.4或5次迭代就足够了.这意味着我们只需要从网络或硬盘读取设备4-5次.

一些结果:

N2=10000
HISTN=100

Median with our algorithm: 0.497143 - 4 iterations of the algorithm
Time: 0.000787
Median after sorting: 0.497143
Time: 0.001626

(Algorithm is 2 times faster)

N2=1000000
HISTN=1000

Median with our algorithm: 0.500665 - 4 iterations of the algorithm
Time: 0.028874
Median after sorting: 0.500665
Time: 0.097498

(Algorithm is ~3 times faster)

如果要并行化算法,每台机器可以有N个元素并计算直方图.一旦计算出来,他们就会将它发送到主机,这将对所有直方图求和(简单,它可能非常小......算法甚至适用于2个间隔的直方图).然后它将向从机发送新指令(即新间隔)以便计算新的直方图.请注意,每台机器不需要了解其他机器拥有的N个元素.