基于汉明重量的索引

时间:2012-11-24 15:53:20

标签: c sorting cuda indexing hammingweight

假设我们有一个bitsize n=4;的整数 我所描述的问题是如何根据汉明权重及其知道bitsize的值将数字索引到数组位置。例如。 具有16个用于bitsize 4的元素的数组将看起来像这样:

|0|1|2|4|8|3|5|6|9|10|12|7|11|13|14|15|

元素按汉明重量(必要)分组,并根据大小排序(不必要)。 只要您可以进行排序,就不需要排序。 3(0011)做一些操作并返回索引5,5(0101) - > 6等。

n位的所有组合都将存在,并且不会重复。例如。 3的bitsize将具有数组:

|0|1|2|4|3|5|6|7|

我最好有一个没有循环的解决方案。 或任何讨论simillar解决方案的论文。 或者最后只是抛出任何关于如何做到这一点的想法。

2 个答案:

答案 0 :(得分:13)

请注意,您可以使用以下函数使用相同的汉明重量枚举数字(按计数顺序):

int next(int n) { // get the next one with same # of bits set
  int lo = n & -n;       // lowest one bit
  int lz = (n + lo) & ~n;      // lowest zero bit above lo
  n |= lz;                     // add lz to the set
  n &= ~(lz - 1);              // reset bits below lz
  n |= (lz / lo / 2) - 1;      // put back right number of bits at end
  return n;
}

int prev(int n) { // get the prev one with same # of bits set
   int y = ~n;
   y &= -y; // lowest zero bit
   n &= ~(y-1); // reset all bits below y
   int z = n & -n; // lowest set bit
   n &= ~z;        // clear z bit
   n |= (z - z / (2*y)); // add requried number of bits below z
   return n;
 }

例如,在x = 5678上重复应用prev():

0: 00000001011000101110 (5678)
1: 00000001011000101101 (5677)
2: 00000001011000101011 (5675)
3: 00000001011000100111 (5671)
4: 00000001011000011110 (5662)
5: 00000001011000011101 (5661)
6: 00000001011000011011 (5659)
.....

因此理论上你可以通过重复应用来计算数字的索引 这个。然而,这可能需要很长时间。更好的方法是“跳过”某些组合。

有两条规则:

 1. if the number starts with: ..XXX10..01..1 we can replace it by ..XXX0..01..1
adding corresponding number of combinations
 2. if the number starts with: ..XXX1..10..0 again replace it by XXX0..01..1 with corresponding number of combinations 

以下算法使用相同的汉明权重来计算数字中的数字的索引(我没有打扰快速实现二项式):

#define LOG2(x) (__builtin_ffs(x)-1)

int C(int n, int k) { // simple implementation of binomial
 int c = n - k; 
 if(k < c) 
   std::swap(k,c);
 if(c == 0)
  return 1;
 if(k == n-1) 
  return n;
 int b = k+1;
 for(int i = k+2; i <= n; i++) 
    b = b*i;
 for(int i = 2; i <= c; i++)
   b = b / i;
 return b;
}
int position_jumping(unsigned x) {
   int index = 0;
  while(1) {

    if(x & 1) { // rule 1: x is of the form: ..XXX10..01..1
        unsigned y = ~x;
        unsigned lo = y & -y; // lowest zero bit
        unsigned xz = x & ~(lo-1); // reset all bits below lo
        unsigned lz = xz & -xz; // lowest one bit after lo
        if(lz == 0) // we are in the first position!
           return index;

        int nn = LOG2(lz), kk = LOG2(lo)+1;       
        index += C(nn, kk); //   C(n-1,k) where n = log lz and k = log lo + 1

        x &= ~lz; //! clear lz bit
        x |= lo; //! add lo

    } else { // rule 2: x is of the form: ..XXX1..10..0
        int lo = x & -x; // lowest set bit
        int lz = (x + lo) & ~x;  // lowest zero bit above lo  
        x &= ~(lz-1); // clear all bits below lz
        int sh = lz / lo;

        if(lz == 0) // special case meaning that lo is in the last position
            sh=((1<<31) / lo)*2;
        x |= sh-1;

        int nn = LOG2(lz), kk = LOG2(sh);
        if(nn == 0)
           nn = 32;
        index += C(nn, kk);
    }
    std::cout << "x: " << std::bitset<20>(x).to_string() << "; pos: " << index << "\n";
  }
 }

例如,给定数字x = 5678 该算法将在4次迭代中计算其索引:

  x: 00000001011000100111; pos: 4
  x: 00000001011000001111; pos: 9
  x: 00000001010000011111; pos: 135
  x: 00000001000000111111; pos: 345
  x: 00000000000001111111; pos: 1137

请注意,1137是具有相同汉明权重的数字组中的5678的位置。因此,您必须相应地移动此索引以考虑具有较小汉明权重的所有数字

答案 1 :(得分:1)

这是一个概念工作,只是为了开始讨论 第一步是最困难的 - 用近似值求解计算因子 还有更好的想法吗?

Ideone link

#include <stdio.h>
#include <math.h>

//gamma function using Lanczos approximation formula
//output result in log base e
//use exp() to convert back
//has a nice side effect: can store large values in small [power of e] form
double logGamma(double x)
{
    double tmp = (x-0.5) * log(x+4.5) - (x+4.5);
    double ser = 1.0 + 76.18009173     / (x+0) - 86.50532033    / (x+1)
                     + 24.01409822     / (x+2) -  1.231739516   / (x+3)
                     +  0.00120858003  / (x+4) -  0.00000536382 / (x+5);
    return tmp + log(ser * sqrt(2*M_PI) );  
}

//result from logGamma() are actually (n-1)!
double combination(int n, int r)
{
    return exp(logGamma(n+1)-( logGamma(r+1) + logGamma(n-r+1) ));
}

//primitive hamming weight counter
int hWeight(int x)
{
    int count, y;
    for (count=0, y=x; y; count++)
        y &= y-1; 
    return count;
}

//-------------------------------------------------------------------------------------
//recursively find the previous group's "hamming weight member count" and sum them
int rCummGroupCount(int bitsize, int hw)
{
    if (hw <= 0 || hw == bitsize) 
        return 1;
    else
        return round(combination(bitsize, hw)) + rCummGroupCount(bitsize,hw-1);
}
//-------------------------------------------------------------------------------------

int main(int argc, char* argv[])
{
    int bitsize = 4, integer = 14;
    int hw = hWeight(integer);
    int groupStartIndex = rCummGroupCount(bitsize,hw-1);
    printf("bitsize: %d\n", bitsize);
    printf("integer: %d  hamming weight: %d\n", integer, hw);
    printf("group start index: %d\n", groupStartIndex);
}
  

输出:

     

bitsize:4
  整数:14汉明重量:3
  小组开始指数:11