12

I am trying to run a simulation to test the average Levenshtein distance between random binary strings.

My program is in python but I am using this C extension. The function that is relevant and takes most of the time computes the Levenshtein distance between two strings and is this.

lev_edit_distance(size_t len1, const lev_byte *string1,
                  size_t len2, const lev_byte *string2,
                  int xcost)
{
  size_t i;
  size_t *row;  /* we only need to keep one row of costs */
  size_t *end;
  size_t half;

  /* strip common prefix */
  while (len1 > 0 && len2 > 0 && *string1 == *string2) {
    len1--;
    len2--;
    string1++;
    string2++;
  }

  /* strip common suffix */
  while (len1 > 0 && len2 > 0 && string1[len1-1] == string2[len2-1]) {
    len1--;
    len2--;
  }

  /* catch trivial cases */
  if (len1 == 0)
    return len2;
  if (len2 == 0)
    return len1;

  /* make the inner cycle (i.e. string2) the longer one */
  if (len1 > len2) {
    size_t nx = len1;
    const lev_byte *sx = string1;
    len1 = len2;
    len2 = nx;
    string1 = string2;
    string2 = sx;
  }
  /* check len1 == 1 separately */
  if (len1 == 1) {
    if (xcost)
      return len2 + 1 - 2*(memchr(string2, *string1, len2) != NULL);
    else
      return len2 - (memchr(string2, *string1, len2) != NULL);
  }
  len1++;
  len2++;
  half = len1 >> 1;
  /* initalize first row */
  row = (size_t*)malloc(len2*sizeof(size_t));
  if (!row)
    return (size_t)(-1);
  end = row + len2 - 1;
  for (i = 0; i < len2 - (xcost ? 0 : half); i++)
    row[i] = i;

  /* go through the matrix and compute the costs.  yes, this is an extremely
   * obfuscated version, but also extremely memory-conservative and relatively
   * fast.  */
  if (xcost) {
    for (i = 1; i < len1; i++) {
      size_t *p = row + 1;
      const lev_byte char1 = string1[i - 1];
      const lev_byte *char2p = string2;
      size_t D = i;
      size_t x = i;
      while (p <= end) {
        if (char1 == *(char2p++))
          x = --D;
        else
          x++;
        D = *p;
        D++;
        if (x > D)
          x = D;
        *(p++) = x;
      }
    }
  }
  else {
    /* in this case we don't have to scan two corner triangles (of size len1/2)
     * in the matrix because no best path can go throught them. note this
     * breaks when len1 == len2 == 2 so the memchr() special case above is
     * necessary */
    row[0] = len1 - half - 1;
    for (i = 1; i < len1; i++) {
      size_t *p;
      const lev_byte char1 = string1[i - 1];
      const lev_byte *char2p;
      size_t D, x;
      /* skip the upper triangle */
      if (i >= len1 - half) {
        size_t offset = i - (len1 - half);
        size_t c3;

        char2p = string2 + offset;
        p = row + offset;
        c3 = *(p++) + (char1 != *(char2p++));
        x = *p;
        x++;
        D = x;
        if (x > c3)
          x = c3;
        *(p++) = x;
      }
      else {
        p = row + 1;
        char2p = string2;
        D = x = i;
      }
      /* skip the lower triangle */
      if (i <= half + 1)
        end = row + len2 + i - half - 2;
      /* main */
      while (p <= end) {
        size_t c3 = --D + (char1 != *(char2p++));
        x++;
        if (x > c3)
          x = c3;
        D = *p;
        D++;
        if (x > D)
          x = D;
        *(p++) = x;
      }
      /* lower triangle sentinel */
      if (i <= half) {
        size_t c3 = --D + (char1 != *char2p);
        x++;
        if (x > c3)
          x = c3;
        *p = x;
      }
    }
  }

  i = *end;
  free(row);
  return i;
}

Can this be sped up?

I will be running the code in 32 bit ubuntu on an AMD FX(tm)-8350 Eight-Core Processor.

Here is the python code that calls it.

from Levenshtein import distance
import random
for i in xrange(16):
    sum = 0
    for j in xrange(1000):
        str1 = bin(random.getrandbits(2**i))[2:].zfill(2**i)
        str2 = bin(random.getrandbits(2**i))[2:].zfill(2**i)
        sum += distance(str1,str2)
    print i,sum/(1000*2**i)
4

4 回答 4

3

你也许可以并行运行。在开始时生成一个巨大的随机列表，然后在您的循环中，一次生成线程（8 个线程），每个进程处理列表的一部分，并将其最终结果添加到 sum 变量。或者一次生成 8 个列表，一次做 8 个。

openmp 建议的问题是“由于大量数据依赖性，该算法的并行性很差” - 维基百科

from threading import Thread

sum = 0

def calc_distance(offset) :
    sum += distance(randoms[offset][0], randoms[offset][1]) #use whatever addressing scheme is best

threads = []
for i in xrange(8) :
    t = new Thread(target=calc_distance, args=(i))
    t.start()
    threads.append(t)

之后....

for t in threads :
     t.join()

如果 levenshtein 距离内核可用（或可编码），我认为这种方法稍后也可以很好地移植到 opencl。

这只是记忆中的一个快速帖子，所以可能有一些问题需要解决。

于 2013-05-06T16:29:14.297 回答
1

我会做什么：

1）非常小的优化：一劳永逸地分配row以避免内存管理开销。或者您可以尝试realloc()，或者您可以在静态变量中跟踪row's 的大小（并且也具有静态变量row）。然而，即使安装成本很低，这也能节省很少的钱。

2）您正在尝试计算平均值。也可以在 C 中进行平均计算。这应该在通话中节省一些东西。同样，零钱，但它很便宜。

3）既然你对实际计算不感兴趣，只对结果感兴趣，那么，假设你有三台PC，每台都是四核机器。然后在它们中的每一个上运行程序的四个实例，循环缩短十二倍。你将在十二分之一的时间内得到十二个结果：平均这些结果，鲍勃是你的叔叔。

选项 #3 除了循环之外根本不需要修改，您可能希望将其设置为命令行参数，以便您可以在可变数量的计算机上部署程序。实际上，您可能希望同时输出结果及其“权重”，以尽量减少将结果相加时出错的可能性。

for j in xrange(N):
    str1 = bin(random.getrandbits(2**i))[2:].zfill(2**i)
    str2 = bin(random.getrandbits(2**i))[2:].zfill(2**i)
    sum += distance(str1,str2)
print N,i,sum/(N*2**i)

但是，如果您对通用的Levenshtein 统计数据感兴趣，我不太确定仅使用 0 和 1 符号进行计算是否适合您的目的。从字符串 01010101 中，您可以通过翻转八个字符或删除第一个字符并在末尾添加一个零来获得 10101010，这有两种不同的成本。如果您拥有字母表中的所有字母，则第二种可能性的可能性会大大降低，这应该会改变平均成本方案中的某些内容。还是我错过了什么？

于 2013-05-01T20:58:46.290 回答
1

您可以从以下站点学习一些 OpenMP 概念和指令开始：A 初学者的 OpenMP 入门

您需要一个与 OpenMP 兼容的编译器。这是一个可以工作的编译器列表。您将希望-fopenmp在编译代码时使用该选项。

我只是将编译器指令添加#pragma omp parallel for到您的代码中，以告诉编译器以下代码块可以并行运行。通过将 while 循环更改为 for 循环，或者在整个函数中应用 OpenMP 模式，您可以看到额外的性能提升。omp_set_num_threads()您可以通过在这些块之前使用函数来调整用于执行 for 循环的线程数来调整性能。一个好的数字是 8，因为您将在 8 核处理器上运行。

lev_edit_distance(size_t len1, const lev_byte *string1,
              size_t len2, const lev_byte *string2,
              int xcost)
{
  size_t i;
  size_t *row;  /* we only need to keep one row of costs */
  size_t *end;
  size_t half;

 // Set the number of threads the OpenMP framework will use to parallelize the for loops
 omp_set_num_threads(8);

  /* strip common prefix */
  while (len1 > 0 && len2 > 0 && *string1 == *string2) {
    len1--;
    len2--;
    string1++;
    string2++;
  }

  /* strip common suffix */
  while (len1 > 0 && len2 > 0 && string1[len1-1] == string2[len2-1]) {
    len1--;
    len2--;
  }

  /* catch trivial cases */
  if (len1 == 0)
    return len2;
  if (len2 == 0)
    return len1;

  /* make the inner cycle (i.e. string2) the longer one */
  if (len1 > len2) {
    size_t nx = len1;
    const lev_byte *sx = string1;
    len1 = len2;
    len2 = nx;
    string1 = string2;
    string2 = sx;
  }
  /* check len1 == 1 separately */
  if (len1 == 1) {
    if (xcost)
      return len2 + 1 - 2*(memchr(string2, *string1, len2) != NULL);
    else
      return len2 - (memchr(string2, *string1, len2) != NULL);
  }
  len1++;
  len2++;
  half = len1 >> 1;
  /* initalize first row */
  row = (size_t*)malloc(len2*sizeof(size_t));
  if (!row)
    return (size_t)(-1);
  end = row + len2 - 1;

  #pragma omp parallel for
  for (i = 0; i < len2 - (xcost ? 0 : half); i++)
    row[i] = i;

  /* go through the matrix and compute the costs.  yes, this is an extremely
   * obfuscated version, but also extremely memory-conservative and relatively
   * fast.  */
  if (xcost) {
   #pragma omp parallel for
   for (i = 1; i < len1; i++) {
      size_t *p = row + 1;
      const lev_byte char1 = string1[i - 1];
      const lev_byte *char2p = string2;
      size_t D = i;
      size_t x = i;
      while (p <= end) {
        if (char1 == *(char2p++))
          x = --D;
        else
          x++;
        D = *p;
        D++;
        if (x > D)
          x = D;
        *(p++) = x;
      }
    }
  }
  else {
    /* in this case we don't have to scan two corner triangles (of size len1/2)
     * in the matrix because no best path can go throught them. note this
     * breaks when len1 == len2 == 2 so the memchr() special case above is
     * necessary */
    row[0] = len1 - half - 1;
    #pragma omp parallel for
    for (i = 1; i < len1; i++) {
      size_t *p;
      const lev_byte char1 = string1[i - 1];
      const lev_byte *char2p;
      size_t D, x;
      /* skip the upper triangle */
      if (i >= len1 - half) {
        size_t offset = i - (len1 - half);
        size_t c3;

        char2p = string2 + offset;
        p = row + offset;
        c3 = *(p++) + (char1 != *(char2p++));
        x = *p;
        x++;
        D = x;
        if (x > c3)
          x = c3;
        *(p++) = x;
      }
      else {
        p = row + 1;
        char2p = string2;
        D = x = i;
      }
      /* skip the lower triangle */
      if (i <= half + 1)
        end = row + len2 + i - half - 2;
      /* main */
      while (p <= end) {
        size_t c3 = --D + (char1 != *(char2p++));
        x++;
        if (x > c3)
          x = c3;
        D = *p;
        D++;
        if (x > D)
          x = D;
        *(p++) = x;
      }
      /* lower triangle sentinel */
       if (i <= half) {
        size_t c3 = --D + (char1 != *char2p);
        x++;
        if (x > c3)
          x = c3;
        *p = x;
      }
    }
  }

  i = *end;
  free(row);
  return i;
}

您还可以对 for 循环中正在操作的变量进行归约操作，以提供简单的并行计算，如求和、乘法等。

int main()
{
    int i = 0,
        j = 0,
        sum = 0;
    char str1[30]; // Change size to fit your specifications
    char str2[30];

    #pragma omp parallel for
    for(i=0;i<16;i++)
    {
        sum = 0;
            // Could do a reduction on sum across all threads
        for(j=0;j<1000;j++)
        {
            // Calls will have to be changed
            // I don't know much Python so I'll leave that to the experts 
            str1 = bin(random.getrandbits(2**i))[2:].zfill(2**i)
            str2 = bin(random.getrandbits(2**i))[2:].zfill(2**i)
            sum += distance(str1,str2)
        }
        printf("%d %d",i,(sum/(1000*2*i)));
    }
}
于 2013-05-01T18:31:37.363 回答
0

Someone else did a great deal of research a year or two ago and did run-time testing as well.

He came up with this and basicly used a solution tree to speed things up.

于 2013-05-08T12:58:31.017 回答