Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
526 views
in Technique[技术] by (71.8m points)

matrix multiplication - Cache friendly method to multiply two matrices

I intend to multiply 2 matrices using the cache-friendly method ( that would lead to less number of misses)

I found out that this can be done with a cache friendly transpose function.

But I am not able to find this algorithm. Can I know how to achieve this?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

The word you are looking for is thrashing. Searching for thrashing matrix multiplication in Google yields more results.

A standard multiplication algorithm for c = a*b would look like

void multiply(double[,] a, double[,] b, double[,] c)
{
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            for (int k = 0; k < n; k++)
                C[i, j] += a[i, k] * b[k, j]; 
}

Basically, navigating the memory fastly in large steps is detrimental to performance. The access pattern for k in B[k, j] is doing exactly that. So instead of jumping around in the memory, we may rearrange the operations such that the most inner loops operate only on the second access index of the matrices:

void multiply(double[,] a, double[,] B, double[,] c)
{  
   for (i = 0; i < n; i++)
   {  
      double t = a[i, 0];
      for (int j = 0; j < n; j++)
         c[i, j] = t * b[0, j];

      for (int k = 1; k < n; k++)
      {
         double s = 0;
         for (int j = 0; j < n; j++ )
            s += a[i, k] * b[k, j];
         c[i, j] = s;
      }
   }
}

This was the example given on that page. However, another option is to copy the contents of B[k, *] into an array beforehand and use this array in the inner loop calculations. This approach is usually much faster than the alternatives, even if it involves copying data around. Even if this might seem counter-intuitive, please feel free to try for yourself.

void multiply(double[,] a, double[,] b, double[,] c)
{
    double[] Bcolj = new double[n];
    for (int j = 0; j < n; j++)
    {
        for (int k = 0; k < n; k++)
            Bcolj[k] = b[k, j];

        for (int i = 0; i < n; i++)
        {
            double s = 0;
            for (int k = 0; k < n; k++)
                s += a[i,k] * Bcolj[k];
            c[j, i] = s;
        }
   }
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...