5
\$\begingroup\$

I have tried to obtain speed gains by transferring Matlab calculations to C and call it using .mex. My goal is to operate on a matrix \$B\$ which has roughly the dimension 10000x1000

$$ \frac{1}{\text{size(B,1)}}\sum_{i=1}^{2+l*t} \text{mean}( (B(:,i)-B(:,i+1))^2)$$

for \$ l*t<\text{size(B,2) and chosen } t \in \mathbb{N}\$.

My matlab code is as follows:

A=randn(10000,1000);

tic  
RVV_Da_path=ones(size(A,2)-1,1);
i=2;
j=1;
while i <= size(A,2)+1-t_step
   RVV_Da_path(j)=    sum(mean(diff((A(:,1:i)),1,2).^2,1));   
   i=i+t_step;
   j=j+1;
end
RVV_Da_path(j)=    sum(mean(diff((A(:,1:end)),1,2).^2,1));
toc

My C code goes as follows:

#include <math.h>
#include "mex.h" 

void mexFunction(int nlhs, double *plhs[], int nrhs, const mxArray *prhs[])
{

  // calculate 1/size(A,1)*sum_{i=1}^{2+l*t} mean(A(:,i)-A(:,i+1))^2, for l*t < size(A,2), t \in N

#define RV_Out  plhs[0]
#define A_In    prhs[0]
#define t_In    prhs[1]

  double *A,*RV,p,Num_Of_Rows;
  int M, N,Mt,t,i,j,m,k;

  A=mxGetPr(A_In);

  t=mxGetScalar(t_In);
  M=mxGetM(A_In);

  Num_Of_Rows=mxGetM(A_In);
  N=mxGetN(A_In);  

  RV_Out=mxCreateDoubleMatrix(1, N, mxREAL); // init vektor to save the sum of squared first differences
  RV = mxGetPr(RV_Out);
  memset(RV, 1, N);

  i=0;    
  k=0;
  while(i < N-t)
  {
    for(j=0;j<=i;j++)
    {
      for(m=0;m<M;m++)        
      {
        RV[k]+=1/Num_Of_Rows*(pow(A[m + M * j]  -  A[m + M * (j+1)],2.0)) ; // calculate the sum of squared first differences until last possible step
      }   
    }

    i+=t;    
    k++;    
  }

  for(j=0;j<N-1;j++)
  {
    for(m=0;m<M;m++)        
    {
      RV[k]+=1/Num_Of_Rows*(pow(A[m + M * j]  -  A[m + M * (j+1)],2.0)) ; // calculate the sum of squared first differences until end
    }       
  }

  for(i=k+1;i<N;i++) RV[i]=1;  // set the remaining entries equal to 1  
  return;      

}

Now when I translate the C code to a .mex file and compare the speed of the Matlab and the C code the C code is roughly 4-8 times slower for big \$ B\$'s. Why is this the case? Where is the bottleneck in my C-code and how can I work around it?

\$\endgroup\$
  • 1
    \$\begingroup\$ Did you use a profiler to determine where your C code is spending most of its time? Are you running on a multiprocessor machine? Is Matlab set up to take advantage of the multiple processors? What optimization level are you using when you compile? Are MMX and SIMD turned on? \$\endgroup\$ – Snowbody Mar 18 '15 at 14:55
  • \$\begingroup\$ I work on a multi core machine and I didn't parallelize the matlab code actively. I use the -mex filename.c function of matlab to compile. My compiler and the compiler ist from Microsoft Windows SDK 7.1. Further, I don't have any options which I can enable/disable when I use the compiler via matlab. Unfortunately, I'm new to C/C++ and never heard of MMX or SIMD and couldn't find any reference to it in connection with the matlab compile routine. \$\endgroup\$ – Phun Mar 19 '15 at 11:15
  • \$\begingroup\$ Yesterday, I worked on my matlab code and wrote it more efficiently . It is now ~230 times faster on system with fast CPU and ~30 times fast on a system with slow CPU (with no multithreading e.g.). So I think that matlab does some parallization in the background on default. I updated the code in the entry post. \$\endgroup\$ – Phun Mar 19 '15 at 11:15
  • \$\begingroup\$ Please don't do that. It's against the policy of codereview.se to change the code after you post it. Please put it back the way it was, and post another question with the new code. \$\endgroup\$ – Snowbody Mar 19 '15 at 15:25
  • \$\begingroup\$ Ok, I changed it back again. \$\endgroup\$ – Phun Mar 19 '15 at 19:06
4
\$\begingroup\$

The overall Matlab algorithm can be simplified. The key is to realize that:

  • Inter-column differences never change
  • You can use cumulative sums on the means of each column

First let us remove the variable step factor. So tstep = 1. I will reincorporate the variable step factor later.

Let B be a \$m \times n\$ matrix. B(:,1:i) yields a \$m \times i\$ submatrix of B.
Let D = diff(B(:,1:i),1,2). This yields the column differences of B(:,1:i).

  • The first column of D, aka D(:,1), is equal to the second column of B minus the first column of B, aka B(:,2) - B(:,1)
  • The second column of D, aka D(:,2) is equal to the third column of B minus the second column of B, aka B(:,3) - B(:,2)
  • ...
  • The last column of D, aka D(:,i-1) is equal to the \$i^{th}\$ column of B minus the \$(i-1)^{th}\$ column of B, aka B(:,i) - B(:,i-1)

You'll notice that if i increases by 1, these column differences remain the same. The only thing that changes is that we obtain one additional column difference in D. Hence you do not need to perform the diff() function multiple times. Simply perform it once on the entire \$m \times n\$ matrix to yield a \$m \times (n-1)\$ matrix of column differences and take submatrices of this difference matrix.

So the original code you posted:

B=(20*randn(10000,1000));
RV_B=ones(size(B,2),1);
t_step = 1;

i=2;
j=1;

while i <= size(B,2)+1-t_step
    RV_B(j)=    sum(mean(diff((B(:,1:i)),1,2).^2,1));   
    i=i+t_step;
    j=j+1;
end

RV_B(j)=    sum(mean(diff((B(:,1:end)),1,2).^2,1));
M_1=RV_B';

Can be simplified to:

RV_B=ones(size(B,2),1);
t_step = 1;
D = diff(B,1,2);

i=2;
j=1;

while i <= size(B,2)+1-t_step
    RV_B(j)=    sum(mean(D(:,1:i-1).^2,1));   
    i=i+t_step;
    j=j+1;
end

RV_B(j)=    sum(mean(D(:,1:end).^2,1));
M_1=RV_B';

The next step is to realize that if D is our \$m \times (n-1)\$ difference matrix, mean(D) yields a \$1 \times (n-1)\$ matrix of column means. If we truncate columns of this matrix as we do when we take submatrix D(:,1:i-1) it does not affect the means of the columns that are still there. So instead of performing mean() over and over on submatrices, we can again pull it out the loop, perform mean() once on the entire matrix D, and take submatrices of this mean matrix.

So now the code could look like this:

RV_B=ones(size(B,2),1);

t_step = 1;
D = diff(B,1,2);
MD = mean(D.^2);

i=2;
j=1;

while i <= size(B,2)+1-t_step
    RV_B(j)=    sum(MD(:, 1:i-1));   
    i=i+t_step;
    j=j+1;
end

RV_B(j)=    sum(MD(:, 1:end));
M_1 = RV_B';

Similarly, instead of performing a sum on the rows of the submatrices, we can perform a cumulative sum along the rows (or rather row): $$\begin{align}CSMD[i][j] &= \sum_{k=1}^{j} MD[i][k] \\ &= CSMD[i][j-1] + MD[i][j]\end{align}\\$$

Incorporating Matlab's built-in cumsum() along with changing some intermediary variable names to better illustrate their purpose yields the following code:

diffB = diff(B,1,2);
meanSquaredDiff = mean(diffB.^2);
M_1 = cumsum(meanSquaredDiff, 2);

Now we no longer have any explicit loops (in the matlab code at least). This algorithm is much faster and simpler. Consequently, it is much easier to port to C.

You will find that for tstep=1, M_1 for your algorithm has dimension size(B,2) and for the simplified code I provided it has dimension size(B,2)-1. However, in your original algorithm, M_1(end) and M_1(end-1) are always the same if tstep=1, hence you were performing one calculation too many.

Now let us reincorporate the variable step factor. Realize that if we increase tstep it is equivalent to taking every \$\textrm{tstep}^{th}\$ output after the cumulative sum. So if you want tstep=5 you perform the same three-line code above but at the end sample it via M_1 = M_1(1:5:end) or in general M_1 = M_1(1:step:end). You can use isequal to compare the outputs of your original approach to this approach. Remember that you one-padded your output so if you use \$\textrm{tstep} \gt 1\$, you need to remove the extra padding when comparing the matrices.

Finally the last index of the non-padded output will differ as well (if \$\textrm{tstep} \gt 1\$) since your computation computes the last index differently than the rest of the indices. You can fix this by performing tmp = M_1(end); M_1 = M_1(1:step:end); M_1(end) = tmp; instead of simply M_1 = M_1(1:step:end) however the latter approach is more consistent in my opinion.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.