Interpreter. Computational time

Hello, All! We have found some interesting moment. 1. If we use to call next function:

function [result,time] = Func(x,y,u)
N = size(x,1);
M = size(y,1);
result = zeros(M,1);  
tic;
for j=1:M
  for i=1:N
      result(j) = result(j) + u(i) / ( y(j) - x(i) );
  end;
end;
time = toc;

we obtain time = 1.9sec.

2. If we use

N = length(x,1);
M = length(y,1);

or use to call next function:

function [result,time] = Func(x,y,u,N,M)
result = zeros(M,1);  
tic;
for j=1:M
  for i=1:N
      result(j) = result(j) + u(i) / ( y(j) - x(i) );
  end;
end;
time = toc;

we obtain time = 3.0sec.

3. But if we place the function body into the program body we obtain time = 3.7sec.

4. Besides, if we call this function in a cycle then computational time of each function increased.

Time data averaged by 10 runs for random x,y,u. If we measure time outside the functions (or time of the functions executions) we obtain the same time. Can anybody explain why it is happening?

P.s. We use Matlab 2009b and 2010b and test this on different workstations.

Answer 1

[Matt Fig]

SIZE and LENGTH are not the same thing!

size(rand(2,10000),1)
length(rand(2,10000))

You should show what your data looks like, i.e., what does:

size(x)
size(y)

show?

[Dmitry Maryin]

Yes, you are absolutely right! I cited this (length) as an example of the problem.

Data (x,y,u) looks like:

x = sort(rand(N,1));

And I used:

size(x,1)

[Dmitry Maryin]

size(x,1) show value of N (eg 10000);

size(y,1) show value of M (eg 9999).

[Sean de Wolski]

N = length(x,1);

Should throw an error unless you've overwritten function length

[Dmitry Maryin]

Yes, I did mistake when typed. In the code uses: N = length(x);