Willem-Jan de Goeij

Actually, I made a mistake. I've mixed two examples I was looking at.

The average of 2 and 0 is of course 1.
So I would pick 1 as a solution.

What I see in your code is that the problem above is solved by the following set of equations:

2 x = 1+3 = 4
  x = 10+6+4 - 4*5 = 0

The least squares solution for this set of equations, is indeed x=1.6.
But why is the first equation multiplied by 2?

If you keep the factor for x in both equations equal to one, you get

x = (1+3)/2 = 2
x = 10+6+4-4*5 = 0

This gives x=1 as least squares solution.

Dear Jan,

I've tested calling your function from inside a parfor loop. The results show that repetitions occur.

>> matlabpool local
Starting matlabpool using the 'local' configuration ... connected to 3 labs.
>> shuffleparfor
output =

8 1 3 7 5 2 6 4
2 3 8 7 1 5 6 4
8 1 3 7 5 2 6 4
2 3 8 7 1 5 6 4
8 1 3 7 5 2 6 4
2 3 8 7 1 5 6 4
4 8 7 3 6 2 5 1
4 8 7 3 6 2 5 1
4 8 7 3 6 2 5 1

output = zeros(9,8);
Shuffle(1234567890, 'seed');
    
parfor ii=1:9
    output(ii, :) = Shuffle(8, 'index');
end

I use parfor loops on my 64 bit computer with Matlab 64 bit.

I see there are only 32 bit mex compilations on your website.

Does anyone have a 64 bit mex compilation of this function?

Works perfectly in my application, which is a deployed parallel function.

When deploying a function using this progress monitor note that "pctRunOnAll javaaddpath java" will have no effect, you need to embed the folder containing the java classes into the CTF archive when you compile... then it runs like a charm :)

Thanks

After figuring out I had to reduce "progressStepSize" for my needs - it really works!