quick calculation in one second

Hello, if there is a sequence x(i) (more than 1 million number points), how to optimize the algorithm and calculate the equation below in about one second? Thanks!

N = length(x);            

J=[1 2 4 10 20 40 100 200 400 1000 2000 4000 10000 20000 40000 100000 200000 400000 1000000];

for h=1:1:19       
    q = J(h);      
    if (q<=N/3)             
        p(1:1:N-3*q+1)=0;      
        for j=1:1:(N-3*q+1)  
            p(j)=sum(x(j+2*q:1:j+3*q-1))-2*sum(x(j+q:1:j+2*q-1))+sum(x(j:1:j+q-1));
        end        
        time(h)=q;
        Stability(h) = sqrt(sum(p.*p)/(6*q*q*(N-3*q+1)));  
    end       
end

Answer 1

[Walter Roberson]

This appears to be the same equation as your previous question http://www.mathworks.com/matlabcentral/answers/270756-quick-numerical-calculation-without-for-loop

Florian was able to suggest a method there that took about 11 seconds in a parallel environment. Are you running a parallel environment?

Do you have a reason to expect that your formula can be calculated in about 1 second on whatever machine you are using?

[Hao Zhang]

Yes, even it is in a parallel environment, it does not work as we expected. indeed, it can be done within one second. Calculation in a vector or matrix form , not using for loop, may be a possible way, i think.

[Walter Roberson]

You will not be able to evaluate that within 1 second by executing that formula in a straight-forward way. Each term is summed multiple times, with the overall number of summations on the order of magnitude of 6*m^2