Could I see a significant time boost using int16 instead of double on large array operations?

I am using the genetic algorithm with a custom fitness function that is necessarily very complicated.

Many addition/subtraction operations are performed on a large (300,000x10x10) 3D array to get the final result. All of the operations are conducted on complex integers.

This causes difficulty as the generations can take >10 minutes each with a population size of 500. The solution space is VERY large (~10^100) so it's important to get as many generations as possible during a fixed time window (a couple hours).

Would converting all of the objects to int16 or uint16 save a significant amount of time here? My guess is yes, but I don't know if Matlab does some smart things with doubles when they are all actually integers - and it would take quite a bit of work to recast all the objects.

Answer 1

[Walter Roberson]

No, no time would be saved, as the operation is not permitted.

>> foo = complex(randi(1000,3000,10,10,'uint16'), randi(1000,3000,10,10,'uint16'));

>> bar = complex(randi(1000,3000,10,10,'uint16'), randi(1000,3000,10,10,'uint16'));

>> for K = 1 : 20; t(K) = timeit(@() foo-bar, 0); end

Error using -

Complex integer arithmetic is not supported.

Error in @()foo-bar

Error in timeit>roughEstimate (line 182)

f();

Error in timeit (line 61)

t_rough = roughEstimate(f, num_outputs);

>> for K = 1 : 20; t(K) = timeit(@() foo+bar, 0); end

Error using +

Complex integer arithmetic is not supported.

Error in @()foo+bar

Error in timeit>roughEstimate (line 182)

f();

Error in timeit (line 61)

t_rough = roughEstimate(f, num_outputs);

However:

>> foo = complex(randi(1000,3000,10,10,'uint16'), randi(1000,3000,10,10,'uint16'));

>> bar = complex(randi(1000,3000,10,10,'uint16'), randi(1000,3000,10,10,'uint16'));

>> food = double(foo); bard = double(bar);

>> for K = 1 : 20; t(K) = timeit(@() complex(real(foo)-real(bar),imag(foo)-imag(bar)), 0); end

>> for K = 1 : 20; td(K) = timeit(@() food-bard, 0); end

>> mean(td)/mean(t)

ans =

3.37810543407242

so there are gains to be made if you are willing to do all of the real() and imag() yourself, which should be valid for addition and subtraction.