Purposefully Slow MATLAB Function?

I want to write a really, really, slow program for MATLAB. I'm talking like, O(2^n) or worse. It has to finish, and it has to be deterministically slow, so no "if rand() = 123,123, exit!" This sounds crazy, but it's actually for a distributed systems test. I need to create a .m file, compile it (with MCC), and then run it on my distributed system to perform some debugging operations.

The program must constantly be doing work, so sleep() is not a valid option.

I tried making a random large matrix and finding its inverse, but this was completing too quickly. Any ideas?

Upvotes: 2

Answers (11)

Ofri Raviv

Reputation: 24823

this will run 100% cpu for WANTED_TIME seconds

WANTED_TIME = 2^n; % seconds
t0=cputime; 
t=cputime; 
while (t-t0 < WANTED_TIME)
    t=cputime; 
end;

Upvotes: 1

Jason S

Reputation: 189776

Easy! Go back to your Turing machine roots and think of processes that are O(2^n) or worse.

Here's a fairly simple one (warning, untested but you get the point)

N = 12; radix = 10;
odometer = zeros(N, 1);
done = false;
while (~done)
    done = true;
    for i = 1:N
        odometer(i) = odometer(i) + 1;
        if (odometer(i) >= radix)
            odometer(i) = 0;
        else
            done = false;
            break;
        end
    end
end

Even better, how about calculating Fibonacci numbers recursively? Runtime is O(2^N), since fib(N) has to make two function calls fib(N-1) and fib(N-2), but stack depth is O(N), since only one of those function calls happens at a time.

function y = fib(n)
   if (n <= 1)
      y = 1;
   else
      y = fib(n-1) + fib(n-2);
   end
end

Upvotes: 2

Vinko Vrsalovic

Reputation: 340236

I don't speak MATLAB but something equivalent to the following might work.

loops = 0
counter = 0
while (loops < MAX_INT) {
    counter = counter + 1;
    if (counter == MAX_INT) {
        loops = loops + 1;
        counter = 0;
    }
}

This will iterate MAX_INT*MAX_INT times. You can put some computationally heavy thing in the loop for it to take longer if this is not enough.

Upvotes: 2

Amro

Reputation: 124563

Try this one:

tic
isprime( primes(99999999) );
toc

EDIT:

For a more extensive set of tests, use these benchmarks (perhaps for multiple repetitions even):

disp(repmat('-',1,85))
disp(['MATLAB Version ' version])
disp(['Operating System: ' system_dependent('getos')])
disp(['Java VM Version: ' version('-java')]);
disp(['Date: ' date])
disp(repmat('-',1,85))

N = 3000;   % matrix size
A = rand(N,N);  
A = A*A;

tic;  A*A; t=toc;
fprintf('A*A \t\t\t%f sec\n', t)

tic; [L,U,P] = lu(A); t=toc; clear L U P
fprintf('LU(A)\t\t\t%f sec\n', t)

tic; inv(A); t=toc;
fprintf('INV(A)\t\t\t%f sec\n', t)

tic; [U,S,V] = svd(A); t=toc; clear U S V
fprintf('SVD(A)\t\t\t%f sec\n', t)

tic; [Q,R,P] = qr(A); t=toc; clear Q R P
fprintf('QR(A)\t\t\t%f sec\n', t)

tic; [V,D] = eig(A); t=toc; clear V D
fprintf('EIG(A)\t\t\t%f sec\n', t)

tic; det(A); t=toc;
fprintf('DET(A)\t\t\t%f sec\n', t)

tic; rank(A); t=toc;
fprintf('RANK(A)\t\t\t%f sec\n', t)

tic; cond(A); t=toc;
fprintf('COND(A)\t\t\t%f sec\n', t)

tic; sqrtm(A); t=toc;
fprintf('SQRTM(A)\t\t%f sec\n', t)

tic; fft(A(:)); t=toc;
fprintf('FFT\t\t\t%f sec\n', t)

tic; isprime(primes(10^7)); t=toc;
fprintf('Primes\t\t\t%f sec\n', t)

The following are the results on my machine using N=1000 for one iteration only (note primes is using as upper bound 10^7 NOT 10^8 [takes way more time!])

A*A             0.178329 sec
LU(A)           0.118864 sec
INV(A)          0.319275 sec
SVD(A)          15.236875 sec
QR(A)           0.841982 sec
EIG(A)          3.967812 sec
DET(A)          0.121882 sec
RANK(A)         1.813042 sec
COND(A)         1.809365 sec
SQRTM(A)        22.750331 sec
FFT             0.113233 sec
Primes          27.080918 sec

Upvotes: 1

mtrw

Reputation: 35098

This naive implementation of the Discrete Fourier Transform takes ~ 9 seconds for a 2048 long input vector x on my 1.86 GHz single core machine. Going to 4096 inputs extends the time to ~ 35 seconds, close to the 4x I would expect for O(N^2). I don't have the patience to try longer inputs :)

function y = SlowDFT(x)

t = cputime;
y = zeros(size(x));
for c1=1:length(x)
    for c2=1:length(x)
        y(c1) = y(c1) + x(c2)*(cos((c1-1)*(c2-1)*2*pi/length(x)) - ...
                            1j*sin((c1-1)*(c2-1)*2*pi/length(x)));
    end
end
disp(cputime-t);

EDIT: Or if you're looking to stress memory more than CPU:

function y = SlowDFT_MemLookup(x)

t = cputime;
y = zeros(size(x));
cosbuf = cos((0:1:(length(x)-1))*2*pi/length(x));
for c1=1:length(x)
    cosctr = 1;
    sinctr = round(3*length(x)/4)+1;
    for c2=1:length(x)
         y(c1) = y(c1) + x(c2)*(cosbuf(cosctr) ...
                            -1j*cosbuf(sinctr));
         cosctr = cosctr + (c1-1);
         if cosctr > length(x), cosctr = cosctr - length(x); end
         sinctr = sinctr + (c1-1);
         if sinctr > length(x), sinctr = sinctr - length(x); end
    end
end
disp(cputime-t);

This is faster than calculating sin and cos on each iteration. A 2048 long input took ~ 3 seconds, and a 16384 long input took ~ 180 seconds.

Upvotes: 4