Fast calculation of short cumulative product vector

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I want to calculate a reasonable short vector many times with variable inputs a and len
%y = [a a^2 a^3...a^(len)]
My methods so far are the following:
y1 = cumprod(a(ones(len,1)))
y2 = a.^(1:len)
y1 seems to be faster than y2 for len>11 or so. Any suggestions for doing this even faster? The knowledge that y(n) = y(n-1)*y(1) should be exploitable somehow, I just dont see how.

Answers (6)

James Tursa
James Tursa on 6 Sep 2011
Or this variation, which I don't see in the current posts:
y = cumprod(a.*ones(1,n));
I think the MATLAB JIT may recognize this formulation and not actually do the multiply, but just populate the result ala repmat.
And if this calculation is dominating your overall runtime, you can always turn to a simple mex function to do it and avoid the imtermediate array formulations.

Oleg Komarov
Oleg Komarov on 6 Sep 2011
time = zeros(100,3);
for len = 1:100
a = rand(1);
o = ones(1,len);
v = 1:len;
tic
for s = 1:1000; y1 = cumprod(a(o)); end
time(len,1) = toc;
tic
for s = 1:1000; y2 = a.^(v); end
time(len,2) = toc;
tic
for s = 1:1000; y3 = bsxfun(@power,a,v); end
time(len,3) = toc;
end
plot(time)
legend('cumprod','.^','bsxfun')
EDIT cumprod, power, explog, cumprodx, cumprodpow
James' cumprodpow is the fastest.
num_iter = 3000;
time = zeros(100,5);
for len = 1:100
a = rand(1);
tic
for s = 1:num_iter
y1 = cumprod(a(ones(1,len)));
end
time(len,1) = toc;
pause(0.001)
tic
for s = 1:num_iter
y2 = a.^(1:len);
end
time(len,2) = toc;
pause(0.001)
tic
for s = 1:num_iter
y3 = exp(log(a)*(1:len));
end
time(len,3) = toc;
pause(0.001)
tic
for s = 1:num_iter
y4 = cumprodx(a,len);
end
time(len,4) = toc;
pause(0.001)
tic
for s = 1:num_iter
y5 = cumprod(a.*ones(1,len));
end
time(len,5) = toc;
end
plot(time)
legend('cumprod','power','explog','cumprodx','cumprodpow')
  1 Comment
Andrei Bobrov
Andrei Bobrov on 6 Sep 2011
Hi friends!
Firstly, +1;
secondly, small additions:
time = zeros(100,5);
for len = 1:100
a = rand(1);
o = ones(1,len);
v = 1:len;
tic
for s = 1:1000; y1 = cumprod(a(o)); end
time(len,1) = toc;
tic
for s = 1:1000; y2 = a.^(v); end
time(len,2) = toc;
tic
for s = 1:1000; y3 = bsxfun(@power,a,v); end
time(len,3) = toc;
tic
for s = 1:1000; y4 = arrayfun(@(x)a^x,v); end
time(len,4) = toc;
tic
for s = 1:1000, for v = len:-1:1, y5(v) = a^v; end; end
time(len,5) = toc;
end
plot(time)
legend('cumprod','power','bsxfun','arrayfun','for..end')

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Grzegorz Knor
Grzegorz Knor on 6 Sep 2011
My suggestion:
y3 = bsxfun(@power,a,1:len)

Derek O'Connor
Derek O'Connor on 6 Sep 2011
Nobody has followed Knut's suggestion:
y(n) = y(n-1)*y(1)
It is in fact a special case of Horner's Method for evaluating a polynomial
yn(x) = a0 + a1*x + a2*x^2 + ... + an*x^n.
The time behavior of the two methods below is very erratic but the loop or Horner method always seems to beat the CumProd method by 30% to 50% up to about n = 150. After that, CumProd gets better and better. At n = 1000, Horner's time is about 5*CumProd time.
If you have the time try [yc,yh,time] = CumProdvsHorner(1000, 10000);
Has anyone got an explanation for the spikes in the execution times?
function [yc,yh,yx,time] = CumProdvsHorner(nmax,ns);
% Calculation of y = [a a^2 ... a^n]
% USE: [yc,yh,yx,time] = CumProdvsHorner(300,1000);
nmeths=3;
time = zeros(nmax,nmeths);
ntime = zeros(nmax,nmeths);
for n = 2:nmax
x = rand; xsave = x;
yc = zeros(1,n);
yh = zeros(1,n);
yx = zeros(1,n);
tc = tic;
for s = 1:ns
yc = cumprod(x.*ones(1,n));
end
time(n,1) = toc(tc);
tr = tic;
x = xsave;
for s = 1:ns
yh(1) = x;
for k = 2:n
yh(k) = x*yh(k-1);
end
end
time(n,2) = toc(tr);
tx = tic;
for s = 1:ns
yx = cumprodx(x,n);
end
time(n,3) = toc(tx);
end % for n
figure
plot(20:nmax,time(20:nmax,:))
legend('CumProd','Horner','CumProdx')
xlabel('n');
ylabel('Time (secs)')
title('Calculate y(x,n) = [x x^2 x^3 ... x^n]')
for m = 1:nmeths
ntime(:,m) = time(:,m)./(1:nmax)';
end
figure
plot(20:nmax,ntime(20:nmax,:))
legend('CumProd','Horner','CumProdx')
xlabel('n');
ylabel('Normalized Time T(n)/n')
title('Calculate y(x,n) = [x x^2 x^3 ... x^n]')
%
% END CumProdvsHorner
EDIT: Added plots for normalized times, renamed some variables.
  5 Comments
Derek O'Connor
Derek O'Connor on 7 Sep 2011
@proecsm
Take a look at the normalized times with the updated function using:
[yc,yh,time] = CumProdvsHorner(1000,1000);
CumProd's behavior is very strange. As Walter says, this may be due to switching in and out of BLAS.

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James Tursa
James Tursa on 7 Sep 2011
Here is the quick and dirty mex solution:
// File cumprodx.c
// c = cumprodx(a,n) does equivalent of c = cumprod(a.*ones(1,n))
// a = real double scalar
// n = real double scalar with integral value > 0
// Caution: No argument checking
#include "mex.h"
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[])
{
double a = mxGetScalar(prhs[0]);
mwSize i, n = mxGetScalar(prhs[1]);
double *pr;
plhs[0] = mxCreateDoubleMatrix(1,n,mxREAL);
pr = mxGetPr(plhs[0]);
pr[0] = a;
for( i=1; i<n; i++ ) {
pr[i] = pr[i-1] * a;
}
}
  3 Comments
Derek O'Connor
Derek O'Connor on 7 Sep 2011
A good point. I had overlooked the function call overhead, which
can be a significant part of the time to do the necessary (n-1)
multiplications.
However, the timings serve as a warning not to forget the simpler,
loopy methods, especially with (an improving?) JIT compiler.
Vectorization has become something of a fetish with some Matlab
programmers, which blinds them to simpler methods.

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Knut
Knut on 7 Sep 2011
Thank you for all your help. There seems to be many different ways to accomplish this, each with different speed-tradeoffs. The mex-file might run faster with single-presicion and optimization flags, but I did not bother.
For lengths < ~80 elements, 'Horners Method' seems to be the best choice. I did notice some strange "warm-up" effects: increasing the number of iterations caused the relative performance to shift quite a lot. Is this the JIT playing games on me?
num_iter = 100;
len_span = [1:50 100:100:2000 1e4 3e4 1e5 3e5 1e6];
num_iter_span = 1e6./len_span;
time = zeros(length(len_span),7);
for len_idx = 1:length(len_span)
num_iter = num_iter_span(len_idx);
len = len_span(len_idx);
eps = 1e-7;
a = rand(1);
tic
for s = 1:num_iter;
y1 = cumprod(a*ones(1, len));
end
time(len_idx,1) = toc;
assert(max(abs(y1-y1))<eps);
tic
for s = 1:num_iter;
y2 = a.^(1:len);
end
time(len_idx,2) = toc;
assert(max(abs(y1-y2))<eps);
tic
for s = 1:num_iter;
y3 = bsxfun(@power,a,1:len);
end
time(len_idx,3) = toc;
assert(max(abs(y1-y3))<eps);
tic
for s = 1:num_iter,
for v = len:-1:1
y4(v) = a^v;
end;
end
time(len_idx,4) = toc;
assert(max(abs(y1-y4))<eps);
tic
for s = 1:num_iter,
y5(1) = a;
for v = 2:len
y5(v) = a*y5(v-1);
end;
end
time(len_idx,5) = toc;
assert(max(abs(y1-y5))<eps);
tic
for s = 1:num_iter,
y6 = cumprodx(a,len);
end
time(len_idx,6) = toc;
assert(max(abs(y1-y6))<eps);
tic
for s = 1:num_iter,
y7 = exp(log(a)*(1:len));
end
time(len_idx,7) = toc;
assert(max(abs(y1-y7))<eps);
end
figure
semilogx(len_span, time)
xlabel('vector length')
ylabel('time/vector length')
grid on
axis tight
legend('cumprod(a([1 1,...]))','power','bsxfun','for..end', 'Horners Method', 'mex', 'log')
  3 Comments
Oleg Komarov
Oleg Komarov on 7 Sep 2011
You didn't mention that in your original post.
Anyways, in this case I find that James' cumprod(a.*ones(1,len)) is the fastest, always.

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