How can I optimize this recursive for loop?

5 vues (au cours des 30 derniers jours)
LeChat
LeChat le 1 Mai 2020
Modifié(e) : LeChat le 5 Avr 2021
Hi,
I have this recursive fo loop and I would like to know if there is some way in Matlab by which I could optimize the computation time:
tic
a = 2;
Tmax = 1e2;% which I want to increase by 1 order of magnitude
dt = 1e-2;% which I want to decrease by 1 order of magnitude
Nt = Tmax/dt;
t = (0:Nt-1)'*dt;
Nr = 1e1;
x = zeros(Nt,Nr);
y = zeros(Nt,Nr);
eps = 1e-3;
edt = 1e-2;
adt = 1e-4;
for i = 1:Nt-1
tt = (1+t(i)-t(1:i))*ones(1,Nr);
e1 = ones(i,1);
ex = e1*x(i,:);
ey = e1*y(i,:);
exp1 = exp(-.25*((ex-x(1:i,:)).^2+(ey-y(1:i,:)).^2)./tt);
x(i+1,:) = x(i,:) + adt*sum((ex-x(1:i,:))./tt.^2.*exp1) + edt*randn(1,Nr);
y(i+1,:) = y(i,:) + adt*sum((ey-y(1:i,:))./tt.^2.*exp1) + edt*randn(1,Nr);
end
toc
Obviously I cannot use a parfor (because of the recursivity of this loop, and I tried to make all the vectors gpuArrays but the computation time seems ~20% longer (even without the gather step).
I was wondering if there were any other way to either make this vectorial or using the pdist function from the Statistics toolbox (or another one called dist but I forgot which toolbox is was from and I don't know the difference with pdist), or something on the GPU...?
Thank you for your help and ideas.
  9 commentaires
LeChat
LeChat le 2 Avr 2021
This has improved the performances of the computation. Thank you!
Jan
Jan le 2 Avr 2021
Modifié(e) : Jan le 2 Avr 2021
The results between the originak version and the faster version of Sindar are different:
% Original:
exp1 = exp(-.25*((ex-x(1:i,:)).^2 + (ey-y(1:i,:)).^2) ./ tt);
% Faster:
exp1 = exp((-.25*(ex).^2 + (ey).^2).*tti)
In the 2nd version, the -0.25 is multiplied to the first term only. This reduces the runtime significantly, because the values are growing faster and EXP(Inf) is reached soon, which is much cheaper than finite calculation.
You need:
exp1 = exp((-0.25 * (ex.^2 + ey.^2) .* tti)
The last row of exp1 is 0, so it does not matter in the sum. Omitting it does not reduce the runtime significantly, although the number of EXP calls is reduced. See my answer.

Connectez-vous pour commenter.

Réponse acceptée

Jan
Jan le 2 Avr 2021
Modifié(e) : Jan le 2 Avr 2021
tic
Tmax = 1e2;% which I want to increase by 1 order of magnitude
dt = 1e-2;% which I want to decrease by 1 order of magnitude
Nt = Tmax/dt;
t = (0:Nt-1)'*dt;
Nr = 1e1;
x = zeros(Nt,Nr);
y = zeros(Nt,Nr);
edt = 1e-2;
adt = 1e-4;
for i = 1:Nt-1
tt = 1 ./ (1 + t(i) - t(1:i-1));
ex = x(i, :) - x(1:i-1, :);
ey = y(i, :) - y(1:i-1, :);
exp1 = exp((-0.25 * tt) .* (ex.^2 + ey.^2)) .* tt.^2;
x(i+1, :) = x(i, :) + adt * sum(ex .* exp1, 1) + edt * randn(1, Nr);
y(i+1, :) = y(i, :) + adt * sum(ey .* exp1, 1) + edt * randn(1, Nr);
end
toc
This runs in 10 sec instead of the original 14 sec.
  1 commentaire
LeChat
LeChat le 5 Avr 2021
Modifié(e) : LeChat le 5 Avr 2021
Actually I ended up doing the following , following previous comments (and Sindar's advice):
tic
Tmax = 1e2;% which I want to increase by 1 order of magnitude
dt = 1e-2;% which I want to decrease by 1 order of magnitude
Nt = Tmax/dt;
t = (0:Nt-1)'*dt;
Nr = 1e1;
x = zeros(Nt,Nr);
y = zeros(Nt,Nr);
%
Fx = zeros(Nt,Nr);
Fy = zeros(Nt,Nr);
%
edt = 1e-2;
adt = 1e-4;
for i = 1:Nt-1
tt = (1 + t(i) - t(1:i-1));
ex = x(i, :) - x(1:i-1, :);
ey = y(i, :) - y(1:i-1, :);
exp1 = exp((-0.25./tt) .* (ex.^2 + ey.^2)) ./ tt.^2;
Fx(i+1,:)=adt*sum(ex.*exp1,1);
Fy(i+1,:)=adt*sum(ey.*exp1,1);
x(i+1, :) = x(i, :) + Fx(i+1,:) + edt * randn(1, Nr);
y(i+1, :) = y(i, :) + Fy(i+1,:) + edt * randn(1, Nr);
end
toc
It gives very similar performances as what you posted Jan, I don't know if doing the sum separately is very interesting (performance-wise)...?

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Application Deployment dans Help Center et File Exchange

Produits


Version

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by