Struggling to Vectorize the following Code.

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This code is running really slow would like to vectorize it, however i don't really know how to do so in this case. The code is below, it does work but the N value really should be in the 1000's I've managed to vectorize code before for a much simplier system but this one i have no idea :( any help or suggestions are appreciated. Thanks
hold on
% Number of iterations
N = 100;
x = 1; y = 1;z = 1;
for w21 = linspace(-12,-3,N)
for i = 1:N-1
y = y_iterate(x,z,w21);
z = z_iterate(y);
x = x_iterate(y);
if i >= (N - 200)
p = plot(w21,x,'.k','MarkerSize',3);
end
end
end
function val = x_iterate(y)
val = -3 + 8.*(1 ./ (1 + exp(-y)));
end
function val = z_iterate(y)
val = -7 + 8.*(1 ./ (1 + exp(-y)));
end
function val = y_iterate(x,z,w21)
val = 4 + w21.*(1 ./ (1 + exp(-x))) + 6.*(1 ./ (1 + exp(-z)));
end

Answers (2)

Fabio Freschi
Fabio Freschi on 9 Nov 2019
Edited: Fabio Freschi on 9 Nov 2019
Almost all execution time is due to the plot function as highlighted by the profiler:
You can call the plot only once at the end of the two loops. One way is to load the data to plot in two vectors and call plot only once
hold on
% Number of iterations
N = 100;
x = 1; y = 1;z = 1;
% initialization
aa = [];
bb = [];
for w21 = linspace(-12,-3,N)
for i = 1:N-1
y = y_iterate(x,z,w21);
z = z_iterate(y);
x = x_iterate(y);
if i >= (N - 200)
% update
aa = [aa; w21];
bb = [bb; x];
end
end
end
p = plot(aa,bb,'.k','MarkerSize',3);
function val = x_iterate(y)
val = -3 + 8.*(1 ./ (1 + exp(-y)));
end
function val = z_iterate(y)
val = -7 + 8.*(1 ./ (1 + exp(-y)));
end
function val = y_iterate(x,z,w21)
val = 4 + w21.*(1 ./ (1 + exp(-x))) + 6.*(1 ./ (1 + exp(-z)));
end
  8 Comments
Fabio Freschi
Fabio Freschi on 9 Nov 2019
Edited: Fabio Freschi on 9 Nov 2019
Try this: for N = 1000 it takes about 4 secs
hold on
% Number of iterations
N = 1000;
x = 1; y = 1;z = 1;
% initialization
aa = zeros(200*N,1);
bb = zeros(200*N,1);
k = 1;
for w21 = linspace(-12,-3,N)
for i = 1:N-1
y = y_iterate(x,z,w21);
z = z_iterate(y);
x = x_iterate(y);
if length(x) ~= 1
p = 1;
end
if i >= (N - 200)
% update
aa(k) = w21;
bb(k) = x;
k = k+1;
end
end
end
p = plot(aa(1:k-1),bb(1:k-1),'.k','MarkerSize',3);
function val = x_iterate(y)
val = -3 + 8.*(1 ./ (1 + exp(-y)));
end
function val = z_iterate(y)
val = -7 + 8.*(1 ./ (1 + exp(-y)));
end
function val = y_iterate(x,z,w21)
val = 4 + w21.*(1 ./ (1 + exp(-x))) + 6.*(1 ./ (1 + exp(-z)));
end

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Walter Roberson
Walter Roberson on 9 Nov 2019
No, you have a chaotic system. Each iteration depends on the bitwise details of the previous iteration.
It is possible to calculate the algebraic iteration to any given number of steps in terms of the original inputs, getting out more complicated formulas each step, but the rounding that takes place at each individual step turns out to be important to the long-term outcome, and the algebraic formulas for multiple steps will not reproduce that round-off the same way.
Example: after two iterations, the new y is
w21/(exp(3 - 8/(exp(- w21/(exp(-x) + 1) - 6/(exp(-z) + 1) - 4) + 1)) + 1) + 6/(exp(3 - 8/(exp(- w21/(exp(-x) + 1) - 6/(exp(-z) + 1) - 4) + 1)) + 1) + 4
where those are the initial x, z values in the formula.
Your system immediately loses its original y coordinate, by the way: the original y coordinate plays no role in the system.
  3 Comments
Fabio Freschi
Fabio Freschi on 9 Nov 2019
The bottleneck of the code is the plot. In this sense, plotting all data in a single call can be seen as vectorization.

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