Speed up the code

9 views (last 30 days)
Pooneh Shah Malekpoor
Pooneh Shah Malekpoor on 5 Mar 2021
Edited: Geoff Hayes on 5 Mar 2021
Hello
This code is part of a total code. It calculates the geometrical features of the main problem geometry. However, it takes so much time due to mupadmex. How can I improve it? Similar code is written for the solution of x2 and x3 which I have not copied (due to lack of space).
BTW, if i omit the if imaginary condition, would there be any problem? I mean is it possible that double command generate imaginary numbers?
any clever way of speeding up the code is highly appriciated.
%Slopes of three lines
m1=(P(2,2)-P(1,2))/(P(2,1)-P(1,1)); % line 1
m2=(P(3,2)-P(2,2))/(P(3,1)-P(2,1)); % line 2
m3=(P(4,2)-P(3,2))/(P(4,1)-P(3,1)); % line 3
syms x1 %introducing the variables x1,x2,x3
syms x2
syms x3
int=0; % counting the points of intersection - it has to be 2 at the end, in case there is intersection
Points(2,2)=0; %for preallocation
% intersection of circle with line 3
y3=P(3,2)+m3*(x3-P(3,1));
c3 = (x3-xc)^2+(y3-yc)^2-R^2;
x_sol3 = vpasolve(c3);
y_sol3 = subs(y3,'x3',x_sol3); %convert sym to numeric value in y3
x_sol3 = double(x_sol3);
y_sol3 = double(y_sol3);
if imag(x_sol3(1))==0 && imag(x_sol3(2))==0 % This is to guarantee there is real intersection (not imaginary)
for v=1:2
if x_sol3(v)>P(3,1) && x_sol3(v)<P(4,1)
if v==2 && abs(x_sol3(1)-x_sol3(2))<1e-6 % means if two roots are the same do not count it
int;
else
int=int+1;
Points(int,:)=[x_sol3(v) y_sol3(v)];
end
end
end
end
  1 Comment
Pooneh Shah Malekpoor
Pooneh Shah Malekpoor on 5 Mar 2021
I have now replaced symbolic solver with fsolve to speed up the code; however, it only gives out one of the solution and not the other one...the is much more quick now, but does not printout one of the solutions. what should i do?
m1=(P(2,2)-P(1,2))/(P(2,1)-P(1,1)); % line 1
m2=(P(3,2)-P(2,2))/(P(3,1)-P(2,1)); % line 2
m3=(P(4,2)-P(3,2))/(P(4,1)-P(3,1)); % line 3
int=0; % counting the points of intersection - it has to be 2 at the end, in case there is intersection
Points(2,2)=0;
% intersection of circle with line 3
F = @(x)[(x(1)-xc)^2+((P(3,2)+m3*(x(1)-P(3,1)))-yc)^2-R^2;(x(2)-xc)^2+((P(3,2)+m3*(x(2)-P(3,1)))-yc)^2-R^2];
opts = optimoptions('fsolve', 'Display', 'off');
x = fsolve(@(x) F(x), zeros(1,2), opts);
y3=P(3,2)+m3*(x-P(3,1));

Sign in to comment.

Answers (0)

Community Treasure Hunt

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

Start Hunting!