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How can I find intersection points without mouse?

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Hossein Alishahi
Hossein Alishahi 2021 年 3 月 7 日
コメント済み: Jan 2021 年 3 月 8 日
Hi,
I hope you are keeping safe.
Couls you please help me regarding my matlab code, It is urgent and I would appreciate if you help me. I want to find intersections in this plot without mouse(Using code).
%% Input Parameters
clear all
clc
alpha1=input('Enter positive "alpha1"?');
alpha2=input('Enter positive "alpha2"?');
beta1=input('Enter positive "beta1"?');
beta2=input('Enter positive "beta2"?');
lambda=input('Enter positive "lambda"?');
eta1=input('Enter positive "eta1"?');
eta2=input('Enter positive "eta2"?');
P=input('Enter positive "P"?');
%% Feasible Area
x = -1:P;m = 0; c = eta2; y = m * x + c;
plot(x, y, 'black')
y = -1:P; m = 0; c = eta1; y = m * x + c;
plot(y, x, 'black')
eta1=min (eta1,P); eta2=min (eta2,P);
if eta1+eta2<P
X=[0 0 eta1 eta1]; Y=[0 eta2 eta2 0];
fill(X,Y,[0.85 0.85 0.85]);
else
X=[0 0 P-eta2 eta1 eta1]; Y=[0 eta2 eta2 P-eta1 0];
fill(X,Y,[0.85 0.85 0.85]); axis([-1 P -1 P])
end
X=[0 0]; Y=[0 P];
line(X,Y,'Color','black')
hold on
line(Y,X,'Color','black')
g = @(x,y) x+y-P;
gp =fimplicit(g,[0 P 0 P], 'black')
f = @(x,y) log((1+alpha1*x./(1+alpha2*y))) -lambda*log((1+beta1*x./(1+beta2*y)));
fp =fimplicit(f,[-1 P -1 P], '-')
hold on
grid on
For alpha1=1 , alpha2=0.2, beta1=1,beta2=2, lambda=2, eta1=3,eta2=4, P=2
Many Thanks in advance
  3 件のコメント
Hossein Alishahi
Hossein Alishahi 2021 年 3 月 8 日
Firstly , If you read my request carefully, I have not claimed that my question is more unrgent than others. Secondly, you can run ginput function and then select your desired coordinate from lines by using mouse.
Thanks,
Jan
Jan 2021 年 3 月 8 日
Some members of this forum react allergic to the term "urgent", see e.g. https://www.mathworks.com/matlabcentral/answers/29922-why-your-question-is-not-urgent-or-an-emergency .
You are drawing some objects. I cannot guess between which of these obejcts you want to locate the interection. The more the readers have to guess, the longer takes it to answer.

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