How to create a function to define a square

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TOKEN
TOKEN 2019년 4월 21일
댓글: Walter Roberson 2019년 4월 21일
How to create a function to define a square
  댓글 수: 3
TOKEN
TOKEN 2019년 4월 21일
I am actually doing the shapes recoginition using optimization tools, I have to define a square first.
here is the code
clc
clear
M0=zeros(1024,1024); % Background of zeros
x=1:1024;
y=1:1024;
[X,Y]=meshgrid(x,y); % grid of pixel positions
for in=11
A=double(imread([num2str(in,'%4.3d'),'.jpg']))/255; % Load image and scale to 1
M=@(p) M0+double((X-p(1)).^2+(Y-p(2)).^2<(p(3))^2); % define your teplate
% You may need a separate funtion file
fun=@(p) sum(sum(abs(A-M(p)).^2)); % The objective funtion to minimize
% find the avergae center position and ditribution of the nonzero
% pixels in image
pos=find(A>0);
x0=mean(X(pos));
y0=mean(Y(pos));
d=(range(X(pos))+range(Y(pos)))/2;
trl=[x0 y0 d/2]; % The trial solution % number of parameters depend on the shape
lb=[0 0 0];% lower bounds of the prameters
ub=[1024 1024 512];% upper bounds of the prameters
% Optimization process
%You dont have to change the below lines
opts=optimoptions('ga');
opts.InitialPopulationMatrix=trl;
opts.Display='iter';
[sol,fval]=ga(fun,length(trl),[],[],[],[],lb,ub,[],opts);
if fval<100
'circle'
else
'unknown'
end
figure(1)
clf
surf(M(sol),'linestyle','none')
view(2)
daspect([1 1 1])
figure(2)
clf
surf(A,'linestyle','none')
view(2)
daspect([1 1 1])
end
Walter Roberson
Walter Roberson 2019년 4월 21일
square: noun. A planar figure with four equal sides and four right angles.

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