Does anyone know how to use the matlab to calculate the minimu distance between a point outside oval and the oval surface？

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huazai2020 2020-6-28

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编辑： huazai2020 2020-7-18

Does anyone know how to use the matlab to calculate the minimu distance between a point outside oval and the oval surface？

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Image Analyst 2020-6-28

编辑：Image Analyst 2020-6-28

bwdist() perhaps?

huazai2020 2020-6-29

The oual can be both given in the form of an equation or data points. How to use the bwdist(), could you share me the code?

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Matt J 2020-6-30

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编辑：Matt J 2020-6-30

在 MATLAB Online 中打开

You can use trustregprob from the File Exchange

https://www.mathworks.com/matlabcentral/fileexchange/53191-quadratic-minimization-with-norm-constraint?focused=6974391&tab=function

For example, consider the following ellipse and external point y,

A=[2 1;1,2]; %Ellipse equation matrix
y=[1;0.5]; %External point
z=[0.584808315593597 ; 0.201052451754066]; %Closest point
ellipsefun=@(p,q) sum([p;q].*(A*[p;q]))-1 ;
hold on
 fimplicit(ellipsefun, [-1 1.3  -1 1.3])
 plot(y(1),y(2),'rx',z(1),z(2),'bo');
 axis equal
 
hold off

I found the closest point z using the code below,

Aroot=chol(A);
L=Aroot\eye(2);
z=Aroot\trustregprob(L.'*L, L.'*y,1);

and the minimum distance is just,

>> distance=norm(z-y)
distance =
    0.5116

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huazai2020 2020-7-7

What is FEX AND H1? Are they these?

Find the projection of point P in R^n on the ellipsoid

E = { x = x0 + U*(z.*radii) : |z| = 1 }, where U is orthogonal matrix of the orientation of E, radii are the axis lengths, and x0 is the center.

Or on generalized conic E = { x : x'*A*x + b'*x + c = 0 }.

The projection is the minimization problem:

min | x - P | (or max | x - P|) for x in E.

Method: solve the Euler Lagrange equation with respect to the Lagrange multiplier, which can be written as polynomial equation (from an idea by Roger Stafford)

Bruno Luong 2020-7-8

FEX short word for File Exchange. Click on it I put the link on my message.

H1: https://www.mathworks.com/help/matlab/matlab_prog/add-help-for-your-program.html

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Image Analyst 2020-7-4

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在 MATLAB Online 中打开

"The oual can be both given in the form of an equation or data points." <== if you have data points (xb, yb) on the boundary of an ellipse/oval, you can use sqrt() to find the distances to some other point (xp, yp). Then use in() to find the minimum distance.

distances = sqrt(xb-xp).^2 + (yb-yp).^2);
minDistance = min(distances)

If you have an image then you need to get the boundary points first:

boundaries = bwboundaries(binaryImageOfEllipse);
boundaries = boundaries{1}; % Extract double array from cell array.
xb = boundaries(:, 2);
yb = boundaries(:, 1);

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Image Analyst 2020-7-18

在 MATLAB Online 中打开

No, you shouldn't use an inner loop because your minDistance will just be the distance between (x1,y1) and xwoint(Num2) and ywoint(Num2) instead of the whole array. Do it like this:

clc;
clear all;
A = xlsread('data1.xlsx');
xpoint=A(:,1);
ypoint=A(:,2);
Num1=length(xpoint);
xwoint=A(:,3);
ywoint=A(:,4);
Num2=length(xwoint);
for k=1:1:Num1
	x1 = xpoint(k);
	y1 = ypoint(k);
	distances = sqrt((xwoint-x1).^2 + (ywoint-y1).^2);
	minDistance(k) = min(distances)*1000;
end
plot(minDistance, 'b-');
% xlswrite('thick.xlsx',minDistance,'sheet1');