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Simplex optimisation using fminsearch

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Ebube Ezi
Ebube Ezi 2018-8-23
编辑： Bruno Luong 2023-3-23
Please how can i use the simplex optimisation syntax called 'fminsearch' to optimise a circle fitted to a set of random points? I've tried using the equation of a circle as the function for the syntax but am not making any head way. I presume am not getting the function right.
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John D'Errico
John D'Errico 2023-3-22
Why do you need to use fminsearch? This is a problem simply solved without recourse to fminsearch.

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回答（2 个）

Walter Roberson
Walter Roberson 2018-8-24
x = vector_of_x_coordinates;
y = vector_of_y_coordinates;
obj = @(xyr) sum( (x-xyr(1)).^2 + (y-xyr(2)).^2 - xyr(3).^2 );
xyr0 = [guess_x, guess_y, guess_r];
[XYR, residue] = fminsearch(obj, xyr0);
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Biraj Khanal
Biraj Khanal 2023-3-22
编辑：Biraj Khanal 2023-3-22
Shouldn't that be resolved when the absolute value of the sum is used?
Biraj Khanal
Biraj Khanal 2023-3-23
So, this is my adaptation of a tutorial I found. The idea is to minimize the error as distance between the coordinates of guessed circle and the actual data points.
obj = @(hkr) sumerror(hkr,x,y);
[sol ,residue] = fminsearch(obj,[Guess_h, Guess_k, Guess_r]);
function E = sumerror(hkr,x,y)
t=linspace(0,2*pi,length(x));
xx=hkr(1) + hkr(3)*cos(t); % create x and y for the guess iteration
yy=hkr(2) + hkr(3)*sin(t);
Ex=sum((xx-x).^2);
Ey=sum((yy-y).^2);
E=(Ex+Ey)/2; %average of the eror in x and y coordinates
end
It seems to work for me. If anyone gets a simpler idea to do this, it would be great.

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John D'Errico
John D'Errico 2023-3-23
编辑：John D'Errico 2023-3-23
Ran in:
You are still looking for help on this?
There is ABSOLUTELY no need to use a simplex optimizer for this problem. NONE AT ALL. Let me make up some data.
n = 50;
XY = normalize(randn(n,2),2,'norm') + 2 + randn(n,2)/5;
plot(XY(:,1),XY(:,2),'.')
axis equal
The data is a circle, with center at [2,2], and a radius of 1.
[C,R,rmse] = circlefit(XY)
C = 1×2
2.0039 2.0386
R = 1.0565
rmse = 0.2043
theta = linspace(0,2*pi)';
XYhat = C + R*[cos(theta),sin(theta)];
hold on
plot(XYhat(:,1),XYhat(:,2),'r-')
plot(C(1),C(2),'gx')
hold off
So a circle at center pretty near the point [2,2], with radius estimated as also very close to 1.
Even if you have only half an arc of a circle, it still works.
k = XY(:,1) > 2;
plot(XY(k,1),XY(k,2),'.')
axis equal
[C,R,rmse] = circlefit(XY(k,:))
C = 1×2
2.1750 2.0898
R = 0.9377
rmse = 0.2168
theta = linspace(-pi/2,pi/2)';
XYhat = C + R*[cos(theta),sin(theta)];
hold on
plot(XYhat(:,1),XYhat(:,2),'r-')
plot(C(1),C(2),'gx')
hold off
As expected, the estimates are less good there, but that data is seriously noisy. Honestly, if someone gave me this data and asked to fit a circle to it, I would have guessed the data is almost useless to fit a circle.
I've attached circlefit to this answer. (circlefit also works on higher dimensional data, where it can fit a sphere to data in 3-d, etc.)
Circlefit uses an algebraic trick to remove the quadratic terms in the circle equations. This reduces the problem to finding the best intersection of many straight lines. But that is a problem easily solved using linear algebra. It can also use robustfit from the stats toolbox, if you have it.
Don't write your own code to solve problems like this.

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