Non linear least square fitting

Hi,

I am really stuck trying to figure out how to fit a circle to some data points. I have watched plenty of videos on the topic but do not really understand. Could you please explain to me in the most simple way how to plot some random data points, for example x = [2; 3; -2; 1]; and y = [3; 2; 4; -1];, in matlab?

I have only come up with the following but do not know how to proceed:

x = [2; 3; -2; 1];
y = [3; 2; 4; -1];
N = length(x);


% circle's equation x^2+y^2 = 2xc1+2yc2+c3
X = [ones(N,3),x];
Y=y;
phi = inv(X'*X)*X'*Y
plot(x,y,'*')

Thanks alot in advance!

[Adam Danz]

Is your question how to fit a circle to your data (implied by your question title and first sentence) or is it how to create random coordinates (the question that is asked)? Do the randome data points have to be along the circumference of a cirlce? With or without noise?

[Jenny Andersen]

What I meant was how to fit a circle to the data (non linear regression). With random I meant arbitrary points, and for example the points given above. I really just want to understand how to perform least square fitting in matlab, so the points does not really matter in this case. Without noise I guess.

[Matt J]

Jenny's answer converted to a comment:

I tried the following this time, but it does not seem to be working. Anyone knows why?

x = [2; 3; -2; 1];
y = [3; 2; 4; -1];
N = length(x);
plot (x,y,'*')

% circle's equation x^2+y^2 = 2xc1+2yc2+c3
x = c1+sqrt(-y^2+2ay+a^2+c
y = c1+sqrt(-y^2+2ax+a^2+c
X = [ones(N,3),x];
Y = y;
plot(X,Y)

[Matt J]

I have only come up with the following but do not know how to proceed ... phi = inv(X'*X)*X'*Y

You have tried to set up linear equations X*phi=Y and find its Ordinary Least Squares (OLS) solution. This is not a good approach here, because X and Y both have noise in them. But even if you were to do it this way, the better Matlab algorithm for OLS would be,

phi=X\Y

Answer 1

[Adam Danz]

Create noisy circular data

This section creates the noisy data we're fitting.

% Create randome points along a circle
r = 22.52;                          % radius
cnt = [9.923, -2.42];               % (x,y) center
theta = rand(1,1000)*2*pi;          % Theta
xData = r*cos(theta)+cnt(1);        % x coordinates
yData = r*sin(theta)+cnt(2);        % y coordinates

% Add noise to coordinates
xData = xData + (rand(size(xData))*10-5); 
yData = yData + (rand(size(yData))*10-5); 

Plot the input data

The only inputs are xData and yData. The rest of the variables above are unknowns.

% Plot the data with noise
clf()
plot(xData, yData, 'ko')
axis equal
grid on

Fit the xData,yData coordinates to a circle

This section uses nonlinear least squares fitting x = lsqnonlin(fun,x0). The first line defines the function to fit and is the equation for a circle. The second line are estimated starting points. See the link for more info on this function. The output circFit is a 1x3 vector defining the [x_center, y_center, radius] of the fitted circle.

f = @(a) (xData-a(1)).^2 + (yData-a(2)).^2 - a(3).^2;
a0 = [mean(xData),mean(yData),max(xData)-mean(xData)]; 
circFit = lsqnonlin(f,a0);

Add the fitted circle to the noisy data

% use the circFit parameters to create the fitted circle
theta = linspace(0,2*pi,100);  % arbitrary spacing
xFit = circFit(3)*cos(theta) + circFit(1); 
yFit = circFit(3)*sin(theta) + circFit(2);

hold on
plot(xFit,yFit,'r-','LineWidth',3)

Compare the fitted values to the actual (noisy) values

Of course these values may differ when this code is repeated due to random noise.

This creates a table comparing the known and fitted results.

T = array2table([[cnt,r];circFit],'VariableNames',{'xCnt','yCnt','radius'},...
    'RowNames',{'Real values','FitValues'})

T =

2×3 table

xCnt yCnt radius

______ ______ ______

Real values 9.923 -2.42 22.52

FitValues 9.8052 -2.548 22.926

[Matt J]

I really just want to understand how to perform least square fitting in matlab, so the points does not really matter in this case.

But bear in mind, there are analytical alternatives to Adam's solution, for the particular case of circle fitting.

[Jenny Andersen]

Ok. Could you possibly show me the analytical alternatives to Adam's solution?

[Jenny Andersen]

Thanks alot Adam btw.

[Adam Danz]

Thanks, @Matt J. I meant to mention alternative approaches to the special case when fitting a circle but by the time I finished typing out the answer, I had forgotten.

@Jenny Andersen, there are several option on the file exchange.

• https://www.mathworks.com/matlabcentral/fileexchange/5557-circle-fit explained in https://blogs.mathworks.com/pick/2008/03/14/fitting-a-circle-easily/
• https://www.mathworks.com/matlabcentral/fileexchange/36361-circle_fit
• https://www.mathworks.com/matlabcentral/fileexchange/44219-fast-circle-fitting-using-landau-method
• https://www.mathworks.com/matlabcentral/fileexchange/22678-circle-fit-taubin-method

But Matt J's answer is even better since you won't need to use a function external to what Matlab already provides.

Answer 2

[Matt J]

Ok. Could you possibly show me the analytical alternatives to Adam's solution?

See attached. Note that my method does not minimize the same fitting error metric as Adam's. If you prefer his, though (and maybe you should), it might be worth mentioning that I do observe somewhat better convergence when initialized with mine. Notice the smaller final "Norm of step" and "First-order optimality" when the iterations terminate.

f = @(a) (xData-a(1)).^2 + (yData-a(2)).^2 - a(3).^2;
opts=optimoptions('lsqnonlin','Display','iter');


a0 = [mean(xData),mean(yData),max(xData)-mean(xData)]; 
lsqnonlin(f,a0,[],[],opts);

                                                 Norm of      First-order 
         Iteration  Func-count     f(x)          step          optimality
             0          4     6.04793e+07                      1.13e+07
             1          8     1.69499e+07          3.881       6.82e+05      
             2         12     1.67336e+07       0.317084        4.6e+03      
             3         16     1.67336e+07     0.00219778          0.221      


[c,R]=circfitAlgebraic([xData;yData]);
lsqnonlin(f,[c,R],[],[],opts);

                                                 Norm of      First-order 
         Iteration  Func-count     f(x)          step          optimality
             0          4     1.74094e+07                      1.22e+06
             1          8     1.67337e+07       0.555282       1.41e+04      
             2         12     1.67336e+07     0.00672023           2.07      
             3         16     1.67336e+07    9.87177e-07        0.00653     

[Jenny Andersen]

Thanks alot!

Answer 3

[cui,xingxing]

I offer here a robust least squares fit for reference purposes only.

%% Create randome points along a circle
r = 1;
numPts = 100;
theta = linspace(0,2*pi,numPts);
xdata = r*cos(theta)+0.1*randn(1,numPts);
ydata = r*sin(theta)+0.1*randn(1,numPts);

% add noise
xdata(1) = 5;xdata(2) = 4;xdata(3) = 5;
ydata(1) = 4;ydata(2) = 5;ydata(3) = 5;
h1 = scatter(xdata,ydata); % sample data

% noise least square fit circle
fun = @(x)(xdata-x(1)).^2+(ydata-x(2)).^2-x(3).^2;
x = lsqnonlin(fun,[0,0,1]);

Local minimum possible.

lsqnonlin stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

h2 = viscircles(x(1:2),x(3),color="red");hold on; % noise least square fit
h3 = viscircles([0,0],1,color="blue");% ground truth


%% robust fit
fitFcn = @(points)robustLeastSquareCircle(points);
distFcn = @(model, points)((xdata-model(1)).^2+(ydata-model(2)).^2-model(3).^2).^2;
sampleSize = 3;
maxDistance = 0.5;
data = [xdata(:),ydata(:)];

[modelRansac,inlierIdx] = ransac(data,fitFcn,distFcn,sampleSize,maxDistance);

Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.


Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.

h4 = viscircles(modelRansac(1:2),modelRansac(3),color="black");

modelRobust = fitFcn(data(inlierIdx,:));

Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.

h5 = viscircles(modelRobust(1:2),modelRobust(3),color="green");% robust fit'

legend([h1,h2,h3,h4,h5],{'point samples','least square fit','ground truth','Ransac samples fit','robust fit'},Location="northwest")
grid on;axis equal

%% support function
function model = robustLeastSquareCircle(points)
% least square fit circle
xdata = points(:,1);
ydata = points(:,2);
fun = @(x)(xdata-x(1)).^2+(ydata-x(2)).^2-x(3).^2;
model = lsqnonlin(fun,[0,0,1]);
end

-------------------------Off-topic interlude, 2024-------------------------------

I am currently looking for a job in the field of CV algorithm development, based in Shenzhen, Guangdong, China,or a remote support position. I would be very grateful if anyone is willing to offer me a job or make a recommendation. My preliminary resume can be found at: https://cuixing158.github.io/about/ . Thank you!

Email: [e-mail]

[Matt J]

This looks pretty much the same as @Adam Danz's solution.

[cui,xingxing]

I have improved it slightly in the back