Asked by Jan Jan
on 14 Jun 2019

Hello,

there is unknown nonlinear function transformating points of the 2D surface. We have set of known points in the format:

[X,Y,X',Y'] %X coordinate before transfomation, Y coordinate before transformation, X coordinate after transfomation, Y coordinate after transformation

So we know coordinates of point before and after transfomation. What we need is a method to calculate coordinates after transformation for any X and Y. I tried two methods already. First one is using Lagrange polynomial interpolation using complex numbers in a way that

From there there was 10 random known points chosen to form 10-th order polynomial. Process was repeated 50 times and average interpolating polynomial was received. Unfortunately this method was giving very inacurrate results. I am posting this section of code.

A = importdata('basegridX.txt'); %121base grid X values

B = importdata('basegridY.txt'); %121base grid Y values

C = importdata('transgridX.txt');%121 shifted grid X values

D = importdata('transgridY.txt');%121 shifted grid Y values

E=complex(A,B);

F=complex(C,D);

syms z

M=0;

for i=1:50

msize = numel(E);

rng(i)

G=E(randperm(msize, 10));

rng(i)

H=F(randperm(msize, 10)); %choose semi random element 50 times

z1=G(1,1);

z2=G(2,1);

z3=G(3,1);

z4=G(4,1);

z5=G(5,1);

z6=G(6,1);

z7=G(7,1);

z8=G(8,1);

z9=G(9,1);

z10=G(10,1);

f1=H(1,1);

f2=H(2,1);

f3=H(3,1);

f4=H(4,1);

f5=H(5,1);

f6=H(6,1);

f7=H(7,1);

f8=H(8,1);

f9=H(9,1);

f10=H(10,1);

L1=((z-z2)*(z-z3)*(z-z4)*(z-z5)*(z-z6)*(z-z7)*(z-z8)*(z-z9)*(z-z10))/((z1-z2)*(z1-z3)*(z1-z4)*(z1-z5)*(z1-z6)*(z1-z7)*(z1-z8)*(z1-z9)*(z1-z10))*f1;

L2=((z-z1)*(z-z3)*(z-z4)*(z-z5)*(z-z6)*(z-z7)*(z-z8)*(z-z9)*(z-z10))/((z2-z1)*(z2-z3)*(z2-z4)*(z2-z5)*(z2-z6)*(z2-z7)*(z2-z8)*(z2-z9)*(z2-z10))*f2;

L3=((z-z1)*(z-z2)*(z-z4)*(z-z5)*(z-z6)*(z-z7)*(z-z8)*(z-z9)*(z-z10))/((z3-z1)*(z3-z2)*(z3-z4)*(z3-z5)*(z3-z6)*(z3-z7)*(z3-z8)*(z3-z9)*(z3-z10))*f3;

L4=((z-z1)*(z-z2)*(z-z3)*(z-z5)*(z-z6)*(z-z7)*(z-z8)*(z-z9)*(z-z10))/((z4-z1)*(z4-z2)*(z4-z3)*(z4-z5)*(z4-z6)*(z4-z7)*(z4-z8)*(z4-z9)*(z4-z10))*f4;

L5=((z-z1)*(z-z2)*(z-z3)*(z-z4)*(z-z6)*(z-z7)*(z-z8)*(z-z9)*(z-z10))/((z5-z1)*(z5-z2)*(z5-z3)*(z5-z4)*(z5-z6)*(z5-z7)*(z5-z8)*(z5-z9)*(z5-z10))*f5;

L6=((z-z1)*(z-z2)*(z-z3)*(z-z4)*(z-z5)*(z-z7)*(z-z8)*(z-z9)*(z-z10))/((z6-z1)*(z6-z2)*(z6-z3)*(z6-z4)*(z6-z5)*(z6-z7)*(z6-z8)*(z6-z9)*(z6-z10))*f6;

L7=((z-z1)*(z-z2)*(z-z3)*(z-z4)*(z-z5)*(z-z6)*(z-z8)*(z-z9)*(z-z10))/((z7-z1)*(z7-z2)*(z7-z3)*(z7-z4)*(z7-z5)*(z7-z6)*(z7-z8)*(z7-z9)*(z7-z10))*f7;

L8=((z-z1)*(z-z2)*(z-z3)*(z-z4)*(z-z5)*(z-z6)*(z-z7)*(z-z9)*(z-z10))/((z8-z1)*(z8-z2)*(z8-z3)*(z8-z4)*(z8-z5)*(z8-z6)*(z8-z7)*(z8-z9)*(z8-z10))*f8;

L9=((z-z1)*(z-z2)*(z-z3)*(z-z4)*(z-z5)*(z-z6)*(z-z7)*(z-z8)*(z-z10))/((z9-z1)*(z9-z2)*(z9-z3)*(z9-z4)*(z9-z5)*(z9-z6)*(z9-z7)*(z9-z8)*(z9-z10))*f9;

L10=((z-z1)*(z-z2)*(z-z3)*(z-z4)*(z-z5)*(z-z6)*(z-z7)*(z-z8)*(z-z9))/((z10-z1)*(z10-z2)*(z10-z3)*(z10-z4)*(z10-z5)*(z10-z6)*(z10-z7)*(z10-z8)*(z10-z9))*f10;

L=L1+L2+L3+L4+L5+L6+L7+L8+L9+L10;

M=M+L;

i=i+1;

end

M=M/50; %average polynomials sum

Second idea was to divide surface to smallest possible triangles and treat every transformation as linear transformation and use corresponding linear function to transform points placed inside this elementary triangle. This method yielded better results yet still not precise enough. This is section of code calculating linear functions for elementary triangle.

if mod(i,2)==1

f(i,1)=0.1*(x2-x1)*w+0.1*(x3-x1)*z+x1-0.1*(x2-x1)*w1-0.1*(x3-x1)*z1;

f(i,2)=0.1*(y2-y1)*w+0.1*(y3-y1)*z+y1-0.1*(y2-y1)*w1-0.1*(y3-y1)*z1;

else

f(i,1)=-0.1*(xa2-xa1)*w-0.1*(xa3-xa1)*z+xa1+0.1*(xa2-xa1)*wa1+0.1*(xa3-xa1)*za1;

f(i,2)=-0.1*(ya2-ya1)*w-0.1*(ya3-ya1)*z+ya1+0.1*(ya2-ya1)*wa1+0.1*(ya3-ya1)*za1;

%if checks triangle type, there are two types of triangles with slightly different equation

%x1,x2,x3,y1,y2,y3 are coordinates of triangle vertexes after transfomation - so X' Y' from lagrange example

%w1,z1 are coordinates of triangle right angle vertex before transformation - so X, Y from lagrange example

%w,z are unknowns - coordinates of point we are looking for

%xa1,ya1,wa1 are corresponding coordinates in triangles of second type

%0.1 are hypotenuses length in triangles before transfomation so basically w2-w1, w3-w1 etc

I would be very thankfull if someone sees holes in my reasoning or can propose more accurate methods of interpolation that would work with that type of data.

Answer by Matt J
on 15 Jun 2019

If you have a model of the transformation the best approach would be to do a surface fit.

Jan Jan
on 15 Jun 2019

I do not really understand what you mean by 'model of transformation'.

Anyway I just tried this approach using cftool I set X and Y values as coordinates of red dots from picture above, but now it doesn't allow me to add green dots in complex form as Z values. ("Using only the real component of complex data.")

I worked around the issue and tried to do it separately for X and Y values, I tried every fit method and all of them are giving too inaccurate methods. Linear fit gave exactly the same results as one of my previous version, best seemed to be spline interpolant but still errors above 10%.

Matt J
on 16 Jun 2019 at 12:18

I do not really understand what you mean by 'model of transformation'.

I mean you haven't said what the mathematical form of the final surface is. If you know the mathematical form, you should try to fit the unknown parameters in that model using cftool or similar.

but still errors above 10%.

How are you making that assessment? You would have to know the correct interpolated values in advance to be able to say how big the errors are.

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## 2 Comments

## Matt J (view profile)

Direct link to this comment:https://in.mathworks.com/matlabcentral/answers/467167-every-point-on-2d-surface-is-being-transformed-by-unknown-function-how-to-interpolate-coordinates-o#comment_714931

## Jan Jan (view profile)

Direct link to this comment:https://in.mathworks.com/matlabcentral/answers/467167-every-point-on-2d-surface-is-being-transformed-by-unknown-function-how-to-interpolate-coordinates-o#comment_714952

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