Projective Transform - matlab code - image-processing

Question

i can't use any toolbox function i need to build it from scratch.
% load images
img1 = readImage('roadSign.tif');
img2 = readImage('lena.tif');
% call the main function
mapIntoImage(img1,img2)
function [newImage] = mapIntoImage(imageA,imageB)
% Input: imageA, imageB - a grayscale image in the range [0..255].
%
% Output: newImage – imageA into which image B has been mapped.
%
showImage(imageA)
hold on
% Initially, the list of points is empty.
xy = [];
% Loop, picking up the points.
disp('Please enter corners of place to insert image in clockwise order.')
for j = 1:4
[xi,yi] = ginput(1);
%draw a yellow dot
plot(xi,yi,'yo')
xy(:,j) = [xi;yi];
end
% get x1 y1 cordinates - xy(:, 1)
imgRow = size(imageB,1);
imgCol = size(imageB,2);
[X,Y] = meshgrid(1:imgCol,1:imgRow);
imgBcords = [0 size(imageB, 1) size(imageB,1) 0 ;
0 0 size(imageB,2) size(imageB,2)];
coefs = findCoefficients(xy, imgBcords);
A = [coefs(1) coefs(2) coefs(5);coefs(3) coefs(4) coefs(6); coefs(7) coefs(8) 1];
temp = zeros(size(X,1), size(X,2), 3);
new = ones(256);
for i = 1:size(X,1)
for j = 1:size(X,2)
temp(i,j,:) =A*[X(i,j); Y(i,j); new(i,j)];
end
end
end
function [ result ] = findCoefficients( imageA, imageB )
% finds coefficients for inverse mapping algorithem
% takes 2 X 2d vectors each consists of 4 points x,y
% and returns the coef accroding to reverse mapping function
%
% x y 0 0 1 0 -xx' -yx'
% 0 0 x y 0 1 -xy' -yy'
% y' and x' are in the destenation picture;
A = [imageB(1,1) imageB(2,1) 0 0 1 0 -imageB(1,1)*imageA(1,1) -imageB(2,1)*imageA(1,1);
0 0 imageB(1,1) imageB(2,1) 0 1 -imageB(1,1)*imageA(2,1) -imageB(2,1)*imageA(2,1);
imageB(1,2) imageB(2,2) 0 0 1 0 -imageB(1,2)*imageA(1,2) -imageB(2,2)*imageA(1,2);
0 0 imageB(1,2) imageB(2,2) 0 1 -imageB(1,2)*imageA(2,2) -imageB(2,2)*imageA(2,2);
imageB(1,3) imageB(2,3) 0 0 1 0 -imageB(1,3)*imageA(1,3) -imageB(2,3)*imageA(1,3);
0 0 imageB(1,3) imageB(2,3) 0 1 -imageB(1,3)*imageA(2,3) -imageB(2,3)*imageA(2,3);
imageB(1,4) imageB(2,4) 0 0 1 0 -imageB(1,4)*imageA(1,4) -imageB(2,4)*imageA(1,4);
0 0 imageB(1,4) imageB(2,4) 0 1 -imageB(1,4)*imageA(2,4) -imageB(2,4)*imageA(2,4)];
B = [imageB(1,1); imageB(2,1); imageB(1,2); imageB(2,2); imageB(1,3); imageB(2,3); imageB(1,4); imageB(2,4)];
result = pinv(A)*B;
end
i want to build now the transform
[x' y' 1] = A*[X Y 1];
i have figured out that i would need to use repmat, but i can't seem to get to the real syntax without loops.
what's the most efficient way to do it?

Answer 1

A projective transform has the form of
$ x' = \frac {a_{11}x+a_{12}y+a_{13}}{a_{13}x+a_{23}y+a_{33}} \\
y' = \frac {a_{21}x+a_{22}y+a_{23}}{a_{13}x+a_{23}y+a_{33}}
$
Where the coefficients are defined up to some scale factor. One of the ways to ensure a constant scale factor is to set $a_{33}=1$. One easy way to think about it is to use the homogenous coordinates:
$ \left( \begin{array}{ccc} x'\\y'\\S\end{array} \right) =
\left( \begin{array}{ccc} a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{array} \right)
\left( \begin{array}{ccc} x\\y\\1\end{array} \right)
$
These coordinates are defined up to scale. That is,
$ \left( \begin{array}{ccc} x'/S\\y'/S\\1\end{array} \right) \equiv
\left( \begin{array}{ccc} x'\\y'\\S\end{array} \right)$
Thus, in your case you should do: (Assuming that x and y are column vectors, and A is the transpose of the matrix that I described above:
XY = A * [x y ones(size(x))];
XY(:,1) = XY(:,1)./XY(:,3);
XY(:,2) = XY(:,2)./XY(:,3);