Computing a matrix which transforms a quadrangle to another quadrangle in 2D

In the figure below the goal is to compute the homography matrix H which transforms the points a1 a2 a3 a4 to their counterparts b1 b2 b3 b4. That is:

[b1 b2 b3 b4] = H * [a1 a2 a3 a4]

What way would you suggest to be the best way to calculate H(3x3). a1...b4 are points in 2D which are represented in homogeneous coordinate systems ( that is [a1_x a1_y 1]', ...). EDIT: For these types of problems we use SVD, So i would like to see how this can be simply done in Matlab.

EDIT:

Here is how I initially tried to solve it using svd (H=Q/P) in Maltlab. Cosider the following code for the given example

px=[0 1 1 0];  % a square
py=[1 1 0 0];

qx=[18 18 80 80];    % a random quadrangle
qy=[-20 20 60 -60];
if (DEBUG)
  fill(px,py,'r');
  fill(qx,qy,'r');
end

Q=[qx;qy;ones(size(qx))];
P=[px;py;ones(size(px))];
H=Q/P;
H*P-Q
answer:
   -0.0000         0         0         0         0
  -20.0000   20.0000  -20.0000   20.0000    0.0000
   -0.0000         0         0         0   -0.0000

I am expecting the answer to be a null matrix but it is not!... and that's why I asked this question in StackOverflow. Now, we all know it is a projective transformation not obviously Euclidean. However, it is good to know if in general care calculating such matrix using only 4 points is possible.

Using the data you posted:

P = [px(:) py(:)];
Q = [qx(:) qy(:)];

Compute the transformation:

H = Q/P;

apply the transformation:

Q2 = H*P;

Compare the results:

err = Q2-Q

output:

err =
   7.1054e-15   7.1054e-15
  -3.5527e-15  -3.5527e-15
  -1.4211e-14  -2.1316e-14
   1.4211e-14   1.4211e-14

which is zeros for all intents and purposes..

---

EDIT:

So as you pointed out in the comments, the above method will not compute the 3x3 homography matrix. It simply solves the system of equations with as many equations as points provided:

H * A = B   -->   H = B*inv(A)   -->   H = B/A (mrdivide)

Otherwise, MATLAB has the CP2TFORM function in the image processing toolbox. Here is an example applied to the image shown:

%# read illustration image
img = imread('https://i.sstatic.net/ZvaZK.png');
img = imcomplement(im2bw(img));

%# split into two equal-sized images
imgs{1} = img(:,fix(1:end/2));
imgs{2} = img(:,fix(end/2:end-1));

%# extract the four corner points A and B from both images
C = cell(1,2);
for i=1:2
    %# some processing
    I = imfill(imgs{i}, 'holes');
    I = bwareaopen(imclearborder(I),200);
    I = imfilter(im2double(I), fspecial('gaussian'));

    %# find 4 corners
    C{i} = corner(I, 4);

    %# sort corners in a consistent way (counter-clockwise)
    idx = convhull(C{i}(:,1), C{i}(:,2));
    C{i} = C{i}(idx(1:end-1),:);
end

%# show the two images with the detected corners
figure
for i=1:2
    subplot(1,2,i), imshow(imgs{i})
    line(C{i}(:,1), C{i}(:,2), 'Color','r', 'Marker','*', 'LineStyle','none')
    text(C{i}(:,1), C{i}(:,2), num2str((1:4)'), 'Color','r', ...
        'FontSize',18, 'Horiz','left', 'Vert','bottom')
end

With the corners detected, now we can obtain the spatial transformation:

%# two sets of points
[A,B] = deal(C{:});

%# infer projective transformation using CP2TFORM
T = cp2tform(A, B, 'projective');

%# 3x3 Homography matrix
H = T.tdata.T;
Hinv = T.tdata.Tinv;

%# align points in A into B
X = tformfwd(T, A(:,1), A(:,2));

%# show result of transformation
line(X([1:end 1],1), X([1:end 1],2), 'Color','g', 'LineWidth',2)

The result:

>> H = T.tdata.T
H =
      0.74311    -0.055998    0.0062438
      0.44989      -1.0567   -0.0035331
      -293.31       62.704      -1.1742

>> Hinv = T.tdata.Tinv
Hinv =
       -1.924     -0.42859   -0.0089411
      -2.0585      -1.2615   -0.0071501
       370.68       39.695            1

We can confirm the calculation ourselves:

%# points must be in Homogenous coordinates (x,y,w)
>> Z = [A ones(size(A,1),1)] * H;
>> Z = bsxfun(@rdivide, Z, Z(:,end))   %# divide by w
Z =
          152           57            1
          219          191            1
           62          240            1
           92          109            1

which maps to the points in B:

%# maximum error
>> max(max( abs(Z(:,1:2)-B) ))
ans =
   8.5265e-14