Calculate a 2D homogeneous perspective transformation matrix from 4 points in MATLAB

Question

I\'ve got coordinates of 4 points in 2D that form a rectangle and their coordinates after a perspective transformation has been applied.

The perspective transfo

Answer 1

Should have been an easy question. So how do I get M*A=B*W into a solvable form? It's just matrix multiplications, so we can write this as a system of linear equations. You know like: M11*A11 + M12*A21 + M13*A31 = B11*W11 + B12*W21 + B13*W31 + B14*W41. And every system of linear equations can be written in the form Ax=b, or to avoid confusion with already used variables in my question: N*x=y. That's all.

An example according to my question: I generate some input data with a known M and W:

M = [
    1 2 3;
    4 5 6;
    7 8 1
];
A = [
    0 0 1 1;
    0 1 0 1;
    1 1 1 1
];
W = [
    4 0 0 0;
    0 3 0 0;
    0 0 2 0;
    0 0 0 1
];
B = M*A*(W^-1);

Then I forget about M and W. Meaning I now have 13 variables I'm looking to solve. I rewrite M*A=B*W into a system of linear equations, and from there into the form N*x=y. In N every column has the factors for one variable:

N = [
    A(1,1) A(2,1) A(3,1)      0      0      0      0      0      0 -B(1,1)       0       0       0;
         0      0      0 A(1,1) A(2,1) A(3,1)      0      0      0 -B(2,1)       0       0       0;
         0      0      0      0      0      0 A(1,1) A(2,1) A(3,1) -B(3,1)       0       0       0;
    A(1,2) A(2,2) A(3,2)      0      0      0      0      0      0       0 -B(1,2)       0       0;
         0      0      0 A(1,2) A(2,2) A(3,2)      0      0      0       0 -B(2,2)       0       0;
         0      0      0      0      0      0 A(1,2) A(2,2) A(3,2)       0 -B(3,2)       0       0;
    A(1,3) A(2,3) A(3,3)      0      0      0      0      0      0       0       0 -B(1,3)       0;
         0      0      0 A(1,3) A(2,3) A(3,3)      0      0      0       0       0 -B(2,3)       0;
         0      0      0      0      0      0 A(1,3) A(2,3) A(3,3)       0       0 -B(3,3)       0;
    A(1,4) A(2,4) A(3,4)      0      0      0      0      0      0       0       0       0 -B(1,4);
         0      0      0 A(1,4) A(2,4) A(3,4)      0      0      0       0       0       0 -B(2,4);
         0      0      0      0      0      0 A(1,4) A(2,4) A(3,4)       0       0       0 -B(3,4);
         0      0      0      0      0      0      0      0      1       0       0       0       0
];

And y is:

y = [ 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 1 ];

Notice the equation described by the last row in N whose solution is 1 according to y. That's what I mentioned in my question, you have to fix one of the entries of M to get a single solution. (We can do this because every multiple of M is the same transformation.) And with this equation I'm saying M33 should be 1.

We solve this for x:

x = N\y

and get:

x = [ 1.00000; 2.00000; 3.00000; 4.00000; 5.00000; 6.00000; 7.00000; 8.00000; 1.00000; 4.00000; 3.00000; 2.00000; 1.00000 ]

which are the solutions for [ M11, M12, M13, M21, M22, M23, M31, M32, M33, w1, w2, w3, w4 ]

When doing this in JavaScript I could use the Numeric JavaScript library which has the needed function solve to solve Ax=b.