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 transformation is calculated in homogeneous coordinates and defined by a 3x3 matrix M. If the matrix is not known, how can I calculate it from the given points?

The calculation for one point would be:

| M11 M12 M13 |   | P1.x |   | w*P1'.x |
| M21 M22 M23 | * | P1.y | = | w*P1'.y |
| M31 M32 M33 |   | 1    |   | w*1     |

To calculate all points simultaneously I write them together in one matrix A and analogously for the transformed points in a matrix B:

    | P1.x P2.x P3.x P4.x |
A = | P1.y P2.y P3.y P4.y |
    | 1    1    1    1    |

So the equation is M*A=B and this can be solved for M in MATLAB by M = B/A or M = (A'\B')'.

But it's not that easy. I know the coordinates of the points after transformation, but I don't know the exact B, because there is the factor w and it's not necessary 1 after a homogeneous transformation. Cause in homogeneous coordinates every multiple of a vector is the same point and I don't know which multiple I'll get.

To take account of these unknown factors I write the equation as M*A=B*W where W is a diagonal matrix with the factors w1...w4 for every point in B on the diagonal. So A and B are now completely known and I have to solve this equation for M and W.

If I could rearrange the equation into the form x*A=B or A*x=B where x would be something like M*W I could solve it and knowing the solution for M*W would maybe be enough already. However despite trying every possible rearrangement I didn't managed to do so. Until it hit me that encapsulating (M*W) would not be possible, since one is a 3x3 matrix and the other a 4x4 matrix. And here I'm stuck.

Also M*A=B*W does not have a single solution for M, because every multiple of M is the same transformation. Writing this as a system of linear equations one could simply fix one of the entries of M to get a single solution. Furthermore there might be inputs that have no solution for M at all, but let's not worry about this for now.

What I'm actually trying to achieve is some kind of vector graphics editing program where the user can drag the corners of a shape's bounding box to transform it, while internally the transformation matrix is calculated.

And actually I need this in JavaScript, but if I can't even solve this in MATLAB I'm completely stuck.

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 ]

W is not needed after M has been calculated. For a generic point (x, y), the corresponding w is calculated while solving x' and y'.

| M11 M12 M13 |   | x |   | w * x' |
| M21 M22 M23 | * | y | = | w * y' |
| M31 M32 M33 |   | 1 |   | w * 1  |

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