Question

Hi i am trying to create the affine transform that will allow me to transform a triangle into another one. What i have are the coordinates for the 2 triangles. Can you help me?

Following the answer by Adam Rosenfield i came up with this code in case anyone is bored to solve the equation himself :

public static AffineTransform createTransform(ThreePointSystem source,
            ThreePointSystem dest) {        
    double x11 = source.point1.getX();
    double x12 = source.point1.getY();
    double x21 = source.point2.getX();
    double x22 = source.point2.getY();
    double x31 = source.point3.getX();
    double x32 = source.point3.getY();
    double y11 = dest.point1.getX();
    double y12 = dest.point1.getY();
    double y21 = dest.point2.getX();
    double y22 = dest.point2.getY();
    double y31 = dest.point3.getX();
    double y32 = dest.point3.getY();

    double a1 = ((y11-y21)*(x12-x32)-(y11-y31)*(x12-x22))/
                ((x11-x21)*(x12-x32)-(x11-x31)*(x12-x22));
    double a2 = ((y11-y21)*(x11-x31)-(y11-y31)*(x11-x21))/
                ((x12-x22)*(x11-x31)-(x12-x32)*(x11-x21));
    double a3 = y11-a1*x11-a2*x12;
    double a4 = ((y12-y22)*(x12-x32)-(y12-y32)*(x12-x22))/
                ((x11-x21)*(x12-x32)-(x11-x31)*(x12-x22));
    double a5 = ((y12-y22)*(x11-x31)-(y12-y32)*(x11-x21))/
                ((x12-x22)*(x11-x31)-(x12-x32)*(x11-x21));
    double a6 = y12-a4*x11-a5*x12;
    return new AffineTransform(a1, a4, a2, a5, a3, a6);
}

Answer 1

I’m going to assume you’re talking about 2D here. An affine transformation matrix has 9 values in it:

    | a1 a2 a3 |
A = | a4 a5 a6 |
    | a7 a8 a9 |

There are 3 input vertices x1, x2, and x3, which when transformed should become y1, y2, y3. However, since we’re working in homogeneous coordinates, applying A to x1 doesn’t necessarily give y1 — it gives a multiple of y1. So, we also have the unknown multipliers k1, k2, and k3, with the equations:

A*x1 = k1*y1
A*x2 = k2*y2
A*x3 = k3*y3

Each of those is a vector, so we really have 9 equations in 12 unknowns, so the solution is going to be underconstrained. If we require that a7=0, a8=0, and a9=1, then the solution will be unique (this choice is natural, since it means if the input point is (x, y, 1), then the output point will always have homogeneous coordinate 1, so the resulting transform is just a 2×2 transform plus a translation).

Hence, this reduces the equations to:

a1*x11 + a2*x12 + a3 = k1*y11
a4*x11 + a5*x12 + a6 = k1*y12
                   1 = k1
a1*x21 + a2*x22 + a3 = k2*y21
a4*x21 + a5*x22 + a6 = k2*y22
                   1 = k2
a1*x31 + a2*x32 + a3 = k3*y31
a4*x31 + a5*x32 + a6 = k3*y32
                   1 = k3

So, k1 = k2 = k3 = 1. Plugging these in and converting to matrix form yields:

| x11 x12   1   0   0   0 |   | a1 |   | y11 |
| x21 x22   1   0   0   0 |   | a2 |   | y21 |
| x31 x32   1   0   0   0 | * | a3 | = | y31 |
|   0   0   0 x11 x12   1 |   | a4 |   | y12 |
|   0   0   0 x21 x22   1 |   | a5 |   | y22 |
|   0   0   0 x31 x32   1 |   | a6 |   | y32 |

Solving this 6×6 system of equations yields you your affine transformation matrix A. It will have a unique solution if and only if the 3 points of your source triangle are not collinear.