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

OpenCV has a neat function that does this called getPerspectiveTransform. The source code for this function is available on github with this description:

/* Calculates coefficients of perspective transformation
 * which maps (xi,yi) to (ui,vi), (i=1,2,3,4):
 *
 *      c00*xi + c01*yi + c02
 * ui = ---------------------
 *      c20*xi + c21*yi + c22
 *
 *      c10*xi + c11*yi + c12
 * vi = ---------------------
 *      c20*xi + c21*yi + c22
 *
 * Coefficients are calculated by solving linear system:
 * / x0 y0  1  0  0  0 -x0*u0 -y0*u0 \ /c00\ /u0\
 * | x1 y1  1  0  0  0 -x1*u1 -y1*u1 | |c01| |u1|
 * | x2 y2  1  0  0  0 -x2*u2 -y2*u2 | |c02| |u2|
 * | x3 y3  1  0  0  0 -x3*u3 -y3*u3 |.|c10|=|u3|,
 * |  0  0  0 x0 y0  1 -x0*v0 -y0*v0 | |c11| |v0|
 * |  0  0  0 x1 y1  1 -x1*v1 -y1*v1 | |c12| |v1|
 * |  0  0  0 x2 y2  1 -x2*v2 -y2*v2 | |c20| |v2|
 * \  0  0  0 x3 y3  1 -x3*v3 -y3*v3 / \c21/ \v3/
 *
 * where:
 *   cij - matrix coefficients, c22 = 1
 */

This system of equations is smaller as it avoids solving for W and M33 (called c22 by OpenCV). So how does it work? The linear system can be recreated by the following steps:

Start with the formulas for projection transformation:

     c00*xi + c01*yi + c02
ui = ---------------------
     c20*xi + c21*yi + c22

     c10*xi + c11*yi + c12
vi = ---------------------
     c20*xi + c21*yi + c22

Multiply both sides with the denominator:

(c20*xi + c21*yi + c22) * ui = c00*xi + c01*yi + c02
(c20*xi + c21*yi + c22) * vi = c10*xi + c11*yi + c12

Distribute ui and vi:

c20*xi*ui + c21*yi*ui + c22*ui = c00*xi + c01*yi + c02
c20*xi*vi + c21*yi*vi + c22*vi = c10*xi + c11*yi + c12

Assume c22 = 1:

c20*xi*ui + c21*yi*ui + ui = c00*xi + c01*yi + c02
c20*xi*vi + c21*yi*vi + vi = c10*xi + c11*yi + c12

Collect all cij on the left hand side:

c00*xi + c01*yi + c02 - c20*xi*ui - c21*yi*ui = ui
c10*xi + c11*yi + c12 - c20*xi*vi - c21*yi*vi = vi

And finally convert to matrix form for four pairs of points:

/ x0 y0  1  0  0  0 -x0*u0 -y0*u0 \ /c00\ /u0\
| x1 y1  1  0  0  0 -x1*u1 -y1*u1 | |c01| |u1|
| x2 y2  1  0  0  0 -x2*u2 -y2*u2 | |c02| |u2|
| x3 y3  1  0  0  0 -x3*u3 -y3*u3 |.|c10|=|u3|
|  0  0  0 x0 y0  1 -x0*v0 -y0*v0 | |c11| |v0|
|  0  0  0 x1 y1  1 -x1*v1 -y1*v1 | |c12| |v1|
|  0  0  0 x2 y2  1 -x2*v2 -y2*v2 | |c20| |v2|
\  0  0  0 x3 y3  1 -x3*v3 -y3*v3 / \c21/ \v3/

This is now in the form of Ax=b and the solution can be obtained with x = A\b. Remember that c22 = 1.