Homography

Deriving 2D transforms from 4 points

Source

Given 4 pairs of points of the form $(x,y)\to (x^{\prime },y^{\prime })$, compute the homography matrix

$$h=\left[\begin{array}{ccc}{h}_{11}& h_{12}& tx\\ h_{21}& h_{22}& ty\\ h_{31}& h_{32}& 1\end{array}\right]$$

We write the pre-transformed points as $P$ and the transformed points as $P^{\prime }$

$$P=\left[\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\\ x_3& y_3\\ x_4& y_4\end{array}\right],P^{\prime }=\left[\begin{array}{cc}{x}_1^{\prime }& y_1^{\prime }\\ x_2^{\prime }& y_2^{\prime }\\ x_3^{\prime }& y_3^{\prime }\\ x_4^{\prime }& y_4^{\prime }\end{array}\right]$$

Defining the equation

$$\begin{array}{r}{x}^{\prime }=\frac{h_{11}x+h_{12}y+tx}{h_{31}x+h_{32}y+1}\\ y^{\prime }=\frac{h_{21}x+h_{22}y+ty}{h_{31}x+h_{32}y+1}\end{array}$$

After cross-multiplying and re-arranging:

$$\begin{array}{r}{h}_{11}x+h_{12}y+tx-h_{31}x{x}^{\prime }-h_{32}y{x}^{\prime }=x^{\prime }\\ h_{21}x+h_{22}y+ty-h_{31}x{y}^{\prime }-h_{32}y{y}^{\prime }=y^{\prime }\end{array}$$ $$\left[\begin{array}{cccccccc}{x}_1& y_1& 1& 0& 0& 0& -x_1{x}_1^{\prime }& -y_1{x}_1^{\prime }\\ 0& 0& 0& x_1& y_1& 1& -x_1{y}_1^{\prime }& -y_1{y}_1^{\prime }\\ x_2& y_2& 1& 0& 0& 0& -x_2{x}_2^{\prime }& -y_2{x}_2^{\prime }\\ 0& 0& 0& x_2& y_2& 1& -x_2{y}_2^{\prime }& -y_2{y}_2^{\prime }\\ x_3& y_3& 1& 0& 0& 0& -x_3{x}_3^{\prime }& -y_3{x}_3^{\prime }\\ 0& 0& 0& x_3& y_3& 1& -x_3{y}_3^{\prime }& -y_3{y}_3^{\prime }\\ x_4& y_4& 1& 0& 0& 0& -x_4{x}_4^{\prime }& -y_4{x}_4^{\prime }\\ 0& 0& 0& x_4& y_4& 1& -x_4{y}_4^{\prime }& -y_4{y}_4^{\prime }\end{array}\right]\left[\begin{array}{c}{h}_{11}\\ h_{12}\\ tx\\ h_{21}\\ h_{22}\\ ty\\ h_{31}\\ h_{32}\end{array}\right]=\left[\begin{array}{c}{x}_1^{\prime }\\ y_1^{\prime }\\ x_2^{\prime }\\ y_2^{\prime }\\ x_3^{\prime }\\ y_3^{\prime }\\ x_4^{\prime }\\ y_4^{\prime }\end{array}\right]$$

We solve $Ah=b$ for $h$ (i. e. $h=A^{-1}b$).

This gets us the 3x3 $h$. However, matrix3d takes a 4x4 matrix. Since we don’t care about $z$, we can make it just map back to itself:

$$h=\left[\begin{array}{cccc}{h}_{11}& h_{12}& 0& tx\\ h_{21}& h_{22}& 0& ty\\ 0& 0& 1& 0\\ h_{31}& h_{32}& 0& 1\end{array}\right]$$