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assume we have 3 vertices and a point inside the triangle. with a homography matrix [a,b,c] [d,e,f] [g,h,1] Also the final location of the 3 vertices and the point that is not inside the triangle anymore. how to find out this homography matrix? you are free to set the vertices to any location. The one I thought is after transformation, the triangle becomes a single line, all points lie on this line. However, the matrix I found is not invertible. anyone please help? |
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Suppose we have the four points $p_1,p_2,p_3,p_4$ with homogeneous coordinates $$ p_1=\begin{bmatrix}-1 \\ 0 \\ 1\end{bmatrix}, \quad p_2=\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}, \quad p_3=\begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix}, \quad p_4=\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}, $$ and we want to map $p_1,p_2,p_3$ to themselves but $p_4$ should be mapped to $$q=\begin{bmatrix}0 \\ -1 \\ 1\end{bmatrix}.$$ This results in a linear equation system of the form $$ \left\{ \begin{array}{rcl} Hp_1 & = & \lambda_1 p_1\\ Hp_2 & = & \lambda_2 p_2 \\ Hp_3 & = & \lambda_3 p_3 \\ Hp_4 & = & \lambda_4 q \end{array} \right., $$ in which there are twelve unknowns (eight from $H$ and four $\lambda_i$) and twelve equations. Solve this system and get $$H=\frac{1}{3}\begin{bmatrix}3 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & -2 & 3\end{bmatrix}.$$ Note however that you may not in general assume that a given element in $H$ is one just because the scaling is arbitrary, because it may need to be zero. |
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