Q1: Affine Rectification

1.1 Output Images and Angle Evaluation Results:

Before	After
0.9801935239283572	0.9999413201351588
0.9077385010215981	0.9982815650684259

Before	After
0.9954293641543964	0.9999413201351588
0.9964914336529828	0.9999752424843763

Before	After
0.9928400002003689	0.9999744329310002
0.9958562285955483	0.9995508549532741

Before	After
0.9868327646462675	0.9999251970022865
0.9987191070813484	0.9999383464088631

Before	After
0.7842324298864688	0.999978139517972
0.9999970152053781	0.9999869448826936

1.2 Implementation Description

Use the annotated points to get two pairs of parallel lines. Use $x=l_1$x$l_2$ to get the intersection of two lines. Use $x_1$x$x_2$ to get the line at infinite. Then, we can compute H with:

$l=H^{-T}l'$, l refers to the line at infinite, $l=[l_1,l_2,l_3]^{-T}$.

We get one solution H=$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ l_{1} & l_{2} & l_{3} \end{bmatrix} $$

Q2: Metric Rectification

2.1 Output Images and Angle Evaluation Results:

Before	After
-0.10272308450664172	0.05650143872109491
0.45958021577399355	0.01784973627322923

Before	After
0.43166311591881273	0.09891134391597332
0.416638993307934	0.06254952999238895

Before	After
-0.2526709009836574	0.0018168249234062953
0.08817394212374821	0.008198536236441723

Before	After
0.1677181548716634	-0.008091866263126083
0.03498266075814032	-0.029306932732914674

Before	After
0.6685881410241951	-0.021182064318718698
-0.04479521389935166	-0.009640234139807519

2.2 Implementation Description

According to $$ l' C_\infty^{*'} M' = 0 $$, use SVD to get the solution of AC=0 (image is already affine-rectified)

Then use SVD for $ C_\infty^{*'} $ and get 2 Non-zero eigenvalues.

$$ C_\infty^{*'} = U \begin{bmatrix} \sigma_1 & 0 & 0 \\ 0 & \sigma_2 & 0 \\ 0 & 0 & 0 \end{bmatrix} U^T $$

With $C_\infty^{*} =HC_\infty^{*'} H^{T}$, we get $$ H= \begin{bmatrix} \sqrt{\sigma_1} & 0 & 0 \\ 0 & \sqrt{\sigma_2} & 0 \\ 0 & 0 & 1 \end{bmatrix} U^T $$

Q3: Planar Homography from Point Correspondences

3.1 Output:

3.2 Implementation Description:

With x'=Hx, we know that H can be solved in Ah=0, where A_i=\begin{bmatrix} x_2^i & y_2^i & 1 & 0 & 0 & 0 & -x_1^i x_2^i & -x_1^i y_2^i & -x_1^i \\ 0 & 0 & 0 & x_2^i & y_2^i & 1 & -y_1^i x_2^i & -y_1^i y_2^i & -y_1^i \end{bmatrix}

Then get the solution H with eigenvalues decomposition of A.

Q5: Bonus: More Planar Homography from Point Correspondences

5.1 Output

5.2 Implementation Description:

Expand the situation in Q3 to take 12 points and then perform homography on each of them.