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pages/CSE559A/CSE559A_L25.md

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+# CSE559A Lecture 25
+
+## Geometry and Multiple Views
+
+### Cues for estimating Depth
+
+#### Multiple Views (the strongest depth cue)
+
+Two common settings:
+
+**Stereo vision**: a pair of cameras, usually with some constraints on the relative position of the two cameras.
+
+**Structure from (camera) motion**: cameras observing a scene from different viewpoints
+
+Structure and depth are inherently ambiguous from single views.
+
+Other hints for depth:
+
+- Occlusion
+- Perspective effects
+- Texture
+- Object motion
+- Shading
+- Focus/Defocus
+
+#### Focus on Stereo and Multiple Views
+
+Stereo correspondence: Given a point in one of the images, where could its corresponding points be in the other images?
+
+Structure: Given projections of the same 3D point in two or more images, compute the 3D coordinates of that point
+
+Motion: Given a set of corresponding points in two or more images, compute the camera parameters
+
+#### A simple example of estimating depth with stereo:
+
+Stereo: shape from "motion" between two views
+
+We'll need to consider:
+
+- Info on camera pose ("calibration")
+- Image point correspondences
+
+![Simple stereo system](https://notenextra.trance-0.com/CSE559A/Simple_stereo_system.png)
+
+Assume parallel optical axes, known camera parameters (i.e., calibrated cameras).  What is expression for Z?
+
+Similar triangles $(p_l, P, p_r)$ and $(O_l, P, O_r)$:
+
+$$
+\frac{T-x_l+x_r}{Z-f}=\frac{T}{Z}
+$$
+
+$$
+Z = \frac{f \cdot T}{x_l-x_r}
+$$
+
+### Camera Calibration
+
+Use an scene with known geometry
+
+- Correspond image points to 3d points
+- Get least squares solution (or non-linear solution)
+
+Solving unknown camera parameters:
+
+$$
+\begin{bmatrix}
+su\\
+sv\\
+s
+\end{bmatrix}
+= \begin{bmatrix}
+m_{11} & m_{12} & m_{13} & m_{14}\\
+m_{21} & m_{22} & m_{23} & m_{24}\\
+m_{31} & m_{32} & m_{33} & m_{34}
+\end{bmatrix}
+\begin{bmatrix}
+X\\
+Y\\
+Z\\
+1
+\end{bmatrix}
+$$
+
+Method 1: Homogenous linear system. Solve for m's entries using least squares.
+
+$$
+\begin{bmatrix} 
+X_1 & Y_1 & Z_1 & 1 & 0 & 0 & 0 & 0 & -u_1X_1 & -u_1Y_1 & -u_1Z_1 & -u_1 \\
+0 & 0 & 0 & 0 & X_1 & Y_1 & Z_1 & 1 & -v_1X_1 & -v_1Y_1 & -v_1Z_1 & -v_1 \\
+\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
+X_n & Y_n & Z_n & 1 & 0 & 0 & 0 & 0 & -u_nX_n & -u_nY_n & -u_nZ_n & -u_n \\
+0 & 0 & 0 & 0 & X_n & Y_n & Z_n & 1 & -v_nX_n & -v_nY_n & -v_nZ_n & -v_n
+\end{bmatrix}
+\begin{bmatrix} m_{11} \\ m_{12} \\ m_{13} \\ m_{14} \\ m_{21} \\ m_{22} \\ m_{23} \\ m_{24} \\ m_{31} \\ m_{32} \\ m_{33} \\ m_{34} \end{bmatrix} = 0
+$$
+
+Method 2: Non-homogenous linear system. Solve for m's entries using least squares.
+
+**Advantages**
+
+- Easy to formulate and solve
+- Provides initialization for non-linear methods
+
+**Disadvantages**
+
+- Doesn't directly give you camera parameters
+- Doesn't model radial distortion
+- Can't impose constraints, such as known focal length
+
+**Non-linear methods are preferred**
+
+- Define error as difference between projected points and measured points
+- Minimize error using Newton's method or other non-linear optimization
+
+#### Triangulation
+
+Given projections of a 3D point in two or more images (with known camera matrices), find the coordinates of the point
+
+##### Approaches 1: Geometric approach
+
+Find shortest segment connecting the two viewing rays and let $X$ be the midpoint of that segment
+
+![Triangulation geometric approach](https://notenextra.trance-0.com/CSE559A/Triangulation_geometric_approach.png)
+
+##### Approaches 2: Non-linear optimization
+
+Minimize error between projected point and measured point
+
+$$
+||\operatorname{proj}(P_1 X) - x_1||_2^2 + ||\operatorname{proj}(P_2 X) - x_2||_2^2
+$$
+
+![Triangulation non-linear optimization](https://notenextra.trance-0.com/CSE559A/Triangulation_non_linear_optimization.png)
+
+##### Approaches 3: Linear approach
+
+$x_1\simeq P_1X$ and $x_2\simeq P_2X$
+
+$x_1\times P_1X = 0$ and $x_2\times P_2X = 0$
+
+$[x_{1_{\times}}]P_1X = 0$ and $[x_{2_{\times}}]P_2X = 0$
+
+Rewrite as:
+
+$$
+a\times b=\begin{bmatrix}
+0 & -a_3 & a_2\\
+a_3 & 0 & -a_1\\
+-a_2 & a_1 & 0
+\end{bmatrix}
+\begin{bmatrix}
+b_1\\
+b_2\\
+b_3
+\end{bmatrix}
+=[a_{\times}]b
+$$
+
+Using **singular value decomposition**, we can solve for $X$
+
+### Epipolar Geometry
+
+What constraints must hold between two projections of the same 3D point?
+
+Given a 2D point in one view, where can we find the corresponding point in the other view?
+
+Given only 2D correspondences, how can we calibrate the two cameras, i.e., estimate their relative position and orientation and the intrinsic parameters?
+
+Key ideas:
+
+- We can answer all these questions without knowledge of the 3D scene geometry
+- Important to think about projections of camera centers and visual rays into the other view
+
+#### Epipolar Geometry Setup
+
+![Epipolar geometry setup](https://notenextra.trance-0.com/CSE559A/Epipolar_geometry_setup.png)
+
+Suppose we have two cameras with centers $O,O'$
+
+The baseline is the line connecting the origins
+
+Epipoles $e,e'$ are where the baseline intersects the image planes, or projections of the other camera in each view
+
+Consider a point $X$, which projects to $x$ and $x'$
+
+The plane formed by $X,O,O'$ is called an epipolar plane
+There is a family of planes passing through $O$ and $O'$
+
+Epipolar lines are projections of the baseline into the image planes
+
+**Epipolar lines** connect the epipoles to the projections of $X$
+Equivalently, they are intersections of the epipolar plane with the image planes – thus, they come in matching pairs.
+
+**Application**: This constraint can be used to find correspondences between points in two camera. by the epipolar line in one image, we can find the corresponding feature in the other image.
+
+![Epipolar line for converging cameras](https://notenextra.trance-0.com/CSE559A/Epipolar_line_for_converging_cameras.png)
+
+Epipoles are finite and may be visible in the image.
+
+![Epipolar line for parallel cameras](https://notenextra.trance-0.com/CSE559A/Epipolar_line_for_parallel_cameras.png)
+
+Epipoles are infinite, epipolar lines parallel.
+
+![Epipolar line for perpendicular cameras](https://notenextra.trance-0.com/CSE559A/Epipolar_line_for_perpendicular_cameras.png)
+
+Epipole is "focus of expansion" and coincides with the principal point of the camera
+
+Epipolar lines go out from principal point
+
+Next class:
+
+### The Essential and Fundamental Matrices
+
+### Dense Stereo Matching
+
+

pages/CSE559A/_meta.js

@@ -26,5 +26,7 @@ export default {
     CSE559A_L21: "Computer Vision (Lecture 21)",
     CSE559A_L22: "Computer Vision (Lecture 22)",
     CSE559A_L23: "Computer Vision (Lecture 23)",
-    CSE559A_L24: "Computer Vision (Lecture 24)"
+    CSE559A_L24: "Computer Vision (Lecture 24)",
+    CSE559A_L25: "Computer Vision (Lecture 25)",
+    CSE559A_L26: "Computer Vision (Lecture 26)",
 }

pages/Math416/Math416_L25.md

@@ -94,7 +94,7 @@ $$
 
 QED
 
-## Application ot valuating definite integrals
+## Application to evaluating definite integrals
 
 Idea:
 

pages/Math416/Math416_L26.md

@@ -1 +1,148 @@
+# Math416 Lecture 26
 
+## Continue on Application to evaluating definite integrals
+
+Note: Contour can never go through a singularity.
+
+Recall the semi annulus contour.
+
+Know that $\int_\gamma f(z)dz=0$.
+
+So $\int_A+\int_B+\int_C+\int_D=0$.
+
+From last lecture, we know that $\int_D=0$ and $\int_A+\int_C=2i\int_0^\infty \frac{\sin x}{x}dx$.
+
+### Integrating over $B$
+
+Do $B$, we have $\gamma(t)=\epsilon e^{it}$ for $t\in[0,\pi]$.
+
+$\int_B=-\int_0^\pi f(\epsilon e^{it})\epsilon i e^{it}dt$.
+
+$f(z)=\frac{e^{iz}}{z}=\frac{1}{z}(1+iz-\frac{z^2}{2!}+\cdots)$.
+
+So $z f(z)=1+O(\epsilon)$ and $f(z)=\frac{1}{z}+O(\frac{\epsilon}{z})$.
+
+$$
+\begin{aligned}
+\int_B&=-\int_0^\pi (\frac{1}{\epsilon}e^{it}+O(1))\epsilon i e^{it}dt\\
+&=-i\int_0^\pi 1dt+O(\epsilon)\\
+&=-i\pi+O(\epsilon)
+\end{aligned}
+$$
+
+### Integrating over $D$
+
+#### Method 1: Using estimate
+
+$z=Re^{it}$ for $t\in[0,\pi]$.
+
+$f(z)=\frac{e^{iz}}{z}=\frac{e^{iRe^{it}}}{Re^{it}}$.
+
+$Re^{it}=R(\cos t+i\sin t)$, $iRe^{it}=-R(\sin t-i\cos t)$.
+
+$e^{iRe^{it}}=e^{-R\sin t}e^{iR\cos t}$.
+
+$\max|f(z)|=\max\frac{|e^{iR\cos t}|}{|R e^{it}|}=\frac{1}{R}$.
+
+This only bounds the function $|\int_D|\leq \pi R\frac{1}{R}=\pi$.
+
+This is not a good estimate.
+
+#### Method 2: Hard core integration
+
+$\gamma(t)=Re^{it}$ for $t\in[0,\pi]$.
+
+$$
+\begin{aligned}
+\int_D&=\int_0^\pi \frac{e^{iRe^{it}}}{R e^{it}}iR e^{it}dt\\
+&=i\int_0^\pi e^{iR\cos t}e^{-R\sin t}dt\\
+\end{aligned}
+$$
+
+Notice that we can use $\frac{2}{\pi}t$ to replace $\sin t$.
+
+$$
+\begin{aligned}
+\left|\int_D\right|&\leq\int_0^\pi e^{-R\sin t}dt\\
+&=2\int_0^{\pi/2} e^{-R\sin t}dt\\
+&\leq 2\int_0^{\pi/2} e^{-2Rt/\pi}dt\\
+&=-\frac{2\pi}{R}(e^{-\frac{R\pi}{2}t})|_0^{\pi/2}\\
+&\leq\frac{\pi}{R}
+\end{aligned}
+$$
+
+As $R\to\infty$, $\left|\int_D\right|\to 0$.
+
+So $\int_D=0$.
+
+So we have $\int_A+\int_C=2i\int_0^\infty \frac{\sin x}{x}dx=i\pi$.
+
+So $\int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}$.
+
+## Application to evaluate $\int_{-\infty}^\infty \frac{\cos x}{1+x^4}dx$
+
+$f(z)=\frac{e^{iz}}{1+z^4}=\frac{\cos z+i\sin z}{1+z^4}$.
+
+Our desired integral can be evaluated by $\int_{-R}^R f(z)dz$
+
+To evaluate the singularity, $z^4=-1$ has four roots by the De Moivre's theorem.
+
+$z^4=-1=e^{i\pi+2k\pi i}$ for $k=0,1,2,3$.
+
+So $z=e^{i\theta}$ for $\theta=\frac{\pi}{4}+\frac{k\pi}{2}$ for $k=0,1,2,3$.
+
+So the singularities are $z=e^{i\pi/4},e^{i3\pi/4},e^{i5\pi/4},e^{i7\pi/4}$.
+
+Only $z=e^{i\pi/4},e^{i3\pi/4}$ are in the upper half plane.
+
+So we can use the semi-circle contour to evaluate the integral. Name the path as $\gamma$.
+
+$\int_\gamma f(z)dz=2\pi i\left[\operatorname{Res}_{z=e^{i\pi/4}}(f)+\operatorname{Res}_{z=e^{i3\pi/4}}(f)\right]$.
+
+The two poles are simple poles.
+
+$\operatorname{Res}_{z_0}(f)=\lim_{z\to z_0}(z-z_0)f(z)$.
+
+So 
+
+$$
+\begin{aligned}
+\operatorname{Res}_{z=e^{i\pi/4}}(f)&=\lim_{z\to e^{i\pi/4}}(z-e^{i\pi/4})\frac{e^{iz}}{1+z^4}\\
+&=\frac{(z-e^{i\pi/4})e^{iz}}{(z-e^{i\pi/4})(z-e^{i3\pi/4})(z-e^{i5\pi/4})(z-e^{i7\pi/4})}\\
+&=\frac{e^{ie^{i\pi/4}}}{(e^{i\pi/4}-e^{i3\pi/4})(e^{i\pi/4}-e^{i5\pi/4})(e^{i\pi/4}-e^{i7\pi/4})}
+\end{aligned}
+$$
+
+A short cut goes as follows:
+
+We know $p(z)=1+z^4$ has four roots $z_1,z_2,z_3,z_4$.
+
+$$
+\lim_{z\to z_0}\frac{(z-z_0)}{p(z)}=\frac{1}{p'(z_0)}
+$$
+
+So
+
+$$
+\operatorname{Res}_{z=e^{i\pi/4}}(f)=\frac{e^{ie^{i\pi/4}}}{4e^{i3\pi/4}}
+$$
+
+Similarly,
+
+$$
+\operatorname{Res}_{z=e^{i3\pi/4}}(f)=\frac{e^{ie^{i3\pi/4}}}{4e^{i\pi/4}}
+$$
+
+So the sum of the residues is
+
+$$
+\begin{aligned}
+\operatorname{Res}_{z=e^{i\pi/4}}(f)+\operatorname{Res}_{z=e^{i3\pi/4}}(f)&=\frac{e^{ie^{i\pi/4}}}{4e^{i3\pi/4}}+\frac{e^{ie^{i3\pi/4}}}{4e^{i\pi/4}}\\
+&=\frac{e^{\frac{i}{\sqrt{2}}} e^{-\frac{1}{\sqrt{2}}}}{4[-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}]}+\frac{e^{-\frac{i}{\sqrt{2}}}-e^{-\frac{1}{\sqrt{2}}}}{4[\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}]}\\
+&=\frac{\pi\sqrt{2}}{2}e^{-\frac{1}{\sqrt{2}}}(\cos\frac{1}{\sqrt{2}}+\sin\frac{1}{\sqrt{2}})
+\end{aligned}
+$$
+
+SKIP
+
+Review on next lecture.