CS231A. Review for Problem Set 1. Saumitro Dasgupta

Transcription

1 CS231A Review for Problem Set 1 Saumitro Dasgupta

2 On today's menu Camera Model Rotation Matrices Homogeneous Coordinates Vanishing Points Matrix Calculus Constrained Optimization Camera Calibration Demo

3 Camera Model Review x = K[R t] X x: 2D point in image space (homogeneous) X: 3D point in world space (homogeneous) K: Camera calibration matrix (intrinsics) [R t] : Camera rotation and translation (extrinsics)

4 Intrinsics Visualization Launch Interactive Demo

5 The Extrinsics Quiz P b = [R t] P a P a P b What are the coordinate systems of the points and? World to Camera: P camera = [R t] P world Let R = I, t = [ 0 0 0] What are the world coordinates of the camera center? [ 0 0 0] R t = [ ] For and 0 0 0? Still [ ] [R t] For any arbitrary? R t

6 Rotation Matrices Orthogonality Rotation matrices are orthogonal. Orthogonal matrices satisfy Equivalently: R R = R R R 1 = I = R The rows and columns of a rotation matrix have length 1.

7 Rotation Matrices Determinant For any orthogonal matrix, det (M) 2 M, we have: = det (M) det (M) = det ( M ) det (M) = det ( M M ) det (M) = det (I) = 1 = ±1 However, rotations cannot include reflections. Therefore: det (R) = 1

8 Rotation Matrices Interpreting the rows and columns The columns of a rotation matrix are the original coordinate system's basis vectors represented in the rotated coordinate system. R [ 1 0 0] = [ R 1 R 2 R ] [ ] 3 = R 1 What about the rows? Hint: Rows of R are columns of R. The rotated coordinate system's basis vectors represented in the original coordinate system.

9 To homogeneous: From homogeneous: Homogeneous coordinates R n [ ] Conversion x y becomes [ x y 1] [ ] x y w becomes [ x ] Mapping from (Euclidean) to (Projective) In projective space, all scalar multiples of a point are equivalent. [ ] [ ] [ ] P n and both map back to 2 3 in R 2. Cartesian:Euclidean as Homogeneous:Projective w y w

10 Homogeneous coordinates What's the point? We would like to use the powerful tools of linear algebra Is perspective projection a linear transformation? No. Involves division by depth. Is translation a linear transformation? No. Doesn't preserve the origin. How do we deal with points at infinity? Homogeneous coordinates provide a solution for all of the above.

11 Vanishing Points Perspective Study, Leonardo da Vinci (1481)

12 Vanishing Points Multiple Vanishing Points Original Image Credits: Miquel González, The Louvre, Paris, France

13 Vanishing Points Outside the clipped image Original Image Credits: Mecki Mac on Flickr, Goldau, Switzerland

14 Vanishing Points Under perspective projection, lines that are parallel in the world are no longer parallel in the image. Exception: Lines parallel to the image plane remain parallel. In image space, parallel lines meet at the vanishing point. In projective space, parallel lines meet at points at infinity (also known as an ideal points). The homogeneous coordinate is zero. Points at infinity are imaged as vanishing points.

15 Matrix Calculus Gradient and Hessian The gradient and Hessian are defined for real valued functions Gradient x f (x) = f x 1 f x 2 f x n f : R n 2 xf (x) = R. 2 f x f x 2 x 1 2 f x n x 1 2 f x 1 x 2 2 f x f x n x 2 Hessian 2 f x 1 x n 2 f x 2 x n 2 f x 2 n

16 Matrix Calculus Some properties of the gradient The gradient points in the direction of steepest ascent. The gradient is orthogonal to the level set. Contours and gradients of f (x, y) = x 2 + y 2.

17 Matrix Calculus Useful identities [ a] = [ x] x x x a = a x x A x [ Ax] = ( A + ) x [ x] = 2x x The Matrix Cookbook is your new best friend.

18 Unconstrained Optimization Review Let f : R n R be a continuously differentiable function. What are the necessary and sufficient conditions for minimum? f ( x ) to be a local The gradient at x is zero: f ( x ) = 0 The Hessian is positive semi-definite: f ( x ) 0 2

19 Constrained Optimization Subject to equality constraints Given f : R n and g : R n R m, R Minimize: Subject to: f (x) g(x) = 0 We can solve optimization problems of this form using the method of Lagrange multipliers.

20 Constrained Optimization Lagrange Multilpliers Define the Lagrangian as: where Each λ i λ R m λ Λ(x,λ) = f (x) + λ g(x) is known as a Lagrange Multiplier. The necessary conditions for optimality are: x Λ(x,λ) = 0 Λ(x,λ) = 0 λ

21 Constrained Optimization Feasible Region Minimize: Subject to: f (x) = x [ 1 2] g(x) = x x 9 = 0

22 Constrained Optimization Lagrange Multipliers: Intuition Must remain on the constraint surface. Recall that gradients are orthogonal to level sets Only possible direction of movement is orthogonal to x g(x). f (x) = λ x x f (x) What if g(x)? Cannot move to change while remaining on the constraint surface we're at a stationary point! x x Therefore, f (x) + λ g(x) = 0. x [f (x) + λg(x) ] = 0.

23 Constrained Optimization The Lagrange Optimality Conditions x Λ(x, λ) = 0 f (x) = λ g(x) A consequence of x x Gives us n equations (recall that ) λ Λ(x, λ) = 0 The equality constraints in disguise. x R n λ R m Gives us m equations (recall that ) We have (n + m) unknowns and (n + m) equations. Solve simultaneously.

24 Camera Calibration Live Demo Reference: MATLAB Documentation.