1. Computer Graphics Recitation 11

2. 2 The plan today Least squares approach  General / Polynomial fitting  Linear systems of equations  Local polynomial surface fitting

3. 3 y = f (x) Motivation Given data points, fit a function that is “close” to the points y x P i = (x i, y i )

4. 4 Motivation Local surface fitting to 3D points

5. 5 Line fitting Orthogonal offsets minimization – we learned x y P i = (x i, y i )

6. 6 Line fitting The origin of the line is the center of mass The direction v is the eigenvector corresponding to the largest eigenvalue of the scatter matrix S (2  2) In 2D this requires solving a square characteristic equation In higher dimensions – higher order equations…

7. 7 Line fitting y-offsets minimization x y P i = (x i, y i )

8. 8 Line fitting Find a line y = ax + b that minimizes Err(a, b) = [y i – (ax i + b)] 2 Err is quadratic in the unknown parameters a, b Another option would be, for example: But – it is not differentiable, harder to minimize…

9. 9 Line fitting – LS minimization To find optimal a, b we differentiate Err(a, b) : Err(a, b) = (–2x i )[y i – (ax i + b)] = 0 Err(a, b) = (–2)[y i – (ax i + b)] = 0

10. 10 Line fitting – LS minimization We get two linear equations for a, b : (–2x i )[y i – (ax i + b)] = 0 (–2)[y i – (ax i + b)] = 0

11. 11 Line fitting – LS minimization We get two linear equations for a, b : [x i y i – ax i 2 – bx i ] = 0 [y i – ax i – b] = 0

12. 12 Line fitting – LS minimization We get two linear equations for a, b : ( x i 2 ) a + ( x i ) b = x i y i ( x i ) a + ( 1) b = y i

13. 13 Line fitting – LS minimization Solve for a, b using e.g. Gauss elimination Question: why the solution is minimum for the error function? Err(a, b) = [y i – (ax i + b)] 2

14. 14 Fitting polynomials y x

15. 15 Fitting polynomials Decide on the degree of the polynomial, k Want to fit f(x) = a k x k + a k-1 x k-1 + … a 1 x+ a 0 Minimize: Err(a 0, a 1, …, a k ) = [y i – (a k x k +a k-1 x k-1 + …+a 1 x+a 0 )] 2 Err(a 0,…,a k ) = (– 2x m )[y i – (a k x k +a k-1 x k-1 +…+ a 0 )] = 0

16. 16 Fitting polynomials We get a linear system of k+1 in k+1 variables

17. 17 General parametric fitting We can use this approach to fit any function f(x)  Specified by parameters a, b, c, …  The expression f(x) linearly depends on the parameters a, b, c, …

18. 18 General parametric fitting Want to fit function f abc… (x) to data points (x i, y i )  Define Err(a,b,c,…) = [y i – f abc… (x i )] 2  Solve the linear system

19. 19 General parametric fitting It can even be some crazy function like Or in general:

20. 20 Solving linear systems in LS sense Let’s look at the problem a little differently:  We have data points (x i, y i )  We want the function f(x) to go through the points:  i =1, …, n: y i = f(x i )  Strict interpolation is in general not possible In polynomials: n+1 points define a unique interpolation polynomial of degree n. So, if we have 1000 points and want a cubic polynomial, we probably won’t find it…

21. 21 Solving linear systems in LS sense We have an over-determined linear system n  k : f(x 1 ) = 1 f 1 (x 1 ) + 2 f 2 (x 1 ) + … + k f k (x 1 ) = y 1 f(x 2 ) = 1 f 1 (x 2 ) + 2 f 2 (x 2 ) + … + k f k (x 2 ) = y 2 … f(x n ) = 1 f 1 (x n ) + 2 f 2 (x n ) + … + k f k (x n ) = y n

22. 22 Solving linear systems in LS sense In matrix form:

23. 23 Solving linear systems in LS sense In matrix form: Av = y

24. 24 Solving linear systems in LS sense More constrains than variables – no exact solutions generally exist We want to find something that is an “approximate solution”:

25. 25 Finding the LS solution v  R k Av  R n As we vary v, Av varies over the linear subspace of R n spanned by the columns of A: Av = A2A2 A1A1 AkAk 1 2. k = 1 A1A1 A2A2 AkAk + 2 +…+ k

26. 26 Finding the LS solution We want to find the closest Av to y : Subspace spanned by columns of A y RnRn Av closest to y

27. 27 Finding the LS solution The vector Av closest to y satisfies: (Av – y)  {subspace of A ’s columns}  column A i, = 0  i, A i T (Av – y) = 0 A T (Av – y) = 0 (A T A)v = A T y These are called the normal equations

28. 28 Finding the LS solution We got a square symmetric system (A T A)v = A T y (n  n) If A has full rank (the columns of A are linearly independent) then (A T A) is invertible.

29. 29 Weighted least squares Sometimes the problem also has weights to the constraints:

30. 30 Local surface fitting to 3D points Normals? Lighting? Upsampling?

31. 31 Local surface fitting to 3D points Locally approximate a polynomial surface from points

32. 32 Fitting local polynomial X Y Z Reference plane Fit a local polynomial around a point P P

33. 33 Fitting local polynomial surface Compute a reference plane that fits the points close to P Use the local basis defined by the normal to the plane! z x y

34. 34 Fitting local polynomial surface Fit polynomial z = p(x,y) = ax 2 + bxy + cy 2 + dx + ey + f z x y

35. 35 Fitting local polynomial surface Fit polynomial z = p(x,y) = ax 2 + bxy + cy 2 + dx + ey + f z x y

36. 36 Fitting local polynomial surface Fit polynomial z = p(x,y) = ax 2 + bxy + cy 2 + dx + ey + f z x y

37. 37 Fitting local polynomial surface Again, solve the system in LS sense: ax 1 2 + bx 1 y 1 + cy 1 2 + dx 1 + ey 1 + f = z 1 ax 2 2 + bx 2 y 2 + cy 2 2 + dx 2 + ey 2 + f = z 1... ax n 2 + bx n y n + cy n 2 + dx n + ey n + f = z n Minimize  ||z i – p(x i, y i )|| 2

38. 38 Fitting local polynomial surface Also possible (and better) to add weights:  w i ||z i – p(x i, y i )|| 2, w i > 0 The weights get smaller as the distance from the origin point grows.

39. 39 Geometry compression using relative coordinates Given a mesh:  Connectivity  Geometry – (x, y, z) of each vertex

40. 40 Geometry compression using relative coordinates The size of the geometry is large (compared to connectivity) (x, y, z) coordinates are hard to compress  Floating-point numbers – have to quantize  No correlation

41. 41 Geometry compression using relative coordinates Represent each vertex with relative coordinates: average of the neighbours the relative coordinate vector

42. 42 Geometry compression using relative coordinates We call them  –coordinates: average of the neighbours the relative coordinate vector

43. 43 Geometry compression using relative coordinates When the mesh is smooth, the  –coordinates are small.  –coordinates can be better compressed.

44. 44 Geometry compression using relative coordinates Matrix form to compute the  –coordinates:

45. 45 Geometry compression using relative coordinates Matrix form to compute the  –coordinates: The same L matrix for y and z … L is called the Laplacian of the mesh

46. 46 Geometry compression using relative coordinates How do we restore the (x, y, z) from the  –coordinates? Need to solve the linear system: Lx =  (x)

47. 47 Geometry compression using relative coordinates Lx =  (x) But:  L is singular   (x) contains quantization error

48. 48 Geometry compression using relative coordinates Lx =  (x) Solution: choose some anchor vertices whose (x, y, z) position is known (in addition to  )

49. 49 Geometry compression using relative coordinates We add the anchor vertices to our linear system: constrained anchor vertices L

50. 50 Geometry compression using relative coordinates Now we have more equations than unknowns Solve in least squares sense!

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