Computer Graphics Recitation 11. 2 The plan today Least squares approach General / Polynomial fitting Linear systems of equations Local polynomial

Computer Graphics

Recitation 11

2

The plan today

Least squares approach General / Polynomial fitting Linear systems of equations Local polynomial surface fitting

3

y = f (x)

Motivation

Given data points, fit a function that is “close” to the points

y

x

Pi = (xi, yi)

4

Motivation

Local surface fitting to 3D points

5

Line fitting

Orthogonal offsets minimization – we learned

x

y

Pi = (xi, yi)

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 (22)

In 2D this requires solving a square characteristic equation In higher dimensions – higher order equations…

n

iiP

nm

1

1

dc

ba

yyy

xxx

yyy

xxxS

n

n

n

n

T

21

21

21

21

7

Line fitting

y-offsets minimization

x

y

Pi = (xi, yi)

8

Line fitting

Find a line y = ax + b that minimizes

Err(a, b) = [yi – (axi + 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…

n

i 1

n

iii baxybaAbsErr

1

)(),(

9

Line fitting – LS minimization

To find optimal a, b we differentiate Err(a, b):

Err(a, b) = (–2xi)[yi – (axi + b)] = 0

Err(a, b) = (–2)[yi – (axi + b)] = 0

a

n

i 1

n

i 1b

10


We get two linear equations for a, b:

(–2xi)[yi – (axi + b)] = 0

(–2)[yi – (axi + b)] = 0

n

i 1

n

i 1

11



[xiyi – axi2 – bxi] = 0

[yi – axi – b] = 0

n

i 1

)1(

n

i 1

)2(

12



( xi2) a + ( xi) b = xiyi

( xi) a + ( 1) b = yi

n

i 1

n

i 1

n

i 1

n

i 1

n

i 1

n

i 1

13


Solve for a, b using e.g. Gauss elimination

Question: why the solution is minimum for the error function?

Err(a, b) = [yi – (axi + b)]2

n

i 1

14

Fitting polynomials

y

x

15

Fitting polynomials

Decide on the degree of the polynomial, k Want to fit f(x) = akx

k + ak-1xk-1 + … a1x+ a0

Minimize:

Err(a0, a1, …, ak) = [yi – (akxk+ak-1x

k-1+ …+a1x+a0)]2

Err(a0,…,ak) = (– 2xm)[yi – (akxk+ak-1x

k-1+…+ a0)] = 0

n

i 1

n

i 1ma

16

Fitting polynomials

We get a linear system of k+1 in k+1 variables

n

ii

ki

n

iii

n

ii

k

n

i

ki

n

i

ki

n

i

ki

n

i

ki

n

ii

n

ii

n

i

ki

n

ii

n

i

yx

yx

y

a

a

a

xxx

xxx

xx

1

1

11

0

1

2

1

1

1

1

1

1

2

1

111

11

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


Want to fit function fabc…(x) to data points (xi, yi)

Define Err(a,b,c,…) = [yi – fabc…(xi)]2

Solve the linear system

0)]())[(2(),,,(

0)]())[(2(),,,(

...1

...1

iiiabc

n

i

iiiabc

n

i

xfyxfb

cbaErrb

xfyxfa

cbaErra

n

i 1

19


It can even be some crazy function like

Or in general:

1732

21

2

2

sin)( xxxfx

e

)(...)()()( 2211,...,, 11xfxfxfxf kkk

20

Solving linear systems in LS sense

Let’s look at the problem a little differently: We have data points (xi, yi)

We want the function f(x) to go through the points:

i =1, …, n: yi = f(xi)

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


We have an over-determined linear system nk:

f(x1) = 1 f1(x1) + 2 f2(x1) + … + k fk(x1) = y1

f(x2) = 1 f1(x2) + 2 f2(x2) + … + k fk(x2) = y2

…

…

…

f(xn) = 1 f1(xn) + 2 f2(xn) + … + k fk(xn) = yn

22


In matrix form:

n

k

nknn

k

k

y

y

y

xfxfxf

xfxfxf

xfxfxf

2

1

2

1

21

22221

11211

)(...)()(

...

)(...)()(

)(...)()(

23


In matrix form:

Av = y

Tn

Tk

jiij

yyyy

v

knxfA

),...,,(

),...,,(

kn matrix, r rectangula a is )( where

21

21

.

24


More constrains than variables – no exact solutions generally exist

We want to find something that is an “approximate solution”:

2minarg~ yAvvv

25

Finding the LS solution

v Rk

Av Rn As we vary v, Av varies over the linear subspace

of Rn spanned by the columns of A:

Av = A2A1 Ak

1

2

.

. k

= 1A 1 A 2 A k + 2 +…+ k

26


We want to find the closest Av to y:2

min yAvv

Subspace spannedSubspace spannedby columns of by columns of AA

yy

Rn

Avclosest to y

27


The vector Av closest to y satisfies:

(Av – y) {subspace of A’s columns}

column Ai, <Ai, Av – y> = 0

i, AiT(Av – y) = 0

AT(Av – y) = 0

(ATA)v = ATy

These arecalled the

normal equations

28


We got a square symmetric system (ATA)v = ATy (nn)

If A has full rank (the columns of A are linearly independent) then (ATA) is invertible.

yAAAv

yAv

TT

v

1

2

)(

min

29

Weighted least squares

Sometimes the problem also has weights to the constraints:

system square a is this)(

matrix diagonaln n is W where),()(min

on dependt doesn' and 0,)]([min

ii

i2

,...,,1

,...,, 2121

WyAvWAA

wyAvWyAv

wxfyw

TT

iT

v

iii

n

ii k

k

30


Normals? Lighting? Upsampling?

31


Locally approximate

a polynomial surface

from points

32

Fitting local polynomial

X Y

ZReference plane

Fit a local polynomial around a point P

P

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


Fit polynomial z = p(x,y) = ax2 + bxy + cy2 + dx + ey + f

z

x y

35



z

x

y

36



z

x y

37


Again, solve the system in LS sense:

ax12 + bx1y1 + cy1

2 + dx1 + ey1 + f = z1

ax22 + bx2y2 + cy2

2 + dx2 + ey2 + f = z1

. . .

axn2 + bxnyn + cyn

2 + dxn + eyn + f = zn

Minimize ||zi – p(xi, yi)||2

38


Also possible (and better) to add weights:

wi ||zi – p(xi, yi)||2, wi > 0

The weights get smaller as the distance from the origin point grows.

39

Geometry compression using relative coordinates

Given a mesh: Connectivity Geometry – (x, y, z) of each vertex

40


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


Represent each vertex with relative coordinates:

)(

1

iNjj

iii v

dvv

average of the neighbours

the relative coordinate vector

42


We call them –coordinates:

),,(),,( )()()( zi

yi

xiiii zyx

average of the neighbours

the relative coordinate vector

43


When the mesh is smooth, the –coordinates are small.

–coordinates can be better compressed.

44


Matrix form to compute the –coordinates:

)(

)(1

)(2

)(1

1

2

1

111

111

11

11

1

10

1

1

1

1

10

001

111

22

11

xn

xn

x

x

n

n

ddd

ddd

dd

dd

x

x

x

x

nnn

nnn

45


Matrix form to compute the –coordinates:

The same L matrix for y and z… L is called the Laplacian of the mesh

)( xLx

46


How do we restore the (x, y, z) from the –coordinates?

Need to solve the linear system:

Lx = (x)

47


Lx = (x)

But: L is singular (x) contains quantization error

48


Lx = (x)

Solution: choose some anchor vertices whose (x, y, z) position is known (in addition to )

49


We add the anchor vertices to our linear system:

0100

0010

nx

x

x

2

1

known

known

xn

x

x

x

x

2

1

)(

)(2

)(1

constrainedanchor vertices

L

)(~ xxL

50


Now we have more equations than unknowns Solve in least squares sense!

2)(~

min x

xxL

See you next time

Documents

Computer Graphics Recitation 11. 2 The plan today Least squares approach General / Polynomial fitting Linear systems of equations Local polynomial