Least Square

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# Least Square - PowerPoint PPT Presentation

Least Square. y = f ( x ). Motivation. Given data points, fit a function that is “ close ” to the points Local surface fitting to 3D points. y. P i = ( x i , y i ). x. Line Fitting. y - offsets minimization. y. P i = ( x i , y i ). x. Line Fitting.

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### Least Square

y = f (x)

Motivation
• Given data points, fit a function that is “close” to the points
• Local surface fitting to 3D points

y

Pi = (xi, yi)

x

Line Fitting
• y-offsets minimization

y

Pi = (xi, yi)

x

Line Fitting
• Find a liney = ax + bthat minimizes
• E(a,b)is quadratic in the unknown parametersa, b
• Another option would be, for example:
• But – it is not differentiable, harder to minimize…
Line Fitting – LS minimization
• To find optimal a, b we differentiate E(a, b):

E(a, b) = (–2xi)[yi – (axi + b)] = 0

E(a, b) = (–2)[yi – (axi + b)] = 0

Line Fitting – LS minimization
• We get two linear equations for a, b:

(–2xi)[yi – (axi + b)] = 0

(–2)[yi – (axi + b)] = 0

[xiyi – axi2 – bxi] = 0

[yi – axi – b] = 0

Line Fitting – LS minimization
• We get two linear equations for a, b:

( xi2) a + ( xi) b = xiyi

( xi)a + ( 1)b = yi

Line Fitting – LS minimization
• Solve for a, b using e.g. Gauss elimination
• Question: why the solution is the minimum for the error function?

E(a, b) = [yi – (axi + b)]2

Fitting Polynomials
• Decide on the degree of the polynomial, k
• Want to fit f (x) = akxk + ak-1xk-1 + … + a1x+ a0
• Minimize:

E(a0,a1, …,ak) = [yi – (akxik+ak-1xik-1+ …+a1xi+a0)]2

E(a0,…,ak) = (– 2xm)[yi – (akxik+ak-1xik-1+…+ a0)] = 0

Fitting Polynomials
• We get a linear system of k+1 in k+1 variables
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, …
General Parametric Fitting
• Want to fit function fabc…(x) to data points (xi, yi)
• Define E(a,b,c,…) = [yi – fabc…(xi)]2
• Solve the linear system
General Parametric Fitting
• It can even be some crazy function like
• Or in general:
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…
Solving Linear Systems in LS Sense
• We have an over-determined linear system nk:

f(x1) = 1f1(x1) + 2f2(x1) + … + kfk(x1) = y1

f(x2) = 1f1(x2) + 2f2(x2) + … + kfk(x2) = y2

f(xn) = 1f1(xn) + 2f2(xn) + … + kfk(xn) = yn

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”:
Finding the LS Solution
• v Rk
• Av  Rn
• As we vary v, Av varies over the linear subspace of Rnspanned by the columns of A:

Av =

= 1

+ 2

+…+ k

1

2

.

.

k

A1

A2

Ak

A1

A2

Ak

Av

closest to y

Finding the LS Solution
• We want to find the closest Av to y:

y

Subspace spanned

by columns of A

Rn

Finding the LS Solution
• 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 are

called the

normal equations

Finding the LS Solution
• We got a square symmetric system
• (ATA)v = ATy
• If A has full rank (the columns of A are linearly independent) then (ATA) is invertible.

(kk)

### Laplacian Editing

Handle Based Editing

Handles, they are free to move by user

Model deforms when handles are places at target positions

General Idea
• A visually pleasing deformed mesh should maintain
• local parameterization (triangle shape)
• local geometry information (local shape)
• We decompose the global geometry into
• coefficients of the Laplace operator
• Laplacian coordinates (LCs)
Laplacian Coordinates
• LC represent local geometry
• Coefficients of Laplacian operator represent local parameterization

or

xi

xj

...

Laplacian Editing
• We solve new vertex positions with handle constraints
• In general there is no solution
• We use LS method to find optimal solution

Solve x:

### Shape Matching

Shape Matching
• We have two objects in correspondence
• Want to find the rigid transformation that aligns them
Shape Matching
• When the objects are aligned, the lengths of the connecting lines are small.
Shape Matching – Formalization
• Align two point sets
• Find a translation vector t and rotation matrix R so that:
Shape Matching – Solution
• Turns out we can solve the translation and rotation separately.
• Theorem: if (R, t) is the optimal transformation, then the points {pi} and {Rqi + t} have the same centers of mass.
Finding the Rotation R
• To find the optimal R, we bring the centroids of both point sets to the origin:
• We want to find R that minimizes
Finding the Rotation R
• First we compute optimum linear transformation A that minimizes
• We solve it in LS sense and build the systems

=

Q = q0 q1 … qn

P = p0 p1 … pn

A

QT

PT

AT

=

3x3

Finding the Rotation R
• We can solve columns of AT = {a1, a2, a3} superlatively
• By using LS method

QQT AT = QPT  A = (QQT)-1(QPT)

QT

=

PT

a1

a2

a3

3x1

3x3

3x3

Finding the Rotation R
• Second we extract the pure rotation part R from A
• We use singular value decomposition (SVD)
• A = UWV
• A = RS
• R = UV, S = VTWV
• Since U and V are orthogonal, therefore R does
• S is a positive-semidefinite symetric matrix
• Finally we compute translation t = p - Rq

U, V: orthogonal matricesW: diagonal matrix

Reference
• SVD
• Numerical Recipes in C (section2.6)
• http://www.nrbook.com/a/bookcpdf.php
• A Tutorial on Principal Component Analysis
• http://www.snl.salk.edu/~shlens/pub/notes/pca.pdf
• Laplacian Editing
• Demo Program
• http://www.cse.ust.hk/~oscarau/comp290/290project_demo_code.zip
• Differential Coordinates for Interactive Mesh Editing [PDF]