3D Geometry forComputer Graphics

Class 3

2

Last week - eigendecomposition

A

We want to learn how the transformation A works:

3

Last week - eigendecomposition

A

If we look at arbitrary vectors, it doesn’t tell us much.

4

Spectra and diagonalization

A

If A is symmetric, the eigenvectors are orthogonal (and there’s always an eigenbasis).

A = UUT

n

2

1

= Aui = i ui

5

In real life…

Matrices that have eigenbasis are rare – general transformations involve also rotations, not only scalings.

We want to understand how general transformations behave.

Need a generalization of eigendecomposition

SVD: A = UVT

Before we learn SVD, we’ll see Principal Component Analysis – usage of spectral analysis to analyze the shape and dimensionality of scattered data.

6

The plan for today

First we’ll see some applications of PCA Then look at the theory.

7

PCA finds an orthogonal basis that best represents given data set.

The sum of distances2 from the x’ axis is minimized.

PCA – the general idea

x

y

x’

y’

8

PCA – the general idea

PCA finds an orthogonal basis that best represents given data set.

PCA finds a best approximating plane (again, in terms of distances2)

3D point set instandard basis

x y

z

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PCA – the general idea

PCA finds an orthogonal basis that best represents given data set.

PCA finds a best approximating plane (again, in terms of distances2)

3D point set instandard basis

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Application: finding tight bounding box

An axis-aligned bounding box: agrees with the axes

x

y

minX maxX

maxY

minY

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Usage of bounding boxes (bounding volumes)

Serve as very simple “approximation” of the object Fast collision detection, visibility queries Whenever we need to know the dimensions (size) of the object

The models consist of thousands of polygons

To quickly test that they don’t intersect, the bounding boxes are tested

Sometimes a hierarchy of BB’s is used

The tighter the BB – the less “false alarms” we have

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Finding cluster representatives

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Finding cluster representatives

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Application: finding tight bounding box

Oriented bounding box: we find better axes!

x’y’

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Application: finding tight bounding box

This is not the optimal bounding box

x y

z

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Application: finding tight bounding box

Oriented bounding box: we find better axes!

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Managing high-dimensional data

Data base of face scans (3D geometry + texture)

10,000 points in each scan x, y, z, R, G, B 6 numbers for each point Thus, each scan is a 10,000*6 = 60,000-dimensional vector!

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Managing high-dimensional data

How to find interesting axes is this 60000-dimensional space? axes that measures age, gender, etc… There is hope: the faces are likely to be governed by a small set

of parameters (much less than 60,000…)

age axis gender axis

FaceGen demo

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Notations

Denote our data points by x1, x2, …, xn Rd

11 11 2

22 21 2

1 2

, , ,

n

n

dd dn

xx x

xx x

xx x

1 2 dx x x

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The origin is zero-order approximation of our data set (a point)

It will be the center of mass:

It can be shown that:

The origin of the new axes

1

1

n

ini

m x

2

1

argminn

ii

x

m x - x

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Scatter matrix

Denote yi = xi – m, i = 1, 2, …, n

where Y is dn matrix with yk as columns (k = 1, 2, …,

n)

TS YY

1 1 1 1 21 2 1 1 12 2 2 1 21 2 2 2 2

1 21 2

dn

dn

d d d dn n n n

y y y y y y

y y y y y yS

y y y y y y

Y YT

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Variance of projected points

In a way, S measures variance (= scatterness) of the data in different directions.

Let’s look at a line L through the center of mass m, and project our

points xi onto it. The variance of the projected points x’i is:

Original set Small variance Large variance

21

1

var( ) || ||n

ini

L

x m

L L L L

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Variance of projected points

Given a direction v, ||v|| = 1, the projection of xi onto L = m + vt is:

|| || , || || , Ti i i ix m v x m v v y v y

vm

xi

x’iL

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Variance of projected points

So,

2 2 21 1 1

1 1

21 1 1 1 1

var( ) || || ( ) || ||

|| || , ,

n n

n n ni i

T T Tn n n n n

L Y

Y Y Y YY S S

T Ti i

T T T

x -m v y v

v v v v v v v v v

2 21 1 1 1

1 22 2 2 2

2 1 2 1 2 21 2

1 1

1 2

( ) || ||

i n

n nT d di n

i i

d d d di n

y y y y

y y y yv v v v v v Y

y y y y

Tiv y v

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Directions of maximal variance

So, we have: var(L) = <Sv, v> Theorem:

Let f : {v Rd | ||v|| = 1} R,

f (v) = <Sv, v> (and S is a symmetric matrix).

Then, the extrema of f are attained at the eigenvectors of S.

So, eigenvectors of S are directions of maximal/minimal variance!

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Summary so far

We take the centered data points y1, y2, …, yn Rd

Construct the scatter matrix S measures the variance of the data points Eigenvectors of S are directions of maximal variance.

TS YY

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Scatter matrix - eigendecomposition

S is symmetric

S has eigendecomposition: S = VVT

S = v2v1 vd

1

2

d

v2

v1

vd

The eigenvectors formorthogonal basis

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Principal components

Eigenvectors that correspond to big eigenvalues are the directions in which the data has strong components (= large variance).

If the eigenvalues are more or less the same – there is no preferable direction.

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Principal components

There’s no preferable direction

S looks like this:

Any vector is an eigenvector

There is a clear preferable direction

S looks like this:

is close to zero, much smaller than .

TVV

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How to use what we got

For finding oriented bounding box – we simply compute the bounding box with respect to the axes defined by the eigenvectors. The origin is at the mean point m.

v2v1

v3

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For approximation

x

yv1

v2

x

y

This line segment approximates the original data set

The projected data set approximates the original data set

x

y

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For approximation

In general dimension d, the eigenvalues are sorted in descending order:

1 2 … d The eigenvectors are sorted accordingly. To get an approximation of dimension d’ < d, we

take the d’ first eigenvectors and look at the subspace they span (d’ = 1 is a line, d’ = 2 is a plane…)

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For approximation

To get an approximating set, we project the original data points onto the chosen subspace:

xi = m + 1v1 + 2v2 +…+ d’vd’ +…+dvd

Projection:

xi’ = m + 1v1 + 2v2 +…+ d’vd’ +0vd’+1+…+ 0 vd

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Optimality of approximation

The approximation is optimal in least-squares sense. It gives the minimal of:

The projected points have maximal variance.

2

1

n

k kk

x x

Original set projection on arbitrary line projection on v1 axis

See you next time

36

Technical remarks:

i 0, i = 1,…,d (such matrices are called positive semi-definite). So we can indeed sort by the magnitude of i

Theorem: i 0 <Sv, v> 0 v

Proof:

Therefore, i 0 <Sv, v> 0 v

,

( ) ( ) ,

T T T T

T T T T

S V V S S V V

V V

v v v v v v

v v v v v v

2 2 21 2, ... dS 1 2 dv v u u u