Simple Mathematics Simple Mathematics for graphicsfor graphics

Simple Mathematics Simple Mathematics for graphicsfor graphics

Volumes from Silhouettes

• Start with collection of “calibrated” images

Volumes from Silhouettes

Cup on turntable example

3D Curves from Edges• “Feature-based” stereo matching

3D Curves from Edges• Extract extremal and internal

edges

3D Curves from Edges• Match curves along epipolar lines

viewing rayviewing rayepipolar planeepipolar plane

epipolar lineepipolar line

What is Computer Graphics?

• Using a computer to generate an image from a representation.

Model Imagecomputer

Representations• How do we represent an object?

– Points • p = [x y z]

– Mathematical Functions• X2 + Y2=R2

– Polygons (most commonly used)• Points• Connectivity

Representing Points and Vectors

• A 3D point p = [ x y z ]– Represents a location with respect to

some coordinate system

• A 3D vector v = [ x y z]– Represents a displacement from a

position

Linear Algebra• Why study Linear Algebra?

– Deals with the representation and operations commonly used in Computer Graphics

Vector Spaces• Consists of a set of elements,

called vectors and two operations that are defined on them, addition and multiplication

Vector Addition• Given V = [X Y Z] and W = [A B C]

– V+W = [X+A Y+B Z+C]

• Properties of Vector addition– Commutative: V+W=W+V– Associative (U+V)+W = U+(V+W)– Additive Identity: V+0 = V– Additive Inverse: V+W = 0, W=-V

Parallelogram Rule• To visualize what a vector addition

is doing, here is a 2D example:

V

W

V+W

Vector Multiplication• Given V = [X Y Z] and a Scalar s and t

– sV = [sX sY sZ]

• Properties of Vector multiplication• Associative: (st)V = s(tV)• Multiplicative Identity: 1V = V• Scalar Distribution: (s+t)V = sV+tV• Vector Distribution: s(V+W) = sV+sW

Dot Product and Distances

• Given u = [x y z] and v = [a b c]– v•u = ax+by+cz

• The Euclidean distance of u from the origin is sqrt( x2+y2+z2) and is denoted by ||u||– Notice that ||u|| = sqrt(u•u)

• The Euclidean distance between u and v is sqrt( (x-a) 2+(y-b) 2+(z-c) 2) ) and is denoted by || u-v ||

Properties of the Dot Product

• Given a vector u, v, w and scalar s– The result of a dot product is a

SCALAR value– Commutative: v•w = w•v– Non-degenerate: v•v=0 only when

v=0– Bilinear: v•(u+sw)=v•u+s(v•w)

Angles and Projection• Alternative view of the dot product• v•w=||v|| ||w|| cos() where is the

angle between v and w• If v is a unit vector (||v|| = 1) then if

we perpendicularly project w onto v can call this newly projected vector u then ||u|| = v•w

w

vu

Matrices• A compact way of representing operations on

points and vectors• 3x3 Matrix A looks like

• a(i,j) refers to the element of matrix A

333231

232221

131211

aaa

aaa

aaa

Matrix Multiplication• If A is an nk matrix and B is a kp then

AB is a np matrix with entries c(i,j)where– c(i,j)=a(i,s)b(s,j)

• Alternatively, if we took the rows of A and columns of B as individual vectors then

• c(i,j)=Ai•Bj where the subscript refers to the row and column, respectively

Matrix Multiplication Properties

• Associative: (AB)C = A(BC)• Distributive: A(B+C) = AB+AC• Multiplicative Identity: I= diag(1)

(square matrix• NOT commutative: ABBA

Determinant of a Matrix

• Defined on a square matrix (nxn)

• Where A1i determinant of (n-1)x(n-1) submatrix A gotten by deleting the first row and the ith column

n

i ii AAA

1 11)1(det

Recursive Definition!!• The basis case• det of a 2x2 matrix is defined to

be ad-bc where

db

ca

Uses of the Determinant?

• Linear Independence of columns in a matrix

• Cross Product– Given 2 vectors v=[v1 v2 v3], w=[w1

w2 w3], the cross product is defined to be the determinant of

1221

3113

2332

321

321

wvwv

wvwv

wvwv

www

vvv

kji

Cross Product Properties

• The Cross Product of v and w is denoted by vw

• Is a VECTOR, perpendicular to the plane defined by v and w

• ||vw||=||v|| ||w|| |sin| is the angle between v and w

• vw=-(wv)v

wvw

Matrix Transpose and Inverse

• The Transpose of a matrix A, denoted by AT is defined as aij=aji (exchanging the rows and columns)

• If A and B are nxn matrices and AB=BA=I then B is the inverse of A, denoted by A-1

• (AB)-1=B-1A-1 same applies for transposeT

MM T 11

Methods for finding the Inverse

• Explicit Methods– Gaussian-Jordan Elimination

• Create the Augmented matrix [A|I] and reduce the left side to the identity using elementary row operations and the right hand side will be the inverse. ie. [I|A-1]

– Cramer’s Rule• Solve for A’ where a’

ij=det(submatrix(Aij))

• A-1=(1/det(A))(A’)T

Implicit Methods• Instead of explicitly calculating A-1, there

are techniques that solve equations of the form Ax=b (system of linear equations).

• Clearly x=A-1b but we do not need to explicitly calculate A-1 to calculate x.– LU Decomposition– QR Factorization– Singular Value Decomposition (SVD)– Conjugate Gradient if sparse…

Transformations• Why use transformations?

Create object in convenient coordinates

Reuse basic shape multiple times Hierarchical modeling System independent Virtual cameras

Translation

z

y

x

z

y

x

t

t

t

= +

z

y

x

tttT zyx ),,(

z

y

x

z

y

x

t

t

t

= +

'

'

'

z

y

x

Properties of Translation

v=v)0,0,0(T

=v),,(),,( zyxzyx tttTsssT

=

=v),,(1zyx tttT

v),,(),,( zyxzyx tttTsssT v),,(),,( zyxzyx sssTtttT

v),,( zzyyxx tststsT

v),,( zyx tttT

Rotations (2D)

sin

cos

ry

rx

cos)sin(sin)cos('

sin)sin(cos)cos('

rry

rrx

)sin('

)cos('

ry

rx

cossinsincos)sin(

sinsincoscos)cos(

cossin'

sincos'

yxy

yxx

yx,

',' yx

x

y

Rotations 2D• So in matrix notation

y

x

y

x

cossin

sincos'

'

Rotations (3D)

100

0cossin

0sincos

)(

cos0sin

010

sin0cos

)(

cossin0

sincos0

001

)(

z

y

x

R

R

R

Properties of RotationsIRa )0(

)()()()( aaaa RRRR

)()()( aaa RRR

)()()(1 Taaa RRR

)()()()( abba RRRR order matters!

Combining Translation & Rotation

)1,1(T

)45( R

)45( R

)1,1(T

Combining Translation & Rotation

Tvv'

RTR

TR

R

vv

vv

vv

''

)(''

'''

vv R'

TR

T

vv

vv

''

'''

Scaling

zs

ys

xs

z

y

x

z

y

x

'

'

'

z

y

x

zyx

s

s

s

sssS

00

00

00

),,(

Uniform scaling iff zyx sss

Homogeneous Coordinates

z

y

x

w

Z

Y

X

can be represented as

wherew

Zz

w

Yy

w

Xx ,,

Translation Revisited

11000

100

010

001

),,(z

y

x

t

t

t

z

y

x

tttTz

y

x

zyx

Rotation & Scaling Revisited

11000

000

000

000

),,(z

y

x

s

s

s

z

y

x

sssSz

y

x

zyx

11000

0cossin0

0sincos0

0001

)(z

y

x

z

y

x

Rx

Combining Transformations

vv

vvvv

vvv

vv

M

TRSTRT

RSR

S

'''

''''''

'''

'

where TRSM

Transforming Tangents

t

qp

qp

qpt

qpt

M

M

MM

)(

'''

Transforming Normals

nnn

nn

nn

tntn

tn

tn

tn

TT

T

TT

TT

T

T

T

MM

M

M

M

M

11'

'

'

'

0'

0''

0

Rotations about an arbitrary axis

Rotate by around a unit axis r

r

• We can view the rotation around an arbitrary axis as a set of simpler steps

• We know how to rotate and translate around the world coordinate system

• Can we use this knowledge to perform the rotation?

Rotation about an arbitrary axis

• Translate the space so that the origin of the unit vector is on the world origin

• Rotate such that the extremity of the vector now lies in the xz plane (x-axis rotation)

• Rotate such that the point lies in the z-axis (y-axis rotation)

• Perform the rotation around the z-axis• Undo the previous transformations

Rotation about an arbitrary axis

• Step 1Rotate x-axis

x

y

z

(a,b,c)

x

Closer Look at Y-Z Plane

• Need to rotate degrees around the x-axis

y

z

Equations for

||)1,0,0(||||),,0(||

)1,0,0(),,0()cos(

||),,0(||||)1,0,0(||

||),,0()1,0,0(||)sin(

cb

cb

cb

cb

Rotation about the Y-axis

• Using the same analysis as before, we need to rotate degrees around the Y-axis

y

z

x

Rotation about the Z-axis

• Now, it is aligned with the Z-axis, thus we can simply rotate degrees around the Z-axis.

• Then undo all the transformations we just did

Equation summary

c

b

a

TRRRRRT

c

b

a

rot xyzyxaxis )()()()()(

'

'

'

)( 111

Deformations

Transformations that do not preserve shape Non-uniform scaling Shearing Tapering Twisting Bending

Shearing

11000

01

01

01

1

'

'

'

z

y

x

ss

ss

ss

z

y

x

zyzx

yzyx

xzxy

0

0

0

1

zyzx

yzyx

xz

xy

ss

ss

s

s

x

y

x

y

Tapering

11000

0)(00

00)(0

0001

1

'

'

'

z

y

x

xf

xf

z

y

x

Image courtesy of Watt, 3D Computer Graphics

Twisting

11000

0))(cos(0))(sin(

0010

0))(sin(0))(cos(

1

'

'

'

z

y

x

yy

yy

z

y

x

Image courtesy of Watt, 3D Computer Graphics

Bending

11000

0)()(0

0)()(0

0001

1

'

'

'

z

y

x

ykyh

ygyf

z

y

x

Image courtesy of Watt, 3D Computer Graphics

What have we seen so far?

• Basic representations (point, vector)• Basic operations on points and vectors

(dot product, cross products, etc.)• Transformation – manipulative

operators on the basic representation (translate, rotate, deformations) – 4x4 matrices to “encode” all these.

Why do we need this?• In order to generate a picture from

a model, we need to be able to not only specify a model (representation) but also manipulate the model in order to create more interesting images.

Overview• The next set of slides will deal with

the other half of the process (at least in a simplistic fashion)

• From a model, how do we generate an image

Model Imagecomputer

Accumulation of Transformations

baseT

)()()(

)()(

RRTRMM

RTRMM

TM

basearmhand

basebasearm

basebase

base

arm hand

Example: Robot arm

Orthonormal Coordinates

Iff u and v are orthonormal:

Tyx

yx

Mvv

uu

M

100

0

01

100100

10

01

100

0

011 vr

ur

yx

yx

y

x

yx

yx

vv

uu

r

r

vv

uu

TM

Coordinate Systems Object coordinates World coordinates Camera coordinates Normalized device coordinates Window coordinates

Object CoordinatesConvenient place to model the

object

O

World CoordinatesCommon coordinates for the scene

O

O

W

Camera CoordinatesCoordinate system with the camera

in a convenient pose

u

v

n

1000

nr

vr

ur

zyx

zyx

zyx

nnn

vvv

uuu

M

Normalized Device Coordinates

Device independent coordinatesVisible coordinate usually range

from:

11

11

11

z

y

x

Window CoordinatesAdjusting the NDC to fit the window

0

0

2)1(

2)1(

yheight

yy

xwidth

xx

ndw

ndw

),( 00 yx is the lower left of the window

width

height

Perspective ProjectionTaking the camera coordinates to

NDC

z

x

near

Perspective Projection

z

x

near'p

p

z

xnearx

z

x

near

x

'

'

Rasterization

Array of pixels

0

0 1 2 3 4

1

2

3

Rasterizing LinesGiven two endpoints, find the pixels that make up the line.

),(),,( 1100 yxyx

Rasterizing LinesRequirements

1. No gaps

2. Minimize error (distance to line)

Rasterizing Lines

Line(int x0, int y0, int x1, int y1) float dx = x1 – x0; float dy = y1 – y0; float m = dy/dx; float x = x0, y= y0;

for(x = x0; x <= x1; x++) setPixel(x,round(y)); y = y+m;

Assume –1 < m < 1, x0 < x1

Rasterizing LinesProblems with previous algorithm

1. round takes time2. uses floating point arithmetic

Midpoint Algorithm

NE

E

P=(x,y)

MQ

If Q <= M, choose E. If Q > M, choose NE

Implicit Form of a Line

dxBcdxbdya

dxBydxxdy

Bxdx

dyycbyax

0

0

Implicit form Explicit form

Positive below the lineNegative above the lineZero on the line

Decision Function

cybxayxFd

cybxayxFd

)()1(),1(

),(

2

1

2

1

Choose NE if d > 0Choose E if d <= 0

Incrementing d

cybxayxFdnew )()2(),2(2

1

2

1

If choosing E:

But:

cybxayxFdold )()1(),1(2

1

2

1

So:

Eaddd oldnewinc

Incrementing d

cybxayxFdnew )()2(),2(2

3

2

3

If choosing NE:

But:

cybxayxFdold )()1(),1(2

1

2

1

So:

NEbaddd oldnewinc

Initializing d

2

1

2

1

2

1

2

1

00

0000 )()1(),1(

ba

bacybxa

cybxayxFd

Multiply everything by 2 to remove fractions (doesn’t change the sign)

Midpoint AlgorithmLine(int x0, int y0, int x1, int y1) int dx = x1 – x0, dy = y1 – y0; int d = 2*dy-dx; int delE = 2*dy, delNE = 2*(dy-dx); int x = x0, y = y0; setPixel(x,y);

while(x < x1) if(d<=0) d += delE; x = x+1; else d += delNE; x = x+1; y = y+1; setPixel(x,y);

Assume 0 < m < 1, x0 < x1

See: Bresenham Algorithm

Limitations?• The midpoint line algorithm

assumes that the slope (m) is between 0 and 1

• This implies that this algorithm only applies to lines in region 1

• Extending to other regions left as an a programming assignment

1

23

4

5

67

8

Anti-aliasing LinesLines appear jaggy

Sampling is inadequate

Anti-aliasing LinesTrade intensity resolution for spatial resolution

Anti-aliasing LinesLine(int x0, int y0, int x1, int y1) float dx = x1 – x0; float dy = y1 – y0; float m = dy/dx; float x = x0, y= y0;

for(x = x0; x <= x1; x++) int yi = floor(y); float f = y – yi; setPixel(x,yi, 1-f); setPixel(x,yi+1, f); y = y+m;

Assume 0 < m < 1, x0 < x1

Putting it all together!!• Take your representation (points) and

transform it from Object Space to World Space• Take your World Space point and transform it

to Camera Space• Perform the remapping and projection onto the

image plane in Normalized Device Coordinates• Perform this set of transformations on each

point of the polygonal object• “Connect the dots” through line rasterization

IntuitivelyObjectSpace

WorldSpace

CameraSpace

Rasterization