Transforming Transformations James F. Blinn · CSE590B Lecture 4 More about P1 Transforming Transformations James F. Blinn JimBlinn.Com

CSE590B Lecture 4

More about P1

Transforming Transformations

James F. Blinn JimBlinn.Com

http://courses.cs.washington.edu/courses/cse590b/13au/

Previously On

CSE590b

Transformations

M

A B

C D

M

A B

x w x wC D

x Ax Cw

w Bx Dw

The Function

x Ax Cw

w Bx Dw

x

w

x

w

A

B

D

B

0

0 0

x Aw

w B

x Dw Bx Dw

w B

The “Phase Space” of M

M MM MM M

M MM M d

sing

ular

dM

M

0

M MM

M

0

Nilpotent

2t

d

1 2tan , td

Also a valid matrix

Possible

numeric

signatures

New Coordinate System A B E F G H

C D G H E F

2 2 2 2

214

2 2 214

E H F G

t E

H F Gd

2 2 2 2

AD BC

E F E F G H G H

E H F G

2

2 2 2

4

4

A D AD BC

F G H

d

2t A D E

Plot in EFGH space

2 2 2

0 1H F G

E E E

2 2 2

0H F G

E E Ed

F/E

B/E

0 plane at infinityt

Compare with Q version

H/E

G/E

F/E

2 2 2

2 2 2

0 1

0 plane at infinity

0

H F G

E E E

t

F G H

E E Ed

Roadmap of M

2 2 2 2

2 2 2 2

ˆcos sin

ˆsin cos

ˆ ˆ

E E

H H

E H F G

E H F G

2 2 2

2 2 2

ˆ0 1 0

ˆ ˆ ˆ

ˆ0 1

ˆ

ˆ0 sin cos 0

ˆ ˆ

H F G

E E E

Ht

E

H F G

E E Ed

−3

−2

−1

01

23

y

E H

0

1

1

0

identity

singular

rotations

Involutions

(trace=0)

d = 0

Single eigenvalue

Nilpotent

And Now

IM

M

Finding Real Eigenvalues

det 0 M I

Find such that

singular

Outside cone (red,blue):

Two intersections on line

Inside cone (green):

No intersections on line

Linear combo of M and I is singular

Transformation of Transformations *

A BA B

C DC D

T T

p q A B s qA B

r s C D r pC D

t u A B v uA B

s v C D s tC D

A tvA uvC stB sqD

B utA uuC tB tqD

C svA vvC ssB svD

D usA uvC tsB tvD

ts us ts uv AA

us tv ts uv DD

tu tu tt uu BB

rv rv rr vv CC

Transformation of Transformations

p q A B s qA B

r s C D r pC D

1 1 0 0

1 1 0 0

0 0 1 1

0 0 1 1

A E

D F

B G

C H

ts us ts uv AA

us tv ts uv DD

tu tu tt uu BB

rv rv rr vv CC

1 12 2

1 12 2

0 0 0

0

0

0

tv us EE

tv us ts uv ts uv FF

tu sv tt ss uu vv tt ss uu vv GG

tu sv tt ss uu vv tt ss uu vv HH

A B E F G H

C D G H E F

Distill with Rotation Transform

cos sin

sin cos

t u

s v

1 12 2

1 12 2

0 0 0

0

0

0

tv us EE

tv us ts uv ts uv FF

tu sv tt ss uu vv tt ss uu vv GG

tu sv tt ss uu vv tt ss uu vv HH

1 0 0 0

0 cos 2 sin 2 0

0 sin 2 cos 2 0

0 0 0 1

EE

FF

GG

HH

E

FB

E

F

Similar to

what we did

with Q:

Rotation Transform

1 0 0 0

0 cos 2 sin 2 0

0 sin 2 cos 2 00

0 0 0 1

EE

FF

G

HH

2 2 2E H F

2 214

F H

F H F H

d

Rotate to make G zero

12t E

E

H

Ft 0

= 0

d = 0

d = 0 ~

~

Note: Not dividing by E yet

Rotation Transform

1 0 0 0

0 cos 2 sin 2 0

0 sin 2 cos 2 00

0 0 0 1

EE

FF

G

HH

2 2 2E H F

2 214

F H

F H F H

d

Rotate to make G zero

12t E

E

H

Ft

d

Project onto unit sphere

[E,F,G,H]

Mappings

E

H

F

E

F

Quadratic Polys (3D)

E

FB

F/E

B/E

project

Transformation mtx (4D)

project

rotate

rotate

Sign Flips

1 0 0 0

0 1 0 1 0 0

1 0 0 0 1 0

0 0 0 1

EE

FF

GG

HH

1 0 0 0

1 0 0 1 0 0

0 1 0 0 1 0

0 0 0 1

EE

FF

GG

HH

1 0 0 0

0 1 0 1 0 0

1 0 0 0 1 0

0 0 0 1

EE

FF

GG

HH

Rotating 90 degrees flips (F,G) sign

Scaling by -1 in x flips (E,F) signs

Exchanging x,w flips (E,G) signs E

H

F

t

d

Sign Flips

1 0 0 0

0 1 0 1 0 0

1 0 0 0 1 0

0 0 0 1

EE

FF

GG

HH

Exchanging x,w flips (E,G) signs E

H

F

t

d

Sign Flip effect on angle range

E

H

0o-90

o

-180o

+180o

+90o

E

H

0o-90

o

+90o

+170o

+10o

-10o

10 10/

170 10

Further transform that keeps G=0

12

12

0 0

0

0

0

tv usEE

tv us ts uvFF

tu sv tt ss uu vvGH

tu sv tt ss uu vvH

0

0

tu sv

tt ss uu vv

s u s uor

v t v t

E

H

F

t

d

t u

u t

= Diagonal scale

Effect of Diagonal Scale

0 0

0 2

0 2

E tt uu Et u

F tt uu tu Fu t

H tu tt uu H

E

H

Ft

d

2 2

2 2

4

4

F H

t E

d

22 2

22 2 2 2

E tt uu E

F H tt uu F H

2 2 2

2 2 2 2

E E t

F H F H d

2 2 2 20 E t F Hd

Effect of Diagonal Scale

E

H

Ft=0

d 0

0

2 2 2 2 2

2

2 22

2 2

2 22

2

0 0 plane

0 0 cone along axis

0 0 intersecting planes

0 0 cone along axis

0 0 plane

neg

neg

t E t H F

E

E tH tF F

tH tF

E tH tF H

E

d d

d

d

d

d

sing

ular

dNilpotent

t

2 2

2 2

2 , discrim

2 , discrim

F ttF tuH uuF H F

H ttH tuF uuH F H

Diagonal Scale to get F or H zero

0 0

0 2

0 2

E tt uu Et u

F tt uu tu Fu t

H tu tt uu H

H

F

H

F E

H

F

t

d

If positive can make F=0

If positive can make H=0

d=0 case

H

F

0 0

0 2

0 2

E tt uu E

F tt uu tu F

H tu tt uu H

0

0 2

0

0

H F

tt uu EE

tt tu uu FF

t u Et u

t u F

0

0 2

0

0

H F

tt uu EE

tt tu uu FF

t u Et u

t u F

H

F

Distilled EFGH Space

H

F1 0

0 1

0

0

E F

E F

1 1

1 1

2 1

1 0

2 1

1 0

1 1

1 1

1 0

0 1

0 1

1 0

0 1

1 0

E H

H E

1 0

0 0

Phase Dgm and Distilled

H

F1 0

0 1

0

0

E F

E F

1 1

1 1

2 1

1 0

2 1

1 0

1 1

1 1

1 0

0 1

0 1

1 0

0 1

1 0

E H

H E

sing

ular

d

M

M

0

M MM

M

0

Nilpotent

M MM MM M

M MM M d

Phase Dgm and Distilled

H

F1 0

0 1

0

0

E F

E F

1 1

1 1

2 1

1 0

2 1

1 0

1 1

1 1

1 0

0 1

0 1

1 0

0 1

1 0

sing

ular

d

M

M

0

M MM

M

0

Nilpotent

M MM MM M

M MM M d

M VRM MVR VR

VR

M VR H

Distinguish

between

E

FB

Q

I

0 1

1 0

Phase Dgm and Distilled

H

F1 0

0 1

0

0

E F

E F

1 1

1 1

2 1

1 0

2 1

1 0

1 1

1 1

1 0

0 1

0 1

1 0

0 1

1 0

sing

ular

d

M

M

0

M MM

M

0

Nilpotent

M MM MM M

M MM M d

M VR H Use if d £ 0

Internal Structure

Using outer products to make M

tk unM =

1 0

0 1

A BA B C D

C D

1 0 0 1A B A B

C D C D

Write k,n in terms of t,u

tuk = a + b

tun = c + d

M = t ttua b

c u utu d

tk unM =

Transformation

M

= t ttua b

c u utu d

T* T

T

T

T

T

T*

T*

T*

T*

uT* uT

uT

u ~

M = t ttua b

c u utu d

~ ~ ~

~

~ ~

~~~

=u T u~

=t T t~

Pick nice t,u

t

u

= 0 1

= 1 0

t ttu

u utu

0 0

0 1

0 0

1 0

0 1

0 0

1 0

0 0

t

tt

u D

+ B

C u

ut

u

AA B

C D

t

tt

u D

+ B

C u

ut

u

AM~ ~ ~ ~ ~

~~~~

Nilpotent

M M

Nilpotent

= 0MM

M

H

F1 0

0 1

0

0

E F

E F

1 1

1 1

2 1

1 0

2 1

1 0

1 1

1 1

1 0

0 1

0 1

1 0

0 1

1 0

0 1

0 0

0

0

1

1

A B E F G H

C D G H E F

E

F

G

H

1 1

1 1

0

1

0

1

A B E F G H

C D G H E F

E

F

G

H

Distilled: Can Transform to:

Nilpotent

0 1

0 0

ttN

t

H

F

M M

Nilpotent

= 0MM

M

ttN a a

ttN t t

x

w

x

w

ttN N tt

d

2t

Idempotent 0 0

0 1

tu

H

F

M M

= 0MM

M

sin

gu

lar

tuD

tuD a a

tuD t t

tuD u u 0

x

w

x

w

tuD tuD

d

2t

An Identity

t u t u

M M

= 0MM

M

t u tu

tu

t u tu

tu

H/E

G/E

F/E

t u

t u tu I

General Scales (Eigenvectors)

0

0

E F

E F

u tut(E+F) +(FE)M =

ut(E+F)M =u u

u

tu+ (F-E)M =t t

t

H

FM M

= 0MM

M

x

w

x

w

x

w

x

w

H/E

G/E

F/E

E/H

G/H

F/H

d

2t

Scale Involution (Eigenvectors) 0

, 00

E FE

E F

u tutF +FVS =

utFM =u u

u

tuFM =t t

t

H

FM M

= 0MM

M

x

w

x

w

H/E

G/E

F/E

E/H

G/H

F/Hd

2t

u t

Rotation

E H

H E

u tutE

+ H t uut

H

F

M M

= 0MM

M

x

w

x

w

IE

+ H t uut

^

d

2t

Rotation Involution

, 0E H

EH E

+ H t uut

H

F

M M

= 0MM

M

d

2t

x

w

x

w

A

B

D

B

Single eigenvalue d = 0

1 1

0 1

u

tu

t tt

H

F

MMd0 = 0

M

M M

IdentityI+N

x

w

x

w

I tt ^

d

2t

Pure & Mixed Tensors - A Relation

Q a,b

Q T

Q T

=0

F/E

B/E

T I

x,w a,b Q Q Q1

2a,b

Roots of Q

are

Eigenvalues of T

Exemplary Transformations

Exemplary Transformation

Tp q r r~q~p~

x

wp qr

x

wp q

r

Construct T given Eigenvectors

T = p qpq a b

T = pp p

T = qq q

T = p qpq a bp p p

p

Want

Basic Answer

How it works for p

Construct T given

two different output points

T = p qpq a b~ ~

T = pp p

T = qq q

~

~

Want

Basic Answer

Works for p and q

T = pqa ~p p

T = qpb ~q q

Third point

=r~ p~ q~c +d

T = p qpq a b~ ~

T = rr ~

T = p qpq a br r r~ ~

Pick a and b to make

= p qpq a br r~ ~r~

= +p

r

q r q

p

p r

q~

~ ~

~

~ ~

~

~

The answer

a = pr

rp~ ~qr

qr~ ~

b=

T =p qpq ~ ~+

pr rp~ ~qrqr~ ~

How it works

T =p qpq ~ ~+

pr rp~ ~qrqr~ ~

pp p

T =p qpq ~ ~+

pr rp~ ~qrqr~ ~

qq q

T =p qpq ~ ~+

pr rp~ ~qrqr~ ~

rr r

T =p qpq ~ ~+

pr rp~ ~qrqr~ ~

T =p qpq ~ ~+

pr rp~ ~qrqr~ ~

pp p

T =p qpq ~ ~+

pr rp~ ~qrqr~ ~

qq q

T =qq ~

+pr rp~ ~

rr p~

qr~ ~

=q

pr

r r~

qp~ ~

How it works T =p qpq ~ ~+

pr rp~ ~qrqr~ ~

T = p

q

~p r

q r~~p

p

T = qp

~r p~~q r

qq

=q

pr

rr~qp~ ~Tr

Net T =p qpq ~ ~+

pr rp~ ~qrqr~ ~

Determinant

T =

p qpq ~ ~+

pr rp~ ~qrqr~ ~

T

=T

q

r

q

p

r p

q

r

q

p

r p

~~

~

~ ~

~

p q r

r pq

qr p

r q p

p rq

pr q

q

r

q

p

r p

An invariant of the points p,q,r Geometric meaning of sign

Determinant is negative

exactly when necessary

for order reversal

Higher Dimensions

Tp

q

db

p~ ~q

db~~

Higher Dimensions

Tp

q

sr

p~ ~q

sr~~

T = p q

q

~ ~+ bs~ g

s

s

p

p

qa

Works for p,q,s

Now find a,b,g to make it work for r

Higher Dimensions

T

p

q

q

~

~

s~

s

s

p

p

q

s

p

p

q

r

r

q

s

p

q

r

r

s

p

q

s

r

rr

s

q

r

p

s

r

q

p

T =p q

p

q ~ ~+

pr rp~ ~

qr

qr~ ~

p

p

p

s

s

s

q

q

q

Four Points in P1

A B C D

D B

C

A

A B C

D

x

w

Interleaving of Four Points

D A B C

D A BC

B CD A

A B C D

D B

C

A

B A C D

A B C

D

x

w

Same interleaving

Different interleaving

D A

C

B

AB C

D

x

w

Three possible interleavings

A B D C

A B C

D

x

w

w

x

A BC

D

w

x

A B CD

A B C D

A C B D

D B

C

A

D C

B

A

C B

D

A

Diagrams for Four Points

A A Ax w A

B B Bx w B

C C Cx w C

D D Dx w D A D

A C

A B

B D

B C

C D

A D

A C

A B

D B

B C

C D1V

2V

3V

Diagrams for Four Points

A DA CA B D B B CC D 0

1 2 3 0V V V

A D

A C

A B

D B

B C

C D1V

2V

3V

Cross Ratio

1

2

2 2

3 1 2

3 1 2

1 1

1

1

11

V

V

V V

V V V

V V V

V V

31 2

2 3 1

, ,VV V

V V V

B C

A C

A D

B D2

3

V

V

Could pick any of as “cross ratio”

Relationship

between

“Absolute Invariant”

3D view of invariant space

V3

V2

V1

V1

V2

V3

1 2 3 0V V V

2 2 2

1 2 3 1V V V

1

1cos V

A D

A C

A B

D B

B C

C D1V

2V

3V

Homogeneously

scale to normalize

onto unit circle

3D view of invariant space

V3

V2

V1

V1

V2

V3

V3 V2

V1

V3 V2

V1

1 2 3 0V V V

Determine which sign

indicates which interleaving

1 1

0 1

1 1

x w

A

B

C

D

AB

C

D

x

w

A D

A C

A B

D B

B C

C D1V

2V

3V

w x

2x

w x

Follow in V space

A B C

D

1

2

3

1 0

1

2

1

V

V

V

D

A B CD

1

2

3

1 1

0

2

2

V

V

V

D

V3 V2

V1

V3 V2

V1

V3 V2

V1

A B CD

1

2

3

1 3

2 2

1

1

2

V

V

V

D

Follow in V space

A B C

D

1

2

3

1 0

1

2

1

V

V

V

D

A B CD

1

2

3

1 1

0

2

2

V

V

V

D

V3 V2

V1

V3 V2

V1

V3 V2

V1

A B CD

1

2

3

1 3

2 2

1

1

2

V

V

V

D

A B CD

1

2

3

0 1

1

0

1

V

V

V

D

V3 V2

V1

A B CD

1

2

3

1 3

2 2

2

1

1

V

V

V

D

V3 V2

V1

V3 V2

V1

A B CD

1

2

3

1 1

1

0

1

V

V

V

D

2V1V

3V

…

Follow in V space

A B C

D

1

2

3

1 0

1

2

1

V

V

V

D

A B CD

1

2

3

1 1

0

2

2

V

V

V

D

V3 V2

V1

V3 V2

V1

V3 V2

V1

A B CD

1

2

3

1 3

2 2

1

1

2

V

V

V

D

A B CD

1

2

3

0 1

1

0

1

V

V

V

D

V3 V2

V1

A B CD

1

2

3

1 3

2 2

2

1

1

V

V

V

D

V3 V2

V1

V3 V2

V1

A B CD

1

2

3

1 1

1

0

1

V

V

V

D

Harmonic Set

A B D

C

D B C

A

2V1V

3V 2V

3V

Can

transform

to

Can

transform

to

Determine which sign

indicates which interleaving

1 1

0 1

1 1

x w

A

B

C

D

1

2

3

^ ^

^ ^ 2

^ ^

V w x

V x

V w x

A B C D

A C B D

A D B C

A B C

1V w x

D

A B C

2 2V x

A B C

3V w x

Signs and interleaving Sign(V1,V2,V3) Ordering of points on P1 View in (x,w) plane

D B

C

AA B C

D

A B C

D

C B

D

AA B CD

A B C

D

C D

B

AA B C

D

A B C

D

Better Interleaving Test

1 2 2 2VV w x x x w x

A B C

A B C

2 3 2V V x w x

A B C

2 2

3 1V V w x w x w x

A B C

1V w x

D

A B C

2 2V x

A B C

3V w x

Old way

Better way

Diagrams

A C

A B

D B

C D

V1 V2 = A

C

A B

D

B

C D

A

C

A B

D

B

C D

A D

A C D B

B C

V2 V3 =A

D

A C

D BB

C= A

D

A C

D BB

C

A D

A B

B C

C D

V3V1 =A

D

A B

B

CCD

A

D

A B

B

CCD

Best Interleaving Test

w

x

A B C

D

A

D

A B

C

B

D C

0Û

w

x

A BC

D

A

D

A C

B

C

D B

0Û

A B CD

A

C

A B

D

B

C D

0Û