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COMP 546 Lecture 8 image motion 2 Tues. Feb. 6, 2018 1 Overview of Today Abstract computational problem: how to estimate the local image velocity ? Solution based on V1 motion detectors 2 Intensity changes in XY


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SLIDE 1

1

COMP 546

Lecture 8

image motion 2

  • Tues. Feb. 6, 2018
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SLIDE 2

Overview of Today

  • Abstract computational problem:

how to estimate the local image velocity ?

  • Solution based on V1 motion detectors

2

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SLIDE 3

Intensity changes in XY

(

๐œ– ๐œ–๐‘ฆ ๐ฝ ๐‘ฆ, ๐‘ง , ๐œ– ๐œ–๐‘ง ๐ฝ ๐‘ฆ, ๐‘ง )

is the 2D spatial gradient at ๐‘ฆ, ๐‘ง .

๐‘ฆ ๐‘ง +

  • +

+ +

  • +

+

3

โˆ‡ ๐ฝ

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SLIDE 4

Intensity changes in XYT

(

๐œ– ๐œ–๐‘ฆ ๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข , ๐œ– ๐œ–๐‘ง ๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข , ๐œ– ๐œ–๐‘ข ๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข )

is the 3D spatio-temporal gradient at ๐‘ฆ, ๐‘ง, ๐‘ข .

๐‘ง ๐‘ข ๐‘ฆ

4

โˆ‡ ๐ฝ

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SLIDE 5

Suppose that image โ€œobjectsโ€ are โ€œmovingโ€. (What do these terms mean?)

(๐‘ฆ, ๐‘ง, ๐‘ข) ๐‘ค๐‘ฆ, ๐‘ค๐‘ง

5

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SLIDE 6

Intensity conservation

๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข = ๐ฝ ๐‘ฆ + ๐‘ค๐‘ฆ โˆ†๐‘ข, ๐‘ง + ๐‘ค๐‘งโˆ†๐‘ข, ๐‘ข + โˆ†๐‘ข

Assume a moving pointโ€™s intensity doesnโ€™t change over time. This is similar to the assumption made last lecture, namely that the left and right images are related by a (disparity) shift.

6

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SLIDE 7

๐ฝ ๐‘ฆ + ๐‘ค๐‘ฆ โˆ†๐‘ข, ๐‘ง + ๐‘ค๐‘งโˆ†๐‘ข, ๐‘ข + โˆ†๐‘ข = ๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข +

๐œ– ๐œ–๐‘ฆ ๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข

๐‘ค๐‘ฆโˆ†๐‘ข +

๐œ– ๐œ–๐‘ง ๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข

๐‘ค๐‘งโˆ†๐‘ข +

๐œ– ๐œ–๐‘ข ๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข

โˆ†๐‘ข + H.O.T.

Taylor series

7

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SLIDE 8

8

๐ฝ ๐‘ฆ + โˆ†๐‘ฆ โ‰ˆ ๐ฝ ๐‘ฆ + ๐œ–๐ฝ ๐œ–๐‘ฆ โˆ†๐‘ฆ ๐‘ฆ ๐‘ฆ + โˆ†๐‘ฆ

๐œ–๐ฝ ๐œ–๐‘ฆ

We can estimate by taking

๐ฝ ๐‘ฆ + 1 โˆ’ ๐ฝ(๐‘ฆ โˆ’ 1) 2 .

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SLIDE 9

๐ฝ ๐‘ฆ + ๐‘ค๐‘ฆ โˆ†๐‘ข, ๐‘ง + ๐‘ค๐‘งโˆ†๐‘ข, ๐‘ข + โˆ†๐‘ข = ๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข +

๐œ– ๐œ–๐‘ฆ ๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข

๐‘ค๐‘ฆโˆ†๐‘ข +

๐œ– ๐œ–๐‘ง ๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข

๐‘ค๐‘งโˆ†๐‘ข +

๐œ– ๐œ–๐‘ข ๐ฝ ๐‘ฆ, ๐‘ง, ๐‘ข

โˆ†๐‘ข + H.O.T.

โ‰ˆ 0

9

๐๐›๐จ๐๐Ÿ๐ฆ ๐œ๐Ÿ๐๐›๐ฏ๐ญ๐Ÿ ๐ฉ๐  ๐ฃ๐จ๐ฎ๐Ÿ๐จ๐ญ๐ฃ๐ฎ๐ณ ๐๐ฉ๐จ๐ญ๐Ÿ๐ฌ๐ฐ๐›๐ฎ๐ฃ๐ฉ๐จ

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SLIDE 10

โ€œMotion Constraint Equationโ€

๐œ– ๐ฝ ๐œ–๐‘ฆ

๐‘ค๐‘ฆ +

๐œ– ๐ฝ ๐œ–๐‘ง ๐‘ค๐‘ง + ๐œ– ๐ฝ ๐œ–๐‘ข = 0

10

( ๐œ– ๐ฝ

๐œ–๐‘ฆ , ๐œ– ๐ฝ ๐œ–๐‘ง , ๐œ– ๐ฝ ๐œ–๐‘ข ) โˆ™ (

๐‘ค๐‘ฆ, ๐‘ค๐‘ง, 1 ) = 0

Previous slides (and assumptions) imply:

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SLIDE 11

๐œ– ๐ฝ ๐œ–๐‘ฆ

๐‘ค๐‘ฆ + ๐œ– ๐ฝ

๐œ–๐‘ง ๐‘ค๐‘ง + ๐œ– ๐ฝ ๐œ–๐‘ข = 0

๐‘ค๐‘ฆ ๐‘ค๐‘ง

Motion constraint equation is a line in velocity space

11

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SLIDE 12

๐œ– ๐ฝ ๐œ–๐‘ฆ

๐‘ค๐‘ฆ +

๐œ– ๐ฝ ๐œ–๐‘ง ๐‘ค๐‘ง + ๐œ– ๐ฝ ๐œ–๐‘ข = 0

๐‘ค๐‘ฆ ๐‘ค๐‘ง But many velocities satisfy this equation.

12

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SLIDE 13

๐‘ง ๐‘ข ๐‘ฆ

The black vector in the figure below is the 3D image gradient. It is perpendicular to the gray plane. The red vectors are perpendicular to the 3D image gradient (see equation) and thus the red vectors lie in the grey plane.

13

(

๐œ– ๐ฝ ๐œ–๐‘ฆ , ๐œ– ๐ฝ ๐œ–๐‘ง , ๐œ– ๐ฝ ๐œ–๐‘ข ) โˆ™ (

๐‘ค๐‘ฆ, ๐‘ค๐‘ง, 1 ) = 0

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SLIDE 14

๐œ– ๐ฝ ๐œ–๐‘ฆ

๐‘ค๐‘ฆ +

๐œ– ๐ฝ ๐œ–๐‘ง ๐‘ค๐‘ง + ๐œ– ๐ฝ ๐œ–๐‘ข = 0

โ€œNormal velocityโ€ is the velocity component in the direction of the XY gradient.

Normal velocity

14

๐‘ค๐‘ฆ ๐‘ค๐‘ง

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SLIDE 15

โ€œAperture Problemโ€

The same issue arises with any 1D pattern e.g. single bar, edge, constant gradient.

15

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SLIDE 16

Solution of Aperture Problem

We need more than one line or edge (or gradient

  • rientation).

Could be in the same aperture or neighboring aperture.

normal velocities shown

16

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SLIDE 17

โ€œIntersection of Constraintsโ€ (IOC)

๐‘ค๐‘ฆ ๐‘ค๐‘ง

Suppose two nearby points have two different spatial gradients and the same image velocity.

normal velocities shown

17

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SLIDE 18

๐œ– ๐ฝ ๐œ–๐‘ฆ (๐‘ฆ1, ๐‘ง1, ๐‘ข) ๐‘ค๐‘ฆ + ๐œ– ๐ฝ ๐œ–๐‘ง (๐‘ฆ1, ๐‘ง1, ๐‘ข) ๐‘ค๐‘ง + ๐œ– ๐ฝ ๐œ–๐‘ข (๐‘ฆ1, ๐‘ง1, ๐‘ข) = 0

Intersection of Constraints (IOC)

๐‘ค๐‘ฆ ๐‘ค๐‘ง

๐œ– ๐ฝ ๐œ–๐‘ฆ (๐‘ฆ2, ๐‘ง2, ๐‘ข) ๐‘ค๐‘ฆ + ๐œ– ๐ฝ ๐œ–๐‘ง (๐‘ฆ2, ๐‘ง2, ๐‘ข) ๐‘ค๐‘ง + ๐œ– ๐ฝ ๐œ–๐‘ข (๐‘ฆ2, ๐‘ง2, ๐‘ข) = 0

IOC gives a unique solution.

18

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SLIDE 19

Another Example (counterintuitive)

๐‘ค๐‘ฆ ๐‘ค๐‘ง

19

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SLIDE 20

Overview of Today

  • Abstract computational problem:

how to estimate the local image velocity ?

  • Solution based on V1 motion detectors

20

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SLIDE 21

Recall: XYT Gabor

21

sin 2๐œŒ ๐‘‚ (๐‘™0 ๐‘ฆ + ๐‘™1 ๐‘ง) + 2๐œŒ ๐‘ˆ ๐œ• ๐‘ข

๐‘ฆ ๐‘ง ๐‘ข

sin

2๐œŒ ๐‘‚ (๐‘™0 ๐‘ฆ + ๐‘™1 ๐‘ง) + 2๐œŒ ๐‘ˆ ๐œ• ๐‘ข

๐ป(๐‘ฆ, ๐‘ง, ๐‘ข, ๐œ๐‘ฆ, ๐œ๐‘ง, ๐œ๐‘ข)

๐‘ฆ ๐‘ข ๐‘ง ๐‘ฆ ๐‘ข ๐‘ง

cos

2๐œŒ ๐‘‚ (๐‘™0 ๐‘ฆ + ๐‘™1 ๐‘ง) + 2๐œŒ ๐‘ˆ ๐œ• ๐‘ข

๐ป(๐‘ฆ, ๐‘ง, ๐‘ข, ๐œ๐‘ฆ, ๐œ๐‘ง, ๐œ๐‘ข)

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SLIDE 22

(

๐œ– ๐œ–๐‘ฆ ๐ฝ, ๐œ– ๐œ–๐‘ง ๐ฝ, ๐œ– ๐œ–๐‘ข ๐ฝ )

๐‘ง ๐‘ข ๐‘ฆ

22

(2๐œŒ ๐‘‚ ๐‘™0 , 2๐œŒ ๐‘‚ ๐‘™1 ,

2๐œŒ ๐‘ˆ ๐œ• )

We should get maximum response from a Gabor whose frequency (๐‘™0 , ๐‘™1 , ๐œ• ) is parallel to intensity gradient.

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SLIDE 23

Each motion sensitive V1 cell is sensitive only to the 1D motion component perpendicular to its orientation.

23

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SLIDE 24

24

Suppose true motion vector ๐‘ค๐‘ฆ, ๐‘ค๐‘ง is toward the right and the image contains lots of oriented structure in each local neighborhood. Which V1 motion cells would have a large response ? ๐‘ค๐‘ฆ ๐‘ค๐‘ง

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SLIDE 25

25

Suppose true motion vector ๐‘ค๐‘ฆ, ๐‘ค๐‘ง is toward the right and the image contains lots of oriented structure in each local neighborhood. Which V1 motion cells would have a large response ? ๐‘ค๐‘ฆ ๐‘ค๐‘ง

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SLIDE 26

26

Suppose true motion vector ๐‘ค๐‘ฆ, ๐‘ค๐‘ง is toward the right and the image contains lots of oriented structure in each local neighborhood. Which V1 motion cells would have a large response ? ๐‘ค๐‘ฆ ๐‘ค๐‘ง

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SLIDE 27

27

Suppose true motion vector ๐‘ค๐‘ฆ, ๐‘ค๐‘ง is toward the right and the image contains lots of oriented structure in each local neighborhood. Which V1 motion cells would have a large response ? ๐‘ค๐‘ฆ ๐‘ค๐‘ง

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SLIDE 28

Motion pathway in the brain

28

(contains V1) (temporal lobe contains motion area MT and MST : โ€œmiddle temporalโ€ & โ€œmedial superior temporalโ€)

MST โŸต MT โŸตV1

(next lecture) (today)

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SLIDE 29

V1 โ†’ MT

29

MT cells receive inputs from orientation/motion tuned V1 cells. Many MT cells are velocity tuned. They require multiple

  • rientations in their

receptive field (to avoid the aperture problem). ๐‘ค๐‘ฆ ๐‘ค๐‘ง

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SLIDE 30

V1 โ†’ MT

30

๐‘ค๐‘ฆ ๐‘ค๐‘ง MT cells receive inputs from orientation/motion tuned V1 cells. Many MT cells are velocity tuned. They require multiple

  • rientations in their

receptive field (to avoid the aperture problem).

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SLIDE 31

V1 โ†’ MT

31

๐‘ค๐‘ฆ ๐‘ค๐‘ง For any velocity vector, there is a family of

  • rientation/motion tuned

V1 cells that would respond well, provided the moving image contains the spatial

  • rientaiton that this cell is

sensitive to.

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SLIDE 32

32

For full version of the MT model, see [Simoncelli and Heeger Vision Research 1998]

Gabors can have different sizes and spatial frequencies.

dot pattern and velocity

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SLIDE 33

33

Random dot patterns (such as above) contain oriented structure at all frequencies. Later in the course when we discuss linear systems and Fourier transforms, you will learn a more formal (mathematical) statement of what this means.