CS 6320 Computer Vision Practice Set

Spring 2019Name:UID:

Each question carries 20 points. Show the necessary steps to compute the final solutions. You are

free to use extra papers.

1. Pose Estimation: Let us consider a calibrated camera. Let the origin of the camera be

given by O(0, 0, 0). The image resolution is 640 ⇥ 480 and the principal point is given by

(320, 240). We assume the following parameters for the camera:

✓uv1

◆⇠

✓200 0 320 00 200 240 00 0 1 0

◆⇣I 0

0T 1

⌘0

@Xm

Y m

Zm

1

1

A

where (u, v) correspond to pixel coordinates and I denotes the 3⇥ 3 identity matrix.

K�1 =1

200

0

@1 0 �3200 1 �2400 0 200

1

A

The projections of two 3D points A and B on the image are given by a(120, 240) and

b(320, 240), respectively. Find the coordinates of the two 3D points A(X1, Y1, Z1) and

B(X2, Y2, Z2) that satisfy the 2 conditions: (1) the length of the line segment OA is given

by |OA| = 200, and (2) the line segments OA and AB are perpendicular to each other. [20

points]

1

A -

- x,

K- '

La, ] = xi (To) B -

- Kk' '

(4) = tyg )

I oat = T⇐¥+G = zoo

⇒ 2×2 = 40000 ⇒ X .

-

I look

Since A is in front M cameraits z value has to

te positive ⇒ a

=L:{I ) f' II

,

" →

onto AJ :O

[¥3 . L¥⇒= .

* ( ÷.

2. Model fitting: In Figure ??, you are given a set of 6 2D points (A,B,C,D,E, F ). We can

obtain a line equation by choosing 2 points at a time. Given two 2D points (x1, y1) and (x2, y2),

we can obtain the line equation using the expression (y � y1)(x1 � x2) = (x � x1)(y1 � y2).

The distance of a point (x0, y0) from a line ax + by + c = 0 is given by|ax0+by0+c|p

a2+b2. Let the

threshold in selecting the inliers for a given line equation is given by 0.6.

• (a) Find a pair of points that gives the line equation with the maximum number of

inliers. Show the line equations and inliers.

• (b) Find a point pair that gives the line equation with the minimum number of inliers.

Show the line equations and inliers.

[20 points]

Figure 1:

2

Best line ADy In heirs = { A ,B,E,D3

Worst line BE n=4 Inters = { BE }=

-

3. 3D Modeling: Let us consider a scenario where a 3D point Q is observed by two cameras.

Let the 2 camera matrices be given by:

K1 = K2 =

0

@200 0 3200 200 2400 0 1

1

A

Rotation matrices: R1 = R2 = I.

Translation matrices: t1 = 0, t2 = (100, 0, 0)T.

The corresponding 2D points on the images are given by:

q1 =

0

@5204401

1

Aq2 =

0

@3204401

1

A

Compute the 3D point Q. [20 points]

3

Kikki'=

9. =L !]+akT[orig a- [ '

E]tkkI[ oh]⇒=

BE (E) t' 49 )

II 00

itg= ( If )

Dist ( 9,1925 - Cd,-1005 -124

,-725

⇒j= 26,

- too ) +46,

- k ) = o

⇒X

,- too -127

,-27<=0

2DEr 3,1,

-2×2-100 = -

-⇒

2=4,4 ,- WC - D .

- o X,

-_ too

X. = ) a 4<=100

9=C=

4. Vocabulary tree: We are given three images I1, I2 and I3. Each of these images have two

2D descriptors as given below.

I1 : {(1.5, 0.5), (�1.5,�1.5)}, I2 : {(1.5, 1.5), (1.5,�1.5)}, I3 : {(1.5, 0.5), (1.5,�0.5)}

In Figure ??, we show a vocabulary tree with branch factor 4. Using this vocabulary tree

find the best image match among the 3 possible pairs {(I1, I2), (I1, I3), (I2, I3)}. Assume that

the nodes in the tree has the same weight wi = 1. Show the normalized di↵erence in each of

the three pairs. [20 points]

Figure 2:

4

- -

TL* . "I.

t '

1- I

of ,

= [ 2,

o 0012oooo ]

GE ( 2, lil 4°

o oooooo I

5- ( 3540€ ,

ooo

00003

Norm - Diff Car,Gd =

Iz - Iz = Iz =3

Norm - Diff ( 9, its ) =

§ - Iz = 12=2

Norm - Diff ( Kid -

- Byz- 93£ = Yee

5. Belief Propagation: You have 2 Boolean variables (x1, x2 2 {0, 1}) and 3 equations as

shown below:

x1 + 4x2 = 5

�x1 + 2x2 = 1

�3x1 + 2x2 = �1.2

Show the factor graph and use Belief propagation to solve the equations. Please show the

messages in each iteration till the algorithm terminates. [20 points]

5

y④-

D-④yes

a'E¥ n.IE# n Yeast.

in

absolute values for

⑨ It ⑦ the difference .

Steps All zero messages

mi→a=o

,mz→a=7M→E9Mz→E ,

MI → Eo ,Mz→c=o

- x -

Stere ma→ ,

=min [Mz→aCn£t

Coracoid

k Mua Capt a ached ) =L !]

man -

- It:3^

b - si =

main [ massacred teacher ]Most Cnd tqci.az )

=

Similar 's mz→=f ;) me . ,

-

- ( :3 ) meet :3)

Steps 4=1 ⇒ bit toSteph

u,

←I

,

Nz ← \

-

S :

m ,→a= ( :⇒ meet:)must = ( 2 meet be ⇒Mhc = ( 3) Mate -_ ( I )

Stere :

ma→= main LI I=c⇒

man -

- i.EE: ":D

⇐( hi:)man ⇐min:L :*:⇐⇐C⇒

an = I:c:=L

Steps :

me- ' '

= ( %) mc→z=( to:{ )-

G,

-

- C so:{ ) h -

- ( so:3 )Step se

,← I

,nz←l

, tTERME