Post on 21-Jun-2020
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