Description
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Name: |
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Note the following class changes: |
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Prof. Gennert will be away the week of Nov 4. To make up for lost class time, we will meet as |
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follows: |
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Mon 28 |
Oct. |
4:30-7:20pm with breaks. HW #7 due. |
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Wed 30 |
Oct. |
4:30-5:50pm as usual. |
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Mon 4 Nov. |
No class. HW #8 due. |
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Wed 6 Nov. |
No class. |
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Mon 11 |
Nov. |
4:30-7:20pm with breaks. HW #9 due. |
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Wed 13 |
Nov. |
Resume usual schedule. |
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⃗⃗ |
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1. Singular Value Decomposition (4 pts): This is an important tool that allows us to solve |
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systems of equations |
⃗ ≈ 0, |
| ⃗| |
= 1. This is useful in camera calibration, where we get |
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such a system of equations and want a non-zero parameter vector ⃗. See Szeliski for an |
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introduction to SVD. |
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You can think of the SVD as follows: |
⃗ in the |
direction, scales it by |
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1. It takes the component of input |
and |
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1 |
direction. |
1 |
1 |
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outputs it in the |
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, |
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2. Repeat for all |
, and . |
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3. The output vector is the sum of all the contributions |
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= [ | |
] = |
1 |
[1 | |
1 |
0 |
]=[2 0], =[ |
] = |
1 |
[ |
1 |
1 |
] |
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], =[1 |
1 |
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a. Suppose that 2×2 matrix M has singular value decomposition |
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1 |
2 |
1 −1 |
0 |
0 1 1 |
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2 |
1 −1 |
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√2 |
1√2 |
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2 |
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Without computing M, predict the value of ⃗ for ⃗ = [1] and ⃗ = [−1] based on |
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1 |
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⃗, and finally, so does ⃗. Now compute M and use it to verify [1] and |
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the SVD properties. Hint: Note that in both cases |
⃗ has a simple form, so does |
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1 |
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[−1]. |
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b. Let matrix have singular value decomposition =. Show that matrix |
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has SVD= 2 |
. |
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Camera Calibration (10 pts): One restriction in camera calibration is that the world points
chosen must not lie in a single plane, that is, they cannot be ⃗co-,planar,1≤ ≤otherwise calibration
will fail. To see this, suppose that there are K world points. We know that we
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∙ ̃ = 0 (Szeliski, eqn.
2.7).
need at least 6 points for calibration, ≥ 6. Consider what happens if all world points lie in
Image Plane
Plane containing
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1
1
·
·
·
⃗⃗
The calibration equation is given by ⃗⃗⃗ = 0, written out as
1
0
0
0
0
0
1 1
1 1
1 1
1
1
1
1
1
1
2
0
0
0
1
1
⋮
]
0
⋮
0
⋯
1
1
1 1
⋮1 1
1 1
1 [
0
12
⃗⃗
⃗⃗
⃗⃗⃗ that we
Ideally, there should be a single vector ⃗⃗⃗ such that ⃗⃗⃗ = 0 (or ≈ 0). This is the
hope to find through the singular value decomposition of . The key concept here is the rank
columns. For this
⃗⃗⃗
to exist and be unique,
must have rank = 11 (or 12 with a very small
of a matrix, which is the number of independent rows, also the number of independent
value for the smallest singular value 12. Show that if all world points are co-planar, then
⃗⃗⃗cannot= 0⃗⃗have rank greater than 9 by finding 3 independent non-zero vectors ⃗⃗⃗ such that
. (These independent vectors establish that the nullspace of has rank at least 3;
hence ⃗⃗⃗’s=rank0⃗⃗ cannot exceed 12-3=9.) In this case, it is not possible to find a unique ⃗⃗⃗ such
that and the calibration ⃗⃗⃗ procedure fails.
Hint: The non ̃-zero∙ ̃ =vectors0 are mostly 0s. Use the fact that if all world points are co-planar, then .
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3. Focus of Expansion (8 pts): Suppose that the viewer (camera) is moving. We can model this |
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⃗ |
⃗ |
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in the imaging equations as |
⃗ |
⃗ |
⃗⃗ |
⃗ |
= |
2 |
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= |
+ and |
⃗ |
⃗ |
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⃗⃗ |
⃗⃗ |
⃗⃗ |
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∙ |
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by letting R and depend on time, = ( ), = ( ). Assume that the camera is |
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⃗⃗ |
⃗⃗ |
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translating with velocity |
≡ |
( ) and that there is no rotation, |
( ) = 0 . As we know |
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proving that image points appear to move toward or away from the FOE. What is k?
