CS 4495 Problem Set 3: Geometry solution

$29.99

Original Work ?
Category: You will Instantly receive a download link for .ZIP solution file upon Payment

Description

5/5 - (4 votes)

The past several lectures have dealt with the geometry of imaging. In this project you will epxlore
calibrating a camera with respect to 3D world coordinates as well as esitmating the relationship
between two camera views.
As a reminder as to what you hand in: A Zip le that has
1. Images (either as JPG or PNG or some other easy to recognize format) clearly labeled using
the convention PScounter.jpg
2. Code you used for each question. It should be clear how to run your code to generate the
results. Code should be in dierent folders for each main part with names like PS1-1-code.
For some parts  especially if using Matlab  the entire code might be just a few lines.
3. Finally a PDF le that shows all the results you need for the problem set. This will include
the images appropriately labeled so it is clear which section they are for and the small number
of written responses necessary to answer some of the questions. Also, for each main section,
if it is not obvious how to run your code please provide brief but clear instructions. If there
is no Zip le you will lose at least 50%!
This project uses les stored in the directory http://www.cc.gatech.edu/~afb/classes/CS4495-Fall2013/
ProblemSets/PS3.
1 Calibration
The les pts2d-pic_a.txt and pts3d.txt are a list of twenty 2D and 3D points of the image pic_a.jpg
The goal is to compute the projection matrix that goes from world 3D coordinates to 2D image
coordinates. Recall that using homogeneous coordinates the equation is:


u
v
1

 ‘


s ∗ u
s ∗ v
s

 =


m1,1 m1,2 m1,3 m1,4
m2,1 m2,2 m2,3 m2,4
m3,1 m3,2 m3,3 m3,4






X
Y
Z
1




Recall you solve for the 3×4 matrix M using either SVD to solve the homogeneous version of the
equations or by setting m4,4 to 1 and then using the a normal least squares method. Remember
that M is only known up to a scale factor.
To make sure that your code is correct, we are going to give you a set of normalized points in the
les ./pts2d-norm-pic_a.txt and ./pts3d-norm.txt. If you solve for M using all the points you
should get a matrix that is a scaled equivalent of the following:
1
MnormA =


−0.4583 0.2947 0.0139 −0.0040
0.0509 0.0546 0.5410 0.0524
−0.1090 −0.1784 0.0443 −0.5968


For example, this matrix will take the last normalized 3D point which is < 1.2323, 1.4421, 0.4506, 1.0 >
and will project it to the < u, v > of < 0.1419, −0.4518 > where we converted the homogeneous 2D
point < us, vs, s > to its inhomogeneous version by dividing by s (i.e. the real transformed pixel in
the image).
1.1 Create the least squares function that will solve for the 3×4 matrix MnormA given the normalized
2D and 3D lists, namely ./pts2d-norm-pic_a.txt and ./pts3d-norm.txt. Test it on the
normalized 3D points by multiplying those points by your M matrix and comparing the
resulting the normalized 2D points to the normalized 2D points given in the le. Remember
to divide by the homogeneous value to get an inhomogeneous point. You can do the comparison
by checking the residual between the predicted location of each test point using your equation
and the actual location given by the 2D input data. The residual is just the distance (square
root of the sum of squared dierences in u and v).
Output: code that does the solving, the matrix M you recovered from the normalized points,
the < u, v > projection of the last point given your M matrix, and the residual between that
projected location and the actual one given in the le.
Now you are ready to calibrate the cameras. Using the 3D and 2D point lists for the image, we’re
going to compute the camera projection matrix. To understand the eects of overconstraining the
system, you’re going to try using sets of 8, 12 and 16 points and then look at the residuals. To
debug your code you can used the normalized set from above but for the actual question you’ll need
to use the ./pts2d-pic_b.txt and ./pts3d.txt
1.2 For the three point set sizes k of 8, 12, and 16, repeat 10 times:
1. Randomly choose k points from the 2D list and their corresponding points in the 3D list.
2. Compute the projection matrix M on the chosen points.
3. Pick 4 points not in your set of k and compute the average residual.
4. Save the M that gives the lowest residual.
Output: code that does the computation, and the average residual for each trial of each k (so
that would be 10 x 3 = 30 numbers). Explain any dierence you see between the results for
the dierent k.
Output: Best M
Finally we can solve for the camera center in the world. Let us dene M as being made up of a 3×3
we’ll call Q and a 4th column will call m4:
M = [Q|m4]
2
From class we said that the center of the camera C could be found by:
C = −Q
−1m4
To debug your code: If you use you the normalized 3D points to get the MnormA given above you
would get a camera center of:
CnormA =< −1.5125, −2.3515, 0.2826 >
1.3 Given the best M from the last part, compute C.
Output: code that does the computation, and the location of the camera in real 3D world
coordinates.
2 Fundamental Matrix Estimation
We now wish to estimate the mapping of points in one image to lines in another by means of the
fundamental matrix. This will require you to use similar methods to those in Problem 1. We will
make use of the corresponding point locations listed in pts2d-pic_a.txt and pts2d-pic_b.txt.
Recall that the denition of the Fundamental Matrix is

u
0 v
0 1



f1,1 f1,2 f1,3
f2,1 f2,2 f2,3
f3,1 f3,2 f3,3




u
v
1

 = 0
Given corresponding points you get one equation per point pair. With 8 or more points you can
solve this. With more points (such as the 20 in the les) you solve using the the same least squares
method as in problem 1 above.
2.1 Create the least squares function that will solve for the 3×3 matrix F˜ that satises the epipolar
constraints dened by the sets of corresponding points. Solve this function to create your least
squares estimate of the 3×3 transform F˜.
Output: code that does the solving. The matrix F˜ generated from your least squares function.
2.2 The linear squares estimate of F˜ is full rank; however, the fundamental matrix is a rank 2
matrix. As such we must reduce its rank. In order to do this we can decompose F˜ using
singular value decomposition into the matrices UΣV
T = F˜. We can then estimate a rank 2
matrix by setting the smallest singular value in Σ to zero thus generating Σ
0
. The fundamental
matrix is then easily calculated as F = UΣ
0V
T
. Use the SVD function to do, well, the SVD.
Duh.
Output: Code and fundamental matrix F.
2.3 Now you can use your matrix F to estimate an epipolar line lb in image ‘b’ corresponding to
point pa in image ‘a’:
lb = F pa
3
Similarly, epipolar lines in image a corresponding to points in image b are related by the
transpose of F.
[Below is one way to draw the epipolar lines. The TAs may post additional ways. The key is
to be able to take the projective geometry form of the line and draw it in the image.]
The resulting lines li dened in homogeneous coordinates can not be drawn using standard
line functions, which take as input two points. In order to use such functions, we can nd the
intersection of a given line li with the image boundaries. If we dene the line lL to be the line
corresponding to the left hand side of the image and lR to be the line corresponding to the
right hand side of the image, we can nd the point Pi,L = li×lL and Pi,R = li×lR. We can now
plot the line running through the points Pi,L, Pi,R. However, we must rst have the equations
for lL and lR making use of the point-line duality, we know that lL = PUL ×PBL. Where PUL
is the point dening the upper left-hand corner of the image and PBL is the bottom left-hand
corner of the image.
An example of such an image is:
Output: Code to perform the estimation and line drawing. Images with the estimated epipolar
lines drawn on them.
***EXTRA CREDIT***
If you look at the results of the last section the epipolar lines are close, but not perfect. The problem
is that the oset and scale of the points is large and biased compared to some of the constants. To
x this, we are going to normalize the points.
In the 2D case this normalization is really easy. We want to construct transformations that make
the mean of the points zero and, optionally, scale them so that their magnitude is about 1.0 or some
other not too large number.


u
0
v
0
1

 =


s 0 0
0 s 0
0 0 1




1 0 −cu
0 1 −cv
0 0 1




u
v
1


4
The transform matrix T is the product of the scale and oset matrices. The cu, cv is just the mean.
To compute a scale s you could estimate the standard deviation after substracting the means. Or
you could nd the maximum absolute value. Then the scale factor s would be the reciprocal of
whatever estimate of the scale you are using.
2.4x Create a two matrics Ta and Tb for the set of points dened in the les ./pts2d-pic_a.txt
and ./pts2d-pic_b.txt respectively. Use these matrices to transform the two sets of points.
Then use these normalized points to create a new Fundamental matrix Fˆ. Compute it as
above including making the smaller singular value zero.
Output: The matrixes Ta, Tb and Fˆ
Finally you can create a better F by:
F = T
T
b F Tˆ
a
2.5x Using the new F redraw the epipolar lines of 2.3. They should be better.
Output: The new F and the images with the better epipolar lines drawn.
5