Parameter estimation class 5
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Transcript of Parameter estimation class 5
Parameter estimationclass 5
Multiple View GeometryCPSC 689
Slides modified from Marc Pollefeys’ Comp 290-089
Singular Value Decomposition
XXVT XVΣ T XVUΣ T
Homogeneous least-squares
TVUΣA
1X AXmin subject to nVX solution
Parameter estimation
• 2D homographyGiven a set of (xi,xi’), compute H (xi’=Hxi)
• 3D to 2D camera projectionGiven a set of (Xi,xi), compute P (xi=PXi)
• Fundamental matrixGiven a set of (xi,xi’), compute F (xi’TFxi=0)
• Trifocal tensor Given a set of (xi,xi’,xi”), compute T
Number of measurements required
• At least as many independent equations as degrees of freedom required
• Example:
Hxx'
11
λ
333231
232221
131211
y
x
hhh
hhh
hhh
y
x
2 independent equations / point8 degrees of freedom
4x2≥8
Approximate solutions
• Minimal solution4 points yield an exact solution for H
• More points• No exact solution, because
measurements are inexact (“noise”)• Search for “best” according to some
cost function• Algebraic or geometric/statistical cost
• X, X’ : measured value from the first image and the second image, respectively -- Known values with noise
• : estimated value – algorithm output
• : true value – (unknown).
Notations
'ˆ,ˆ XX
', XX
XHX '
Gold Standard algorithm
• Cost function that is optimal for some assumptions
• Computational algorithm that minimizes it is called “Gold Standard” algorithm
• Other algorithms can then be compared to it
Direct Linear Transformation(DLT)
ii Hxx 0Hxx ii
i
i
i
i
xh
xh
xh
Hx3
2
1
T
T
T
iiii
iiii
iiii
ii
yx
xw
wy
xhxh
xhxh
xhxh
Hxx12
31
23
TT
TT
TT
0
h
h
h
0xx
x0x
xx0
3
2
1
TTT
TTT
TTT
iiii
iiii
iiii
xy
xw
yw
Tiiii wyx ,,x
0hA i
Direct Linear Transformation(DLT)
• Equations are linear in h
0
h
h
h
0xx
x0x
xx0
3
2
1
TTT
TTT
TTT
iiii
iiii
iiii
xy
xw
yw
0AAA 321 iiiiii wyx
0hA i
• Only 2 out of 3 are linearly independent (indeed, 2 eq/pt)
0
h
h
h
x0x
xx0
3
2
1
TTT
TTT
iiii
iiii
xw
yw
(only drop third row if wi’≠0)• Holds for any homogeneous
representation, e.g. (xi’,yi’,1)
Direct Linear Transformation(DLT)
• Solving for H
0Ah 0h
A
A
A
A
4
3
2
1
size A is 8x9 or 12x9, but rank 8
Trivial solution is h=09T is not interesting
1-D null-space yields solution of interestpick for example the one with 1h
Direct Linear Transformation(DLT)
• Over-determined solution
No exact solution because of inexact measurementi.e. “noise”
0Ah 0h
A
A
A
n
2
1
Find approximate solution- Additional constraint needed to avoid 0, e.g.
- not possible, so minimize
1h Ah0Ah
DLT algorithmObjective
Given n≥4 2D to 2D point correspondences {xi↔xi’}, determine the 2D homography matrix H such that xi’=Hxi
Algorithm
(i) For each correspondence xi ↔xi’ compute Ai. Usually only two first rows needed.
(ii) Assemble n 2x9 matrices Ai into a single 2nx9 matrix A
(iii) Obtain SVD of A. Solution for h is last column of V
(iv) Determine H from h
Inhomogeneous solution
'
'h~
''000'''
'''''000
ii
ii
iiiiiiiiii
iiiiiiiiii
xw
yw
xyxxwwwywx
yyyxwwwywx
Since h can only be computed up to scale, pick hj=1, e.g. h9=1, and solve for 8-vector
h~
Solve using Gaussian elimination (4 points) or using linear least-squares (more than 4 points)
However, if h9=0 this approach fails also poor results if h9 close to zeroTherefore, not recommendedNote h9=H33=0 if origin is mapped to infinity
0
1
0
0
H100Hxl 0
T
Degenerate configurations
x4
x1
x3
x2
x4
x1
x3
x2H? H’?
x1
x3
x2
x4
0Hxx iiConstraints: i=1,2,3,4
TlxH 4* Define:
4444* xxlxxH kT
3,2,1 ,0xlxxH 4* iii
TThen,
H* is rank-1 matrix and thus not a homography
(case A) (case B)
If H* is unique solution, then no homography mapping xi→xi’(case B)
If further solution H exist, then also αH*+βH (case A) (2-D null-space in stead of 1-D null-space)
Solutions from lines, etc.
ii lHl T 0Ah
2D homographies from 2D lines
Minimum of 4 lines
Minimum of 5 points or 5 planes
3D Homographies (15 dof)
2D affinities (6 dof)
Minimum of 3 points or lines
Conic provides 5 constraints
Mixed configurations?
Cost functions
• Algebraic distance• Geometric distance• Reprojection error
• Comparison• Geometric interpretation• Sampson error
Algebraic distanceAhDLT minimizes
Ahe residual vector
ie partial vector for each (xi↔xi’)algebraic error vector
2
22alg h
x0x
xx0Hx,x
TTT
TTT
iiii
iiiiiii
xw
ywed
algebraic distance
22
21
221alg x,x aad where 21
T321 xx,,a aaa
2222alg AhHx,x eed
ii
iii
Not geometrically/statistically meaningfull, but given good normalization it works fine and is very fast (use for initialization)
Geometric distancemeasured coordinatesestimated coordinatestrue coordinates
xxx
2
HxH,xargminH ii
i
d Error in one image
e.g. calibration pattern
221-
HHx,xxH,xargminH iiii
i
dd Symmetric transfer error
d(.,.) Euclidean distance (in image)
ii
iiiii
ii ddii
xHx subject to
x,xx,xargminx,x,H 22
x,xH,
Reprojection error
Comparison of geometric and algebraic distances
iiii
iiiiii wxxw
ywwye
ˆˆ
ˆˆhA
Error in one image
Tiiii wyx ,,x xHˆ,ˆ,ˆx Tiiii wyx
3
2
1
h
h
h
x0x
xx0TTT
TTT
iiii
iiii
xw
yw
222iialg ˆˆˆˆx,x iiiiiiii wxxwywwyd
ii
iiiiiiii
wwd
wxwxwywyd
ˆ/x,x
/ˆ/ˆˆ/ˆ/x,x
iialg
2/1222ii
typical, but not , except for affinities 1iw i3xhˆ iw
For affinities DLT can minimize geometric distance
Possibility for iterative algorithm
represents 2 quadrics in 4 (quadratic in X)
Geometric interpretation of reprojection error
Estimating homography~fit surface to points X=(x,y,x’,y’)T in 4
0Hxx ii
Hν
22
22222
ii
x,xx,x
ˆˆˆˆXX
iiii
iiiiiiii
dd
yyxxyyxx
2H
22 ,Xx,xx,x iiiii ddd
Analog to conic fitting
CxxC,x T2alg d
2C,xd
Sampson error
2XX
HνVector that minimizes the geometric error is the closest point on the variety to the measurement
XX
between algebraic and geometric error
XSampson error: 1st order approximation of 0XAh H C
XH
HXH δX
XδX
C
CC XXδX 0XH C
0δX
X XH
H
C
C eXJδ
Find the vector that minimizes subject to
Xδ eXJδXδ
XJ Hwith
C
Find the vector that minimizes subject to
Xδ eXJδXδ
Use Lagrange multipliers:
Lagrange:
derivatives
0Jδ2λ-δδ XXX eT
TT 0J2λ-δ2 X 0Jδ2 X e
λJδ XT
0λJJ T e
e1TJJλ
e1TT
X JJJδ
XδXX ee1TT
XTX
2
X JJδδδ
Sampson error
2XX
HνVector that minimizes the geometric error is the closest point on the variety to the measurement
XX
between algebraic and geometric error
XSampson error: 1st order approximation of 0XAh H C
XH
HXH δX
XδX
C
CC XXδX 0XH C
0δX
X XH
H
C
C eXJδ
Find the vector that minimizes subject to
Xδ eXJδXδ
ee1TT
XTX
2
X JJδδδ
(Sampson error)
Sampson approximation
A few points
(i) For a 2D homography X=(x,y,x’,y’) (ii) is the algebraic error vector(iii) is a 2x4 matrix,
e.g.
(iv) Similar to algebraic error in fact, same as Mahalanobis distance
(v) Sampson error independent of linear reparametrization (cancels out in between e and J)
(vi) Must be summed for all points(vii) Close to geometric error, but much fewer
unknowns
ee1TT2
X JJδ
eee T2
2
JJT e
31213T2T
11 /hxhx hyhwxywJ iiiiii XJ H
C XHCe
ee
1TT JJ
Statistical cost function and Maximum Likelihood
Estimation• Optimal cost function related to noise
model• Assume zero-mean isotropic Gaussian
noise (assume outliers removed) 22 2/xx,2πσ2
1xPr de
22i 2xH,x /
2iπσ2
1H|xPr ide
i
constantxH,xH|xPrlog 2i2i
σ2
1id
Error in one image
Maximum Likelihood Estimate
2i xH,x id
Statistical cost function and Maximum Likelihood
Estimation• Optimal cost function related to noise
model• Assume zero-mean isotropic Gaussian
noise (assume outliers removed) 22 2/xx,2πσ2
1xPr de
22i
2i 2xH,xx,x /
2iπσ2
1H|xPr
ii dd
ei
Error in both images
Maximum Likelihood Estimate
2i
2i x,xx,x ii dd