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Level Set Formulationfor Curve Evolution
Ron Kimmel
www.cs.technion.ac.il/~ron
Computer Science Department Technion-Israel Institute of Technology
Geometric Image Processing Lab
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Consider a closed planar curve
The geometric trace of the curve can be alternatively represented implicitly as
Implicit representation21:)( RS pC
}0),(|),{( yxyxC
0
0
1
1
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Properties of level sets
The level set normal
Proof. Along the level sets we have zero change,
that is , but by the chain rule
So,
||
N
N
0s
Tyxyx sysxs
,),(
T
||||0,
||
NTT
||
T
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Properties of level sets
The level set curvature
Proof. zero change along the level sets, , also
So,
||
div
0ss
NTds
dT
ds
dyx
ds
dyx sysxss
,,,)(),(
...||
,||
,,||
,||
,,,,
||],,[||
||,
yxyx
syysxysxysxx
TT
yxyx
1
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Optical flow
Problem: find the velocity also known as `optical flow’It’s an `inverse’ problem,Given I(t) find
VII t
,
),( yxV
),( yxV
VtxItxI
dtVdtdtVOVdttxItxdtItxItxI
dtVOVdtdttxIdttxItxI
dttVdtxItxI
t
t
),,(),(
0),,(),,(),(),(),(
0)(),,(),(),(
0),(),(
2222
22
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Aperture Problem
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Aperture Problem
`Normal’ vertical flow
Horizontal flow can not be computed differentially.
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Normal flow
Due to the `aperture problem’only the `normal’ velocity can be locally computed
for the normal flow we have
V
||,
I
IVNNVNV NN
||||
,,, IVI
IIVNVIVII NNNt
NVN
|| IVI Nt
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Level Set Formulation
implicit representation of CThen,
Proof. By the chain rule
Then,
Recall that , and
RR 2:, yx 0),(:, yxyxC
dC d
VN Vdt dt
ttytx yxt
tyx
);,(
0
y
x
C(t)
C(t) level set 0
, ,x y t
x
y
NVNVCyx ttytxt
,,,
||
N
||||
,,
VVNV
|| Vt
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Level Set Formulation
Handles changes in topology
Numeric grid points never collide or drift apart.
Natural philosophy for dealing with gray level images.
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Numerical Considerations
Finite difference approximation. Order of approximation, truncation error, stencil. (Differential) conservation laws. Entropy condition and vanishing viscosity. Consistent, monotone, upwind scheme. CFL condition (stability examples)
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Numerical Considerations
Central derivative
Forward derivative
Backward derivative
)(ihuui )(xu
x
h
1 1 iii
h
uuuD ii
x 211
h
uuuD ii
x
1
h
uuuD ii
x1
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Truncation Error
Taylor expansion about x=ih
)()('
)()('
)()('
)()('')(')()(
)()('')(')()(
2
32!2
11
32!2
11
hOihuuD
hOihuuD
hOihuuD
hOihuhihhuihuhihuu
hOihuhihhuihuhihuu
ix
ix
ix
i
i
21
21
1 1
11
Stencils
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Numerical Approximations
2
1,11,11,11,1
4h
uuuuuD jijijiji
xy
211 2
h
uuuuD iii
xx
1 2- 1
41
41
41
41
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Conservation Law
Rate of change of the amount in a fixed domain G =
Flux across the boundaries of G
Differential conservation law
GG
dSnfudxdt
d ,
G
u
0 div fut
nnf
,
f
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Generalized Solution 1D
In 1D
Weak solution satisfies
u
0),(),(),(),(
0
0 div
1
0
1
0
1
0
1
0
1
0
1
0
0101
dttxftxfdxtxutxu
dtdxfu
dtdxfu
t
t
x
x
t
t
x
x
xt
t
t
x
x
t
)),(()),((),( 10
1
0
txuHtxuHdxtxudt
dx
x
t
x
1t
0t
0x 1x
ff u
u
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Hamilton-Jacobi
In 1D: HJ=Hyperbolic conservation lawsIn 2D: just the `flavor’…
Vanishing viscosity, of
The `entropy condition’ selected the `weak solution’
that is the `vanishing viscosity solution’ also known
as `entropy solution’.
xxxt uuHu )(0
lim
Nεκ
NCt
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Numerical Schemes
Conservation form
Numerical flux
The scheme is monotone, if F is non-decreasing.
Theorem: A monotone, consistent scheme, in conservation form converges to the entropy solution.
Yet, up to 1st order accurate ;-( …
x
gg
t
uu nj
nj
nj
nj
2
12
11
),...,( 11 n
qjn
pjnj uuFu
)(),...,( ),,...,( 1121 uHuuguugg qjpj
nj
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Upwind Monotone
Upwind scheme
For we have upwind-monotone schemes
we define Then, and the final scheme is
)()( 2uhuH
0' )(
0' )(
12
1HuH
HuHg
j
jnj
)))0,,((min(),(
)))0,max()0,((min(),(2
11
21
1
nj
nj
nj
njM
nj
nj
nj
njHJ
uuhuug
uuhuug
xxdtxutx ~),~(),(
1
nj
nj
nj
nj
Du
Du
),(1 nj
nj
nj
nj DDtg
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CFL Stability Condition
At the limitFor 3-point scheme of
we need for the numerical domain of dependenceto include the PDE domain of dependence
0)( xt uHu
0 0, tx t
x1x
tx ~,~
domain of dependence
0x
domain of influence
x̂
'1 Hx
t
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CFL Stability Condition
At the limitFor 3-point scheme of
we need for the numerical domain of dependenceto include the PDE domain of dependence
0)( xt uHu
0 0, tx t
x1x
tx ~,~
0x
'1 Hx
t
x
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1D Example
SolutionCharacteristics dx/dt=1CFL condition
Numeric scheme
xt uu
)0,(),( txutxu t
x
tx 0
0x
x
t
1
ni
xni
t uDuD
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1D Example
where
Characteristics
Numeric scheme
CFL condition
xt uxau )(
1 1
1 1)(
x
xxa
t
x
ta
x
)2()(
))0,min()0,(max(
112||
1121 n
ini
ni
ani
ni
axtn
ini
ni
xi
ni
xi
ni
t
uuuuuuu
uDauDauDii
1 1
1 1
x
x
dx
dt
1
xxxxi uauax ||||
Numerical viscosity
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2D Example
Numeric scheme
CFL condition
t
2
1
h
t
2,1,1
12
22
)0,,min(max)0,,max(
)0,,max()0,,max(
nji
nji
nijh
nij
xnij
x
nij
ynij
ynij
xnij
xnij
t
DD
DDDDD
tt NC
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2D Examples
Some flows
Vt
31
31
31 22
22
22
2div
2div
xyyyxxyyxxtt
yx
xyyyxxyyxxtt
tt
NC
NC
NC
require upwind/monotoneschemes
,2
div ,
22
22
gg
gNNggC
yx
xyyyxxyyxx
tt