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Inverse problems in earth observation and cartography
Shape modelling via higher-order active
contours and phase fields
Ian JermynJosiane Zerubia
Marie RocheryPeter Horvath
Ting PengAymen El Ghoul
2
Overview
Problem: entity extraction from (remote sensing) images. Need for prior ‘shape’ knowledge.
Modelling prior shape knowledge: higher-order active contours (HOACs). Two examples: networks (roads) and circles (tree crowns). Difficulties.
Phase fields: What are they and why use them? Phase field HOACs. Two examples: networks (roads) and circles (tree crowns).
Future.
3
Problem: entity extraction
Ubiquitous in image processing and computer vision: Find in the image the region occupied by particular
entities. E.g. for remote sensing: road network, tree crowns,…
4
Calculate a MAP estimate of the region:
In practice, minimize an energy:
Problem formulation
R = argmaxR
P(RjI ;K )
P(RjI ;K ) / P(I jR;K )P(RjK )
R = argminR
E (R;I )
E (R;I ) = ¡ lnP(I jR;K ) ¡ lnP(RjK )
= E I (I ;R) +EG (R) +const
5
Building EG: active contours
A region is represented by its boundary, R = [], the ‘contour’.
Standard prior energies: Length of R and area of R. Single integrals over the
region boundary: Short range dependencies. Boundary smoothness.
Insufficient prior knowledge.
RR
S1
R2
6
Difficulties
(Remote sensing) images are complex.
Regions of interest distinguished by their shape. But topology can be non-trivial, and
unknown a priori. Strong prior information about the
region needed, without constraining the topology.
7
Building a better EG: higher-order active contours
Introduce prior knowledge via long-range dependencies between tuples of points.
How? Multiple integrals over the contour. E.g. Euclidean invariant two-point term:
Dependent°(t); _°(t)°(t0); _°(t0)
E (°) = ¡ZZ
dt dt0 _°(t) ¢_°(t0) ª (j°(t) ¡ °(t0)j)
8
Prior for networks
Gradient descent with this energy. A perturbed circle evolves towards a structure
composed of arms joining at junctions.
EG (R) =¸L(@R) +®A(R)
¡¯2
ZZdt dt0 _°(t) ¢_°(t0) ª (j°(t) ¡ °(t0)j)
r
r)
9
Prior for a gas of circles
The same energy EG can model a ‘gas of circles’ for certain parameter ranges. Which ranges?
A circle should be a stable configuration of the energy (local minimum). Stability analysis.
11
Optimization problem
Minimize E(R, I) = EI(I, R) + EG(R). Algorithm: gradient descent using distance
function level sets. But the gradient of the HOAC term is non-local,
and requires: The extraction of the contour; Many integrations around the contour; Velocity extension.
16
Problems with HOACs
Modelling: Space of regions is complicated to express in the
contour representation. Probabilistic formulation is difficult.
Parameter and model learning are hampered.
Algorithm: Not enough topological freedom. Gradient descent is complex to implement for
higher-order terms. Slow.
Solution: ‘phase fields’.
17
Phase fields
Phase fields are a level set representation (z() = {x : (x) > z}), but the functions, , are unconstrained.
How do we know we are modelling regions?
R
R = 1
R = -1
E0(Á) =Z
d2x
½D2r Á¢r Á+ ¸(
14Á4 ¡
12Á2)
¾
ÁR = arg minÁ: ³ (Á)=R
E0(R)
18
One can show that
R is a minimum for fixed R. Thus gradient descent with E0 mimics
gradient descent with L: ‘valley following’.
Can also add odd potential term to mimic:
Relation to active contours
R1
R2R3
E0(ÁR ) ' ¸C L(@R)
E0(ÁR ) ' ¸C L(@R) +®CA(R)
19
Why use them?
Complex topologies are easily represented. Representation space is linear:
can be expressed, e.g., in wavelet basis for multiscale analysis of shape.
Probabilistic formulation (relatively) simple. Gradient descent is based solely on the PDE
arising from the energy functional: No reinitialization or ad hoc regularization. Implementation is simple and the algorithm is fast.
20
Why use them?
Neutral initialization: No initial region. No bias towards “interior” or
“exterior”. Greater topological freedom:
Can change number of connected components and number of handles without splitting or wrapping.
+
21
How to write HOACs as phase fields?
Use that rR is zero except near R, where it is proportional to the normal vector.
One can show that
EQ (R) = ¡¯C2
ZZdt dt0 _°(t) ¢GC (°(t);°(t0)) ¢_°(t0)
EN L (Á) = ¡¯2
ZZ
2d2x d2x0r Á(x) ¢G(x;x0) ¢r Á(x0)
EN L (ÁR ;¯ ;G) ' EQ (R;¯C ;G)
22
Phase fields : likelihood energies EI
r : normal vector to the contour. r¢r : boundary indicator. (1 § )/2 : characteristic function of the region
(+) or its complement (-). Using these elements, one can construct the
equivalents of active contour and HOAC likelihoods.
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Optimization problem
Minimize
Algorithm: gradient descent, but…
E (R;I ) = E I (I ;R) +EG (R)
= E I (I ;R) +E0(R) +EN L (R)
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Major advantage of phase fields for HOAC energies
Whereas, due to the multiple integrals, HOAC terms require Contour extraction, contour integrations, and force
extension, Phase field HOACs require only a
convolution:
±EN L±Á(x)
= ¯Z
d2x0r 2G(x ¡ x0)Á(x0)
29
Future
New prior models Directed and rectilinear networks (rivers, big
cities…). ‘Gas of rectangles’; Controlled perturbed circle.
Multiscale models; New algorithms (multiscale, stochastic,…); Parameter estimation; Higher-dimensions; …
31
Stability analysis
EC;g(°0+ ±°) =
"
¸C2¼r0+ ®C¼r20 ¡
¯C22¼
Z 2¼
0F00dp
#
+a0
"
¸C2¼+ ®C2¼r0 ¡¯C24¼
Z 2¼
0F10dp
#
+X
kjakj
2"
¸C¼r0k2+ ®C¼
¡¯C24¼
( Ã
2Z 2¼
0F20dp+
Z 2¼
0F21e
ir0kpdp
!
+ k
Ã
2ir0Z 2¼
0F23e
ir0kpdp
!
+k2Ã
r20
Z 2¼
0F24e
ir0kpdp
! ) #
32
HOACs: EI
Linear term that favours large gradients normal to contour.
Quadratic term that favours pairs of points with tangents and image gradients parallel or anti-parallel.
E I ;1(°) =Z
S1dt _°(t) £ r I (°(t))
tangent vectors
gradients
weighting function
E I ;2(°) = ¡ZZ
dt dt0 _° ¢_°0(r I ¢r I 0) ª (j°(t) ¡ °(t0)j)
34
Nonlinearity in the model, not the representation: example
Probability distribution P on R2. P pushes forward to Q on S1. If P is strongly peaked at r0 then
Q() ' P(r0, ). Gradient descent with –ln(P) on R2
mimics gradient descent with -ln(Q) on S1 (‘valley following’).
P (r;µ) = d2x e¡ E (r;µ)
£ exp ¡ ( r4
4 ¡ r22 )
35
Turing stability
To avoid decay of interior and exterior, we require stability of functions § = § 1: 2E/2 must be positive definite (E = E0 + ENL).
For prior terms, this is diagonal in the Fourier basis:
Gives condition on parameters. Better result would be existence and
uniqueness of R for ‘any’ R.
±2E±Á(k0)±Á(k)
(Á§ ) = ±(k;k0)nk2(D ¡ ¯G(k)) +2(¸ ¨ ®)
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