Phase diagram of a higher-order active contour energy Aymen El - PowerPoint PPT Presentation

Outline Phase diagram of a higher-order active contour energy Aymen El Ghoul : Ariana (INRIA/I3S), URISA (Sup’Com Tunis) Josiane Zerubia and Ian Jermyn : ARIANA (INRIA, I3S) Ziad Belhadj : URISA (Sup’Com Tunis) July 26, 2007

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Outline Motivations 1 Region modelling Problem statement 2 Stability analysis of a circle Variational study Stability conditions Stability analysis of a long bar 3 Variational study Stability conditions Experimental results Summary and future work 4

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Outline Motivations 1 Region modelling Problem statement 2 Stability analysis of a circle Variational study Stability conditions Stability analysis of a long bar 3 Variational study Stability conditions Experimental results Summary and future work 4

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Region modelling Probabilistic framework - variational approach Problem Find the region R in the image domain containing a particular entity (road network, tree crowns. . . ) Probabilistic framework MAP estimate: ˆ R = arg max R ∈ R P ( R | I , K ) Using Bayes’s theorem: ˆ R = arg max R ∈ R P ( I | R , K ) P ( R | K ) Variational approach based on energy minimization The probabilities can be written in terms of ‘energies’: exp ( − E ) Energy minimization: � � R = arg min R ∈ R ˆ E image ( I | R , K ) + E prior ( R | K )

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Region modelling HOAC model, [Rochery et al. 06] E g ( γ ) = λ C L ( γ ) + α C A ( γ ) � �� � Snakes [Kass et al. 1988] �� − β C dt dt ′ ˙ γ ( t ′ ) · ˙ γ ( t ) Φ( | γ ( t ) − γ ( t ′ ) | ) 2 1 0.8 Φ (R) ε 0.6 d 0.4 0.2 0 0 1 2 3 4 5 6 7 Distance entre deux points R = | γ (t)− γ (t’)|

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Problem statement Problem statement

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Problem statement Scaling The model has 5 parameters ( λ C , α C , β C , d , ǫ ), so it is difficult to analyse. Scaling: We take ǫ = d . The interaction function can be written as a γ = γ function of ˆ d , α = α C d β = β C d Φ( z ) = Φ( z λ C , ˆ λ C and ˆ Let ˆ d ) . The HOAC model becomes: ˆ �� β dt dt ′ ˙ E g (ˆ γ ) = L (ˆ α A (ˆ γ ( t ′ ) · ˙ γ ( t ) ˆ γ ( t ) − ˆ γ ( t ′ ) | ) , ˆ γ ) + ˆ γ ) − ˆ ˆ Φ( | ˆ 2 E g depends only on the two scaled parameters ˆ ˆ α and ˆ β . Stability analysis (Taylor series expansion): E g ( γ ) = E g ( γ 0 + δγ ) = E g ( γ 0 ) + � δγ | δ E g 2 � δγ | δ 2 E g δγ � γ 0 + 1 δγ 2 | δγ � γ 0

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Outline Motivations 1 Region modelling Problem statement 2 Stability analysis of a circle Variational study Stability conditions Stability analysis of a long bar 3 Variational study Stability conditions Experimental results Summary and future work 4

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Variational study Parametrization Circle parametrization γ 0 ( t ) = ( r 0 ( t ) , θ 0 ( t )) = ( r 0 , t ) where t ∈ [ − π, π [ γ ( t ) = γ 0 ( t ) + δγ ( t ) = ( r ( t ) , θ ( t )) = ( r 0 + δ r ( t ) , θ 0 ( t ) + δθ ( t )) The energy E g is invariant to tangential changes so δθ ( t ) = 0. Perturbation definition The operator F = δ 2 E g δγ 2 is invariant to ‘translation’ on the circle, So the Fourier basis diagonalizes the operator F, So, perturbations are given in terms of Fourier coefficients k a k e ir 0 kt where k = m by : δ r ( t ) = � r 0 and m ∈ Z .

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Variational study Energy of a perturbed circle [Horvath et al. 06] E g ( r ) = E g ( r 0 + δ r ) = e 0 ( r 0 ) + a 0 e 1 ( r 0 ) + 1 � | a k | 2 e 2 ( k , r 0 ) . 2 k

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Variational study Circle energy e 0 20 14 10 18 12 8 16 10 14 6 12 8 e 0 10 e 0 e 0 4 6 8 2 6 4 4 0 2 2 0 0 −2 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 r 0 r 0 r 0 α C = 1, β C = 1 . 54 α C = 1, β C = 1 . 84 α C = 1, β C = 2 . 03 30 35 5 30 0 25 25 −5 20 20 −10 e 0 15 e 0 e 0 15 −15 10 10 −20 5 5 −25 0 0 −30 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 r 0 r 0 r 0 α C = 1, β C = 1 . 10 α C = 1, β C = 0 . 88 α C = − 1, β C = 1

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Stability conditions First stability condition: e 1 = 0 β C ( λ C , α C , r 0 ) = λ C + α C r 0 r 0 α ˆ r 0 ) = 1 + ˆ ⇒ ˆ α, ˆ ⇐ β (ˆ G 10 ( r 0 ) G 10 (ˆ ˆ r 0 ) 0.95 0.9 0.85 0.8 r 0 0.75 i ,r 0 i ) ( β C 0.7 0.65 0.6 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45 β C

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Stability conditions Second stability condition: e 2 positive definite 16 250 14 200 12 150 10 e 0 8 e 2 100 6 50 4 0 2 0 −50 0 0.5 1 1.5 2 2.5 3 0 1 2 3 4 5 6 7 8 r 0 r 0 k α C = 1, r ∗ e 2 ( α C , k , r ∗ 0 = 1 0 ) > 0 e 2 ( α C , k , r 0 ) > 0 , ∀ k � = 1 r 0 , k ) , ∀ k � = 1 α a (ˆ r 0 , k ) > f (ˆ ⇐ ⇒ ˆ . r 0 r 0

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Stability conditions Bounds on the parameter ˆ α 2 3 1.5 2 1 1 0.5 0 0 a f −0.5 −1 −1 −2 −1.5 −3 −2 0 0.5 1 1.5 2 2.5 0 0.5 1 1.5 2 2.5 ˆ ˆ r r 0 0 Graph of a Graph of f 6 8 6 4 4 2 2 0 0 α ˆ fa −2 −2 −4 −4 −6 −6 −8 −8 −10 0 0.5 1 1.5 2 2.5 0 0.5 1 1.5 2 2.5 ˆ r r ˆ 0 0 Graph of f / a Bounds of ˆ α

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Stability conditions Circle phase diagram 7 6 5 asinh ( β C ) 4 3 2 1 0 0 1 2 3 4 5 6 7 asinh ( α C ) r 0 < ∞ 0 . 69 < ˆ

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Outline Motivations 1 Region modelling Problem statement 2 Stability analysis of a circle Variational study Stability conditions Stability analysis of a long bar 3 Variational study Stability conditions Experimental results Summary and future work 4

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Variational study Bar parametrization µ = 1 −1\2 1\2 µ = 2 � x 0 ,µ ( t µ ) = ± µ l t µ , t µ ∈ [ − 0 . 5 , 0 . 5 ] γ 0 ,µ ( t µ ) = w 0 y 0 ,µ ( t µ ) = ± µ 2 where � + 1 si µ = 1 ± µ = − 1 si µ = 2

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Variational study Perturbation definition δγ µ ( t µ ) = ( δ x µ ( t µ ) , δ y µ ( t µ )) . Tangential perturbations make no difference: δ x µ ( t µ ) = 0, Perturbations are given in terms of Fourier coefficients (bar k µ a µ, k µ e ik µ lt µ ‘translation’ invariance) by: δ y µ ( t µ ) = � where k µ = 2 π m µ , m µ ∈ Z , l Contour expression: γ µ ( t µ ) = γ 0 ,µ ( t µ ) + δγ µ ( t µ ) � x µ ( t µ ) = ± µ l t µ = w 0 2 + � k µ a µ, k µ e ik µ lt µ . y µ ( t µ ) = ± µ

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Variational study Energy of a perturbed long bar E g ( γ ) � � = e 0 ( w 0 ) + a 1 , 0 − a 2 , 0 e 1 ( w 0 ) l � | a 1 , k | 2 + | a 2 , k | 2 � + 1 � e 20 + ( a 1 , k a 2 , k + a 1 , − k a 2 , − k ) e 21 2 k + 1 � � � k e 2 a t = e 0 + e 1 a 1 , 0 − a 2 , 0 a ∗ k . 2 k � e 20 e 21 � � � a ∗ a 2 , k a k = , e 2 = . 1 , k e 21 e 20

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Variational study Bar energy e 0 2.8 2.5 2.5 2.7 2 2 2.6 1.5 2.5 1.5 1 e 0 2.4 e 0 e 0 0.5 1 2.3 0 2.2 0.5 −0.5 2.1 2 0 −1 0 0.5 1 1.5 2 0 0.5 1 1.5 2 0 0.5 1 1.5 2 w 0 w 0 w 0 α C = 1, β C = 0 . 66 α C = 1, β C = 1 . 84 α C = 1, β C = 2 . 47 3 3.5 2 2.9 1.9 2.8 2.7 1.8 3 2.6 1.7 e 0 e 0 e 0 2.5 1.6 2.4 2.5 2.3 1.5 2.2 1.4 2.1 2 2 1.3 0 0.5 1 1.5 2 0 0.5 1 1.5 2 0 0.5 1 1.5 2 w 0 w 0 w 0 α C = 1, β C = 0 . 52 α C = 1, β C = 0 . 35 α C = − 1, β C = − 0 . 66

Motivations Stability analysis of a circle Stability analysis of a long bar Summary and future work Stability conditions First stability condition: e 1 = 0 α C α ˆ β C ( λ C , α C , w 0 ) = ⇒ ˆ w 0 ) = α, ˆ G 10 ( w 0 ) ⇐ β (ˆ G 10 ( ˆ w 0 ) ˆ 1.3 1.2 1.1 1 w 0 0.9 i ,w 0 i ) ( β C 0.8 0.7 0.6 0.5 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 β c