-7-11/ XII 2009 1 Statistical Inference for Image Symmetries - PowerPoint PPT Presentation

XXXV Konferencja Statystyka Matematyczna -7-11/ XII 2009 1

Statistical Inference for Image Symmetries Mirek Pawlak pawlak@ee.umanitoba.ca 2

OUTLINE I Problem Statement II Image Representation in the Radial Basis Domain III Semiparametric Inference for Image Symmetry 3

IV Testing for Image Symmetries* • Testing Rotational Symmetries* • Testing Radiality • Testing Reflection and Joint Symmetries V References 4

I Problem Statement 5

• • • • ( ) + ε ij Z ij = f x i , y j 6

D Δ p ij • ( ) + ε ij Z ij = f x i , y j # of data points ∝ n 2 ; Δ ∝ n − 1 7

( ) . Given Problem 1 : Let f ∈ L 2 D ( ) + ε ij , 1 ≤ i , j ≤ n Z ij = f x i , y j and knowing that ( ) ( ) + y sin 2 β ∗ ( ) , x sin 2 β ∗ ( ) − y cos 2 β ∗ ( ) ( ) = f x cos 2 β ∗ f x , y some β ∗ ∈ 0, π [ ) , estimate β ∗ . 8

( ) . Given Problem 1 : Let f ∈ L 2 D ( ) + ε ij , 1 ≤ i , j ≤ n Z ij = f x i , y j and knowing that ( ) ( ) + y sin 2 β ∗ ( ) , x sin 2 β ∗ ( ) − y cos 2 β ∗ ( ) ( ) = f x cos 2 β ∗ f x , y some β ∗ ∈ 0, π [ ) , estimate β ∗ . 9

( ) . Given Problem 2 : Let f ∈ L 2 D ( ) + ε ij , 1 ≤ i , j ≤ n Z ij = f x i , y j verify whether the null hypothesis ( ) ≡ Sf ( ) x , y ( ) H S : f x , y is true or not ( ) x , y ( ) : Reflectional Symmetry, Rotational Symmetry Sf 10

d = 3 d = 8 d = 4 d = 2 An image becomes invariant under rotations through an angle 2 π d d = ∞ Radially Symmetric Objects ( ) x 2 + y 2 ( ) = g f x , y 11

Symmetry → Hermann Weyl, Symmetry, Princenton Univ. Press, 1952. → J.Rosen, Symmetry in Science: An Introduction to the General Theory , Springer, 1995. → J.H. Conway, H. Burgiel, and C. Goodman-Strauss, The Symmetry of Things , A K Peters, 2008. → M. Livio, The Equation That Couldn’t Be Solved: How Mathematical Genius Discovered the Language of Symmetry , Simon & Schuster, 2006. “...Livio writes passionately about the role of symmetry in human perception, arts,....” 12

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II Image Representation in the Radial Basis Domain 16

Radial Moments and Expansions (Zernike Basis) ( ) = ( ) ( ) dxdy ∫∫ ∗ x , y • A pq f f x , y V pq D Degree 2 π ( ) ( ) e − iq θ ρ d ρ d θ 1 ∫ ∫ = f ρ , θ R pq ρ 0 0 Angular dependence Bhatia & Born: “On circle polynomials of Zernike and related orthogonal sets”. Proc. Cambr. Phil. Soc . 1954 ∞ p + 1 ∑ ∑ ( ) ∍ f x , y ( )  ( ) V pq x , y ( ) L 2 D A pq f • π p = 0 q ≤ p 17

Invariant Properties → Reflection ( ) x , y ( ) τ β f ( ) f x , y ( ) = A pq ( ) e − i 2 q β ( ) ∗ A pq τ β f f A pq f 18

→ Rotation ( ) x , y ( ) ( ) r α f f x , y ( ) = A pq f ( ) ( ) e − iq α A pq f A pq r α f 19

( ) from noisy data A pq f D Δ p ij • ( ) + ε ij Z ij = f x i , y j ( ) Z ij ∑ ( ) = ˆ A pq f w pq x i , y j ( ) ∈ D x i , y j ( ) = ( ) ( ) dxdy ∫∫ ∗ x , y ∗ x i , y j ≈ Δ 2 V pq w pq x i , y j V pq p ij 20

( ) + D pq Δ ( ) + G pq Δ ( ) E ˆ A pq = A pq f ( ) = O Δ ( ) D pq Δ ( ) ( ) = O Δ γ G pq Δ 1 < γ = 285 208 *Gauss lattice points problem of a circle * 21

III Semiparametric Inference for Image Symmetry ( ) , Estimation of symmetry parameters: the axis of symmetry β the degree of rotational symmetry ( d ) of nonparametric image ( ) , f ∈ L 2 D ( ) . function: θ = θ f d ∗ β ∗ 22

Semiparametric Inference on the Axis of Reflectional Symmetry ( ) = e − 2 iq β A p , − q f ( ) A pq τ β f β • 23

• Contrast Function p + 1 p N ∑ ∑ ( ) = ( ) − e − 2 iq β A p , − q f ( ) 2 M N β , f A pq f π p = 0 q = − p N ∞ p + 1 p ∑ ∑ ( ) = ( ) − e − 2 iq β A p , − q f ( ) 2 M β , f A pq f π p = 0 q = − p 2 = f − τ β f 24

• Assumption ( ) is invariant under some unique reflection Suppose that f ∈ L 2 D τ β ∗ Hence, if τ β ∗ f = f then we have ( ) = 0 M β ∗ , f or ( ) = 0 for all N M N β ∗ , f 25

• Uniqueness of β ∗ Lemma U : Under Assumption the solution of ( ) = 0 M N β , f is uniquely determined and must equal β ∗ if we choose N so { } for which the gcd of ( ) contains A pq f ( ) ≠ 0 large that M N β , f q ’s is 1. { } , p i ≤ N , i = 1,.., r ( ) ≠ 0,..., A p r q r ( ) ≠ 0 A p 1 q 1 f f ( ) = 1 . gcd q 1 ,..., q r 26

β * = 120  27

( ) based only on A 42 f ( ) , A 86 f ( ) M N β , f 28

( ) based only on A 51 f ( ) , A 86 f ( ) M N β , f 29

p + 1 p 14 ∑ ∑ ( ) = ( ) − e − 2 iq β A p , − q f ( ) 2 M 14 β , f A pq f π p = 0 q = − p 30

( ) = A pq f ( ) e ( ) ir pq f A pq f p + 1 ( ) p ( ) N 2 1 − cos 2 r pq f ∑ ∑ ( ) = ( ) ( ) + 2 q β ⇒ M N β , f 4 A pq f π p = 0 q = − p ⇒ β ∗ : ( ) + q β ∗ = l π r pq f ( ) with A pq f ( ) ≠ 0 . for all p , q 31

• Estimated Contrast Function p + 1 p N ∑ ∑ ( ) = 2 A pq − e − 2 iq β ˆ ˆ ˆ M N β A p , − q π p = 0 q = − p p + 1 ( ) p ( ) N 2 1 − cos 2ˆ ∑ ∑ 4 ˆ r pq + 2 q β A pq = π p = 0 q = − p ( ) ˆ β Δ , N = argmin β ∈ 0, π ) ˆ M N β [ → van der Vaart: Asymptotic Statistics , 1998. 32

• Accuracy of ˆ β Δ , N Theorem 3.1 ( ) and be reflection invariant w.r.t. a unique axis Let f ∈ BV D β ∗ ∈ 0, π [ ) . Let N be so large that β ∗ is determined as the unique ( ) . Then zero of M N β , f ˆ β Δ , N → β ∗ (P) as Δ → 0 ( ) − M N β , f ( ) → 0 (P) M N β ˆ Proof : sup β ∈ 0, π [ ) 33

Theorem 3.2 ( ) β Δ , N = β ∗ + O P Δ ( ) → n 2 rate ← ˆ Proof : ( ) = ˆ ( ) ( ) ˆ ( ) + ˆ ( ) β ∗ ( )  ( ) ˆ β Δ , N − β ∗ 0 = ˆ β Δ , N β 1 1 2 • M N M N M N ( ) ( ) β ∗ ˆ 1 M N β Δ , N − β ∗ = − ˆ ( ) • ( )  ˆ β 2 M N 34

• From Theorem 3.1 : ( ) β ( ) β , f ( ) − M N ( ) → 0 (P) ˆ 2 2 sup β ∈ 0, π M N [ )  β → β ∗ (P) ( ) → M N ( ) ( ) β ∗ , f ( )  ˆ β 2 2 M N p + 1 p N ∑ ∑ ( ) 16 q 2 A pq f 2 ≠ 0 = π p = 0 q = − p 35

( ) ( ) β ∗ Δ − 1 ˆ ( ) ≈ − 1 M N • Δ − 1 ˆ β Δ , N − β ∗ ( ) ( ) β ∗ , f 2 M N p + 1 ( ) = p ( ) N 2 sin 2 ˆ ( ) β ∗ ∑ ∑ • ˆ 8 q ˆ r pq + 2 q β ∗ 1 M N A pq π p = 0 q = − p ( ) ( ) + q β ∗ = l π ← → r pq f p + 1 ( ) p ( ) N 2 sin 2 ˆ ∑ ∑ ( ) 8 q ˆ r pq − r pq f A pq = π p = 0 q = − p 36

p + 1 ( ) p ( ) N 2 sin 2 ˆ ∑ ∑ ( ) ( ) ≈ r pq − r pq f 8 q A pq f π p = 0 q = − p ( ) + O P Δ ( ) • ˆ A pq = A pq f ( ) = O P Δ ( ) r pq − r pq f • ˆ 37

Theorem 3.3 Let f be Lip(1). E ε 4 < ∞ . ⎛ ⎞ ( ) ⇒ N 0, 8 σ 2 Δ − 1 ˆ β Δ , N − β ∗ ⎜ ⎟ ( ) ( ) β ∗ , f 2 ⎝ ⎠ M N p + 1 ( ) = p N ( ) β ∗ , f ∑ ∑ ( ) 16 q 2 A pq f 2 2 M N π p = 0 q = − p 38

⇒ Confidence Interval for β ∗ ⎡ ⎤ ) 2 2 Δ ˆ σ ) 2 2 Δ ˆ σ ( ( β Δ , N − Φ − 1 1 − α β Δ , N + Φ − 1 1 − α ˆ ( ) , ˆ ⎢ ⎥ ( ) ⎢ ⎥ ˆ ˆ 2 2 M N M N ⎣ ⎦ p + 1 p N ( ) = ∑ ∑ 2 16 q 2 ˆ • ˆ 2 M N A pq π p = 0 q = − p { } 2 + Z i , j − Z i , j + 1 ( ) ( ) σ 2 = 1 ∑ 2 Z i , j − Z i + 1, j • ˆ ( ) 4 C Δ ( ) , x i + 1 , y j ( ) , x i , y j + 1 ( ) ∈ D x i , y j 39

Remark 1: f is not reflection symmetric β Δ , N → β  = argmin β f − τ β f 2 ˆ 40

( ) for images invariant under { } Remark 2: Estimation d ∈ 2,3,4,.... rotations through an angle 2 π d ? 41

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IV Testing for Image Symmetries Testing Rotational Symmetries d = 2 - testing image symmetry w.r.t. rotation through π ( ) = e 2 π iq / d A pq f ( ) → A pq r ( ) = e π iq A pq f ( ) A pq f d f 44

The orthogonal projection ∞ p + 1 2 = 1 − e 2 π iq / d 2 A pq f ∑ ∑ ( ) 2 f − r d f π p = 0 q ≤ p The test statistic (d=2) p + 1 N r 2 = ∑ ∑ 2 ˆ T N A pq π p = 0 q ≤ p p = odd 45

Theorem 4.1: A: Under H r 2 : r 2 f = f let N → ∞ , N 7 Δ → 0 , as Δ → 0 Then for p + 1 N r 2 = ∑ ∑ 2 ˆ T N A pq π p = 0 q ≤ p p = odd we have r 2 − σ 2 Δ 2 a N ( ) ( ) ( ) / 4 − N even T N ⎧ N N + 2 ⇒ N 0,2 σ 4 ( ) = ⎨ ( ) a N Δ 2 a N ( ) N + 3 ( ) / 4 − N odd N + 1 ⎩ 46

( ) , s ≥ 2 and let 2 f ≠ f let f ∈ C s D B: Under H r 2 : r N → ∞ , N 3/2 Δ γ − 1 → 0 , N 2 s + 1 Δ → ∞ as Δ → 0 Then we have r 2 − f − r 2 / 4 ( ) T N 2 f ⇒ N 0, σ 2 f − r 2 2 f Δ 47

Proof : • Under H r 2 : T N r 2 is a quadratic form of iid random variables (*de Jong: A CLT for generalized quadratic forms, Pr.Th.Related Fields , 1987*) • Under H r 2 : T N r 2 is dominated by a linear term 48

Similar results exist for • d = 4 - testing image symmetry w.r.t. rotation by π / 2 ( ) = g ρ ( ) • d = ∞ - testing image radiality, i.e., f x , y • Testing image mirror symmetry • Joint hypothesis τ y , r 2 : r 2 f = f AND τ y f = f H τ y , r 2 : r 2 f ≠ f OR τ y f ≠ f H 49