Dense Correspondence and Its Applications - Computer Vision for Visual Effects - page 158

Without any other information, this is the only part of the optical flowfield that can
be obtained. This makes sense because the problem is inherently underconstrained:
wewant to estimate two unknowns ( u and v ) at eachpixel, but only have one equation
( 5.16 ). Therefore, we must make additional assumptions to resolve the remaining
degree of freedom at each pixel.
The most natural assumption is that the optical flow field varies smoothly across
the image; that is, neighboring pixels should have similar flow vectors. Horn and
Schunck phrased this constraint by requiring that the gradient magnitude of the flow
field, namely the quantity

∂ 2 ∂ 2 ∂ 2 ∂ 2

u u v v

+ + + (5.20)

∂ x ∂ y ∂ x ∂ y

should be small. Overall, this leads to an energy function to be minimized over the
flow fields u ( x , y ) and v ( x , y ) :

∂

2

∂ 2 2

+

∂ 2

+

∂ 2

+

∂

I

+ ∂

I

+ ∂

I u u v v

E HS (

) = + λ

u , v

x u y v ∂ ∂ ∂ t ∂ x ∂ y ∂ x ∂ y

(5.21)

x , y

= E data ( u , v ) + λ E smoothness ( u , v )

where is a parameter that specifies the influence of the smoothness term, also

known as a regularization parameter. The larger the value of

λ

, the smoother the
optical flowfield. A largeweight on the regularizer (whichdepends on the domain and
range of I ) is often used to enforce a smooth flowfield. As in Section 3.2.1 , minimizing
a function like Equation ( 5.21 ) can be accomplished by solving the Euler-Lagrange
equations, which in this case are:

λ

∂ 2

I

+ ∂

I

x ∂

I

+ ∂

I

x ∂

I

2 u

λ ∇ = u

y v

∂ x ∂ ∂ ∂ ∂ t

(5.22)

∂ 2

= ∂

x ∂

I I I

+ ∂

y ∂

I I

2 v

λ ∇

y u

+ v

∂ ∂ ∂ y ∂ ∂ t

evaluated simultaneously for all the pixels in the image. Computationally, the solu-
tion proceeds in a similar way to the method described in Section 3.2.1 . The partial
derivatives of the spatiotemporal function I are approximated using finite differences
between the two given images. That is, at pixel

( x , y ) ,

4 I 1

∂ I 1

x ≈

( x + 1, y ) − I 1 ( x , y ) + I 1 ( x + 1, y + 1 ) − I 1 ( x , y + 1 )

∂

)

+ I 2 ( x + 1, y ) − I 2 ( x , y ) + I 2 ( x + 1, y + 1 ) − I 2 ( x , y + 1

4 I 1

∂ I 1

y ≈

( x , y + 1 ) − I 1 ( x , y ) + I 1 ( x + 1, y + 1 ) − I 1 ( x + 1, y )

∂

(5.23)

) + I 2 ( x , y + 1 ) − I 2 ( x , y ) + I 2 ( x + 1, y + 1 ) − I 2 ( x + 1, y

4 I 2

∂ I 1

≈ ( x , y ) − I 1 ( x , y ) + I 2 ( x + 1, y ) − I 1 ( x + 1, y )

∂ t

) + I 2 ( x , y + 1 ) − I 1 ( x , y + 1 ) + I 2 ( x + 1, y + 1 ) − I 1 ( x + 1, y + 1