\underline{Assignment 6}

6.1 Warping and SOR

a. If we rearrange the two equations, we get

αΔu^{k} − J_{13}^{k} = J_{11}^{k}du^{k} + J_{12}^{k}dv^{k} − αΔdu^{k}
αΔv^{k} − J_{23}^{k} = J_{12}^{k}du^{k} + J_{22}^{k}dv^{k} − αΔdv^{k}

0 = [J_{11}^{k}]_{i,j}du_{i,j}^{k} + [J_{12}^{k}]_{i,j}dv_{i,j}^{k} + [J_{13}^{k}]_{i,j} − α ∑ ∑ \frac{du_{ĩ,j̃}^{k} − du_{i,j}^{k}}{h_{l}^{2}}

−α ∑ ∑ \frac{u_{ĩ,j̃}^{k} − u_{i,j}^{k}}{h_{l}^{2}}

0 = [J_{12}^{k}]_{i,j}du_{i,j}^{k} + [J_{22}^{k}]_{i,j}dv_{i,j}^{k} + [J_{23}^{k}]_{i,j} − α ∑ ∑ \frac{dv_{ĩ,j̃}^{k} − dv_{i,j}^{k}}{h_{l}^{2}}

−α ∑ ∑ \frac{v_{ĩ,j̃}^{k} − v_{i,j}^{k}}{h_{l}^{2}}

where the double-summation has the following arguments (they are omitted due to limited space):

∑_{l∈x,y} ∑_{(ĩ,j̃)∈N_{l}(i,j)} .

From this we have to go to an equation system of the form Ax = b. Let’s define the following:

x := ( \begin{matrix}du^{k} \\ du^{k} \\ du^{k} \\ dv^{k} \\ dv^{k} \\ dv^{k} \end{matrix} )

b := ( \begin{matrix}J_{13}^{k} \\ J_{13}^{k} \\ J_{13}^{k} \\ J_{23}^{k} \\ J_{23}^{k} \\ J_{23}^{k} \end{matrix} ) − αD ( \begin{matrix}u^{k} \\ u^{k} \\ u^{k} \\ v^{k} \\ v^{k} \\ v^{k} \end{matrix} )

A := ( \begin{matrix}J_{11}^{k} & & & J_{12}^{k} \\ & J_{11}^{k} & & & J_{12}^{k} \\ & & J_{11}^{k} & & & J_{12}^{k} \\ J_{12}^{k} & & & J_{22}^{k} \\ & J_{12}^{k} & & & J_{22}^{k} \\ & & J_{12}^{k} & & & J_{22}^{k} \end{matrix} )

+α \underbrace{( \begin{matrix}1 & −1 \\ −1 & 2 & −1 \\ & −1 & 1 \\ & & & 1 & −1 \\ & & & −1 & 2 & −1 \\ & & & & −1 & 1 \end{matrix} )}_{=:D}

b. The structure of the double-summation is slightly changed here, since one sometimes needs N_{l}^{−} (newly computed) and N_{l}^{+} (old ones). First, we compute the Gauß-Seidel method. After that, the step to SOR is very easy.

d̃u_{i,j}^{m+1,k} = \frac{α ∑ ∑ \frac{u_{ĩ,j̃}^{k}−u_{i,j}^{k}}{h_{l}^{2}} − [J_{13}^{k}]_{i,j} − ([J_{12}^{k}]_{i,j}dv_{i,j}^{m,k} − α ∑ ∑_{N_{l}^{−}} \frac{du_{ĩ,j̃}^{m+1,k}}{h_{l}^{2}} − α ∑ ∑_{N_{l}^{+}} \frac{du_{ĩ,j̃}^{m,k}}{h_{l}^{2}} )}{[J_{11}]_{i,j} + α ∑ ∑ _{\frac{1}{h_{l}^{2}}}}

d̃v_{i,j}^{m+1,k} = \frac{α ∑ ∑ \frac{v_{ĩ,j̃}^{k}−v_{i,j}^{k}}{h_{l}^{2}} − [J_{23}^{k}]_{i,j} − ([J_{12}^{k}]_{i,j}du_{i,j}^{m+1,k} − α ∑ ∑_{N_{l}^{−}} \frac{dv_{ĩ,j̃}^{m+1,k}}{h_{l}^{2}} − α ∑ ∑_{N_{l}^{+}} \frac{dv_{ĩ,j̃}^{m,k}}{h_{l}^{2}} )}{[J_{22}]_{i,j} + α ∑ ∑ _{\frac{1}{h_{l}^{2}}}}

where m is the iteration step and k the warping level. Remark: In the formula for d̃v_{i,j}^{m+1,k} appears du_{i,j} with m + 1, k. This is due to the fact that du is already computed, assuming the ordering du → dv for the computation. Let’s state the SOR step:

du_{i,j}^{m+1,k} = (1 − ω)du_{i,j}^{m,k} + ωd̃u_{i,j}^{m+1,k}
dv_{i,j}^{m+1,k} = (1 − ω)dv_{i,j}^{m,k} + ωd̃v_{i,j}^{m+1,k}

6.2 Photometric Invariants and Robustification

a. Let’s first give an overview over the different properties where the following assumptions are considered: global multiplicative illumination changes shadow and shading (local multiplicative changes) highlights and specular reflections (local additive / local multi-

plicative changes)

p_{1}
p_{2}
p_{3}
p_{4}
p_{5}
p_{6}

For the decision, what classes do apply, we consider Lecture 5, slide 21. 1.

p_{1} = R − 3B + G
= e (m_{i}(x)w_{i} + m_{b}(x)R̂_{b}(x)) − 3e (m_{i}(x)w_{i} + m_{b}(x)B̂_{b}(x))

+e (m_{i}(x)w_{i} + m_{b}(x)Ĝ_{b}(x)

= e (−m_{i}(x)w_{i} + m_{b}(x)(R̂_{b}(x) − 3B̂_{b}(x) + Ĝ_{b}(x)))

2. Here we have an invariance under local additive changes only.

p_{2} = R^{2} + B^{2} − 2BR = (R − B)^{2}
= [e (m_{i}(x)w_{i} + m_{b}(x)R̂_{b}(x) − m_{i}(x)w_{i} − m_{b}(x)B̂_{b}(x))]^{2}
= [em_{b}(x)(R̂_{b}(x) − B̂_{b}(x))]^{2}
= e^{2}m_{b}(x)^{2}(R̂_{b}(x) − B̂_{b}(x))^{2}

3.

p_{3} = \frac{R − B}{R + G}

= \frac{e (m_{i}(x)w_{i} + m_{b}(x)R̂_{b}(x) − m_{i}(x)w_{i} − m_{b}(x)B̂_{b}(x))}{e (m_{i}(x)w_{i} + m_{b}(x)R̂_{b}(x) + m_{i}(x)w_{i} + m_{b}(x)Ĝ_{b}(x))}

= \frac{m_{b}(x) (R̂_{b}(x) − B̂_{b}(x))}{2m_{i}(x)w_{i} + m_{b}(x) (R̂_{b}(x) + B̂_{b}(x))}

= \frac{R̂_{b}(x) − B̂_{b}(x)}{R̂_{b}(x) + B̂_{b}(x)} if m_{i}(x) = 0.

4.

p_{4} = 2\frac{BG + RG}{B^{2} − R^{2}} − \frac{B + R}{B − R}
= 2\frac{G(B + R)}{(B − R)(B + R)} − \frac{B + R}{B − R} = \frac{2G − B − R}{B − R}

= \frac{m_{b}(x) (2Ĝ_{b}(x) − B̂_{b}(x) − R̂_{b}(x))}{m_{b}(x) (B̂_{b}(x) − R̂_{b}(x))}

= \frac{2Ĝ_{b}(x) − B̂_{b}(x) − R̂_{b}(x)}{B̂_{b}(x) − R̂_{b}(x)}

5. Let’s consider the following fact

ln(\underbrace{c}_{factor} B)_{x} = ( \underbrace{ln c}_{constant} + ln B)_{x} = (ln B)_{x}

p_{5} = (ln B)_{x} + (ln R)_{x}
= \frac{1}{[e(m_{i}(x)w_{i} + m_{b}(x)B̂_{b}(x))]_{x}} + \frac{1}{[e(m_{i}(x)w_{i} + m_{b}(x)R̂_{b}(x))]_{x}}

6. Invariant under global additive only.

p_{6} = ln B_{x} + ln R_{x} = ln(B_{x} · R_{x})
= ln [(e(m_{i}(x)w_{i} + m_{b}(x)B̂_{b}(x)))_{x} · (e(m_{i}(x)w_{i} + m_{b}(x)R̂_{b}(x)))_{x}]

\underline{Assignment 6}

b. We consider here

Ψ(s^{2}) = \sqrt{s^{2} + ε^{2}} with ε = 10^{−3}.

Let’s give the joint robustification

E_{joint}(u, v) = ∫_{Ω} Ψ (∑_{i=1}^{6} (p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t))^{2})
+α (|∇u|^{2} + |∇v|^{2}) dxdy

c. Joint robustification was given in part b.. Separate robustification is

given by

E_{sep}(u, v) = ∫_{Ω} ∑_{i=1}^{6} Ψ((p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t))^{2})
+α (|∇u|^{2} + |∇v|^{2}) dxdy

The Euler-Lagrange equations that have to be satisfied are

0 = F_{u} − \frac{∂}{∂x}F_{u_{x}} − \frac{∂}{∂y} F_{u_{y}}
0 = F_{v} − \frac{∂}{∂x}F_{v_{x}} − \frac{∂}{∂y} F_{v_{y}}

• joint robustification:

F = Ψ (∑_{i=1}^{6} (p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t))^{2})

+α (|∇u|^{2} + |∇v|^{2})

F_{u} = 2 · Ψ′(...)

· ∑_{i=1}^{6}(p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t))

·p_{i_{x}}(x + u, y + v, t + 1)

F_{v} = 2 · Ψ′(...)

· ∑_{i=1}^{6}(p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t))

·p_{i_{y}}(x + u, y + v, t + 1)

F_{u_{x}} = 2αu_{x}
F_{u_{y}} = 2αu_{y}
F_{v_{x}} = 2αv_{x}
F_{v_{y}} = 2αv_{y}
COPCV  Page 5  SS 2009

The Euler-Lagrange equations now read as (divided by 2):

0 = Ψ′(...)( ∑_{i=1}^{6}(p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t))p_{i_{x}}(x + u, y + v, t + 1)) − αΔu
0 = Ψ′(...)( ∑_{i=1}^{6}(p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t))p_{i_{y}} (x + u, y + v, t + 1)) − αΔv

• separate robustification:

F = ∑_{i=1}^{6} Ψ((p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t))^{2})
+α (|∇u|^{2} + |∇v|^{2})
F_{u} = 2 · ∑_{i=1}^{6} Ψ′(...) · (p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t))
·p_{i_{x}}(x + u, y + v, t + 1)
F_{v} = 2 · ∑_{i=1}^{6} Ψ′(...) · (p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t))
·p_{i_{y}}(x + u, y + v, t + 1)
F_{u_{x}} = 2αu_{x}
F_{u_{y}} = 2αu_{y}
F_{v_{x}} = 2αv_{x}
F_{v_{y}} = 2αv_{y}

The Euler-Lagrange equations read as (again divided by 2):

0 = ∑_{i=1}^{6} (Ψ′(...)(p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t)) · p_{i_{x}}(x + u, y + v, t + 1)) − αΔu
0 = ∑_{i=1}^{6} (Ψ′(...)(p_{i}(x + u, y + v, t + 1) − p_{i}(x, y, t)) · p_{i_{y}} (x + u, y + v, t + 1)) − αΔv