Más contenido relacionado La actualidad más candente (20) Similar a Signal Processing Course : Inverse Problems Regularization (20) Más de Gabriel Peyré (16) Signal Processing Course : Inverse Problems Regularization3. J(f) = ||f||2
W 1,2 =
Z
R2
||rf(x)||dxSobolev semi-norm:
Smooth and Cartoon Priors
| f|2
4. J(f) = ||f||2
W 1,2 =
Z
R2
||rf(x)||dxSobolev semi-norm:
Total variation semi-norm: J(f) = ||f||TV =
Z
R2
||rf(x)||dx
Smooth and Cartoon Priors
| f|| f|2
5. J(f) = ||f||2
W 1,2 =
Z
R2
||rf(x)||dxSobolev semi-norm:
Total variation semi-norm: J(f) = ||f||TV =
Z
R2
||rf(x)||dx
Smooth and Cartoon Priors
| f|| f|2
13. Function: ˜f : x 2 R2
7! f(x) 2 R
˜f(x + ") = ˜f(x) + hrf(x), "iR2 + O(||"||2
R2 )
r ˜f(x) = (@1
˜f(x), @2
˜f(x)) 2 R2
Gradient: Images vs. Functionals
14. Function: ˜f : x 2 R2
7! f(x) 2 R
Discrete image: f 2 RN
, N = n2
f[i1, i2] = ˜f(i1/n, i2/n) rf[i] ⇡ r ˜f(i/n)
˜f(x + ") = ˜f(x) + hrf(x), "iR2 + O(||"||2
R2 )
r ˜f(x) = (@1
˜f(x), @2
˜f(x)) 2 R2
Gradient: Images vs. Functionals
15. Function: ˜f : x 2 R2
7! f(x) 2 R
Discrete image: f 2 RN
, N = n2
f[i1, i2] = ˜f(i1/n, i2/n)
Functional: J : f 2 RN
7! J(f) 2 R
J(f + ⌘) = J(f) + hrJ(f), ⌘iRN + O(||⌘||2
RN )
rf[i] ⇡ r ˜f(i/n)
˜f(x + ") = ˜f(x) + hrf(x), "iR2 + O(||"||2
R2 )
r ˜f(x) = (@1
˜f(x), @2
˜f(x)) 2 R2
rJ : RN
7! RN
Gradient: Images vs. Functionals
16. Function: ˜f : x 2 R2
7! f(x) 2 R
Discrete image: f 2 RN
, N = n2
f[i1, i2] = ˜f(i1/n, i2/n)
Functional: J : f 2 RN
7! J(f) 2 R
Sobolev:
rJ(f) = (r⇤
r)f = f
J(f) =
1
2
||rf||2
J(f + ⌘) = J(f) + hrJ(f), ⌘iRN + O(||⌘||2
RN )
rf[i] ⇡ r ˜f(i/n)
˜f(x + ") = ˜f(x) + hrf(x), "iR2 + O(||"||2
R2 )
r ˜f(x) = (@1
˜f(x), @2
˜f(x)) 2 R2
rJ : RN
7! RN
Gradient: Images vs. Functionals
18. rJ(f) = div
✓
rf
||rf||
◆
If 8 n, rf[n] 6= 0,
Sub-di↵erential:
If 9n, rf[n] = 0, J not di↵erentiable at f.
Cu = ↵ 2 R2⇥N
(u[n] = 0) ) (↵[n] = u[n]/||u[n]||)
@J(f) = { div(↵) ; ||↵[n]|| 6 1 and ↵ 2 Crf }
Total Variation Gradient
||rf||
rJ(f)
19. −10 −8 −6 −4 −2 0 2 4 6 8 10
−2
0
2
4
6
8
10
12
−10 −8 −6 −4 −2 0 2 4 6 8 10
−2
0
2
4
6
8
10
12
p
x2 + "2
|x|
Regularized Total Variation
||u||" =
p
||u||2 + "2 J"(f) =
P
n ||rf[n]||"
20. −10 −8 −6 −4 −2 0 2 4 6 8 10
−2
0
2
4
6
8
10
12
−10 −8 −6 −4 −2 0 2 4 6 8 10
−2
0
2
4
6
8
10
12
rJ"(f) = div
✓
rf
||rf||"
◆
p
x2 + "2
|x|
rJ" ⇠ /" when " ! +1
Regularized Total Variation
||u||" =
p
||u||2 + "2 J"(f) =
P
n ||rf[n]||"
rJ"(f)
23. and 0 < ⌧ < 2/L, then f(k) k!+1
! f?
a solution of min
f
J(f).
If f is convex, C1
, rf is L-Lipschitz,Theorem:
f(k+1)
= f(k)
⌧krJ(f(k)
) f(0)
is given.
Gradient Descent
24. and 0 < ⌧ < 2/L, then f(k) k!+1
! f?
a solution of min
f
J(f).
If f is convex, C1
, rf is L-Lipschitz,Theorem:
f(k+1)
= f(k)
⌧krJ(f(k)
) f(0)
is given.
Optimal step size: ⌧k = argmin
⌧2R+
J(f(k)
⌧rJ(f(k)
))
Proposition: One has
hrJ(f(k+1)
), rJ(f(k)
)i = 0
Gradient Descent
26. Gradient Flows and PDE’s
f(k+1)
f(k)
⌧
= rJ(f(k)
)Fixed step size ⌧k = ⌧:
Denote ft = f(k)
for t = k⌧, one obtains formally as ⌧ ! 0:
8 t > 0,
@ft
@t
= rJ(ft) and f0 = f(0)
27. J(f) =
R
||rf(x)||dxSobolev flow:
@ft
@t
= ftHeat equation:
Explicit solution:
Gradient Flows and PDE’s
f(k+1)
f(k)
⌧
= rJ(f(k)
)Fixed step size ⌧k = ⌧:
Denote ft = f(k)
for t = k⌧, one obtains formally as ⌧ ! 0:
8 t > 0,
@ft
@t
= rJ(ft) and f0 = f(0)
30. Optimal choice of t: minimize ||ft f||
! not accessible in practice.
SNR(ft, f) = 20 log10
✓
||f ft||
||f||
◆
Optimal Parameter Selection
t t
40. Sobolev prior: J(f) = 1
2 ||rf||2
f?
= argmin
f2RN
E(f) = ||y f||2
+ ||rf||2
(assuming 1 /2 ker( ))
Sobolev Regularization
41. Sobolev prior: J(f) = 1
2 ||rf||2
f?
= argmin
f2RN
E(f) = ||y f||2
+ ||rf||2
rE(f?
) = 0 () ( ⇤
)f?
= ⇤
yProposition:
! Large scale linear system.
(assuming 1 /2 ker( ))
Sobolev Regularization
42. Sobolev prior: J(f) = 1
2 ||rf||2
f?
= argmin
f2RN
E(f) = ||y f||2
+ ||rf||2
rE(f?
) = 0 () ( ⇤
)f?
= ⇤
yProposition:
! Large scale linear system.
Gradient descent:
(assuming 1 /2 ker( ))
where ||A|| = max(A)
! Slow convergence.
Sobolev Regularization
Convergence: ⇥ < 2/||⇥ ⇥ ||
43. Mask M, = diagi(1i2M )
Example: InpaintingFigure 3 shows iterations of the algorithm 1 to solve the inpainting problem
on a smooth image using a manifold prior with 2D linear patches, as defined in
16. This manifold together with the overlapping of the patches allow a smooth
interpolation of the missing pixels.
Measurements y Iter. #1 Iter. #3 Iter. #50
Fig. 3. Iterations of the inpainting algorithm on an uniformly regular image.
5 Manifold of Step Discontinuities
In order to introduce some non-linearity in the manifold M, one needs to go
log10(||f(k)
f( )
||/||f0||)
k k
E(f(k)
)
M
( f)[i] =
⇢
0 if i 2 M,
f[i] otherwise.
45. Symmetric linear system:
x(k+1)
= argmin E(x)
s.t. x x(k)
2 span(rE(x(0)
), . . . , rE(x(k)
))
Intuition:
Conjugate Gradient
Ax = b () min
x2Rn
E(x) =
1
2
hAx, xi hx, bi
Proposition: 8 ` < k, hrE(xk
), rE(x`
)i = 0
46. Symmetric linear system:
Initialization: x(0)
2 RN
, r(0)
= b Ax(0)
, p(0)
= r(0)
r(k)
=
hrE(x(k)
), d(k)
i
hAd(k), d(k)i
d(k)
= rE(x(k)
) +
||v(k)
||
||v(k 1)||
d(k 1)
v(k)
= rE(x(k)
) = Ax(k)
b
x(k+1)
= x(k)
r(k)
d(k)
Iterations:
x(k+1)
= argmin E(x)
s.t. x x(k)
2 span(rE(x(0)
), . . . , rE(x(k)
))
Intuition:
Conjugate Gradient
Ax = b () min
x2Rn
E(x) =
1
2
hAx, xi hx, bi
Proposition: 8 ` < k, hrE(xk
), rE(x`
)i = 0
47. TV" regularization: (assuming 1 /2 ker( ))
f?
= argmin
f2RN
E(f) =
1
2
|| f y|| + J"
(f)
Total Variation Regularization
||u||" =
p
||u||2 + "2 J"(f) =
P
n ||rf[n]||"
48. TV" regularization: (assuming 1 /2 ker( ))
f(k+1)
= f(k)
⌧krE(f(k)
)
rE(f) = ⇤
( f y) + rJ"(f)
rJ"(f) = div
✓
rf
||rf||"
◆
Convergence: requires ⌧ ⇠ ".
Gradient descent:
f?
= argmin
f2RN
E(f) =
1
2
|| f y|| + J"
(f)
Total Variation Regularization
||u||" =
p
||u||2 + "2 J"(f) =
P
n ||rf[n]||"
49. TV" regularization: (assuming 1 /2 ker( ))
f(k+1)
= f(k)
⌧krE(f(k)
)
rE(f) = ⇤
( f y) + rJ"(f)
rJ"(f) = div
✓
rf
||rf||"
◆
Convergence: requires ⌧ ⇠ ".
Gradient descent:
f?
= argmin
f2RN
E(f) =
1
2
|| f y|| + J"
(f)
Newton descent:
f(k+1)
= f(k)
H 1
k rE(f(k)
) where Hk = @2
E"(f(k)
)
Total Variation Regularization
||u||" =
p
||u||2 + "2 J"(f) =
P
n ||rf[n]||"
52. Noiseless problem: f?
2 argmin
f
J"
(f) s.t. f 2 H
Contraint: H = {f ; f = y}.
f(k+1)
= ProjH
⇣
f(k)
⌧krJ"(f(k)
)
⌘
ProjH(f) = argmin
g=y
||g f||2
= f + ⇤
( ⇤
) 1
(y f)
Inpainting: ProjH(f)[i] =
⇢
f[i] if i 2 M,
y[i] otherwise.
Projected gradient descent:
f(k) k!+1
! f?
a solution of (?).
(?)
Projected Gradient Descent
Proposition: If rJ" is L-Lipschitz and 0 < ⌧k < 2/L,
54. TV
Priors: Non-quadratic
better edge recovery.
=)
Optimization
Variational regularization:
()
– Gradient descent. – Newton.
– Projected gradient. – Conjugate gradient.
Non-smooth optimization ?
Conclusion
Sobolev