Robust Uncertainty Principles: Exact Signal Reconstruction ...

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Robust Uncertainty Principles:Exact Signal Reconstruction from Highly Incomplete

Frequency Information

Emmanuel Candes, California Institute of Technology

SIAM Conference on Imaging Science, Salt Lake City, Utah, May 2004

Collaborators : Justin Romberg (Caltech), Terence Tao (UCLA)

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Incomplete Fourier Information

Observe Fourier samples f(ω) on a domain Ω.

22 radial lines, ≈ 8% coverage

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Classical ReconstructionBackprojection: essentially reconstruct g∗ with

g∗(ω) =

f(ω) ω ∈ Ω

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Original Phantom (Logan−Shepp)

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Naive Reconstruction

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original g∗

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Interpolation?

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original g∗

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Total Variation Reconstruction

Reconstruct g∗ with

ming

‖g‖T V s.t. g(ω) = f(ω), ω ∈ Ω

Original Phantom (Logan−Shepp)

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Reconstruction: min BV + nonnegativity constraint

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original g∗ = original — perfect reconstruction!

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Sparse Spike Train

Sparse sequence of NT spikes Observe NΩ Fourier coefficients

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Interpolation?

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`1 Reconstruction

Reconstruct by solving

ming

∑t

|gt| s.t. g(ω) = f(ω), ω ∈ Ω

For NT ∼ NΩ/2, we recover f perfectly.

original recovered from 30 Fourier samples

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Extension to TV

‖g‖T V =∑

i

|gi+1 − gi| = `1-norm of finite differences

Given frequency observations on Ω, using

min ‖g‖T V s.t. g(ω) = f(ω), ω ∈ Ω

we can perfectly reconstruct signals with a small number of jumps.

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Reconstructed perfectly from 30 Fourier samples

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Model Problem

• Signal made out of T spikes

• Observed at only |Ω| frequency locations

• Extensions

– Piecewise constant signal

– Spikes in higher-dimensions; 2D, 3D, etc.

– Piecewise constant images

– Many others

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Sharp Uncertainty Principles

• Signal is sparse in time, only |T | spikes

• Solve combinatorial optimization problem

(P0) ming

‖g‖`0 := #t, g(t) 6= 0, g|Ω = f|Ω

Theorem 1 N (sample size) is prime

(i) Assume that |T | ≤ |Ω|/2, then (P0) reconstructs exactly.

(ii) Assume that |T | > |Ω|/2, then (P0) fails at exactly reconstructing f ;∃f1, f2 with ‖f1‖`0 + ‖f2‖`0 = |Ω| + 1 and

f1(ω) = f2(ω), ∀ω ∈ Ω

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`1 Relaxation?

Solve convex optimization problem (LP for real-valued signals)

(P1) ming

‖g‖`1 :=∑

t

|g(t)|, g|Ω = f|Ω

• Example: Dirac’s comb

–√

N equispaced spikes (N perfect square).

– Invariant through Fourier transform f = f

– Can find |Ω| = N −√

N with f(ω) = 0, ∀ω ∈ Ω.

– Can’t reconstruct

• More dramatic examples exist

• But all these examples are very special

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Dirac’s Comb

t

f(t)

N

ω

f(ω)

N

f f

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Main Result

Theorem 2 Suppose

|T | ≤ α(M) ·|Ω|

log N

Then min-`1 reconstructs exactly with prob. greater than 1 − O(N−M). (n.b.one can choose α(M) ∼ [29.6(M + 1)]−1.

Extensions

• |T |, number of jump discontinuities (TV reconstruction)

• |T |, number of 2D, 3D spikes.

• |T |, number of 2D jump discontinuities (2D TV reconstruction)

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Heuristics: Robust Uncertainty Principles

f unique minimizer of (P1) iff∑t

|f(t) + h(t)| >∑

t

|f(t)|, ∀h, h|Ω = 0

Triangle inequality∑|f(t)+h(t)| =

∑T

|f(t)+h(t)|+∑T c

|ht| ≥∑T

|f(t)|−|h(t)|+∑T c

|ht|

Sufficient condition∑T

|h(t)| ≤∑T c

|h(t)| ⇔∑T

|h(t)| ≤1

2‖h‖`1

Conclusion: f unique minimizer if for all h, s.t. h|Ω = 0, it is impossible to‘concentrate’ h on T

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Connections:

• Donoho & Stark (88)

• Donoho & Huo (01)

• Gribonval & Nielsen (03)

• Tropp (03) and (04)

• Donoho & Elad (03)

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Dual Viewpoint

• Convex problem has a dual

• Dual polynomial

P (t) =∑ω∈Ω

P (ω)eiωt

– P (t) = sgn(f)(t), ∀t ∈ T

– |P (t)| < 1, ∀t ∈ T c

– P supported on set Ω of visible frequencies

Theorem 3

(i) If FT →Ω and there exits a dual polynomial, then the (P1) minimizer (P1)is unique and is equal to f .

(ii) Conversely, if f is the unique minimizer of (P1), then there exists a dualpolynomial.

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Dual Polynomial

t

P(t) P(ω)

ω

^

Space Frequency

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Construction of the Dual Polynomial

P (t) =∑ω∈Ω

P (ω)eiωt

• P interpolates sgn(f) on T

• P has minimum energy

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Auxilary matrices

Hf(t) := −∑ω∈Ω

∑t′∈E:t′ 6=t

eiω(t−t′) f(t′).

Restriction:

• ι∗ is the restriction map, ι∗f := f |T

• ι is the obvious embedding obtained by extending by zero outside of T

• Identity ι∗ι is simply the identity operator on T .

P := (ι −1

|Ω|H)(ι∗ι −

1

|Ω|ι∗H)−1ι∗sgn(f).

• Frequency support. P has Fourier transform supported in Ω

• Spatial interpolation. P obeys

ι∗P = (ι∗ι −1

|Ω|ι∗T )(ι∗ι −

1

|Ω|ι∗T )−1ι∗sgn(f) = ι∗sgn(f),

and so P agrees with sgn(f) on T .

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Hard Things

P := (ι −1

|Ω|H)(ι∗ι −

1

|Ω|ι∗H)−1ι∗sgn(f).

• (ι∗ι − 1|Ω|ι

∗H) invertible

• |P (t)| < 1, t /∈ T

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Invertibility

(ι∗ι−1

|Ω|ι∗H) = IT −

1

|Ω|H0, H0(t, t′) =

0 t = t′

−∑

ω∈Ω eiω(t−t′). t 6= t′

Fact: |H0(t, t′)| ∼√

|Ω|

‖H0‖2 ≤ Tr(H∗0H0) =

∑t,t′

|H0(t, t′)|2 ∼ |T |2 · |Ω|

Want ‖H0‖ ≤ |Ω|, and therefore

|T |2 · |Ω| = O(|Ω|2) ⇔ |T | = O(√

|Ω|)

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Key Estimates

• Want to show largest eigenvalue of H0 (self-adjoint) is less than Ω.

• Take large powers of random matrices

Tr(H2n0 ) = λ2n

1 + . . . + λ2nT

• Key estimate: develop bounds on E[Tr(H2n0 )]

• Key intermediate result:

‖H0‖ ≤ γ√

log |T |√

|T | |Ω|

with large-probability

• A lot of combinatorics!

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Numerical Results• Signal length N = 1024

• Randomly place Nt spikes, observe Nw random frequencies

• Measure % recovered perfectly

• red = always recovered, blue = never recovered

Nw

Nt/Nw

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Other Phantoms, IOriginal Phantom

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Classical Reconstruction

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original g∗ = classical reconstruction

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Original Phantom

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Total Variation Reconstruction

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original g∗ = TV reconstruction = Exact!

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Other Phantoms, IIOriginal Phantom

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Classical Reconstruction

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Original Phantom

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Total Variation Reconstruction

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original g∗ = TV reconstruction = Exact!

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Scanlines

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2A Scanline of the Original Phantom

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2Classical (Black) and TV (Red) Reconstructions

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Summary

• Exact reconstruction

• Tied to new uncertainty principles

• Stability

• Robustness

• Optimality

• Many extensions: e.g. arbitrary synthesis/measurement pairs

Contact: emmanuel@acm.caltech.edu