Sparse Reconstruction with Compressed Sensing

Author: Ngoc Ly (SysEn 5800 Fall 2021)

Compressed Sensing (CS)

^{Compressed Sensing summary here}

Compression is synonymous with sparsity. So when we talk about compression we are actually referring to the sparsity. We introduce Compressed Sensing and then focus on reconstruction.

Three big groups of algorithms are:

Optimization methods: includes $l_{1}$ minimization i.e. Basis Pursuit, and quadratically constraint $l_{1}$ minimization i.e. basis pursuit denoising.

greedy include Orthogonal matching pursuit and Compressive Sampling Matching Pursuit (CoSaMP)

thresholding-based methods such as Iterative Hard Thresholding(IHT) and Iterative Soft Thresholding, Approximate IHT or AM-IHT, and many more.

More cutting-edge methods include dynamic programming.

We will cover one, i.e. IHT. WHY IHT THEN? Basis pursuit, matching pursuit type algorithms belong to a more general class of iterative thresholding algorithms. So IHT seems like the ideal place to start. If everything compliment with RIP, then IHT has fast convergence.

Introduction

$x\in \mathbb {R} ^{N}$often not really sparse but approximately sparse

$\mathbf {x} \in \mathbb {R} ^{N}$

$\Phi \in \mathbb {R} ^{M\times N}$for $M\ll N$ a Random Gaussian or Bernoulli matrix

$y\in \mathbb {R} ^{M}$are the observed y samples

$e\in \mathbb {R} ^{M}$noise vector $\|e\|_{2}\leq \eta$

sub modual 1

s.t.

How can we reconstruct x from $y=\Phi x+e$ The goal is to reconstruct $x\in \mathbb {R} ^{N}$given $y$ and $\Phi$

Sensing matrix $\Phi$must satisfy RIP i.e. Random Gaussian or Bernoulli matrixies satisfies A $M={\mathcal {O}}(K/log(N/K))$ which is the number of measurements required for standard compressive sensing to recover with high probability.

let $[N]=\{1,\dots ,N\}$be an index set $[N]$ enumerates the columns of $\Phi$ and $x$. $\Phi$ is an under determined systems with infinite solutions since $M\ll N$. Why $l_{2}$ norm won't give sparse solutions, where asl $l_{1}$ norm will return a sparse solution.

submodual 2

The problem formulation is to recover sparse data $\mathbf {x} \in \mathbb {R} ^{N}$

The support of $\mathbf {x}$ is $supp(\mathbf {x} )=\{i\in [N]:\mathbf {x} _{i}\neq 0\}$ we say $\mathbf {x}$ is $k$ sparse when $|supp(x)|\leq k$

We are interested in the smallest $supp(x)$ , i.e. $min(supp(x))$

sub modual 3

Before we get into RIP lets talk about RIC

Restricted Isometry Constant (RIC) is the smallest $\delta _{|s}\geq 0\ s.t.\ s\subseteq [N]$that satisfies the RIP condition introduced by Candes, Tao

Random Gaussian and Bernoulli satisfies RIP

RIP defined as

$(1-\delta _{s})\|x\|_{2}^{2}\leq \|\Phi x\|_{2}^{2}\leq (1+\delta _{s})\|x\|_{2}^{2}$

Let $\Phi \in \mathbb {R} ^{M\times N}$ satisfy RIP, Let $[N]$ be an index set For $s$ is a restriction on $\mathbf {x}$ denoted by $x_{|s}$ $x\in \mathbb {R} ^{N}$ to $s$ k-sparse $\mathbf {x}$ s.t. RIP is satisfied the $s=supp(\mathbf {x} )$ i.e. $s\subseteq [N]$and $\Phi _{|s}\subseteq \Phi$ where the columns of $\Phi _{|s}$ is indexed by $i\in S$

In search for a unique solution we have the following $l_{0}=|supp(x)|$ optimization problem.

sub modual 4

$\mathbf {\hat {s}} ={\underset {s}{argmin}}\|\mathbf {s} \|_{0}\quad s.t.\quad \mathbf {y} =\Phi \mathbf {s}$, which is an combinatorial NP-Hard problem. Hence, if noise is presence the recovery is not stable. [Buraniuk "compressed sensing"]

From Results of Candes, Romberg, Tao, and Donoho

If $\Phi$ satisfies RIP and $\mathbf {y}$ is sparse the $l_{0}$ gives sparse solutions and is a unique. It is equivalent to $l_{1}$ convex optimization problem and can solve by Linear Program.

$\mathbf {\hat {s}} ={\underset {s}{argmin}}\|\mathbf {s} \|_{1}\quad s.t.\quad \mathbf {y} =\Phi \mathbf {s}$

�

x i , i f i ∈ S

( x | S ) i =

0 otherwise

RIP defined as

( 1 − δ s )k x k 22 ≤ k Φx k 22 ≤ ( 1 + δ s )k x k 22

3 Lemmas Page 267 Blumensath Davies IHT for CS

Lemma 1(Blumensath, Davis 2009 Iterative hard thresholding for compressed

sensing), For all index sets Γ and all Φ for which RIP holds with s = | Γ | that is

s = supp ( x )

1k Φ Γ T k 2 ≤

q

1 + δ | Γ | k y k 2

( 1 − δ | Γ | )k x Γ k 22 ≤ k Φ Γ T Φ Γ x Γ k 22 ≤ ( 1 + δ | Γ | )k x Γ k 22

and

k( I − Φ Γ T Φ Γ )k 2 ≤ δ | Γ | k x Γ k 2

SupposeΓ ∩ Λ = ∅

k Φ Γ T Φ Λ ) x Λ k 2 ≤ δ s k x Λ k 2

Lemma 2 (Needell Tropp, Prop 3.5 in CoSaMP: Iterative signal recovery

from incomplete and inaccurate √ samples)

If Φ satisfies RIP k Φx s k 2 ≤ 1 + δ s k x s k 2 , ∀ x s : k x s k 0 ≤ s, Then ∀ x

k Φx k 2 ≤

p

1 + δ s k x k 2 +

p

1 + δ s

k x k 1

sqrts

Lemma 3 (Needell Tropp, Prop 3.5 in CoSaMP: Iterative signal recovery

from incomplete and inaccurate samples)

Let x s be the best s-term approximation to x. Let x r = x − x s Let

y = Φx + e = Φx s + Φx r + e = Φx s + ẽ

If Φ satisfies RIP for sparsity s, then the norm of error ẽ is bounded by

k ẽ k 2 ≤

p

1 + δ s k x − x s k 2 +

p

1 + δ s

k x − x s k 1

√

+ k e k 2

s

∀ x

Theory

Verification of the Sensing matrix

Gel’fand n-width

Errors E ( S, Φ, D )

Definition Mutual Coherence

Let $\Phi \in R^{M\times N}$, the mutual coherence $\mu _{\Phi }$ is defined by:</math>

$\mu _{\Phi }={\underset {i\neq j}{\frac {|\langle a_{i},a_{j}\rangle |}{\|a_{i}\|\|a_{j}\|}}}$

Welch bound $\mu _{\Phi }\geq {\sqrt {\frac {n}{m(n-m)}}}$> $\mu \geq {\sqrt {\frac {N-M}{M(N-1)}}}$>

We want a small $\mu _{\Phi }$ because it will be close to the normal matrix, which satisfies RIP. Also, $\mu _{\Phi }$ will be needed for the step size for the following IHT.

Need to make the connection of Coherence to RIP and RIC.

Algorithm IHT

The $l_{1}$ convex program mentioned in introduction has an equivalent nonconstraint optimization program.

${\underset {y}{min}}\|\mathbf {y} -\Phi \mathbf {x} \|_{2}^{2}+\lambda \|\mathbf {y} \|_{0}$ (cite IT for sparse approximations)  ??? ${\hat {\mathbf {x} }}=arg{\underset {s}{min}}{\frac {1}{n}}\|\mathbf {y} -\Phi \mathbf {x} \|_{2}^{2}+\lambda \|\mathbf {x} \|_{1}$ . In statistics we call the $l_{1}$ regularization LASSO with $\lambda$ as the regularization parameter. This is the closest convex relaxation to $l_{0}$ the first program menttioned in the introduction.[The Benefit of Group Sparsity]

$z_{v}^{(n)}=\nabla f_{v}(x^{(n)})=-\Phi _{v}^{T}(\mathbf {y} -\Phi \mathbf {x} )$ Then $x^{n+1}={\mathcal {H}}\left(\mathbf {x} ^{(n)}-\tau \sum _{j\in N}^{N}z_{v}^{(n)}\right)$

Define the threashholding operators as: ${\mathcal {H}}_{s}[\mathbf {x} ]={\underset {z\in \sum _{s}}{argmin}}\|x-\Phi \mathbf {x} \|_{2}$ selects the best-k term approximation for some k

Stopping criterion is $\|y-\Phi \mathbf {x} ^{(n)}\|_{2}\leq \epsilon$ iff RIC $\delta _{3s}<{\frac {1}{\sqrt {32}}}$

• Initialize $\Phi ,\mathbf {y} ,\mathbf {e} \ {\mbox{with}}\ \mathbf {y} =\mathbf {\Phi } \mathbf {x} |\mathbf {e} and{\mathfrak {M}}$
• output $IHT(\mathbf {y} ,\mathbf {\Phi } ,{\mathcal {S}})$
• Set $x^{(0)}=\mathbf {0}$
• While Stopping criterion false do

• • $x^{(n+1)}\leftarrow {\mathcal {H}}_{|s}\left[x^{(n)}+\Phi ^{T}(\mathbf {y} -\mathbf {\Phi x} ^{(n)})\right]$

• • $n\leftarrow n+1$
  • end while
• return: $IHT(\mathbf {y} ,\mathbf {\Phi } ,{\mathfrak {M}})\leftarrow \mathbf {x} ^{(n)}$

Numerical Example

Check if $\Phi$ satisfies RIP Checking $\Phi$ satisfies RIP is NP-complete in general, must use shortcuts.

We will do some hacking to make use it works. If $\Phi$ is a gaussian random matrix then we know that it satisfies RIP with high probability and IHT will reconstruct the the true signal find $x$ with $l_{1}$ minimization. (cite Ca, Ro, ta, Robust uncertainty exact signal) (cite Blu, Davies IHT for CS) Other words we don't really know in general if $\Phi$ satisfies RIP in general unless we solve an NP-complete problem; however, we can cross our fingers that $\Phi$ satisfies RIP with a high probability because it's Gaussian and not go through all the work for total verification of RIP.

Iterative Hard Thresholding IHT

Applications

Distributed Systems peer-to-peer network

Collaborative sensor networks for energy savings.

Distributed Systems where Energy needs to be preserved.

MRIs of children and pets because they will move around.

Netflix problem