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.

Introduction

Notation =

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

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

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

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

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

put defn of p norm here

$x=\Psi \alpha$ where $\Psi$ is the sparsifying matrix and $\alpha$ are coeficients

sub module goal

The goal of compressed sensing is to being with the under determined linear system $y=\Phi x+e$, Where $\Phi \in \mathbb {R} ^{M\times N}$ for $M\ll N$ How can we reconstruct x from The goal is to reconstruct $x\in \mathbb {R} ^{N}$given $y$ and $\Phi$ Considerably fewer random measurements and reconstruct the original $x$ with high probability from its sparse representation instead of taking a large number of high-resolution measurements and discarding the majority of them. being a random matrix

sub modual

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.

sub module sparsity

A vector $x$ is said to be $k$ sparse in $\mathbb {R} ^{N}$ if it has at most $k$ nonzero coefficients. The set of all k-sparse vectors is denoted by $\Sigma _{k}=\{x:\|x\|_{0}\leq k\}$. Consequently there is ${\binom {N}{k}}$ different subsets of k-sparse vectors. If we draw uniformaly a random k-sparse $x$ from $\Sigma _{k}$ has the entropy$log{\binom {N}{k}}\approx klog(N/k)$ bits are needed for compression of $\Sigma _{k}$ ~cite(Measurements vs Bits)

The goal is to search for the sparsest $x\in \Sigma _{k}$ given the meassurment $y$ and the constraint matrix $\Phi$.

This is antiquated to find the $\min \|x\|_{0}$ in the set $\{x\in \mathbb {R} ^{N}:\Phi x=y\}$

This searching problem can be formulated to the following $l_{0}$ program

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 module

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 module zero norm program

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

sub modual RIP

RIP defined as

$\Phi$ satisfies RIP of order $k$ if for $\forall x\in \Sigma _{k}$ $\delta _{k}\in [0,1)$ satisfies for inequalaty

1. TODO switch s to k

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

If $\Phi$ satisfies RIP then $\Phi$ doesn't send two distinct k-sparse $x\in \Sigma _{k}$ to the measurment vector $y$.

sub module RIP matracies

If a sensing matrix $\Phi$must satisfy RIP, then the number of measurements $M={\mathcal {O}}(K/log(N/K))$ is required recover $x$ with high probability.

sub module RIC

Restricted Isometry Constant (RIC) is the smallest $\delta _{|s}$ in $\{\delta _{k}\in [0,1):(1-\delta _{s})\|x\|_{2}^{2}\leq \|\Phi x\|_{2}^{2}\leq (1+\delta _{s})\|x\|_{2}^{2}\}$

if $M\geq 2k$ then there exists an unique $x$ such that $y=\Phi x$.

for $x$ is k-sparse and and $\Phi$ satisfies RIP of order $2k$ RIC $\delta _{2k}<{\sqrt {2}}-1$ then the $l_{0}$ program can have to relaxed convex form $l_{1}$ program.

sub problem 1 norm program

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.

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

Theory

Verification of the Sensing matrix

Check if $\Phi$ satisfies RIP Checking $\Phi$ satisfies RIP is NP-complete in general so it's unreasonable to ask a computer to verify a matrix satisfies RIP. In order to get around this combinatorial hard problem, we need an understanding of what matrices satisfy RIP.

Random Sensing matrices: Gaussian, Bernoulli, Rademacher Deterministic Sensing Matrices: binary, bipolar, ternary, Vandermond Structural Sensing Matrices: Toeplitz, Circulant, Hadamard Optimized Sensing Matrices (Parkale, Nalbalwar, Sensing Matrices in Compressed Sensing)

Are some examples. Different sensing matrices are more suited for different problems, but in general, we want to use an alternative to Gaussian because it reduces the computational complexity.

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 $\mu _{\Phi }\geq {\sqrt {\frac {n}{m(n-m)}}}$> $\mu \geq {\sqrt {\frac {N-M}{M(N-1)}}}$> is the coherence between $\Phi$ and $\Psi$ 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 $x^{n+1}={\mathcal {H}}\left(\mathbf {x} ^{(n)}-\tau \sum _{j\in N}^{N}z_{v}^{(n)}\right)$

sub modual

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 $\delta _{3s}<{\frac {1}{\sqrt {32}}}$

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

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

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

$\Phi ^{*}$ is a Adjoint matrix i.e. the transpost of it's cofactor.

Numerical Example

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. Donoho proves that nearly all matrices are sensing matrices.

Iterative Hard Thresholding IHT

Applications

Netflix problem