Foundations of Smart Sensing - Inriapeople.rennes.inria.fr/Aline.Roumy/teaching/CompressiveSensing_Sli… · Foundations of Smart Sensing. Teachers: Nancy Bertin CNRS/IRISA, Panama

Foundations of Smart SensingCompressive Sensing

MSc in Statistics for Smart Data - ENSAI

Aline RoumyInria Rennes (France)

[email protected]

December 2018

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Outline1 Part 1 - Why compressive sensing?

2 Part 2 - Compressive sensing: how it works?Notations (Reminder)Sensing processReconstruction processGood sensing matrices? First insightsCompressive sensing vs other schemes

3 Part 3 - Compressive sensing: why it works?A magic tool: concentration inequalityConditions for exact reconstruction with P1

Restricted isometry propertyRIP ⇒ NSP ⇔ recoveryConcentration ⇒ RIP and condition on mGaussian ⇒ Concentration

4 Part 4 - Compressive sensing: what it is good for?

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Organization

• This course is the second and last part of the courseFoundations of Smart Sensing.

• Teachers:

Nancy BertinCNRS/IRISA, Panama

audio and acoustics

Cedric HerzetInria, Fluminance

geophysics

Aline Roumy∗

Inria, Siroccoimage and video

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Course schedule (tentative)

Compressive sensing (CS): a self-sufficient course with a lot ofconnections to sparse approximations

• Dec 5: Lecture (CS: how it works + why it works )

• Dec 6: Lecture (CS: what for -2)+ Lab

• Dec 7: Lab, summaries

Quiz: (within course Quiz ), no grade.

• connect to socrative.com and login as student

• room name ALINER

• you can put your name but if you prefer you can use any othername

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http://www.socrative.com/

Course grading

• Quiz: (within course), no grade.

• Project:I Computer lab

using Matlab, R, Python...I write a report on the lab activities (max 2 pages for the

comments (excluding proofs, figures):deadline TO BE DISCUSSEDsend the pdf file + matlab (or sthg else) code files via email [email protected]

I You will get a grade from the evaluation of your report,and during the lab session.

• Final Exam:I written exam (Dec 13th) for the whole courseI 2 hours

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Course material

• G. Kutyniok, Theory and Applications of Compressed Sensing,GAMM Mitteilungen 36 (2013), 79-101.

• S. Foucart, Notes on compressed sensing, 2009. (pdf)

For more information:

• S. Foucart, H. Rauhut, A mathematical introduction tocompressive sensing, Birkhauser, 2013.Chapters 6 (RIP), 9 (RIP probability).

• M.A. Davenport, M.F. Duarte, Y.C. Eldar, G. KutyniokIntroduction to compressed sensing.in Compressed Sensing: Theory and Applications, Y.C. Eldar andG. Kutyniok eds., Dekker, 2012. (pdf)

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http://www.math.tamu.edu/~foucart/teaching/notes/CS.pdf

http://www-stat.stanford.edu/~markad/publications/ddek-chapter1-2011.pdf






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Part 1 - Why compressive sensing?

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Classical:

sampling + compression

Compressive sensing

sampling AND compression

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Classical: Sampling + compression

Acquisition (sampling/sensing)

• For instance HD video1920x1080=2.07 M pixels/image25Hz: images/s,12(=8+2+2) bits/pixel

→ 0.6 Gbit/s

Compression

• For instance, MPEG-5 (2013)0.6 Gbit/s → 2Mbit/s i.e. compression ratio 300:10.6 Gbit/s → 0.04 bit/pixel

Compressive sensing: can we acquire less data in the first place?To answer the question: let us first look at the classicalsampling+compression technique.

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Sampling: (1) Fourier representation

The frequency viewpoint (Fourier 1822)Signals can be built from the sum of harmonic functions

Wikipedia.

irenevigueguix.wordpress.com

x(t) signalX (f ) Fourier coefficiente2πift harmonic functionAnalysis

X (f ) =

∫ +∞

−∞x(t)e−2πiftdt

Synthesis

x(t) =

∫ +∞

−∞X (f )e2πiftdt

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Sampling: (2) optimal sampling rate

Nyquist–Shannon sampling theorem:“Exact reconstruction of a continuous-timesignal from discrete samples is possible if thesignal is bandlimited and the samplingfrequency is greater than twice the highestfrequency.”

Wikipedia.

IDCOM, University of Edinburgh

Classical Sampling Theory

The Whittaker–Kotelnikov–Shannon

Sampling Theorem states:

“Exact reconstruction of a continuous-time

signal from discrete samples is possible if

the signal is bandlimited and the sampling

frequency is greater than twice the highest

frequency.”

Sampling below this rate introduces aliasing

Subspace of

bandlimited

signals

Sampling

Operator

S i g n a l s p a c e

Reconstruction

Operator (linear)

Observation space

Mike Davies.

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Sampling: (3) optimal sampling rate

Sampling below this rate introduces

aliasing

signal ambiguity

SVI.nl

wikipedia

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Compression: (1) image decomposition

with 2D-discrete cosine transform (DCT) (orthogonal basis)1- Split image into blocks of size N1 × N2 each2- For each N1 × N2 image block (xn1,n2) compute the N1 × N2 blockof transformed image (ck1,k2) with:

ck1,k2 =

N1−1∑n1=0

N2−1∑n2=0

xn1,n2 cos

[π

N1

(n1 +

1

2

)k1

]cos

[π

N2

(n2 +

1

2

)k2

]︸︷︷︸

Φn1,n2(k1, k2)

Example: 8x8 DCT transformTop-left matrix is(Φn1,n2(k1 = 0, k2 = 0))n1,n2

Quiz 1, 2, 3

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Compression: (1) image decomposition

Left: image; Right: discrete cosine transform of image

Key concept: few degrees of freedom in the transform domain

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Compression: (2) image approximation

s-term approximation:Keep the s coefficients cs with largest absolute value and x = Φ−1csLeft: 1% kept; Right:5% kept

Quiz 4

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Can we sample signals at the “Information Rate”?

Yes, we can: compressive sensing

Wikipedia.

E. J. Candes and T. Tao, 2005“Decoding by linear programming”

Wikipedia.

D. L. Donoho, 2006“Compressed sensing”

When compressing a signal we typically take lots of samples, move toa transform domain, and then throw most of the coefficients away!Can we just sample what we need?

Yes! . . . and more surprisingly we can do this non-adaptively.

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Compressive sensing overviewObserve x ∈ Rn via m measurements, with m� nMore precisely, y = Mx where y ∈ Rm

Assumptions:

- signal approximately s-sparse

- use approximately m ≥ O(

s logn

s

)random projections formeasurements

- reconstruct by a non linear mapping

IDCOM, University of Edinburgh

Compressed sensing Overview

Compressed Sensing assumes a

compressible set of signals, i.e.

approximately k-sparse.

Using approximately

𝑚 ≥ 𝒪 𝑘 log2

𝑁

𝑘

random projections for measurements

we have little or no information loss.

Signal reconstruction by a nonlinear

mapping.

Many practical algorithms with

guaranteed performance e.g. 𝐿1 min.,

OMP, CoSaMP, IHT.

Compressible

set of interest

random projection

(observation)

nonlinear

approximation

(reconstruction)

Observe 𝒙 ∈ ℝ𝑁 via 𝑚 ≪ 𝑁 measurements, 𝒚 ∈ ℝ𝑚 where 𝒚 = Φ𝒙

Mike Davies.

THIS is the course SUMMARY!!!

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Part 2 - Maths of compressive sensing -how it works?

Notations (Reminder)

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NormsDefinition (lp-norm)

The lp-norm of x ∈ Rn, p > 1 is defined as

||x ||p =

(

n∑i=1

|xi |p)1/p

p ∈ [1,∞)

maxi|xi | p =∞

If p < 1, definition still valid, but triangle inequality not satisfied⇒ quasi-norm.

Definition (inner product)

〈x , z〉 = zTx =n∑

i=1

xizi

See textbook F.R. for extension to Cn.21/ 66

Definition (support and l0-norm)

The support of a vector x is the index set of its non-zero entries, i.e.

supp (x) = {j ∈ [n] : xj 6= 0}, where [n] = {1, 2, ..., n}

The l0-norm of x is defined as

||x ||0 = card ( supp (x) )

||x ||0 counts the number of non-zero entries of x .||.||0 is not even a quasi-norm.

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Sparsity definition

Definition (s-sparse)

A signal x ∈ Rn is said to be s-sparse if it has at most s non-zeroentries, i.e. ||x ||0 ≤ s.

Definition (Σs)

We define Σs as the set containing all s-sparse signals, i.e.Σs = {x ∈ Rn : ||x ||0 ≤ s}.

Note 1: Sparsity is a highly nonlinear model (Σs is not a linear space)

Quiz 1

Note 2: in many practical cases, x is not sparse itself, but it has asparse representation in some basis Φ. We still say that x is s-sparse,with the understanding that we can write x = Φu, and ||u||0 ≤ s.

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Approximate sparsity

• A sparse signal can be represented exactly giving the positionsand values of its s nonzero components

• Real-world signals are rarely exactly sparse.We need to

I generalize the def: from “sparse’’ to “compressible” signals,I describe the representation error i.e. the error incurred

representing just s components of the signal.

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Best s-term approximation

The best s-term approximation picks the s components thatminimize the representation error

Definition (best s-term approximation)

For p > 0, the lp-error incurred by the best s-term approximation toa vector x ∈ Rn is given by

σs(x)p = minx∈Σs

||x − x ||p

• If x ∈ Σs , then σs(x)p = 0 for any p.

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Compressible signal

Optimal strategy to compute the best s-term approximation:thresholding

• Reorder the elements of x by decreasing magnitude

• Pick the first s elements, set all others to zero.

Definition (compressible signal)

a signal x ∈ Rn is said to be compressible if the error of its bests-term approximation decays quickly in si.e. if ∃C1, q > 0 such that |xi | ≤ C1i−q., when the coefficients havebeen ordered such that |x1| ≥ |x2|... ≥ |xn|.

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Sparsity support

Suppose x ∈ Rn. Let S ⊂ [n] and Sc ⊂ [n]\S• S : sparsity support of x , i.e. the locations of the nonzero

coefficients of x

• Sc : set of locations of the 0 coefficients

• S for compressible signal: set of locations of the coefficientsbelonging to the best s-term approximation of x .

Notation

xS vector obtained by setting the entries of x indexed by Sc to 0.

MS matrix obtained by setting the columns of M indexed by Sc to 0.

• Same notation to denote vectors/matrices where theelements/columns have been removed, instead of being set to 0

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Sensing process

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Compressive sensing

Goal of Compressive sensing (CS): achieve results similar to bests-term approximation using a nonadaptive acquisition/encoder.

Definition (CS)

Let x ∈ Rnx1 be a signal.Let M ∈ Rmxn, with m < n, be a sensing matrix.We take linear measurements of the signal as

y = Mx

with y ∈ Rmx1. Also, yj = Mjx , where Mj is the j-th row of M.

FIRST, assume that x is sparse.

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Sensing process Sensing process

= .

y M

x

• How should we choose a “good” matrix A with m π n??• How do we reconstruct x from y , given A?

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• How do we reconstruct x from y , given M?

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Reconstruction process

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Reconstruction

Now we have taken measurements

y = Mx ,

and want to reconstruct x from y , given knowledge of M .

• M ∈ Rmxn is not square ⇒ not invertible

• Underdetermined system of equations,⇒ infinitely many solutions

• Which one should we pick out of all z’s satisfying y = Mz?

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Least square solution and pseudo inverse

s

sss

Tall matrix + full column rank: A injective (one-to-one)overdetermined system, (at most 1 solution)

x = argminz||y − Az ||2

= A+y = (ATA)−1AT y

PIm(A)y = y = AA+y

Recall: used in OMPs

s

ssss

Fat matrix + full row rank: A surjective (onto)underdetermined/overcomplete system, (∞ nb. solutions)

x = arg minz:y=Az

||z ||2

= A+y = AT (AAT )−1y

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What’s wrong with the pseudoinverse?Example: x ∈ R128x1, s = 3,m = 10,M ∈ R10x128 i.i.d. Gaussian

What’s wrong with the pseudoinverse?Example: x œ R128◊1, K = 3, M = 10, A œ R10◊128

pseudorandom Gaussian

20 40 60 80 100 120-5

0

5

10

15

20x

20 40 60 80 100 120-6

-4

-2

0

2

4

6M+ y

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Minimum l0-norm solution

x = arg minz∈Rn||z ||0 subject to Mz = y

Complexity?

• Problem is non-convex

• Problem is NP-hard:for a given s, try all possible

(ns

)supports, estimate the s

nonzero values of x , check if constraint is satisfied⇒ infeasible for practical problem sizes

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Practical philosophies


Greedyalgorithms

Focus on ||x ||0

Thresholdingalgorithms

Focus on y ∼ Mx

Convex relaxationalgorithms

Solve a nicer problem

Courtesy N. Bertin

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Convex relaxation: minimum l1-norm

Basis Pursuit (BP)


• This problem has polynomial complexity, O(n3)

• It can be solved using linear programming techniques⇒ efficient solvers available

• Solutions to the l1 problem still tend to be rather sparse• Equivalence of solutions between l1 and l0 problems? (See N.

Bertin course)

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Signal sparse in transform domain

Real signals are rarely directly sparse...but rather sparse in a transform domain

original image DCT coefficients of the imageoriginal image in the transform domain

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Signal sparse in transform domain

Let us assume that

• x is not sparse

• x sparse in a transform domain: ∃Φ s.t. x = Φu, u sparse

Then sensing problem is exactly the same.

• We sense y = Mx , with x nonsparse

Since y = Mx = MΦu, with u sparse, reconstruction becomes

• “Solve” y = MΦu for u:

u = arg minz∈Rn||z ||1 subject to MΦz = y

• We obtain x = Φu

• Make sure that MΦ (and not M) is a “good” sensing matrix

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Signal sparsevs signal sparse in transform domain

x sparse

SENSINGy = Mx

RECONSTRUCTIONx = argminz∈Rn ||z ||1

subject to Mz = y.

x = Φu, u sparse

SENSINGy = Mx

RECONSTRUCTIONu = argminz∈Rn ||z ||1

subject to MΦz = yx = Φu

In conclusion: sparse vs sparse in the transform domain: the same!!

• same sensing

• similar reconstruction problem

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Good sensing matrices? First insights

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Sensing process Sensing process

= .

y M

x

• How should we choose a “good” matrix A with m π n??• How do we reconstruct x from y , given A?

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• How should we choose a “good” matrix M with m� n?

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Sensing matrices that are not good

Sensing matrices that are not good

0 5 10 15 20 25 30 35−1

−0.5

0

0.5

1

1.5

2

2.5

3

xMk

• Vector y is all zero!I We need A to be di�erent from x (in this case non-sparse)

∆ incoherence

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Vector y is all zero!→ If x sparse, M must be non-sparse→ We need M to be different from x

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Compressive sensing vs other schemes

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Linear approximation

• Linear transform (DCT, wavelet, ...)

• Approximate the coefficients in a linear way:retain certain fixed coefficients

• Send the non 0 coefficients

+ Send only transform coefficients (support is known)

- Reconstruction not efficient (not all vectors have same supports)

||x − x ||2 = D requires R = 64sLA (if 64 bits used per coefficient)

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Nonlinear approximation

• Linear transform (DCT, wavelet, ...)

• Approximate the coefficients in a nonlinear way: retaincoefficients that best approximate x (Best s-term approximation)

• Send support and values of transform coefficients

+ Adaptive representation, Φ(S)

- Computation not efficient (acquire a lot of coeff and set many to 0)

||x − x ||2 = D requires R = 64sNLA + sNLA log2 n, with sNLA ≤ sLA

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Sparse approximation

• Approximate the coefficients in a nonlinear way:(Best s-term approximation)

• Send support and values of transform coefficients

+ Adaptive representation, D(S)

+/- Less non 0 coeff but higher cost for position (D fat)

- High complexity at encoder

||x − x ||2 = D requires R = 64sSA + sSA log2 d ,with sSA << sNLA ≤ sLA and d >> n

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Compressive sensing

• Acquire directly “few” numbers in a nonadaptive way

+ Universal acquisition

+ Low complexity at encoder

- High complexity at decoder

? Can universal acquisition perform as well as an adaptive model?

→ Surprisingly yes, to some extent

||x − x ||2 = D requires R = 64m

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More on Compressive sensing (CS)vs Sparse approximation (SA)

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CS vs SA (con’t)

Non-linear solvers:

CS Given y and M , find xsparse such that Mx ≈ y .

Return x with guarantee that||x − x || small

SA Given x and D, find csparse such that x = Dc ≈ x .

Return x with guarantee that||x − x || = ||D(c − c)|| small

CS: proximity to the true root

SA: proximity to zero in the range of the function

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CS vs SA (con’t)

Non-linear solvers:





Same decompositionalgorithms

Different criteria

CS: proximity to the true root

SA: proximity to zero in the range of the function

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CS vs SA (con’t)

Non-linear solvers:





Root-finding algorithm:

CS Given y = 0 and f , find xsuch that y = 0 ≈ f (x).


SA Given y = 0 and f , find xsuch that y = 0 ≈ y = f (x).

Return y with guarantee that||f (x)− 0|| small

CS: proximity to the true rootSA: proximity to zero in the range of the function

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CS vs SA (con’t)

𝜖𝜖𝜖𝜖 x

yy=f(x)

solutions of SAsolutions of CS

Root-finding algorithm:

CS Given y = 0 and f , find xsuch that y = 0 ≈ f (x).Return x with guarantee that

||x − x || small

SA Given y = 0 and f , find xsuch that y = 0 ≈ y = f (x).Return y with guarantee that

||f (x)− 0|| small

CS: proximity to the true rootSA: proximity to zero in the range of the function

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Part 3 - Compressive sensing -why it works?

.

Note: analysis performed for the more general inverse problemwith generic matrix A instead of M

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The problem: invert y = Ax

A square

∃ a reconstruction map:

Rm → Rn

y 7→ x = A−1y

mcondition on the matrix

rank(A) = m = n

ker(A) = {z : Az = 0} = {0}

.

.

.51/ 66

The problem: invert y = Ax

A square


Rm → Rn

y 7→ x = A−1y

mcondition on the matrix

rank(A) = m = n

ker(A) = {z : Az = 0} = {0}

.

.

.

A fat


Rm → Rn

y 7→ x =??

mcondition on the matrix ???NEW Reduce the domain of

definition of A: s-sparse

.51/ 66

Part 3 - Maths of CS: why it works?

A magic tool: concentration inequality

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Part 3 - Maths of CS: why it works?

Conditions for exact reconstruction with P1

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Null space property (NSP): definition (1/3)

Definition (NSP(s))

Let A be an m× n matrix. Then A has the null space property (NSP)of order s if, for all v ∈ ker(A) \ 0 and for all index sets S s.t. |S | ≤ s

||vS ||1 < ||vSc ||1 or equivalently ||vS ||1 <1

2||v ||1

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NSP: Uniqueness of P1 (2/3)

Definition (NSP(s))



2||v ||1

Theorem (F&R Th4.5 NSP ⇔ recovery P1)

Let A be an m× n matrix, and let s ≤ m. Then, these conditions areequivalent:

(i) If a solution x of (P1) satisfies ||x ||0 ≤ s, then this is the uniquesolution.

(ii) A satisfies the null space property of order s.

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NSP: Stronger NSP (3/3)

Definition (NSP(s))



2||v ||1

Definition (mixed-norm NSP(s))

Let A be an m × n matrix. Then A has the strong l2 null spaceproperty (NSP) of order s if, for all v ∈ ker(A) \ 0 and for all indexsets S s.t. |S | ≤ s

||vS ||2 <1

2√

s||v ||1

Theorem (mixed-norm NSP(s) ⇒ NSP(s))

The standard NSP(s) is implied by the mixed-norm NSP(s).56/ 66

The restricted isometry property (RIP) (1/2)

Change of notation: s → k , since reference is now Foucart CS-Notes

Definition (Foucart CS-Notes Def. 8.1. RI Ratio)

The k-th order Restricted Isometry Ratio of the measurement matrixA is

γk(A) =βk(A)

αk(A)≥ 1

where αk(A) and βk(A) are the largest and smallest positiveconstants α and β s.t.

∀v ∈ Σk , α||v ||22 ≤ ||Av ||22 ≤ β||v ||22

• A preserves the Euclidean norm of k-sparse vectors

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RIP: alternative definition RIP (2/2)

Definition (RIP and RI constant)

A satisfies the k-th order RIP with constant δ ∈ (0, 1) if

∀v ∈ Σk , (1− δ)||v ||22 ≤ ||Av ||22 ≤ (1 + δ)||v ||22

The smallest such constant δk(A) is the RI constant of the matrix A.

Note that

γk(A) =1 + δk(A)

1− δk(A)

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RIP ⇒ NSP ⇔ recoveryTheorem (RIP ⇒ mixed norm NSP(s) ⇒ NSP(s) ⇔ recovery)

If A satisfies RIP with RI ratio γ2s =β2s

α2s, i.e.

∀v ∈ Σ2s , α2s ||v ||22 ≤ ||Av ||22 ≤ β2s ||v ||22

then

∀v ∈ ker(A)/{0},∀S s.t. |S | ≤ s, ||vS ||2 ≤η√s||v ||1 (1)

where η =γ2s − 1

2=

1

2

β2s − α2s

α2s.

So that if γ2s < 2 then A satisfies the mixed norm NSPi.e. every s-sparse vector x ∈ Rn is the unique solution of (P1) withy = Ax .

Foucart CS-Notes Th. 8.2. Exercise.59/ 66

Concentration ⇒ RIP and condition on mMatrices that satisfy the RIP

Theorem (Foucart CS-Notes Th. 9.1. concentration ⇒ RIP)

Let A be an m × n random matrix drawn according to a probabilitydistribution satisfying the concentration inequality

∀x ∈ Rn,∀ε ∈ (0, 1),P(∣∣∣||Ax ||22 − ||x ||22

∣∣∣ > ε||x ||22)≤ 2 exp(−c(ε)m)

(2)where c(ε) is a constant depending only on ε. Then, for eachδ ∈ (0, 1), there exist constants c0(δ), c1(δ) > 0 depending on δ andon the probability distribution of A such that

P(∣∣∣||Ax ||22 − ||x ||22

∣∣∣ > δ||x ||22,∃x ∈ Σk

)≤ 2 exp(−c0(δ)m) (3)

provided thatm ≥ c1(δ)k ln(e n/k).

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Matrices that satisfy the concentrationinequality

Theorem (Gaussian ⇒ concentration around the mean)

Suppose that the aij entries of the m × n random matrix A areindependent realizations of Gaussian random variables, centered withvariance σ2. For each x ∈ Rn, ε ∈ (0, 1)

P(∣∣∣||Ax ||22 −mσ2||x ||22

∣∣∣ > εmσ2||x ||22)≤ 2 exp

((− ε2

2+ε3

6

)m

).

Foucart CS-Notes Section 9.3.It also holds for

• any subgaussian iid matrices.

• subsampled Fourier matrices (cf. ComputerLab).

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Summary

• if A follows a subgaussian distribution with m = O (s ln(n/s)),

[easy construction / easy to verify...]

then A will satisfy the RIP of order 2s

with probability at least 1− 2e−c0(δ)m)

and hence the NSP

and hence exact reconstruction under P1

• Gaussian, Bernoulli (Rademacher entries) matrices ...,subsampled Fourier matrices satisfy the RIP.

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Part 4 - Compressive sensing -what it is good for?

.

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How to spot a compressive sensing system?

Case 1

• Think about systems that use a raster mode for samplingthen think of physical ways to perform multiplexing instead

• Once you perform the multiplexing,use compressive sensing solvers to reconstruct signal

• Does it work better or as well with fewer measurements ?

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Example: single pixel camera

Classical i=1

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Classical i=2

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Classical i=3

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Classical i=4

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Classical i=5

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Classical i=1000

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Classical i=10000000

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Compressive sensing i=1

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Compressive sensing

if image is 3-sparse, then ∼ 3 measurements are enough

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How to spot a compressive sensing system?

Case 2

• Look for acquisition schemes that multiplexes a signal already

• Is the signal produced by this system sparse in some basis?

• If yes, subsample the acquisition,use compressive sensing solvers to reconstruct signal

• Does it work better or as well with fewer measurements ?

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Documents

Foundations of Smart Sensing - Inriapeople.rennes.inria.fr/Aline.Roumy/teaching/CompressiveSensing_Sli… · Foundations of Smart Sensing. Teachers: Nancy Bertin CNRS/IRISA, Panama