Sparse Approximations

Sparse Approximations

Nick Harvey University of British Columbia

Approximating Dense Objectsby Sparse Objects

Floor joists

Wood Joists Engineered Joists


Bridges

Masonry Arch Truss Arch


Bones

Human Femur Robin Bone

Mathematically• Can an object with many pieces be

approximately represented by fewer pieces?

• Independent random sampling usually does well

• Theme of this talk: When can we beat random sampling?

Dense GraphSparse Graph

6 -1 -1 -1 -1 -1-1 4 -1 -1 -1-1 -1 6 -1 -1 -1

-1 5 -1 -1-1 -1 -1 7 -1 -1 -1-1 -1 -1 5-1 -1 -1 5 -1

-1 -1 -1 -1 6

6 -15 -1 -3

-1 28

-1 2 -11

-3 -1 52

Dense MatrixSparse Matrix

Talk Outline• Vignette #1: Discrepancy theory

• Vignette #2: Singular values and eigenvalues

• Vignette #3: Graphs

• Theorem on “Spectrally Thin Trees”

Discrepancy• Given vectors v1,…,vn2Rd with kvikp bounded.

Want y2{-1,1}n with ki yivikq small.• Eg1: If kvik1·1 then E ki yi vik1 ·• Eg2: If kvik1·1 then 9y s.t. ki yi vik1 ·

Spencer ‘85: Partial Coloring + Entropy MethodGluskin ‘89: Sidak’s LemmaGiannopoulos ‘97: Partial Coloring + SidakBansal ‘10: Brownian Motion + Semidefinite ProgramBansal-Spencer ‘11: Brownian Motion + Potential functionLovett-Meka ‘12: Brownian Motion

Non-algorithmic

Algorithmic

Discrepancy• Given vectors v1,…,vn2Rd with kvikp

bounded.Want y2{-1,1}n with ki yivikq small.

• Eg1: If kvik1·1 then E ki yi vik1 ·• Eg2: If kvik1·1 then 9y s.t. ki yi vik1 · • Eg3: If kvik1·¯, kvik1·±, and ki vik1·1, then

9y with ki yi vik1 · Harvey ’13: Using Lovasz Local Lemma.Question: Can log(±/¯2) factor be improved?

Partitioning sums of rank-1 matrices• Let v1,…,vn2Rd satisfy i vivi

T=I and kvik2·±.Want y2{-1,1}n with ki yivivi

Tk2 small.• Random sampling: E ki yivivi

Tk2 · .Rudelson ’96: Proofs using majorizing measures, then nc-Khintchine

• Marcus-Spielman-Srivastava ’13:9y2{-1,1}n with ki yivivi

Tk2 · .

2

Partitioning sums of matrices• Given dxd symmetric matrices M1,

…,Mn2Rd withi Mi=I and kMik2·±.Want y2{-1,1}n with ki yiMik2 small.

• Random sampling: E ki yiMik2 · Also follows from nc-Khintchine.Ahlswede-Winter ’02: Using matrix moment generating function.Tropp ‘12: Using matrix cumulant generating function.

Partitioning sums of matrices

• Given dxd symmetric matrices M1,…,Mn2Rd withi Mi=I and kMik2·±.Want y2{-1,1}n with ki yiMik2 small.

• Random sampling: E ki yiMik2 · • Question: 9y2{-1,1}n with ki yiMik2 · ?• Conjecture: Suppose i Mi=I and kMikSch-1·±.

9y2{-1,1}n with ki yiMik2 · ?–MSS ’13: Rank-one case is true– Harvey ’13: Diagonal case is true (ignoring

log(¢) factor)

False!

Partitioning sums of matrices

• Given dxd symmetric matrices M1,…,Mn2Rd withi Mi=I and kMik2·±.Want y2{-1,1}n with ki yiMik2 small.

• Random sampling: E ki yiMik2 · • Question: Suppose only that kMik2·1.

9y2{-1,1}n with ki yiMik2 · ?– Spencer/Gluskin: Diagonal case is true

Column-subset selection• Given vectors v1,…,vn2Rd with kvik2=1.

Let st.rank=n/ki viviTk2. Let .

9y2{0,1}n s.t. i yi=k and (1-²)2 · ¸k( i yivivi

T ).

Spielman-Srivastava ’09: Potential function argumentYoussef ’12: Let . 9y2{0,1}n

s.t. i yi=k, (1-²)2 · ¸k( i yivivi

T ) and ¸1( i yiviviT ) ·

(1+²)2.

Column-subset selectionup to the stable rank

• Given vectors v1,…,vn2Rd with kvik2=1.Let st.rank=n/ki vivi

Tk2. Let .For y2{0,1}n s.t. i yi=k, can we control ¸k( i yivivi

T ) and ¸1( i yiviviT ) ?

– ¸k can be very small, say O(1/d).– Rudelson’s theorem: can get ¸1 · O(log d) and

¸k>0.– Harvey-Olver ’13: ¸1 · O(log d / log log d) and ¸k>0.–MSS ‘13: If i vivi

T =I, can get ¸1 · O(1) and ¸k>0.

Graph Laplacian

Lu = D-A =

7 -2 -5-2 3 -1-5 -1 16 -

10-

1010

abcd

a b c d

weighted degree of node

c

negative of u(ac)

Graph with weights u: 5 102 1

Laplacian Matrix:

ab

dc

Effective Resistance from s to t: voltage difference when each edge e is a (1/ue)-ohm resistor and a 1-amp current source placed between s and t= (es-et)T Lu

y (es-et)Effective Conductance: cst = 1 / (effective resistance from s

to t)

Spectral approximation of graphs

®-spectral sparsifier: Lu ¹ Lw ¹ ®¢Lu

5 -1 -1 -1 -1 -14 -1 -1 -1 -1

-1 -1 6 -1 -1 -1 -1-1 5 -1 -1 -1 -1

-1 -1 -1 7 -1 -1 -1 -1-1 -1 -1 5 -1 -1-1 -1 -1 5 -1 -1

-1 -1 -1 -1 6 -1 -1-1 -1 -1 -1 -1 5

-1 -1 -1 -1 -1 -1 6

6 -1 -55 -1 -3 -1

-1 2 -18 -8

-1 2 -11 -1

-3 -1 5 -12 -1 -1

-5 -1 -1 -1 8-1 -8 -1 10

Edge weights u

Edge weights w

Lu = Lw =

Ramanujan Graphs• Suppose Lu is complete graph on n vertices

(ue=1 8e).• Lubotzky-Phillips-Sarnak ’86:

For infinitely many d and n, 9w2{0,1}E such that e we=dn/2 (actually Lw is d-regular)and

• MSS ‘13: Holds for all d¸3, and all n=c¢2k.• Friedman ‘04: If Lw is a random d-regular graph,

then 8²>0

with high probability.

Spectrally-thin trees• Question: Let G be an unweighted graph with n

vertices. Let C = mine (effective conductance of edge e).Want a subtree T of G with .

• Equivalent to

• Goddyn’s Conjecture ‘85: There is a subtree T with

– Relates to conjectures of Tutte (‘54) on nowhere-zero flows,and to approximations of the traveling salesman problem.

Spectrally-thin trees• Question: Let G be an unweighted graph with n

vertices. Let C = mine (effective conductance of edge e).Want a subtree T of G with .

• Rudelson’s theorem: Easily gives ® = O(log n).• Harvey-Olver ‘13: ® = O(log n / log log n).

Moreover, there is an efficient algorithm to find such a tree.

• MSS ’13: ® = O(1), but not algorithmic.

Given an (unweighted) graph G with eff. conductances ¸ C.Can find an unweighted tree T with

Spectrally Thin Trees

Proof overview:1. Show independent sampling gives

spectral thinness, but not a tree.► Sample every edge e independently with

prob. xe=1/ce

2. Show dependent sampling gives a tree, and spectral thinness still works.

Matrix ConcentrationTheorem: [Tropp ‘12]Let Y1,…,Ym be independent, PSD matrices of size nxn.Let Y=i Yi and Z=E [ Y ]. Suppose Yi ¹ R¢Z a.s. Then

Define sampling probabilities xe = 1/ce. It is known that e xe

= n–1.Claim: Independent sampling gives T µ E with E [|T|]=n–1 and

Theorem [Tropp ‘12]: Let M1,…,Mm be nxn PSD matrices.Let D(x) be a product distribution on {0,1}m with marginals x.Let Suppose Mi ¹ Z.ThenDefine Me = ce¢Le. Then Z = LG and Me ¹ Z holds.Setting ®=6 log n / log log n, we get whp.But T is not a tree!

Independent sampling

Laplacian of the single edge eProperties of conductances used



Proof overview:1. Show independent sampling gives spectral thinness,

but not a tree.► Sample every edge e independently with prob.

xe=1/ce

2. Show dependent sampling gives a tree, and spectral thinness still works.► Run pipage rounding to get tree T with Pr[ e2T ] = xe =

1/ce

Pipage rounding[Ageev-Svirideno ‘04, Srinivasan ‘01, Calinescu et al. ‘07, Chekuri et al. ‘09]

Let P be any matroid polytope.E.g., convex hull of characteristic vectors of spanning trees.Given fractional x

Find coordinates a and b s.t. linez x + z ( ea – eb ) stays in current faceFind two points where line leaves PRandomly choose one of thosepoints s.t. expectation is x

Repeat until x = ÂT is integral

x is a martingale: expectation of final ÂT is original fractional x.

ÂT1ÂT2

ÂT3

ÂT4

ÂT5

ÂT6

x

Say f : Rm ! R is concave under swaps if z ! f( x + z(ea-eb) ) is concave 8x2P, 8a, b2[m].Let X0 be initial point and ÂT be final point visited by pipage rounding.Claim: If f concave under swaps then E[f(ÂT)] · f(X0). [Jensen]

Let E µ {0,1}m be an event.Let g : [0,1]m ! R be a pessimistic estimator for E, i.e.,

Claim: Suppose g is concave under swaps. Then Pr[ ÂT 2 E ] · g(X0).

Pipage rounding and concavity

Chernoff BoundChernoff Bound: Fix any w, x 2 [0,1]m and let ¹ = wTx.Define . Then,

Claim: gt,µ is concave under swaps. [Elementary calculus]

Let X0 be initial point and ÂT be final point visited by pipage rounding.Let ¹ = wTX0. Then Bound achieved by independent sampling also achieved by pipage rounding

Matrix Pessimistic Estimators

Main Theorem: gt,µ is concave under swaps.

Theorem [Tropp ‘12]: Let M1,…,Mm be nxn PSD matrices.Let D(x) be a product distribution on {0,1}m with marginals x.Let Suppose Mi ¹ Z.LetThen and .

Bound achieved by independent sampling also achieved by pipage rounding

Pessimistic estimator



Proof overview:1. Show independent sampling gives spectral thinness,

but not a tree.► Sample every edge e independently with prob. xe=1/ce

2. Show dependent sampling gives a tree, and spectral thinness still works.► Run pipage rounding to get tree T with Pr[ e2T ] = xe =

1/ce

Matrix AnalysisMatrix concentration inequalities are usually proven via sophisticated inequalities in matrix analysisRudelson: non-commutative Khinchine inequalityAhlswede-Winter: Golden-Thompson inequalityif A, B symmetric, then tr(eA+B) · tr(eA eB).Tropp: Lieb’s concavity inequality [1973]if A, B Hermitian and C is PD, then z ! tr exp( A + log(C+zB) ) is concave.Key technical result: new variant of Lieb’s theoremif A Hermitian, B1, B2 are PSD, and C1, C2 are PD, then z ! tr exp( A + log(C1+zB1) + log(C2–zB2) ) is concave.

QuestionsCan Spencer/Gluskin theorem be

extended to matrices?Can MSS’13 be made algorithmic?Can MSS’13 be extended to large-rank

matrices?O(1)-spectrally thin trees exist. Can one

be found algorithmically?Are O(1)-spectrally thin trees helpful for

Goddyn’s conjecture?

Sparse Approximations

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