A Robust Form of Kruskal's Identifiability Theorem · 2020. 1. 3. · randomly sampling rwords...

A Robust Form of Kruskal’s IdentifiabilityTheorem

Aditya Bhaskara (Google NYC)

Joint work withMoses Charikar (Princeton University)

Aravindan Vijayaraghavan (NYU → Northwestern)

Background: “understanding” data

Data from various sources:

QUESTION (“simple explanation”): can we think of data asbeing generated from a model with a small number of

parameters?

Very successful; e.g. topic models for documents, hiddenmarkov models for speech, . . .

Background: “understanding” data

Data from various sources:

QUESTION (“simple explanation”): can we think of data asbeing generated from a model with a small number of

parameters?

Very successful; e.g. topic models for documents, hiddenmarkov models for speech, . . .

Topic model for documents

Large collection of documents

Carricature model: each document is about a topic, and eachtopic is a distribution over n words;

aardvark apple ball lion . . . zebra

sports 0 0.0002 0.005 0.0001 . . .

wildlife 0.0006 0 0.0001 0.005 . . .

Model parameters:

I probability that document is on topic i : wi (sum to 1)

I word probability vector for topic i : pi ∈ Rn

Carricature model: each document is about a topic, and eachtopic is a distribution over n words; e.g.

sports 0 0.0002 0.005 0.0001 . . .

wildlife 0.0006 0 0.0001 0.005 . . .

Model parameters:

Carricature model: each document is about a topic, and eachtopic is a distribution over n words; e.g.

sports 0 0.0002 0.005 0.0001 . . .

wildlife 0.0006 0 0.0001 0.005 . . .

Model parameters:

Generating a document: say r-word document

w1 w2wR

word ∼ p2word ∼ p2

word ∼ p2

r words

ASSUMPTION: picking a document from corpus at random,randomly sampling r words ≡ picking r word document as

Goal: find the {wr, pr}

w1 w2wR

word ∼ p2

r words

w1 w2wR

word ∼ p2

r words

What about tensors?

Experiment: Pick three random words from random document

QUESTION: What is

Pr[word1 = i,word2 = j,word3 = k]?

Precisely equal to∑R

r=1wrpr(i)pr(j)pr(k).

(i, j, k)

r=1wr(pr ⊗ pr ⊗ pr).

∴ Finding parameters ≡ tensor decomposition!

What about tensors?

QUESTION: What is

(i, j, k)

What about tensors?

QUESTION: What is

(i, j, k)

Recipe for tensor methods in mixture models

Algebraic statistics literature: [Allman, Mathias, Rhodes 09], ...

1. Estimate a tensor whose decomposition allows reading offthe model parameters

2. Use tensor decomposition

Many applications: mixtures of gaussians, hidden markovmodels, communities, crabs, ...

Are we done?

Two caveats

Efficiency

We need polynomial time algorithms for decomposition!Given T =

∑Rr=1 pr ⊗ pr ⊗ pr, can we find {pr}Rr=1?

RobustnessAlgorithm needs to work with noisy tensorIn applications, we estimate the tensor from samples. Nsamples =⇒ 1/

√N error per entry, typically.

GOAL: Given, target accuracy ε, and

T =R∑r=1

pr ⊗ pr ⊗ pr +N , with ‖N‖ < ε/poly(n),

recover {pr} up to error ε.

Two caveats

Efficiency

T =R∑r=1

Two caveats

Efficiency

R∑r=1

A success story: the full rank case

Given T =

R∑r=1

pr ⊗ pr ⊗ pr +N , ‖N‖ < ε/poly(n)

Define P : matrix (n×R) with columns {pr}.

TheoremIf σR(P ) > 1/poly′(n), then can find {pr} up to error ε in polynomialtime.

Discovered many times: [Harshman 72], [Leurgans, et al. 93], [Chang 96], [Anandkumar, et al.10], [Goyal et al. 13], ...

Given T =

R∑r=1

Given T =

R∑r=1

Beyond non degeneracy

Question: Can we approximately recover parameters underweaker conditions? What about the case R > n?

This talk: easier problem of identifiability.

Theorem (Informal theorem.)

If the Kruskal rank condition holds, then it is possible to recoverthe decomposition up to error ε.

I Not efficient /; open problem to do it efficiently

I Can be done if the components are nearly orthogonal[Anandkumar et al. 14]

Kruskal’s theorem (1977)

R∑r=1

ai ⊗ bi ⊗ ci

GOAL: Find conditions under which the decomposition unique.

Kruskal rankFor matrix A(n×R), the rank is the largest integer k(A) s.t.every k(A) columns of A are linearly independent.

Note: Much stronger than the usual notion of rank; reminiscent of restricted isometry

Theorem (Kruskal)

Suppose T is defined as above, and A,B,C are n×R matriceswith columns ar, br, cr, respectively.Then a sufficient condition for uniqueness of decomposition is

k(A) + k(B) + k(C) ≥ 2R+ 2

R∑r=1

ai ⊗ bi ⊗ ci

Theorem (Kruskal)

k(A) + k(B) + k(C) ≥ 2R+ 2

R∑r=1

ai ⊗ bi ⊗ ci

Theorem (Kruskal)

k(A) + k(B) + k(C) ≥ 2R+ 2

R∑r=1

ai ⊗ bi ⊗ ci

Theorem (Kruskal)

k(A) + k(B) + k(C) ≥ 2R+ 2

Our result

R∑r=1

ai ⊗ bi ⊗ ci +N

GOAL: Find conditions under which decomposition is unique∗.

Robust Kruskal rankFor matrix A(n×R) and parameter τ > 0, the rank is thelargest integer kτ (A) s.t. every n× kτ (A) submatrix hascondition number < τ .

Note: Recall that condition number of B is σmax(B)/σmin(B).

Theorem (Rough)

Let T be as above, and A,B,C be n×R matrices as before.Then the decomposition is robustly unique if

kτ (A) + kτ (B) + kτ (C) ≥ 2R+ 2

Our result

R∑r=1

ai ⊗ bi ⊗ ci +N

Theorem (Rough)

kτ (A) + kτ (B) + kτ (C) ≥ 2R+ 2

Our result

R∑r=1

ai ⊗ bi ⊗ ci +N

Theorem (Rough)

kτ (A) + kτ (B) + kτ (C) ≥ 2R+ 2

Our result

rank k-tensors

We show: For any ε > 0, there is an ε′ = ε/poly(n), such thatT =ε′ T

′ implies the decompositions are ε-close, up to apermutation.

Our result

R∑r=1

ai ⊗ bi ⊗ ci ; T ′ =

R∑r=1

a′i ⊗ b′i ⊗ c′i

Theorem (Formal)Suppose T, T ′ are defined as above, and A,B,C,A′, B′, C ′ are n×Rmatrices. Further, suppose for some τ > 0, that

kτ (A) + kτ (B) + kτ (C) ≥ 2R+ 2.

Then for any ε > 0, there is an ε′ = ε/poly(n, τ) such that if‖T − T ′‖ < ε′, then there exist diagonal matrices ΓA,ΓB ,ΓC , and apermutation Π such that ΓAΓBΓC = I, and

A′ =ε ΓAΠA, B′ =ε ΓBΠB, C ′ =ε ΓCΠC.

Remarks

I Conditions only about kτ (A), kτ (B), kτ (C); nothing isneeded about A′, B′, C ′ (except an upper bound on columnlengths)

I Implies that no tensor close to T has rank < R

I Naturally generalizes to higher order tensors (similar to DeLathauwer’s extension of Kruskal’s theorem)

Main difficulty in proof: handling 1/poly(n) noise

(If ε′ = ε/exp(n), much easier to show that decompositions are ε-close)

Remarks

Proof overview

Suppose A,B,C,A′, B′, C ′ are n×R, column lengths ≤ ρ, and

kτ (A) = kτ (B) = kτ (C) = n ; R = 4n/3.

(I.e., any n columns of A,B,C are well conditioned (thus lin.ind.))

1. Show that A′ is a scaled permutation of A, B′ of B, C ′ of C

2. Show that permutations are equal, and scalings multiply toidentity

Crux: first part – “permutation lemma”

Proof overview

kτ (A) = kτ (B) = kτ (C) = n ; R = 4n/3.

Proof overview

kτ (A) = kτ (B) = kτ (C) = n ; R = 4n/3.

Permutation lemma

Sufficient conditions for A, A′ having same columns (up topermutation)

(Suppose for now, that no two cols of A′ are parallel)∗∗

Key idea. Look at the spaces spannedby subsets of columns of A,A′.

Definition: 〈AS〉 := span of columnsindexed by S ⊆ [R]

Informal: Suppose for all S of size (n− 1), 〈A′S〉 “contains” atleast |S| columns of A. Then the same is true for all S.

(If true for |S| = 1, then every column of A′ is parallel to some column of A)

Permutation lemma

Downward induction: base case |S| = (n− 1)

A A′

a1 ⊗ b1 ⊗ c1 + a2 ⊗ b2 ⊗ c2 + · · · ≈ a′1 ⊗ b′1 ⊗ c′1 + a′2 ⊗ b′2 ⊗ c′2 + . . .

For any w,∑4n/3r=1 〈w, ar〉(br ⊗ cr) ≈

∑4n/3r=1 〈w, a′r〉(b′r ⊗ c′r).

1. Pick w to be orthogonal to 〈A′S〉2. Only n/3 + 1 terms remain on RHS =⇒ at most this

number must remain on LHS! (by Kruskal condition)

3. Implies at least (n− 1) cols {ar} belong to 〈A′S〉

A A′

a1 ⊗ b1 ⊗ c1 + a2 ⊗ b2 ⊗ c2 + · · · ≈ a′1 ⊗ b′1 ⊗ c′1 + a′2 ⊗ b′2 ⊗ c′2 + . . .

∑4n/3r=1 〈w, a′r〉(b′r ⊗ c′r).

1. Pick w to be orthogonal to 〈A′S〉

2. Only n/3 + 1 terms remain on RHS =⇒ at most thisnumber must remain on LHS! (by Kruskal condition)

A A′

a1 ⊗ b1 ⊗ c1 + a2 ⊗ b2 ⊗ c2 + · · · ≈ a′1 ⊗ b′1 ⊗ c′1 + a′2 ⊗ b′2 ⊗ c′2 + . . .

∑4n/3r=1 〈w, a′r〉(b′r ⊗ c′r).

A A′

a1 ⊗ b1 ⊗ c1 + a2 ⊗ b2 ⊗ c2 + · · · ≈ a′1 ⊗ b′1 ⊗ c′1 + a′2 ⊗ b′2 ⊗ c′2 + . . .

∑4n/3r=1 〈w, a′r〉(b′r ⊗ c′r).

A A′

a1 ⊗ b1 ⊗ c1 + a2 ⊗ b2 ⊗ c2 + · · · ≈ a′1 ⊗ b′1 ⊗ c′1 + a′2 ⊗ b′2 ⊗ c′2 + . . .

∑4n/3r=1 〈w, a′r〉(b′r ⊗ c′r).

Downward induction

Show for |S| = (n− 2), . . . , 1:

I start with some S of size (n− 2)

I define Si = S ∪ {i}, for various i 6∈ S — (n/3) + 2 such..

I let Ti be indices of cols of A that lie in 〈A′Si〉; |Ti| ≥ (n− 1)

Main idea. Any col in Ti ∩ Tj must be contained in 〈A′S〉1. Say we have: a =

∑r∈S αra

′r + αia

′i =

∑r∈S βra

′r + βja

2. Since a′i, a′j are not parallel, this means αi = βj = 0

Downward induction

Show for |S| = (n− 2), . . . , 1:

∑r∈S αra

′r + αia

′i =

∑r∈S βra

′r + βja

Downward induction

Show for |S| = (n− 2), . . . , 1:

∑r∈S αra

′r + αia

′i =

∑r∈S βra

′r + βja

Downward induction

Show for |S| = (n− 2), . . . , 1:

Main idea. Any col in Ti ∩ Tj must be contained in 〈A′S〉

1. Say we have: a =∑

r∈S αra′r + αia

′i =

∑r∈S βra

′r + βja

Downward induction

Show for |S| = (n− 2), . . . , 1:

∑r∈S αra

′r + αia

′i =

∑r∈S βra

′r + βja

Downward induction

Show for |S| = (n− 2), . . . , 1:

∑r∈S αra

′r + αia

′i =

∑r∈S βra

′r + βja

Downward induction

Show for |S| = (n− 2), . . . , 1:

Thus if T is the set of cols contained in 〈A′S〉,then for any i, j, Ti ∩ Tj = T

Thus Ti form a sunflower family

Downward induction

Show for |S| = (n− 2), . . . , 1:

Thus if T is the set of cols contained in 〈A′S〉,then for any i, j, Ti ∩ Tj = TThus Ti form a sunflower family

Downward induction

Claim. Implies that the core has size ≥ (n− 2)

Proof.Basic counting argument. We know |Ti| ≥ n− 1 for all i.

Say |T | = n− 3, then |Ti \ T | ≥ 2 for all i.

These don’t intersect for different i (sunflower!)

Thus the #(cols) of A is > (n− 3) + 2× (n/3 + 2) > (4n/3)..

Downward induction

Making things robust – permutation lemma

Sufficient conditions for A, A′ having approximately samecolumns (up to permutation)

Informal: Suppose for all S of size (n− 1), 〈A′S〉 contains atleast |S| columns of A, up to error ε. Then the same is true forall S, with error ε′.

(If true for singletons, then every col of A′ is ε′-close to a scaling of a col of A)

Making things robust – permutation lemma

Sufficient conditions for A, A′ having approximately samecolumns (up to permutation)

Informal: Suppose for all S of size (n− 1), 〈A′S〉 contains atleast |S| columns of A, up to error ε. Then the same is true forall S, with error ε′.

(If true for singletons, then every col of A′ is ε′-close to a scaling of a col of A)

Issue with inductive proof

Suppose we have claim for |S| = (n− 1) with error ε; then canprove claim for |S| = (n− 2) with error only ε · n.

Continuing this way, we only get the claim for |S| = 1 witherror ε · exp(n) /

Means the initial error should have been exponentially small..

The fix

I More combinatorial invariant (different definition of sets T )

I Stronger sunflower argument

I Better “base case”

The fix

Directions

Are there polynomial time algorithms under Kruskal’sconditions?

What about other tensor based algorithms in the case R > n?

(a) Higher order tensors: [Cardoso 92], next talk (also [Goyal etal. 13])(b) Recent work assuming incoherence [Anandkumar, et al. 14]

Can we show “robust” uniqueness under more algebraicconditions?

Can we show that a generic n× n× n tensor has a “stable”unique decomposition up to rank n2/4?

Directions

What about other tensor based algorithms in the case R > n?(a) Higher order tensors: [Cardoso 92], next talk (also [Goyal etal. 13])

(b) Recent work assuming incoherence [Anandkumar, et al. 14]

Directions

What about other tensor based algorithms in the case R > n?(a) Higher order tensors: [Cardoso 92], next talk (also [Goyal etal. 13])(b) Recent work assuming incoherence [Anandkumar, et al. 14]

Directions

Thank you!

Questions?

A Robust Form of Kruskal's Identifiability Theorem · 2020. 1. 3. · randomly sampling rwords...

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