Download Sketching as a Tool for Numerical Linear Algebra

Sketching as a Tool for Numerical
Linear Algebra
All Lectures
David Woodruff
IBM Almaden
Massive data sets
Examples
 Internet traffic logs
 Financial data
 etc.
Algorithms
 Want nearly linear time or less
 Usually at the cost of a randomized approximation
2
Regression analysis
Regression
 Statistical method to study dependencies between
variables in the presence of noise.
3
Regression analysis
Linear Regression
 Statistical method to study linear dependencies
between variables in the presence of noise.
4
Regression analysis
Linear Regression
 Statistical method to study linear dependencies
between variables in the presence of noise.
Example Regression
Example
 Ohm's law V = R ∙ I
250
200
150
Example Regression
100
50
0
0
50
100
150
5
Regression analysis
Linear Regression
 Statistical method to study linear dependencies
between variables in the presence of noise.
Example Regression
Example
 Ohm's law V = R ∙ I
 Find linear function that
best fits the data
250
200
150
Example Regression
100
50
0
0
50
100
150
6
Regression analysis
Linear Regression
 Statistical method to study linear dependencies between
variables in the presence of noise.
Standard Setting
 One measured variable b
 A set of predictor variables a1 ,…, a d
 Assumption:
b = x0 + a 1 x1 + … + a d xd + e
 e is assumed to be noise and the xi are model
parameters we want to learn
 Can assume x0 = 0
 Now consider n observations of b
7
Regression analysis
Matrix form
Input: nd-matrix A and a vector b=(b1,…, bn)
n is the number of observations; d is the number of
predictor variables
Output: x* so that Ax* and b are close
 Consider the over-constrained case, when n À d
 Can assume that A has full column rank
8
Regression analysis
Least Squares Method
 Find x* that minimizes |Ax-b|22 = S (bi – <Ai*, x>)²
 Ai* is i-th row of A
 Certain desirable statistical properties
9
Regression analysis
Geometry of regression
 We want to find an x that minimizes |Ax-b|2
 The product Ax can be written as
A*1x1 + A*2x2 + ... + A*dxd
where A*i is the i-th column of A
 This is a linear d-dimensional subspace
 The problem is equivalent to computing the point of the
column space of A nearest to b in l2-norm
10
Regression analysis
Solving least squares regression via the normal equations
 How to find the solution x to minx |Ax-b|2 ?
 Equivalent problem: minx |Ax-b |22
 Write b = Ax’ + b’, where b’ orthogonal to columns of A
 Cost is |A(x-x’)|22 + |b’|22 by Pythagorean theorem
 Optimal solution x if and only if AT(Ax-b) = AT(Ax-Ax’) = 0
 Normal Equation: ATAx = ATb for any optimal x
 x = (ATA)-1 AT b
 If the columns of A are not linearly independent, the MoorePenrose pseudoinverse gives a minimum norm solution x
11
Moore-Penrose Pseudoinverse
• Any optimal solution x has the form A− b + I − V′V′T z, where
V’ corresponds to the rows i of V T for which Σ𝑖,𝑖 > 0
•
Why?
•
Because A I − V′V′T z = 0, so A− b + I − V′V′T z is a
solution. This is a d-rank(A) dimensional affine space so it
spans all optimal solutions
•
Since A− b is in column span of V’, by Pythagorean theorem,
|A− b + I − V′V′T z|22 = A− b 22 + |(I − V ′ V ′T )z|22 ≥ A− b 22
12
Time Complexity
Solving least squares regression via the normal equations
 Need to compute x = A-b
 Naively this takes nd2 time
 Can do nd1.376 using fast matrix multiplication
 But we want much better running time!
13
Sketching to solve least squares regression
 How to find an approximate solution x to minx |Ax-b|2 ?
 Goal: output x‘ for which |Ax‘-b|2 · (1+ε) minx |Ax-b|2
with high probability
 Draw S from a k x n random family of matrices, for a
value k << n
 Compute S*A and S*b
 Output the solution x‘ to minx‘ |(SA)x-(Sb)|2
 x’ = (SA)-Sb
14
How to choose the right sketching matrix S?
 Recall: output the solution x‘ to minx‘ |(SA)x-(Sb)|2
 Lots of matrices work
 S is d/ε2 x n matrix of i.i.d. Normal random variables
 To see why this works, we
introduce the notion of a
subspace embedding
15
Subspace Embeddings
• Let k = O(d/ε2)
• Let S be a k x n matrix of i.i.d. normal
N(0,1/k) random variables
• For any fixed d-dimensional subspace, i.e.,
the column space of an n x d matrix A
– W.h.p., for all x in Rd, |SAx|2 = (1±ε)|Ax|2
• Entire column space of A is preserved
Why is this true?
Subspace Embeddings – A Proof
• Want to show |SAx|2 = (1±ε)|Ax|2 for all x
• Can assume columns of A are orthonormal
(since we prove this for all x)
• Claim: SA is a k x d matrix of i.i.d. N(0,1/k)
random variables
– First property: for two independent random variables X
and Y, with X drawn from N(0,a2 ) and Y drawn from
N(0,b2 ), we have X+Y is drawn from N(0, a2 + b2 )
X+Y is drawn from N(0,
2
𝑎
+
2
𝑏 )
2
• Probability density function 𝑓𝑧 of Z = X+Y𝑏2is
𝑧
𝑥− 2 2
2
𝑎 +𝑏 𝑓 and 𝑓
𝑧
convolution of probability density
𝑋
𝑌
− 2 functions
−
2
−
•
𝑧−𝑥 2
2𝑎2
2 𝑎 +𝑏
𝑥2
− 2
2𝑏
2
Calculation: 𝑒
= 𝑒
𝑓𝑍 𝑧 = ∫ 𝑓𝑌 𝑧 − 𝑥 𝑓𝑋 𝑥 𝑑𝑥
• 𝑓𝑥 𝑥 =
•Density
𝑓𝑍 𝑧
2 /2𝑎 2
1
−𝑥
𝑒
𝑎 2𝜋 .5
, 𝑓𝑦 𝑦 =
2 /2𝑏2
1
−𝑥
𝑒
2𝑧 2
𝑏
.5
𝑏 2𝜋
𝑥− 2 2
𝑎 +𝑏
−
𝑎𝑏 2
2
/2𝑏2 𝑎2 +𝑏2
.5
𝑎2 +𝑏2 2
2 /2𝑎 2
1
1
−(𝑧−𝑥)
−𝑥
of
distribution:
= Gaussian
∫
𝑒
𝑒
∫
2𝜋 .5 𝑎𝑏
𝑎 2𝜋 .5
𝑏 2𝜋 .5
−
=
1
−𝑧 2 /2 𝑎2 +𝑏2
𝑒
2𝜋 .5 𝑎2 +𝑏2 .5
𝑎𝑏 2
𝑎2 +𝑏2
∫
.5
𝑎2 +𝑏2
2𝜋 .5 𝑎𝑏
𝑒
𝑒 𝑑𝑥
2
𝑏2 𝑧
𝑥− 2 2
𝑎 +𝑏
2
𝑎𝑏 2
𝑎2 +𝑏2
dx
dx = 1
Rotational Invariance
• Second property: if u, v are vectors with <u, v> = 0,
then <g,u> and <g,v> are independent, where g is a
vector of i.i.d. N(0,1/k) random variables
• Why?
• If g is an n-dimensional vector of i.i.d. N(0,1)
random variables, and R is a fixed matrix, then
the probability density function of Rg is
𝑥𝑇 RRT
𝑓
𝑥 =
−
1
𝑒
det(RRT ) 2𝜋 𝑑/2
−1
𝑥
2
– RRT is the covariance matrix
– For a rotation matrix R, the distribution of Rg
and of g are the same
Orthogonal Implies Independent
• Want to show: if u, v are vectors with <u, v> = 0, then
<g,u> and <g,v> are independent, where g is a vector of
i.i.d. N(0,1/k) random variables
• Choose a rotation R which sends u to αe1 , and sends v
to βe2
• < g, u > = < gR, RT u > = < h, αe1 > = αh1
• < g, v > = < gR, RT v > = < h, βe2 > = βh2
where h is a vector of i.i.d. N(0, 1/k) random variables
• Then h1 and h2 are independent by definition
Where were we?
• Claim: SA is a k x d matrix of i.i.d. N(0,1/k) random
variables
• Proof: The rows of SA are independent
– Each row is: < g, A1 >, < g, A2 > , … , < g, Ad >
– First property implies the entries in each row are
N(0,1/k) since the columns Ai have unit norm
– Since the columns Ai are orthonormal, the entries in a
row are independent by our second property
Back to Subspace Embeddings
•
•
•
•
Want to show |SAx|2 = (1±ε)|Ax|2 for all x
Can assume columns of A are orthonormal
Can also assume x is a unit vector
SA is a k x d matrix of i.i.d. N(0,1/k) random variables
• Consider any fixed unit vector 𝑥 ∈ 𝑅𝑑
• SAx 22 = i∈ k < g i , x >2 , where g i is i-th row of SA
• Each < g i , x
>2
is distributed as N
1 2
0,
k
• E[< g i , x >2 ] = 1/k, and so E[ SAx 22 ] = 1
How concentrated is SAx 22 about its expectation?
Johnson-Lindenstrauss Theorem
• Suppose h1 , … , hk are i.i.d. N(0,1) random variables
• Then G = i h2i is a 𝜒 2 -random variable
• Apply known tail bounds to G:
– (Upper) Pr G ≥ k + 2 kx .5 + 2x ≤ e−x
– (Lower) Pr G ≤ k − 2 kx .5 ≤ e−x
• If x =
ϵ2 k
,
16
then Pr G ∈ k(1 ± ϵ) ≥ 1 −
2 k/16
−ϵ
2e
1
δ
• If k = Θ(ϵ−2 log( )), this probability is 1-δ
• Pr SAx 22 ∈ 1 ± ϵ ≥ 1 − 2−Θ d
This only holds for a fixed x, how to argue for all x?
Net for Sphere
• Consider the sphere Sd−1
• Subset N is a γ-net if for all x ∈ Sd−1 , there is a y ∈ N,
such that x − y 2 ≤ γ
• Greedy construction of N
– While there is a point x ∈ Sd−1 of distance larger than
γ from every point in N, include x in N
• The sphere of radius γ/2 around ever point in N is
contained in the sphere of radius 1+ γ around 0d
• Further, all such spheres are disjoint
• Ratio of volume of d-dimensional sphere of radius 1+ γ
to dimensional sphere of radius 𝛾 is 1 + γ d /(γ/2)d , so
N ≤ 1 + γ d /(γ/2)d
Net for Subspace
• Let M = {Ax | x in N}, so M ≤ 1 + γ d /(γ/2)d
• Claim: For every x in Sd−1 , there is a y in M for which
Ax − y 2 ≤ γ
• Proof: Let x’ in S d−1 be such that x − x ′ 2 ≤ γ
Then Ax − Ax ′ 2 = x − x ′ 2 ≤ γ, using that the
columns of A are orthonormal. Set y = Ax’
Net Argument
• For a fixed unit x, Pr SAx 22 ∈ 1 ± ϵ ≥ 1 − 2−Θ d
• For a fixed pair of unit x, x’, SAx 22 , SAx′ 22 , SA x − x ′
are all 1 ± ϵ with probability 1 − 2−Θ d
• SA x − x ′ 22 = SAx 22 + SAx ′ 22 − 2 < SAx, SAx ′ >
• A x − x ′ 22 = Ax 22 + Ax ′ 22 − 2 < Ax, Ax ′ >
– So Pr < Ax, Ax ′ > = < SAx, SAx ′ > ± O ϵ
2
2
= 1 − 2−Θ(d)
• Choose a ½-net M = {Ax | x in N} of size 5𝑑
• By a union bound, for all pairs y, y’ in M,
< y, y′ > = < Sy, Sy′ > ± O ϵ
• Condition on this event
• By linearity, if this holds for y, y’ in M, for αy, βy′ we have
< αy, βy′ > = αβ < Sy, Sy′ > ± O ϵ αβ
Finishing the Net Argument
• Let y = Ax for an arbitrary x ∈ Sd−1
• Let y1 ∈ M be such that y − y1 2 ≤ γ
• Let α be such that α(y − y1 ) 2 = 1
– α ≥ 1/γ (could be infinite)
• Let y2′ ∈ M be such that α y − y1 − y2 ′
• Then y − y1 −
y′2
.
α
y2 ′
α 2
≤
γ
α
2
≤γ
≤ γ2
• Set y2 =
Repeat, obtaining y1 , y2 , y3 , … such that for
all integers i,
y − y1 − y2 − … − yi 2 ≤ γi
• Implies yi 2 ≤ γi−1 + γi ≤ 2γi−1
Finishing the Net Argument
• Have y1 , y2 , y3 , … such that y =
• Sy
2
2
= |S
i yi
and yi
2
≤ 2γi−1
2
|
y
i i 2
=
i
Syi
=
i
yi
2
2
2
2
+2
+2
i,j
i,j
< Syi , Syj >
< yi , yj > ± O ϵ
i,j
yi
2
yj
= | i yi |22 ± O ϵ
= y 22 ± O ϵ
=1±O ϵ
• Since this held for an arbitrary y = Ax for unit x, by
linearity it follows that for all x, |SAx|2 = (1±ε)|Ax|2
2
Back to Regression
• We showed that S is a subspace
embedding, that is, simultaneously for all x,
|SAx|2 = (1±ε)|Ax|2
What does this have to do with regression?
Subspace Embeddings for
Regression
• Want x so that |Ax-b|2 · (1+ε) miny |Ay-b|2
• Consider subspace L spanned by columns of A
together with b
• Then for all y in L, |Sy|2 = (1± ε) |y|2
• Hence, |S(Ax-b)|2 = (1± ε) |Ax-b|2 for all x
• Solve argminy |(SA)y – (Sb)|2
• Given SA, Sb, can solve in poly(d/ε) time
Only problem is computing SA takes O(nd2) time
How to choose the right sketching matrix S? [S]
 S is a Subsampled Randomized Hadamard Transform
 S = P*H*D
 D is a diagonal matrix with +1, -1 on diagonals
 H is the Hadamard transform
 P just chooses a random (small) subset of rows of H*D
 S*A can be computed in O(nd log n) time
Why does it work?
31
Why does this work?
 We can again assume columns of A are orthonormal
 It suffices to show SAx
2
2
= PHDAx
 HD is a rotation matrix, so HDAx
2
2
2
2
= 1 ± ϵ for all x
= Ax
2
2
= 1 for any x
 Notation: let y = Ax
 Flattening Lemma: For any fixed y,
Pr [ HDy
∞
≥C
log.5 nd/δ
]
n.5
≤
δ
2d
32
Proving the Flattening Lemma
 Flattening Lemma: Pr [ HDy
∞
≥C
log.5 nd/δ
]
n.5
≤
δ
2d
 Let C > 0 be a constant. We will show for a fixed i in [n],
Pr [ HDy
≥C
i
log.5 nd/δ
]
n.5
≤
δ
2nd
 If we show this, we can apply a union bound over all i

HDy i = j Hi,j Dj,j yj
 (Azuma-Hoeffding) Pr |
probability 1
t2
)
2 j β2
j
−(
j Zj |
> t ≤ 2e
, where |Zj | ≤ βj with
 Zj = Hi,j Dj,j yj has 0 mean
 |Zj | ≤

|yj |
n.5
1
2
β
=
j j
n
= βj with probability 1
nd
 Pr |
j Zj | >
C log.5 δ
n.5
δ
C2 log
nd
−
2
≤ 2e
≤
δ
2nd
33
Consequence of the Flattening Lemma
 Recall columns of A are orthonormal
 HDA has orthonormal columns
 Flattening Lemma implies HDAei
probability 1 −
δ
2d
 With probability 1
∞
≤C
log.5 nd/δ
n.5
with
for a fixed i ∈ d
δ
− ,
2
 Given this, ej HDA
2
ej HDAei
≤C
≤C
d.5 log.5 nd/δ
n.5
log.5 nd/δ
n.5
for all i,j
for all j
(Can be optimized further)
34
Matrix Chernoff Bound

Let X1 , … , Xs be independent copies of a symmetric random matrix X ∈ Rdxd
1
with E X = 0, X 2 ≤ γ, and E X T X 2 ≤ 𝜎 2 . Let W = s i∈[s] Xi . For any ϵ > 0,
Pr W
2
> ϵ ≤ 2d ⋅ e
(here W
γϵ
3
−sϵ2 /(𝜎2 + )
2
= sup Wx 2 / x 2 )

Let V = HDA, and recall V has orthonormal columns

Suppose P in the S = PHD definition samples uniformly with replacement. If
row i is sampled in the j-th sample, then Pj,i = n, and is 0 otherwise

Let Yi be the i-th sampled row of V = HDA

Let Xi = Id − n ⋅ YiT Yi
 E X i = Id − n ⋅

Xi
2
≤ Id
1
j n
ViT Vi = Id − V T V = 0d
2 + n ⋅ max ej HDA
2
2
= 1 + n ⋅ C 2 log
nd
δ
d
⋅ n = Θ(d log
nd
δ
)
35
Matrix Chernoff Bound

Recall: let Yi be the i-th sampled row of V = HDA

Let Xi = Id − n ⋅ YiT Yi

E X T X + Id = Id + Id − 2n E YiT Yi + n2 E[YiT Yi YiT Yi ]
1
i n
nd
T
2
v
v
C
log
i
i i
δ
= 2Id − 2Id + n2
⋅ viT vi viT vi = n
d
⋅ n = C 2 dlog
T
i vi vi
nd
𝛿
⋅ 𝑣𝑖
2
2

Define Z = n

Note that 𝑋 𝑇 𝑋 + 𝐼𝑑 and Z are real symmetric, with non-negative
eigenvalues
Claim: for all vectors y, we have: 𝑦 𝑇 𝑋 𝑇 𝑋𝑦 + 𝑦 𝑇 𝑦 ≤ 𝑦 𝑇 𝑍𝑦
Proof: 𝑦 𝑇 𝑋 𝑇 𝑋𝑦 + 𝑦 𝑇 𝑦 = 𝑛 𝑖 𝑦 𝑇 𝑣𝑖𝑇 𝑣𝑖 𝑦 𝑣𝑖 22 = 𝑛 𝑖 < 𝑣𝑖 , 𝑦 >2 𝑣𝑖 22 and
𝑛𝑑 𝑑
𝑛𝑑
𝑇
𝑇
𝑇
2
2
2
𝑦 𝑍𝑦 = 𝑛
𝑦 𝑣𝑖 𝑣𝑖 𝑦 𝐶 log
⋅ =𝑛
< 𝑣𝑖 , 𝑦 > 𝐶 log
𝛿
𝑛
𝛿


𝑖
𝑋𝑇 𝑋

Hence, 𝐸

Hence, E X T X
𝐼𝑑
𝑖
2
2
≤ 𝐸 𝑋 𝑇 𝑋 + 𝐼𝑑
= |𝐸 𝑋 𝑇 𝑋 + 𝐼𝑑 |2 + 1
𝑛𝑑
2
≤ 𝑍 2 + 1 ≤ 𝐶 𝑑 log
+1
𝛿
= O d log
nd
δ
2
+ 𝐼𝑑
2
36
Matrix Chernoff Bound
nd
δ

Hence, E X T X

Recall: (Matrix Chernoff) Let X1 , … , Xs be independent copies of a
symmetric random matrix X ∈ Rdxd with E X = 0, X 2 ≤ γ, and E X T X
≤ 𝜎 2 . Let W =
2
1
s
= O d log
i∈[s] X i .
Pr |Id − PHDA
d

Set s = d log
nd log δ
δ
ϵ2
T
For any ϵ > 0, Pr W
PHDA
2
> ϵ ≤ 2d ⋅ e
−s ϵ2 /(Θ(d log
2
2
γϵ
−sϵ2 /(𝜎2 + 3 )
> ϵ ≤ 2d ⋅ e
nd
)
δ
δ
, to make this probability less than 2
37
SRHT Wrapup
 Have shown |Id − PHDA
T
|2 < ϵ using Matrix
PHDA
Chernoff Bound and with s = d log
d
nd log δ
δ
ϵ2
samples
 Implies for every unit vector x,
|1− PHDAx 22 | = x T x − x PHDA
so PHDAx
2
2
T
PHDA x < ϵ ,
∈ 1 ± ϵ for all unit vectors x
 Considering the column span of A adjoined with b, we can
again solve the regression problem
 The time for regression is now only O(nd log n) +
𝑑 log 𝑛
poly(
𝜖
38
). Nearly optimal in matrix dimensions (n >> d)
Faster Subspace Embeddings S [CW,MM,NN]
 CountSketch matrix
 Define k x n matrix S, for k = O(d2/ε2)
 S is really sparse: single randomly chosen non-zero
entry per column
[
[
0010 01 00
1000 00 00
0 0 0 -1 1 0 -1 0
0-1 0 0 0 0 0 1
 nnz(A) is number of non-zero entries of A
Can compute
S ⋅ A in nnz(A)
time!
39
Simple Proof [Nguyen]
 Can assume columns of A are orthonormal
 Suffices to show |SAx|2 = 1 ± ε for all unit x
 For regression, apply S to [A, b]
 SA is a 2d2/ε2 x d matrix
 Suffices to show | ATST SA – I|2 · |ATST SA – I|F · ε
 Matrix product result shown below:
Pr[|CSTSD – CD|F2 · [3/(𝛿(# rows of S))] * |C|F2 |D|F2] ≥ 1 − 𝛿
 Set C = AT and D = A.
 Then |A|2F = d and (# rows of S) = 3 d2/(δε2)
40
Matrix Product Result [Kane, Nelson]
 Show: Pr[|CSTSD – CD|F2 · [3/(𝛿(# rows of S))] * |C|F2 |D|F2] ≥ 1 − 𝛿
 (JL Property) A distribution on matrices S ∈ Rkx n has the (ϵ, δ, ℓ)-JL
moment property if for all x ∈ Rn with x 2 = 1,
ES Sx 22 − 1
ℓ
≤ ϵℓ ⋅ δ
 (From vectors to matrices) For ϵ, δ ∈ 0,
1
2
, let D be a distribution on
matrices S with k rows and n columns that satisfies the (ϵ, δ, ℓ)-JL
moment property for some ℓ ≥ 2. Then for A, B matrices with n rows,
Pr AT S T SB − AT B
S
F
≥3ϵ A
F
B
F
≤δ
41
From Vectors to Matrices

1
(From vectors to matrices) For ϵ, δ ∈ 0, 2 , let D be a distribution on matrices
S with k rows and n columns that satisfies the (ϵ, δ, ℓ)-JL moment property
for some ℓ ≥ 2. Then for A, B matrices with n rows,
Pr AT S T SB − AT B
S

Proof: For a random scalar X, let X
 Sometimes consider X = T
 | T

F
|p = E T
p
F
F
p
≥3ϵ A
= EX
F
B
F
≤δ
p 1/p
for a random matrix T
1/p
Can show |. |p is a norm
 Minkowski’s Inequality: X + Y

F
p
≤ X
p
+ Y
p
For unit vectors x, y, need to bound |〈Sx, Sy〉 - 〈x, y〉|ℓ
42
From Vectors to Matrices

For unit vectors x, y, |〈Sx, Sy〉 - 〈x, y〉|ℓ
=
≤
≤
1
(
2
1
(
2
1
2
Sx 22 −1 + Sy
Sx
2
2
2
2
− 1 ℓ + Sy
1
ℓ
−1 − S x−y
2
2
1
ℓ
−1 ℓ+ S x−y
(ϵ ⋅ δ +ϵ ⋅ δ + x − y
≤3ϵ⋅δ
2
2
2
2
− x−y
2
2
2
2
− x−y
|ℓ
2
2 ℓ)
1
ℓ
ϵ⋅δ )
1
ℓ
Sx,Sy − x,y ℓ
x2y2
1
ℓ

By linearity, for arbitrary x, y,

Suppose A has d columns and B has e columns. Let the columns of A be
A1 , … , Ad and the columns of B be B1 , … , Be

Define Xi,j =

AT S T SB
−
1
Ai 2 Bj
2
2
AT B F
=
≤3ϵ⋅δ
⋅ ( SAi , SBj − Ai , Bj )
i
2
j Ai 2
⋅
2 2
Bj Xi,j
2
43
From Vectors to Matrices

Have shown: for arbitrary x, y,

For Xi,j =

| AT ST SB − A B F |ℓ/2 =
1
Sx,Sy − x,y ℓ
x2y2
SAi , SBj − Ai , Bj : AT S T SB − AT B
⋅
Ai 2 Bj
2
T 2
i
≤
j
2
2
2
⋅ Bj Xi,j
j
Ai
2
2
2 |
⋅ Bj |Xi,j
ℓ/2
j
Ai
2
2
⋅ Bj
i
≤ 3ϵδ
1
ℓ
= 3 ϵδ

T T
T
Since E A S SB − A B
Pr
AT S T SB
−
AT B
F
ℓ
F
1
ℓ
2
F
=
i
j
Ai
2
2
2
2
⋅ Bj Xi,j
2
ℓ/2
2
2
2
2
Xi,j
Ai
2
2
2
i
2
A
j
2
F
B
2
ℓ
2
Bj
2
2
F
T T
T
= A S SB − A B
> 3ϵ A
2
2
Ai
i
=

≤3ϵ⋅δ
1
ℓ
ℓ/2
2
F ℓ
, by Markov’s inequality,
2
F
B
F
≤
1
3ϵ A F B F
ℓ
E |AT S T SB − AT B|ℓF ≤ δ
44
Result for Vectors

Show: Pr[|CSTSD – CD|F2 · [3/(𝛿(# rows of S))] * |C|F2 |D|F2] ≥ 1 − 𝛿

(JL Property) A distribution on matrices S ∈ Rkx n has the (ϵ, δ, ℓ)-JL moment
property if for all x ∈ Rn with x 2 = 1,
ES Sx

2
2
−1
ℓ
≤ ϵℓ ⋅ δ
1
(From vectors to matrices) For ϵ, δ ∈ 0, 2 , let D be a distribution on matrices
S with k rows and n columns that satisfies the (ϵ, δ, ℓ)-JL moment property
for some ℓ ≥ 2. Then for A, B matrices with n rows,
Pr AT S T SB − AT B
S

F
≥3ϵ A
F
B
F
≤δ
Just need to show that the CountSketch matrix S satisfies JL property and
bound the number k of rows
45
CountSketch Satisfies the JL Property

(JL Property) A distribution on matrices S ∈ Rkx n has the (ϵ, δ, ℓ)-JL moment
property if for all x ∈ Rn with x 2 = 1,
ES Sx
2
2
−1
ℓ
≤ ϵℓ ⋅ δ

We show this property holds with ℓ = 2. First, let us consider ℓ = 1

For CountSketch matrix S, let
 h:[n] -> [k] be a 2-wise independent hash function
 σ: n → {−1,1} be a 4-wise independent hash function

Let δ E = 1 if event E holds, and δ E = 0 otherwise

E[ Sx 22 ] =
j∈ k
E[
i∈ n
δ h i = j σi x i
=
j∈ k
i1,i2∈[n] E[δ
=
j∈ k
i∈[n] E[δ
=
1
k
j∈[k[
i∈ n
2
]
h i1 = j δ h i2 = j σi1 σi2 ]xi1 xi2
h i = j 2 ]xi2
xi2 = x
2
2
46
CountSketch Satisfies the JL Property


E[ Sx 42 ] = 𝐸[
𝑗1 ,𝑗2 ,𝑖1 .𝑖2 ,𝑖3 ,𝑖4
𝑗′∈
j∈ k
i∈ n
𝑘
δ h i = j σi xi
2
2
i′∈ n
δ h i′ = j′ σi′ xi′ ] =
𝐸[𝜎𝑖1 𝜎𝑖2 𝜎𝑖3 𝜎𝑖4 𝛿 ℎ 𝑖1 = 𝑗1 𝛿 ℎ 𝑖2 = 𝑗1 𝛿 ℎ 𝑖3 = 𝑗2 𝛿 ℎ 𝑖4 = 𝑗2 ]𝑥𝑖1 𝑥𝑖2 𝑥𝑖3 𝑥𝑖4

We must be able to partition {𝑖1 , 𝑖2 , 𝑖3 , 𝑖4 } into equal pairs

Suppose 𝑖1 = 𝑖2 = 𝑖3 = 𝑖4 . Then necessarily 𝑗1 = 𝑗2 . Obtain

Suppose 𝑖1 = 𝑖2 and 𝑖3 = 𝑖4 but 𝑖1 ≠ 𝑖3 . Then get

Suppose 𝑖1 = 𝑖3 and 𝑖2 = 𝑖4 but 𝑖1 ≠ 𝑖2 . Then necessarily 𝑗1 = 𝑗2 . Obtain
1
1
2 2
4
𝑥
𝑥
≤
𝑥
2 . Obtain same bound if 𝑖1 = 𝑖4 and 𝑖2 = 𝑖3 .
𝑗 𝑘 2 𝑖1 ,𝑖2 𝑖1 𝑖2
𝑘

Hence, E[ Sx 42 ] ∈ [ x 42 , 𝑥 42 (1 + 𝑘)] = [1, 1 + k]

So, ES Sx
2
2
2
−1
2
2
1
𝑗𝑘
4
𝑖 𝑥𝑖
1 2 2
𝑗1 ,𝑗2 ,𝑖1 ,𝑖3 𝑘 2 𝑥𝑖1 𝑥𝑖3
= 𝑥
= 𝑥
4
2
4
4
− 𝑥
4
4
2
2
1
≤ 1 + 𝑘 − 2 + 1 = 𝑘 . Setting k = 2𝜖2 𝛿 finishes the proof
47
Where are we?

(JL Property) A distribution on matrices S ∈ Rkx n has the (ϵ, δ, ℓ)-JL
moment property if for all x ∈ Rn with x 2 = 1,
ES Sx

2
2
−1
ℓ
≤ ϵℓ ⋅ δ
1
(From vectors to matrices) For ϵ, δ ∈ 0, 2 , let D be a distribution on
matrices S with k rows and n columns that satisfies the (ϵ, δ, ℓ)-JL moment
property for some ℓ ≥ 2. Then for A, B matrices with n rows,
Pr AT S T SB − AT B
S
2
F
≥ 3 ϵ2 A
2
F
B
2
F
≤δ
2

We showed CountSketch has the JL property with ℓ = 2, 𝑎𝑛𝑑 𝑘 = 𝜖2 𝛿

Matrix product result we wanted was:
Pr[|CSTSD – CD|F2 · (3/(𝛿k)) * |C|F2 |D|F2] ≥ 1 − 𝛿
We are now down with the proof CountSketch is a subspace embedding

48
Course Outline
 Subspace embeddings and least squares regression
 Gaussian matrices
 Subsampled Randomized Hadamard Transform
 CountSketch
 Affine embeddings
 Application to low rank approximation
 High precision regression
 Leverage score sampling
49
Affine Embeddings

Want to solve min 𝐴𝑋 − 𝐵 2𝐹 , A is tall and thin with d columns, but B has a large
𝑋
number of columns

Can’t directly apply subspace embeddings

Let’s try to show SAX − SB
we need of S

Can assume A has orthonormal columns

Let 𝐵 ∗ = 𝐴𝑋 ∗ − 𝐵, where 𝑋 ∗ is the optimum

F
= 1 ± ϵ AX − B
F
for all X and see what properties
𝑆 𝐴𝑋 − 𝐵 2𝐹 − 𝑆𝐵 ∗ 2𝐹 = 𝑆𝐴 𝑋 − 𝑋 ∗ + 𝑆 𝐴𝑋 ∗ − 𝐵 2𝐹 − 𝑆𝐵 ∗ 2𝐹
= 𝑆𝐴 𝑋 − 𝑋 ∗ 2𝐹 − 2𝑡𝑟[ 𝑋 − 𝑋 ∗ 𝑇 𝐴𝑇 𝑆 𝑇 𝑆𝐵∗ ]
∈ 𝑆𝐴 𝑋 − 𝑋 ∗ 2𝐹 ± 2 𝑋 − 𝑋 ∗ 𝐹 𝐴𝑇 𝑆 𝑇 𝑆𝐵 ∗ 𝐹 (use 𝑡𝑟 𝐶𝐷 ≤ 𝐶 𝐹 𝐷 𝐹 )
∈ 𝑆𝐴 𝑋 − 𝑋 ∗ 2𝐹 ±2𝜖 𝑋 − 𝑋 ∗ 𝐹 B ∗ F (if we have approx. matrix product)
∈ 𝐴 𝑋 − 𝑋 ∗ 2𝐹 ± 𝜖( 𝐴 𝑋 − 𝑋 ∗ 2𝐹 + 2 𝑋 − 𝑋 ∗ 𝐹 𝐵 ∗ ) (subspace embedding for50A)
Affine Embeddings
 We have 𝑆 𝐴𝑋 − 𝐵
2
𝐹
− 𝑆𝐵∗
2
𝐹
∈ 𝐴 𝑋 − 𝑋∗
2
𝐹
± 𝜖(|𝐴(𝑋
51
Affine Embeddings
 Know: 𝑆 𝐴𝑋 − 𝐵 2𝐹 − 𝑆𝐵∗ 2𝐹 − 𝐴𝑋 − 𝐵
±2𝜖 𝐴𝑋 − 𝐵 2𝐹
 Know: 𝑆𝐵∗ 2𝐹 = 1 ± 𝜖 𝐵∗ 2𝐹
 𝑆 𝐴𝑋 − 𝐵
2
𝐹
= (1 ± 2𝜖) 𝐴𝑋 − 𝐵 2𝐹 +𝜖 𝐵∗
= 1 ± 3𝜖 𝐴𝑋 − 𝐵 2𝐹
2
𝐹
− 𝐵∗
2
𝐹
∈
2
𝐹
 Completes proof of affine embedding!
52
Affine Embeddings: Missing Proofs
 Claim: A + B
2
F
= A
2
F
 Proof: A + B
2
F
=
Ai + Bi
2
2
2
2
Bi
=
i
Ai
+ B
+
i
= A
2
F
+ 2Tr(AT B)
2
2
+ 2〈Ai , Bi 〉
i
2
F
+ B
2
F
+ 2Tr(AT B)
53
Affine Embeddings: Missing Proofs
 Claim: Tr AB ≤ A
i
Ai
2
≤
2
i
i A 2
= A
F
B
B
F
i
i〈A , Bi 〉
 Proof: Tr AB =
≤
F
Bi
1
2
2
for rows Ai and columns Bi
by Cauchy-Schwarz for each i
1
2 2
B
i i 2
another Cauchy-Schwarz
F
54
Affine Embeddings: Homework Proof
 Claim: SB ∗ 2F = 1 ± ϵ B ∗ 2F with constant probability if
1
CountSketch matrix S has k = O( 2 ) rows
ϵ
 Proof:
 SB ∗ 2F =
i
SBi∗
2
2
 By our analysis for CountSketch and linearity of expectation,
E SB ∗ 2F = i E SBi∗ 22 = B ∗ 2F
 E[ SB ∗ 4F ] =
i,j
E[ SBi∗
2
2
2
SBj∗ ]
2
 By our CountSketch analysis,E[
SBi∗ 42 ]]
≤
 For cross terms see Lemma 40 in [CW13]
2
∗ 4
Bi 2 (1 + )
k
55
Low rank approximation
 A is an n x d matrix
 Think of n points in Rd
 E.g., A is a customer-product matrix
 Ai,j = how many times customer i purchased item j
 A is typically well-approximated by low rank matrix
 E.g., high rank because of noise
 Goal: find a low rank matrix approximating A
 Easy to store, data more interpretable
56
What is a good low rank approximation?
Singular Value Decomposition (SVD)
Any matrixAAk == argmin
U ¢ Σ ¢ rank
V k matrices B |A-B|F
 U has orthonormal columns
 Σ is diagonal with non-increasing
positive
1/2
2
(|C|
)
F = (Σ
i,jVC
i,j )
The
rows
of
are
k
entries down the diagonal
the orthonormal
top k principal
 V has
rows
components
Computing
Ak exactly is
expensive
 Rank-k approximation:
Ak = Uk ¢ Σk ¢ Vk
57
Low rank approximation
 Goal: output a rank k matrix A’, so that
|A-A’|F · (1+ε) |A-Ak|F
 Can do this in nnz(A) + (n+d)*poly(k/ε) time [S,CW]
 nnz(A) is number of non-zero entries of A
58
Solution to low-rank approximation [S]
 Given n x d input matrix A
 Compute S*A using a random matrix S with k/ε << n
rows. S*A takes random linear combinations of rows of A
A
SA
 Project rows of A onto SA, then find best rank-k
approximation to points inside of SA.
59
What is the matrix S?
 S can be a k/ε x n matrix of i.i.d. normal random
variables
 [S] S can be a k/ε x n Fast Johnson Lindenstrauss
Matrix
 Uses Fast Fourier Transform
 [CW] S can be a poly(k/ε) x n CountSketch matrix
[
[
0010 01 00
1000 00 00
0 0 0 -1 1 0 -1 0
0-1 0 0 0 0 0 1
S ¢ A can be
computed in
nnz(A) time
60
Why do these Matrices Work?

Consider the regression problem min Ak X − A

Write Ak = WY, where W is n x k and Y is k x d

Let S be an affine embedding

Then SAk X − SA

By normal equations, argmin SAk X − SA
X
F
= 1 ± ϵ Ak X − A
F
X
− SA −
for all X
F
= SAk

So, Ak SAk

But Ak SAk

Let’s formalize why the algorithm works now…
− SA
A
F
≤ 1 + ϵ Ak − A
F
− SA
F
is a rank-k matrix in the row span of SA!
61
Why do these Matrices Work?



min
rank−k X
XSA − A
2
F
≤ Ak SAk
−
SA − A|2F ≤ 1 + ϵ A − A_k|2F
By the normal equations,
XSA − A 2F = XSA − A SA − SA
Hence,
min
rank−k X
XSA − A
2
F
= A SA − SA − A
2
F
2
F
+
+ A SA − SA − A
min
rank−k X
2
F
XSA − A SA − SA

Can write SA = U ΣV T in its SVD, where U, Σ are k x k and V T is k x d

Then, min
rank−k X
XSA − A SA − SA
2
F
=
=


min
XUΣ − A SA − UΣ
min
Y − A SA − UΣ
rank−k X
rank−k Y
2
F
2
F
2
F
Hence, we can just compute the SVD of A SA − UΣ
But how do we compute A SA
− UΣ
quickly?
62
Caveat: projecting the points onto SA is slow
 Current algorithm:
1. Compute S*A
2. Project each of the rows onto S*A
3. Find best rank-k approximation of projected points
inside of rowspace of S*A
minrank-k X |X(SA)R-AR|F2
 Bottleneck is step 2
Can solve with affine embeddings
 [CW] Approximate the projection
 Fast algorithm for approximate regression
minrank-k X |X(SA)-A|F2
 Want nnz(A) + (n+d)*poly(k/ε) time
63
Using Affine Embeddings
2
F

We know we can just output arg min

Choose an affine embedding R:
XSAR − AR 2F = 1 ± ϵ XSA − A
rank−k X
XSA − A

Note: we can compute AR and SAR in nnz(A) time

Can just solve

min
rank−k X
min
rank−k X
XSAR − AR
2
F
XSAR − AR
= AR SAR
−
2
F
for all X
2
F
−
SAR − AR
2
F
2
F
+
min
rank−k X
XSAR − AR SAR
k
ϵ
−
SAR
2
F

Compute

Necessarily, Y = XSAR for some X. Output Y SAR − SA in factored form. We’re done!
min
rank−k Y
Y − AR SAR
SAR
using SVD which is n + d poly
time
64
Low Rank Approximation Summary
1. Compute SA
2. Compute SAR and AR
3. Compute
min
rank−k Y
Y − AR SAR
−
SAR
2
F
using SVD
4. Output Y SAR − SA in factored form
Overall time: nnz(A) + (n+d)poly(k/ε)
65
Course Outline
 Subspace embeddings and least squares regression
 Gaussian matrices
 Subsampled Randomized Hadamard Transform
 CountSketch
 Affine embeddings
 Application to low rank approximation
 High precision regression
 Leverage score sampling
66
High Precision Regression
 Goal: output x‘ for which |Ax‘-b|2 · (1+ε) minx |Ax-b|2
with high probability
 Our algorithms all have running time poly(d/ε)
 Goal: Sometimes we want running time poly(d)*log(1/ε)
 Want to make A well-conditioned
 κ A = sup Ax 2 / inf Ax 2
x 2 =1
x 2 =1
 Lots of algorithms’ time complexity depends on κ A
 Use sketching to reduce κ A to O(1)!
67
Small QR Decomposition
 Let S be a 1 + ϵ0 - subspace embedding for A
 Compute SA
 Compute QR-factorization, SA = QR−1
 Claim: κ AR =
1+ϵ0
1−ϵ0
 For all unit x, 1 − ϵ0 ARx
2
≤ SAR x 2 = 1
 For all unit x, 1 + ϵ0 ARx
2
≥ SARx
 So κ AR = sup ARx 2 / inf ARx
x 2 =1
x 2 =1
2
2
=1
≤
1+ϵ0
1−ϵ0
68
Finding a Constant Factor Solution
 Let S be a 1 + ϵ0 - subspace embedding for AR
 Solve x0 = argmin SARx − Sb
x
2
 Time to compute AR and x0 is nnz(A) + poly(d) for constant ϵ0
 xm+1 ← xm + RT AT b − AR xm
 AR xm+1 − x ∗ = AR (xm + RT AT b − ARxm − x ∗ )
= AR − ARRT AT AR xm − x ∗
= U Σ − Σ 3 V T (xm − x ∗ ),
where AR = U ΣV T is the SVD of AR

AR xm+1 − x ∗
 ARxm − b
2
2
2
=
Σ − Σ 3 V T xm − x ∗
= AR xm −
x∗
2
2
+
2
ARx ∗
= O ϵ0 |AR(xm − x ∗ )|2
−b
2
2
69
Course Outline
 Subspace embeddings and least squares regression
 Gaussian matrices
 Subsampled Randomized Hadamard Transform
 CountSketch
 Affine embeddings
 Application to low rank approximation
 High precision regression
 Leverage score sampling
70
Leverage Score Sampling
 This is another subspace embedding, but it is based on sampling!
 If A has sparse rows, then SA has sparse rows!
 Let A = U ΣV T be an n x d matrix with rank d, written in its SVD
 Define the i-th leverage score ℓ i
 What is
iℓ
of A to be Ui,∗
2
2
i ?
 Let q1 , … , qn be a distribution with qi ≥
βℓ i
d
, where β is a parameter
 Define sampling matrix S = D ⋅ ΩT , where D is k x k and Ω is n x k
 Ω is a sampling matrix, and D is a rescaling matrix
 For each column j of Ω, D, independently, and with replacement, pick a row
index i in [n] with probability qi , and set Ωi,j = 1 and Dj,j = (qi k)^(.5) 71
Leverage Score Sampling
 Note: leverage scores do not depend on choice of orthonormal
basis U for columns of A
 Indeed, let U and U’ be two such orthonormal bases
 Claim: ei U
2
2
= ei U′
2
2
for all i
 Proof: Since both U and U’ have column space equal to that of A,
we have U = U ′ Z for change of basis matrix Z
 Since U and U’ each have orthonormal columns, Z is a rotation
matrix (orthonormal rows and columns)
 Then ei U
2
2
= ei U′ Z
2
2
= ei U′
2
2
72
Leverage Score Sampling gives a Subspace Embedding
2
2
2
2

Want to show for S = D ⋅ ΩT , that SAx

Writing A = U ΣV T in its SVD, this is equivalent to showing
= 1 ± ϵ Uy 22 = 1 ± ϵ y 22 for all y

As usual, we can just show with high probability, U T S T SU − I

How can we analyze U T S T SU?

(Matrix Chernoff) Let X1 , … , Xk be independent copies of a symmetric
random matrix X ∈ Rdxd with E X = 0, X 2 ≤ γ, and E X T X 2 ≤ σ2 . Let W
1
=k
j∈[k] X j .
= 1 ± ϵ Ax
for all x
SUy
2
2
2
≤ϵ
For any ϵ > 0,
γϵ
Pr W
(here W
2
= sup
Wx 2
.
x2
2
−kϵ2 /(σ2 + 3 )
> ϵ ≤ 2d ⋅ e
Since W is symmetric, W
2
= sup x T Wx . )
x 2 =1
73
Leverage Score Sampling gives a Subspace Embedding

Let i(j) denote the index of the row of U sampled in the j-th trial
UT
i(j) Ui(j)

Let Xj = Id −

The Xj are independent copies of a symmetric matrix random variable



E
2
≤ Id
XTX
2
+
UT
i j Ui j 2
qi j
= Id − 2E
=
, where Ui(j) is the j-th sampled row of U
UT
i Ui
i qi
qi
E X j = Id −
Xj
qi(j)
= Id − Id = 0d
≤ 1 + max
UT
i j Ui j
qi j
T
UT
i Ui Ui Ui
i
q i
i
+E
− Id ≤
Ui 22
qi
d
≤1+β
T
UT
i j Ui j Ui j Ui j
q2i j
d
β
T
T
i Ui Ui − Id ≤
d
β
− 1 Id ,
where A ≤ B means x T Ax ≤ x Bx for all x

d
Hence, |E X T X |2 ≤ β − 1
74
Applying the Matrix Chernoff Bound

(Matrix Chernoff) Let X1 , … , Xk be independent copies of a symmetric
random matrix X ∈ Rdxd with E X = 0, X 2 ≤ γ, and E X T X 2 ≤ σ2 . Let W
1
=k
j∈[k] X j .
For any ϵ > 0,
Pr W
(here W


2
= sup
d
γ = 1 + β, and σ2 =
X j = Id −
UT
i j Ui j
qi j
Wx 2
.
x2
d
β
2
γϵ
3
−kϵ2 /(σ2 + )
> ϵ ≤ 2d ⋅ e
Since W is symmetric, W
2
= sup x T Wx . )
x 2 =1
−1
, and recall how we generated S = D ⋅ ΩT : For each
column j of Ω, D, independently, and with replacement, pick a row index i in
[n] with probability qi , and set Ωi,j = 1 and Dj,j = (qi k)^(.5)
 Implies W = Id − UT S T SU

Pr Id −
U T S T SU
β
2
> ϵ ≤ 2d ⋅
−kϵ2 Θ d
e
d log d
)
βϵ2
. Set k = Θ(
and we’re done.
75
Fast Computation of Leverage Scores
 Naively, need to do an SVD to compute leverage scores
 Suppose we compute SA for a subspace embedding S
 Let SA = QR−1 be such that Q has orthonormal columns
 Set ℓ′i = ei AR
2
2
 Since AR has the same column span of A, AR = UT −1
 1 − ϵ ARx 2 ≤ SARx 2 = x 2
 1 + ϵ ARx 2 ≥ SARx 2 = x 2
 1 ± O(ϵ ) x 2 = ARx 2 = UT −1 x 2 = T −1 x 2 ,
 ℓi = ei ART
2
2
= 1 ± O(ϵ) ei AR
2
2
= 1 ± O(ϵ) ℓi ′
 But how do we compute AR? We want nnz(A) time
76
Fast Computation of Leverage Scores
 ℓi = 1 ± O(ϵ) ℓi ′
 Suffices to set this ϵ to be a constant
 Set ℓ′i = ei AR 22
 This takes too long
 Let G be a d x O(log n) matrix of i.i.d. normal random variables
 For any vector z, Pr[ zG
2
2
= 1±
1
2
z 2] ≥ 1 −
1
n2
 Instead set ℓ′i = ei ARG 22 .
 Can compute in (nnz(A) + d2 ) log n time
 Can solve regression in nnz(A) log n + poly(d(log n)/ε) time
77
Course Outline
 Subspace embeddings and least squares regression
 Gaussian matrices
 Subsampled Randomized Hadamard Transform
 CountSketch
 Affine embeddings
 Application to low rank approximation
 High precision regression
 Leverage score sampling
 Distributed Low Rank Approximation
78
Distributed low rank approximation
 We have fast algorithms for low rank approximation, but
can they be made to work in a distributed setting?
 Matrix A distributed among s servers
 For t = 1, …, s, we get a customer-product matrix from
the t-th shop stored in server t. Server t’s matrix = At
 Customer-product matrix A = A1 + A2 + … + As
 Model is called the arbitrary partition model
 More general than the row-partition model in which each
customer shops in only one shop
79
The Communication Model
Coordinator
…
Server 1
Server 2
Server s
• Each player talks only to a Coordinator via 2-way communication
• Can simulate arbitrary point-to-point communication up to factor of 2
80
(and an additive O(log s) factor per message)
Communication cost of low rank approximation
 Input: n x d matrix A stored on s servers
 Server t has n x d matrix At
 A = A1 + A2 + … + As
 Assume entries of At are O(log(nd))-bit integers
 Output: Each server outputs the same k-dimensional space W
 C = A1 PW + A2 PW + … + As PW , where PW is the projector onto W
 |A-C|F · (1+ε)|A-Ak|F
 Application: k-means clustering
 Resources: Minimize total communication and computation.
Also want O(1) rounds and input sparsity time
81
Work on Distributed Low Rank Approximation
 [FSS]: First protocol for the row-partition model.
 O(sdk/ε) real numbers of communication
 Don’t analyze bit complexity (can be large)
 SVD Running time, see also [BKLW]
 [KVW]: O(skd/ε) communication in arbitrary partition model
 [BWZ]: O(skd) + poly(sk/ε) words of communication in
arbitrary partition model. Input sparsity time
 Matching Ω(skd) words of communication lower bound
 Variants: kernel low rank approximation [BLSWX], low rank
approximation of an implicit matrix [WZ], sparsity [BWZ]
82
Outline of Distributed Protocols
 [FSS] protocol
 [KVW] protocol
 [BWZ] protocol
83
Constructing a Coreset [FSS]
 Let A = U ΣV T be its SVD
 Let m = k + k/ϵ
 Let Σm agree with Σ on the first m diagonal entries, and be 0
otherwise
 Claim: For all projection matrices Y=I-X onto (n-k)-dimensional
subspaces,
Σm V T Y
where c = A −
2
F
Am 2F
= 1 ± ϵ AY
2
F
+ c,
does not depend on Y
T so that SA = U T UΣV T = Σ V T is a sketch
 We can think of S as Um
m
m
84
Constructing a Coreset

Claim: For all projection matrices Y=I-X onto (n-k)-dimensional subspaces,
Σm V T Y
where c = A − Am

Proof: AY
2
F
= AX
=
2
F
2
TX
−
Σ
V
m
F
− Σm V T X
Σ − Σm V T X
+ U Σ − Σm V T Y
F
+ A − Am
F
+ A − Am
F
= Σm V T
does not depend on Y
2
2
Also, Σm V T Y
2
+
c
=
1
±
ϵ
AY
F,
F
2
= UΣm V T Y
≤ Σm V T Y
2
F
2
2
F
2
− AY
2
F
2
F
= Σm V T Y
2
F
+c
2
F
+ A − Am
F
2
F
− A
2
F
+ AX
2
F
2
F
2
F
2
≤ Σ − Σm V T 2 ⋅
≤ σ2m+1 k ≤ ϵ σ2m+1
X
2
F
m−k+1 ≤ϵ
2
i∈{k+1,..,m+1} σi
≤ ϵ A − Ak
2
F
85
Unions of Coresets

Suppose we have matrices A1 , … , As and construct
1 T,1 2 T,2
s T,s
Σm
V , Σm V , … , Σm
V as in the previous slide, together with c1 , … , cs

i T,i
Then i
Σm
V Y F + ci = 1 ± ϵ AY 2F , where A is the matrix formed by
concatenating the rows of A1 , … , As

Let B be the matrix obtained by concatenating the rows of
1 V T,1 , Σ 2 V T,2 , … , Σ s V T,s
Σm
m
m

Suppose we compute B = U ΣV T and compute Σm V T and B − Bm

Then Σm V T Y

So Σm V T and the constant c +
2
2
+c+
F
i ci
= 1 ± ϵ BY
i ci
2
F
+
i ci
= 1 ± O(ϵ) AY
2
F
2
F
are a coreset for A
86
[FSS] Row-Partition Protocol
[KVW] protocol
will handle 2, 3,
Problems:
Coordinator and 4
1. sdk/ε real numbers of communication
2. bit complexity can be large
3. running time for SVDs [BLKW]
4. doesn’t work in arbitrary partition model
…
P s ∈ Rn s x d
P 2 ∈ Rn2 x d
P1 ∈ Rn1 x d
This is an SVD-based protocol. Maybe
our random matrix techniques can t
 Server t sends the top k/ε + k principal components of P , scaled by the top
improve communication just like they
k/ε + k singular values Σ t , together with 𝑐 𝑡
improved computation?
 Coordinator returns top k principal components of Σ1 V1 ; Σ 2 V 2 ; … ; Σ s V s
87
[KVW] Arbitrary Partition Model Protocol
 Inspired by the sketching algorithm presented earlier
Fix:
Problems:
 Let S be
one ofofthe
k/ε x n random
matrices
Instead
projecting
A onto SA,
recalldiscussed
2 small seed
Can’t
output
projection
of At onto
SA
since
 Scan
be
generated
pseudorandomly
from
T
we can solve min A SA XSA − A F
rank−k X
the rank is too
large
 Coordinator
sends
small seed for S to all servers
 Let 𝑇1 , 𝑇2 be affine embeddings, solve
2
T
 Could
communicate
this
projection
min T1 A SA XSAT2 − T1 AT2 F to the
t and sends it to Coordinator
rank−k
X
 Server
t
computes
SA
coordinator
whoproblem
could find
a k-dimensional
(optimization
is small
and has a
space, form
but communication
depends on n
closed
solution)
s SAt = SA to all servers
 Coordinator
sends
Σ
t=1
 Everyone can then compute XSA and
then output k directions
 There is a good k-dimensional subspace inside of SA. If
we knew it, t-th server could output projection of At onto it88
[KVW] protocol
 Phase 1:
 Learn the row space of SA
optimal k-dimensional
space in SA
SA
cost · (1+ε)|A-Ak|F
89
[KVW] protocol
 Phase 2:
 Find an approximately optimal space W inside of SA
optimal space in SA
approximate
space W in SA
SA
cost · (1+ε)2|A-Ak|F
90
[BWZ] Protocol
 Main Problem: communication is O(skd/ε) + poly(sk/ε)
 We want O(skd) + poly(sk/ε) communication!
 Idea: use projection-cost preserving sketches [CEMMP]
 Let A be an n x d matrix
 If S is a random k/ε2 x n matrix, then there is a constant
𝑐 ≥ 0 so that for all k-dimensional projection matrices P:
SA I − P F + c = 1 ± ϵ A I − P F
91
[BWZ] Protocol
 Let S be a k/ε2 x n projection-cost preserving sketch
 Let T be a d x k/ε2 projection-cost
preserving
sketch
Intuitively,
U looks like
top k
t
 Server t sends SA T to Coordinator
left singular vectors of SA
 Coordinator sends back SAT =
t
t SA T
to servers
Thus, U T SA looks like2 top k
 Each right
server
computes
k/ε of
x kSA
matrix U of top k left singular
singular
vectors
vectors of SAT
 Server t sends U T SAt Top
to Coordinator
k right singular vectors of SA
work because S is a projectioncost preserving sketch!
 Coordinator returns the space U T SA =
tU
T SAt
to output
92
[BWZ] Analysis
 Let W be the row span of UT SA, and P be the projection onto W
 Want to show A − AP
F
≤ 1 + ϵ A − Ak
F
 Since T is a projection-cost preserving sketch,
(*)
SA − SAP
F
≤ SA − UUT SA
F
+ c1 ≤ 1 + ϵ SA − SA
k F
 Since S is a projection-cost preserving sketch, there is a scalar c >
0, so that for all k-dimensional projection matrices P,
SA − SAP
F
+ c = 1 ± ϵ A − AP
 Add c to both sides of (*) to conclude A − AP
F
F
≤ 1 + ϵ A − Ak
F
93
Conclusions for Distributed Low Rank Approximation
 [BWZ] Optimal O(sdk) + poly(sk/ε) communication protocol for low
rank approximation in arbitrary partition model
 Handle bit complexity by adding Tao/Vu noise
 Input sparsity time
 2 rounds, which is optimal [W]
 Optimal data stream algorithms improves [CW, L, GP]
 Communication of other optimization problems?
 Computing the rank of an n x n matrix over the reals
 Linear Programming
 Graph problems: Matching
 etc.
94
Additional Time-Permitting Topics
 Will cover some recent topics at a research-level (many
details omitted)
 Weighted Low Rank Approximation
 Regression and Low Rank Approximation with MEstimator Loss Functions
 Finding Heavy Hitters in a Data Stream optimally
95