Sketching as a Tool for Numerical Linear Algebra
David Woodruff
IBM Almaden
2
Talk Outline
Regression Exact Regression Algorithms Sketching to speed up Least Squares Regression Sketching to speed up Least Absolute Deviation (l1) Regression
Low Rank Approximation Sketching to speed up Low Rank Approximation
Recent Results and Open Questions M-Estimators and robust regression CUR decompositions
3
Regression
Linear Regression Statistical method to study linear dependencies between
variables in the presence of noise.
Example Ohm's law V = R ∙ I Find linear function that
best fits the data
4
Regression
Standard Setting One measured variable b A set of predictor variables a ,…, a Assumption:
b = x + a x + … + a x + 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
1 d
1
1 d
d
0
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Regression analysis
Matrix form
Input: nd-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
Assume that A has full column rank
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Regression analysis
Least Squares Method Find x* that minimizes |Ax-b|22 = (bi – <Ai*, x>)²
Ai* is i-th row of A
Certain desirable statistical properties Closed form solution: x = (ATA)-1 AT b
Method of least absolute deviation (l1 -regression)
Find x* that minimizes |Ax-b|1 = |bi – <Ai*, x>|
Cost is less sensitive to outliers than least squares Can solve via linear programming
Time complexities are at least n*d2, we want better!
7
Talk Outline
Regression Exact Regression Algorithms Sketching to speed up Least Squares Regression Sketching to speed up Least Absolute Deviation (l1) Regression
Low Rank Approximation Sketching to speed up Low Rank Approximation
Recent Results and Open Questions M-Estimators and robust regression CUR decompositions
8
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
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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
Computing S*A may be slow…
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How to choose the right sketching matrix S? [S]
S is a Johnson Lindenstrauss 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 much faster
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Even faster sketching matrices [CW]
[ [0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 00 0 0 -1 1 0 -1 00-1 0 0 0 0 0 1
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
Surprisingly,
this works!
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Simpler and Sharper Proofs [MM, NN, N]
Let B = [A, b] be an n x (d+1) matrix
Let U be an orthonormal basis for the columns of B
Suffices to show |SUx|2 = 1 ± ε for all unit x
Implies |S(Ax-b)|2 = (1 ± ε) |Ax-b|2 for all x
SU is a (d+1)2/ε2 x (d+1) matrix
Suffices to show | UTST SU – I|2 · |UTST SU – I|F · ε
Matrix product result: |CSTSD – CD|F2 · [1/(# rows of S)] * |C|F2 |D|F2
Set C = UT and D = U. Then |U|2F = (d+1) and (# rows of S) = (d+1)2/ε2
|SBx|2 = (1±ε) |Bx|2 for all x
S is called a subspace
embedding
|SBx|2 = (1±ε) |Bx|2 for all x
S is called a subspace
embedding
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Talk Outline
Regression Exact Regression Algorithms Sketching to speed up Least Squares Regression Sketching to speed up Least Absolute Deviation (l1) Regression
Low Rank Approximation Sketching to speed up Low Rank Approximation
Recent Results and Open Questions M-Estimators and robust regression CUR decompositions
14
Sketching to solve l1-regression
How to find an approximate solution x to minx |Ax-b|1 ?
Goal: output x‘ for which |Ax‘-b|1 · (1+ε) minx |Ax-b|1 with high probability
Natural attempt: 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)|1
Turns out this does not work
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Sketching to solve l1-regression [SW]
Why doesn’t outputting the solution x‘ to minx‘ |(SA)x-(Sb)|1
work?
Don‘t know of k x n matrices S with small k for which if x‘ is solution to minx |(SA)x-(Sb)|1 then
|Ax‘-b|1 · (1+ε) minx |Ax-b|1
with high probability
Instead: can find an S so that
|Ax‘-b|1 · (d log d) minx |Ax-b|1
S is a matrix of i.i.d. Cauchy random variables Property: |Ax-b|1 · |S(Ax-b)|1 · (d log d) |Ax-b|1
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Cauchy random variables
Cauchy random variables not as nice as Normal (Gaussian) random variables
They don’t have a mean and have infinite variance
Ratio of two independent Normal random variables is Cauchy
If a and b are scalars and C1 and C2 independent
Cauchys, thena*C1 + b*C2 ~ (|a|+|b|)*C
for a Cauchy C
If a and b are scalars and C1 and C2 independent
Cauchys, thena*C1 + b*C2 ~ (|a|+|b|)*C
for a Cauchy C
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Sketching to solve l1-regression
Main Idea: Let B = [A, b]. Compute a QR-factorization of S*B
Q has orthonormal columns and Q*R = S*B
B*R-1 is a “well-conditioning” of B: Σi=1
d |BR-1ei|1 · Σi=1d |SBR-1ei|1
· (d log d).5 Σi=1d |SBR-1ei|2
· d (d log d).5
|x|1 · |x|2
= |SBR-1x|2
· |SBR-1x|1
· (d log d) |BR-1x|1
These two properties make importance sampling work!
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Want to estimate Σi=1n yi by sampling, for yi ¸ 0
Suppose we sample yi with probability pi
T = Σi=1n δ(yi sampled) yi/pi
E[T] = Σi=1n pi ¢ yi/pi = Σi=1
n yi
Var[T] · E[T2] = Σi=1n pi ¢ yi
2 / pi2 · (Σi=1
n yi ) maxi yi/pi
Bound maxi yi/pi by ε2 (Σi=1n yi )
For us, yi = |(Ax-b)i| and this holds if pi = |eiBR-1|1 * poly(d/ε) !
Importance Sampling
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To get a bound for all x, use Bernstein’s inequality and a net argument
Sample poly(d/ε) rows of B*R-1 where the i-th row is sampled proportional to its 1-norm
T is diagonal matrix with Ti,i = 0 if row i not sampled, otherwise Ti,i = 1/Pr[row i sampled] |TBx|1 = (1 ± ε )|Bx|1 for all x
Solve regression on the (reweighted) samples!
Importance Sampling
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Sketching to solve l1-regression [MM]
Most expensive operation is computing S*A where S is the matrix of i.i.d. Cauchy random variables
All other operations are in the “smaller space”
Can speed this up by choosing S as follows:
[ [0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 00 0 0 -1 1 0 -1 00-1 0 0 0 0 0 1
¢
[
[C1
C2
C3
… Cn
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Further sketching improvements [WZ]
Can show you need a fewer number of sampled rows in later steps if instead choose S as follows
Instead of diagonal of Cauchy random variables, choose diagonal of reciprocals of exponential random variables
[ [0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 00 0 0 -1 1 0 -1 00-1 0 0 0 0 0 1
¢
[
[1/E1
1/E2
1/E3
… 1/En
For recent work on fast
sampling-based algorithms, see Richard’s talk!
For recent work on fast
sampling-based algorithms, see Richard’s talk!
Uses max-stability of exponentials
[Andoni]:max yi/ei ~ |y|1/e
Uses max-stability of exponentials
[Andoni]:max yi/ei ~ |y|1/e
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Talk Outline
Regression Exact Regression Algorithms Sketching to speed up Least Squares Regression Sketching to speed up Least Absolute Deviation (l1) Regression
Low Rank Approximation Sketching to speed up Low Rank Approximation
Recent Results and Open Questions M-Estimators and robust regression CUR decompositions
23
Low rank approximation
A is an n x d matrix
Typically well-approximated by low rank matrix E.g., only high rank because of noise
Want to output a rank k matrix A’, so that
|A-A’|F · (1+ε) |A-Ak|F,
w.h.p., where Ak = argminrank k matrices B |A-B|F
(For matrix C, |C|F = (Σi,j Ci,j2)1/2 )
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Solution to low-rank approximation [S]
Given n x d input matrix A Compute S*A using a sketching matrix S with k << n
rows. S*A takes random linear combinations of rows of A
SA
A
Project rows of A onto SA, then find best rank-k approximation to points inside of SA.
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Low Rank Approximation Idea
Regression problem minX |Ak X – A|F
Solution is X = I, and minimum is |Ak – A|F
This is a generalized regression problem!
If S is a subspace embedding for column space of Ak and also if for any matrices B, C, |BSTSC – BC|F2 · 1/(# rows of S) |B|
F2 |C|F2
Then if X’ is the minimizer to minX |SAkX – SA|F , then
|Ak X’ – A|F · (1+ε) minX |AkX-A|F = (1+ε) |Ak-A|F
But minimizer X’ = (SAk)- SA is in the row span of SA!
S can be matrix of i.i.d. Normals
S can be a Fast Johnson Lindenstrauss Matrix
S can be a CountSketch matrix
S can be matrix of i.i.d. Normals
S can be a Fast Johnson Lindenstrauss Matrix
S can be a CountSketch matrix
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Caveat: projecting the points onto SA is slow
Current algorithm:
1. Compute S*A (easy)
2. Project each of the rows onto S*A
3. Find best rank-k approximation of projected points inside of rowspace of S*A (easy)
Bottleneck is step 2
[CW] Turns out you can approximate the projection Sketching for generalized regression again: minX |X(SA)-A|F
2
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Talk Outline
Regression Exact Regression Algorithms Sketching to speed up Least Squares Regression Sketching to speed up Least Absolute Deviation (l1) Regression
Low Rank Approximation Sketching to speed up Low Rank Approximation
Recent Results and Open Questions M-Estimators and robust regression CUR decompositions
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M-Estimators and Robust Regression
Solve minx |Ax-b|M M: R -> R¸ 0 |y|M = Σi=1
n M(yi)
Least squares and L1-regression are special cases
Huber function, given a parameter c:
M(y) = y2/(2c) for |y| · c
M(y) = |y|-c/2 otherwise
Enjoys smoothness properties of l2and
robustness properties of l1
[CW15] For M-estimators with at least linear and at most quadratic growth, can get O(1)-approximation in nnz(A) + poly(d) time
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CUR Decompositions
[BW14] Can find a CUR decomposition in O(nnz(A) log n) + n*poly(k/ε) time with O(k/ε) columns, O(k/ε) rows, and rank(U) = k
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Open Questions
Recent monograph in NOW Publishers D. Woodruff, “Sketching as a Tool for Numerical Linear Algebra”
Other types of low rank approximation:
(Spectral) How quickly can we find a rank k matrix A’, so that
|A-A’|2 · (1+ε) |A-Ak|2,
w.h.p., where Ak = argminrank k matrices B |A-B|2
(Robust) How quickly can we find a rank k matrix A’, so that
|A-A’|1 · (1+ε) |A-Ak|1,
w.h.p., where Ak = argminrank k matrices B |A-B|1
For other questions regarding Schatten norms and communication-efficiency, see reference above. Thanks!