Modern Optimization Methods for Big Data Problems MATH11146 The University of Edinburgh


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1 Modern Optimization Methods for Big Data Problems MATH11146 The University of Edinburgh Peter Richtárik Week 3 Randomized Coordinate Descent With Arbitrary Sampling January 27, / 30 The Problem We first consider the following problem: minimize f (x) (1) subject to x = (x 1,..., x n ) R n We will assume that f is: smooth (will be made precise later) strongly convex (will be made precise later) So, this is unconstrained minimization of a smooth convex function. 2 / 30
2 Randomized Coordinate Descent with Arbitrary Sampling NSync Algorithm (R. and Takáč 2014, [4]) Input: initial point x 0 R n subset probabilities {p S } for each S [n] = {1, 2,..., n} stepsize parameters v 1,..., v n > 0 for k = 0, 1, 2,... do a) Select a random set of coordinates S k [n] following the law P(S k = S) = p S, S [n] b) Update (possibly in parallel) selected coordinates: x k+1 = x k i S k 1 v i (e T i f (x k ))e i end for (e i is the ith unit coordinate vector) Remark: The NSync algorithm was introduced in The first coordinate descent algorithm using arbitrary sampling. 3 / 30 Two More Ways of Writing the Update Step 1. Coordinatebycoordinate: x k+1 i = { xi k, i / S k, xi k 1 v i ( f (x k )) i, i S k. 2. Via projection to a subset of blocks: If for h R n and S [n] we write h S = h i e i, (2) i S then x k+1 = x k + h Sk for h = (Diag(v)) 1 f (x k ). (3) Depending on context, we shall interchangeably denote the ith partial derivative of f at x by i f (x) = e T i f (x) = ( f (x)) i 4 / 30
3 Samplings Definition 1 (Sampling) By the name sampling we refer to a set valued random mapping with values being subsets of [n] = {1, 2,..., n}. For sampling Ŝ we ine the probability vector p = (p 1,..., p n ) T by We say that Ŝ is proper, if p i > 0 for all i. p i = P(i Ŝ) (4) A sampling Ŝ is uniquely characterized by the probability mass function p S = P(Ŝ = S), S [n]; (5) that is, by assigning probabilities to all subsets of [n]. Later on it will be useful to also consider the probability matrix P = P(Ŝ) = (p ij) given by p ij = P(i Ŝ, j Ŝ) = S:{i,j} S p S. (6) 5 / 30 Samplings: A Basic Identity Lemma 2 ([3]) For any sampling Ŝ we have n p i = E[ Ŝ ]. (7) Proof. n (4)+(5) p i = n p S = p S = p S S = E[ Ŝ ]. S [n]:i S S [n] i:i S S [n] 6 / 30
4 Sampling Zoo  Part I Why consider different samplings? 1. Basic Considerations. It is important that each block i has a positive probability of being chosen, otherwise NSync will not be able to update some blocks and hence will not converge to optimum. For technical/sanity reasons, we ine: Proper sampling. pi = P(i Ŝ) > 0 for all i [n] Nil sampling: P( Ŝ = ) = 1 Vacuous sampling: P( Ŝ = ) > 0 2. Parallelism. Choice of sampling affects the level of parallelism: E[ Ŝ ] is the average number of updates performed in parallel in one iteration; and is hence closely related to the number of iterations. serial sampling: picks one block: P( Ŝ = 1) = 1 We call this sampling serial although nothing prevents us from computing the actual update to the block, and/or to apply he update in parallel. 7 / 30 Sampling Zoo  Part II fully parallel sampling: always picks all blocks: P(Ŝ = {1, 2,..., n}) = 1 3. Processor reliability. Sampling may be induced/informed by the computing environment: Reliable/dedicated processors. If one has reliable processors, it is sensible to choose sampling Ŝ such that P( Ŝ = τ) = 1 for some τ related to the number of processors. Unreliable processors. If processors given a computing task are busy or unreliable, they return answer later or not at all  it is then sensible to ignore such updates and move on. This then means that Ŝ varies from iteration to iteration. 4. Distributed computing. In a distributed computing environment it is sensible: to allow each compute node as much autonomy as possible so as to minimize communication cost, to make sure all nodes are busy at all times 8 / 30
5 Sampling Zoo  Part III This suggests a strategy where the set of blocks is partitioned, with each node owning a partition, and independently picking a chunky subset of blocks at each iteration it will update, ideally from local information. 5. Uniformity. It may or may not make sense to update some blocks more often than others: uniform samplings: P(i Ŝ) = P(j Ŝ) for all i, j [n] doubly uniform (DU): These are samplings characterized by: S = S P(Ŝ = S ) = P(Ŝ = S ) for all S, S [n] τnice: DU sampling with the additional property that P( Ŝ = τ) = 1 distributed τnice: will ine later independent sampling: union of independent uniform serial samplings nonuniform samplings 9 / 30 Sampling Zoo  Part IV 6. Complexity of generating a sampling. Some samplings are computationally more efficient to generate than others: the potential benefits of a sampling may be completely ruined by the difficulty to generate sets according to the sampling s distribution. a τnice sampling can be well approximated by an independent sampling, which is easy to generate... in general, many samplings will be hard to generate 10 / 30
6 Assumption: Strong convexity Assumption 1 (Strong convexity) Function f is differentiable and λstrongly convex (with λ > 0) with respect to the standard Euclidean norm h = ( n ) 1/2 hi 2. That is, we assume that for all x, h R n, f (x + h) f (x) + f (x), h + λ 2 h 2. (8) 11 / 30 Assumption: Expected Separable Overapproximation Assumption 2 (ESO) Assume Ŝ is proper and that for some vector of positive weights v = (v 1,..., v n ) and all x, h R n, E[f (x + hŝ)] f (x) + f (x), h p h 2 p v. (9) Note that the ESO parameters v, p depend on both f and Ŝ. For simplicity, we will often instead of (9) use the compact notation Notation used above: h S = i S (f, Ŝ) ESO(v). h i e i R n (projection of h R n onto coordinates i S) g, h p p v = n p i g i h i R (weighted inner product) = (p 1 v 1,..., p n v n ) R n (Hadamard product) 12 / 30
7 Assumption: Expected Separable Overapproximation Here the ESO inequality again, now without the simplifying notation: E f x + h i e i f (x) + i Ŝ } {{ } complicated n p i i f (x)h i + } {{ } linear in h 1 n p i v i hi 2 2 } {{ } quadratic and separable in h 13 / 30 Complexity of NSync Theorem 3 (R. and Takáč 2013, [4]) Let x be a minimizer of f. Let Assumptions 1 and 2 be satisfied for a proper sampling Ŝ (that is, (f, Ŝ) ESO(v)). Choose starting point x 0 R n, error tolerance 0 < ɛ < f (x 0 ) f (x ) and confidence level 0 < ρ < 1. If {x k } are the random iterates generated by NSync, where the random sets S k are iid following the distribution of Ŝ, then K Ω ( f (x 0 λ log ) f (x ) ) P(f (x K ) f (x ) ɛ) 1 ρ, (10) ɛρ where Ω = max,...,n v i p i n v i E[ Ŝ ]. (11) 14 / 30
8 What does this mean? Linear convergence. NSync converges linearly (i.e., logarithmic dependence on ɛ) High confidence is not a problem. ρ appears inside the logarithm, so it easy to achieve high confidence (by running the method longer; there is no need to restart) Focus on the leading term. The leading term is Ω; and we have a closedform expression for it in terms of parameters v1,..., v n (which depend on f and Ŝ) parameters p1,..., p n (which depend on Ŝ) Parallelization speedup. The lower bound suggests that if it was the case that the parameters v i did not grow with increasing E[ Ŝ ], then we could potentially be getting linear speedup in τ (average number of updates per iteration). So we shall study the dependence of vi on τ (this will depend on f and Ŝ) As we shall see, speedup is often guaranteed for sparse or wellconditioned problems. τ = Question: How to design sampling Ŝ so that Ω is minimized? 15 / 30 Analysis of the Algorithm (Proof of Theorem 3) 16 / 30
9 Tool: Markov s Inequality Theorem 4 (Markov s Inequality) Let X be a nonnegative random variable. Then for any ɛ > 0, P(X ɛ) E[X ]. ɛ Proof. Let 1 X ɛ be the random variable which is equal to 1 if X ɛ and 0 otherwise. Then 1 X ɛ X ɛ. By taking expectations of all terms, we obtain [ ] X P(X ɛ) = E [1 X ɛ ] E ɛ = E[X ]. ɛ 17 / 30 Tool: Tower Property of Expectations (Motivation) Example 5 Consider discrete random variables X and Y: X has 2 outcomes: x 1 and x 2 Y has 3 outcomes: y 1, y 2 and y 3 Their joint probability mass function if given in this table: y 1 y 2 y 3 x x Obviously, E[X] = 0.28x x 2. But we can also write: E[X] = (0.05x x 2 ) + (0.20x x 2 ) + (0.03x x 2 ) = 0.30 }{{} P(Y=y 1 ) 0.30 x x 2) +0.50( } {{ } x x 2) ( x x 2) E[X Y=y 1 ] ( 0.05 = E[E[X Y]]. 18 / 30
10 Tower Property Lemma 6 (Tower Property / Iterated Expectation) For any random variables X and Y, we have E[X ] = E[E[X Y ]]. Proof. We shall only prove this for discrete random variables; the proof is more technical in the continuous case. E[X ] = x xp(x = x) = x x y P(X = x & Y = y) = y = y xp(x = x & Y = y) x P(X = x & Y = y) P(Y = y)x P(Y = y) x = y P(Y = y) xp(x = x Y = y) x } {{ } E[X Y =y] = E[E[X Y ]]. 19 / 30 Proof of Theorem 3  Part I If we let µ = λ/ω, then f (x + h) (8) f (x) + f (x), h + λ 2 h 2 f (x) + f (x), h + µ 2 h 2 v p 1. (12) Indeed, one can easily verify that µ is ined to be the largest number for which λ h 2 µ h 2 v p 1 holds for all h. Hence, f is µstrongly convex with respect to the norm v p 1. Let x be a minimizer of f, i.e., an optimal solution of (1). Minimizing both sides of (12) in h, we get f (x ) f (x) (12) min h R n f (x), h + µ 2 h 2 v p 1 = 1 2µ f (x) 2 p v 1. (13) 20 / 30
11 Proof of Theorem 3  Part II Let h k = v 1 f (x k ). Then in view of (3), we have x k+1 = x k + h k S k. Utilizing Assumption 2, we get E[f (x k+1 ) x k ] = E [ f (x k + h k S k ) x k] (9) f (x k ) + f (x k ), h k p hk 2 p v = f (x k ) 1 2 f (x k ) 2 p v 1 (13) f (x k ) µ(f (x k ) f (x )). Taking expectations in the last inequality, using the tower property, and subtracting f (x ) from both sides of the inequality, we get E[f (x k+1 ) f (x )] (1 µ)e[f (x k ) f (x )]. Unrolling the recurrence, we get E[f (x k ) f (x )] (1 µ) k (f (x 0 ) f (x )). (14) 21 / 30 Proof of Theorem 3  Part III Using Markov s inequality, (14) and the inition of K, we get P(f (x K ) f (x ) ɛ) E[f (x K ) f (x )]/ɛ (14) (1 µ) K (f (x 0 ) f (x ))/ɛ (10) ρ. Finally, let us now establish the lower bound on Ω. Letting { } n = p R n : p 0, p i = E[ Ŝ ], we have Ω (11) = max i v i p i (7) min max p i v i p i = 1 E[ Ŝ ] n v i, where the last equality follows since optimal p i is proportional to v i. 22 / 30
12 How to compute the ESO stepsize parameters v 1,..., v n? 23 / 30 C 1 (A) Functions By inition, the ESO parameters v depend on both f and also highlighted by the notation we use: Ŝ. This is (f, Ŝ) ESO(v). Hence, in order to compute these parameters, we need to specify f and Ŝ. The following inition describes a wide class of functions, often appearing in computational practice, for which we will be able to do this. Definition 7 (C 1 (A) functions) Let A R m n. By C 1 (A) we denote the set of continuously differentiable functions f : R n R satisfying the following inequality for all x, h R n : f (x + h) f (x) + ( f (x)) T h ht A T Ah. (15) Example 8 (Least Squares) The function f (x) = 1 2 Ax b 2 satisfies (15) with an equality. Hence, f C 1 (A). 24 / 30
13 Are There More C 1 (A) Functions? Functions we wish to minimize in machine learning often have a finite sum structure: f (x) = j φ j(m j x), (16) where M j R m n are data matrices and φ j : R m R are loss functions. The next result says that under certain assumptions on the loss functions, f C 1 (A) for some A, and describes what this A is. Theorem 9 ([5]) Assume that for each j, function φ j is γ j smooth: φ j (s) φ j (s ) γ j s s, for all s, s R m. Then f C 1 (A), where A satisfies A T A = J j=1 γ jm T j M j. Remark: The above theorem also says what A T A is. This is important, as will be clear from the next theorem. 25 / 30 ESO for C 1 (A) Functions and an Arbitrary Sampling Theorem 10 ([5]) Assume f C 1 (A) for some real matrix A R m n, let Ŝ be an arbitrary sampling and let P = P(Ŝ) be its probability matrix. If v = (v 1,..., v n ) satisfies P A T A Diag(p v), then (f, Ŝ) ESO(v). 26 / 30
14 Sampling Identity for a Quadratic In in order to prove Theorem 10, we will need the following lemma. Lemma 11 ([5]) Let G be any real n n matrix and Ŝ an arbitrary sampling. Then for any h R n we have E [ h Ṱ Gh ] ( ) S Ŝ = h T P(Ŝ) G h, (17) where denotes the Hadamard (elementwise) product of matrices, and P(Ŝ) is the probability matrix of Ŝ. 27 / 30 Proof of Lemma 11 Proof. Let 1 ij be the indicator random variable of the event i Ŝ & j Ŝ: 1 ij = { 1 if i Ŝ & j Ŝ, 0 otherwise. Note that E[1 ij ] = P(i Ŝ & j Ŝ) = p ij. We now have E [ h Ṱ Gh ] (2) S Ŝ = E n n G ij h i h j = E 1 ij G ij h i h j i Ŝ j Ŝ j=1 n n n n = E[1 ij ]G ij h i h j = p ij G ij h i h j j=1 = h T ( P(Ŝ) G ) h. j=1 28 / 30
15 Proof of Theorem 10 Having established Lemma 11, we are now ready prove Theorem 10. Proof. Fixing any x, h R n, we have E [ f (x + hŝ) ] (15) E [ ] f (x) + f (x), hŝ hṱ S AT AhŜ ] = f (x) + f (x), h p E [ h Ṱ S AT AhŜ (Lemma 11) = f (x) + f (x), h p ht ( P A T A ) h f (x) + f (x), h p ht Diag(p v)h. } {{ } = h 2 p v 29 / 30 References I [1] Yurii Nesterov. Efficiency of coordinate descent methods on hugescale optimization problems. SIAM Journal on Optimization, 22(2): , 2012 [2] Peter Richtárik and Martin Takáč. Iteration complexity of randomized blockcoordinate descent methods for minimizing a composite function. Mathematical Programming, 144(2):1 38, 2014 [3] Peter Richtárik and Martin Takáč. Parallel coordinate descent methods for big data optimization. Mathematical Programming, 1 52, 2015 [4] Peter Richtárik and Martin Takáč. On optimal probabilities in stochastic coordinate descent methods. Optimization Letters, 2015 [5] Zheng Qu and Peter Richtárik. Coordinate descent with arbitrary sampling II: expected separable overapproximation, arxiv: , / 30
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