Lecture 20: Lift and Project, SDP Duality. Today we will study the Lift and Project method. Then we will prove the SDP duality theorem.

Transcription

1 prnceton u. sp 02 cos 598B: algorthms and complexty Lecture 20: Lft and Project, SDP Dualty Lecturer: Sanjeev Arora Scrbe:Yury Makarychev Today we wll study the Lft and Project method. Then we wll prove the SDP dualty theorem. 1 Lft and Project Consder the followng nteger program for the Vertex Cover problem: mn V x x + x j 1 x {0, 1} {, j} E V Replacng the last constrant wth the constrant 0 x 1, we get an LP relaxaton for the problem. Unfortunately, the ntegralty gap of ths LP s 2. However, we can rewrte the nteger program as an equvalent quadratc program: mn V x (1 x )(1 x j ) = 0 {, j} E (1 x )x = 0 V 0 x 1 V The dea of Lft and Project s to smulate quadratc programmng (or degree k programmng) wth an LP/SDP. We ntroduce new varables Y j that represent product terms: Y j ntended = x x j. For brevty, let x 0 = 1, and then Y 0 = x. The condton that (1 x )(1 x j ) = 0 can be now expressed as 1 Y 0 Y j0 + Y j = 0. We also ntroduce new constrants on Y j : 1. Y j = Y j ; 2. Y 0 = Y ; 3. the matrx Y s postve semdefnte; 1

2 2 4. If a T x b was a lnear constrant for the orgnal LP then we add the followng lnear constrants on Y j (for all ): (a T x)x bx (a T x)(1 x ) b(1 x ) Fact: All SDP s we saw (or that have been useful) can be derved usng ths process from the obvous lnear relaxatons 1. The followng secton borrowed from the paper Towards strong nonapproxmablty results n the Lovász-Schrjver herarchy by Mkhal Alekhnovch, Sanjeev Arora and Ianns Tourlaks presents the detals. 2 LS + dervaton of popular SDP relaxatons To llustrate the power of the LS + procedure, we sketch how to use a few rounds of LS + to derve popular SDP relaxatons used n famous approxmaton algorthms. (Ths was suggested by the revewers, who ponted out that ths s not very well-known.) It wll be more convenent to vew LS + as a method for generatng new nequaltes. Gven any relaxaton a T r x b r = 1, 2,...,m (1) (where the trval constrants 0 x 1 are assumed to be ncluded), one round of LS + produces a system of nequaltes n (n + 1) 2 varables Y j for, j = 0, 1,...,n. As mentoned, the ntended meanng s that Y j = x x j and Y 00 = 1, Y 0 = x = x x 0, and Y 00 = 1 so every quadratc expresson n the x s can be vewed as a lnear expresson n the Y j s. Ths s how the quadratc nequaltes below should be nterpreted. (1 x )a T r x (1 x )b = 1,...,n, r = 1,...,m x a T r x x b = 1,...,n, r = 1,...,m x x = x x 0 = 1, 2,...,n (The last constrant corresponds to the fact that x 2 = x for 0/1 varables.) Fnally, one mposes the condton that (Y j ) s postve semdefnte. Obvously, any postve combnaton of the above nequaltes s also mpled, and the dervatons below wll use ths fact. 2.1 Dervng the GW relaxaton The Goemans-Wllamson relaxaton for -cut [GW 94] nvolves fndng unt vectors u 1, u 2,...,u n so as to mnmze 1 4 u u j 2. {,j} E Ths SDP relaxaton can be derved by one round of LS + on the trval lnear relaxaton. Ths relaxaton has 0/1 varables x and d j. In the nteger soluton, x ndcates whch sde of the cut 1 ths process may nvolve several rounds of Lft and Project

3 3 vertex s on, and d j s 1 ff, j are on opposte sdes of the cut. {,j} E d j d j x x j, j = 1, 2,...,n (3) d j x + x j, j = 1, 2,...,n (4) d j 2 (x + x j ), j = 1, 2,...,n (5) Then one round of LS + generates the followng nequaltes on d j : x d j x (x x j ) (6) (1 x )d j (1 x )(x j x ). (7) Addng these and smplfyng usng the fact that x 2 = x for 0/1 varables, one obtans d j (x x j ) 2. Smlarly one can obtan d j (x x j ) 2 whereby t follows d j = (x x j ) 2 = Y +Y jj 2Y j. Now f (Y j ) s any feasble soluton then ts Cholesky decomposton v 0, v 1,...,v n R n+1 are vectors such that Y j =< v, v j >. Then d j = v v j 2. Now defne the set of vectors u 1, u 2,...,u n as u = v 0 2v. These satsfy d j = 1 4 u u j 2 (8) u 2 = v < v 0, v > +4 v 2 = 1. (9) Thus the u s are a feasble soluton to the GW relaxaton. We conclude that one round of LS + produces a relaxaton at least as tght as the GW relaxaton (and n fact one can show that the two relaxatons are the same). 2.2 Dervng the ARV relaxaton Arora, Rao, and Vazran [ARV 04] derve ther log n-approxmaton for sparsest cut usng a smlar SDP relaxaton whose salent feature s the trangle nequalty: u u j 2 + u j u k 2 u u k 2, j, k. (In other words, d j = u u k 2 forms a metrc space.) Ths relaxaton mnus the trangle nequalty s derved smlarly to the GW relaxaton above (detals omtted). The clam s that the trangle nequalty s mpled after three rounds of LS +. As shown n [LS 91], r rounds of LS + mply all nequaltes on subsets of sze r that are true for the nteger soluton. In other words, the nduced soluton on subsets of sze r les n the convex hull of nteger solutons. Thus after three rounds the d j varables restrcted to sets of sze three le n the cut cone. Snce the cut cone s just the set of l 1 (pseudo)metrcs, t follows that the d j varables form a (pseudo)metrc. Thus three rounds of LS + gve a relaxaton that s at least as strong as the ARV relaxaton. (2) 3 Sheral Adams Relaxatons Now let us generalze the Lft and Project method to k-ary programmng [Sheral-Adams 91]. For every subset F of ndces of sze at most k, we ntroduce a new varable Y S. The ntended value of Y S s Y S = S x.

4 4 Frst we requre that for every set S (s.t. S k 2), the matrx (Y S {} {j} ) j s postve semdefnte. Then we add another famly of generc constrants: S, T : S T =, S + T k x ) S(1 x j 0. j T As t often happens n theoretcal CS the case k > 2 s much more complcated than the case k = 2. 4 Dscrepancy 4.1 Two Party Case We are now nterested n applyng the Lft and Project method to analyze the dscrepancy. Recall the defnton of the dscrepancy for two partes. Defnton 1 (2 party case) Dscrepancy of a matrx M s M C = x 1,...,x n {0,1} y 1,...,y n {0,1},j M j x y j. We saw n the last lecture that the cut norm M C s well approxmated by the norm 1 : M 1 = x 1,...,x n { 1,1} y 1,...,y n { 1,1},j M j x y j. Alon and Naor showed that ths norm s approxmated wthn a constant factor by the followng SDP relaxaton: M C = M j u v j. u 1,...,u n S 2n 1 v 1,...,v n S 2n 1,j 4.2 Multparty Case We are nterested n extendng ths method to multparty communcaton complexty. Recall that n the multparty case we have a multdmensonal matrx M j...k (.e. a covarant tensor); our objectve functon s to mze the sum of entres n a set S over all sets S that are cylnder ntersectons. Defnton 2 (3 party case) In the 3 party case, each cylnder ntersecton s determned by ts projectons on the planes O j, O k, and O jk. Let x j, y k, and z jk be ndcator functons of these projectons. Then cylnder ntersecton = {(, j, k) : x j = y k = z jk = 1} = {(, j, k) : 1 x j y k z jk = 0}. So the dscrepency s equal to Dsc(M) = x j {0,1} y k {0,1} z jk {0,1},j,k M,j,k x j y k z jk.

5 5 k j 0 Fgure 1: We mze,j,k M,j,k x j y k z jk over all cylnder ntersectons. We now represent ths optmzaton problem as a Sheral-Adams relaxaton. Denote the optmum value of ths relaxaton by SDP opt (M). Clearly, SDP opt (M) Dsc(M). How well does SDP opt (M) approxmate Dsc(M)? Queston 1: Is SDP opt = O( small Dsc(M))? Queston 2: Show for some natural functon (e.g. clque) that SDP opt s small. Guess: Perhaps, the answer to the frst queston s No. How to solve these problems? The natural way s to use dualty. 5 SDP Dualty A general semdefnte program s an optmzaton program of the followng form: mn C X subject to A 1 X = b 1 A 2 X = b 2 A m X = b m X 0 where C, A 1,..., A m are n n symmetrc matrces; the nner product A B s defned as A B def =,j. A j B j = tr(ab). Note that ths program s a convex program snce the set of all postve semdefnte matrces forms a convex cone. Ths follows from the followng two observatons:

6 6 If A s a postve semdefnte, then so s λa for every λ > 0. If A and B are postve semdefnte matrces, then so s A + B. Indeed, for every vector v, we have v T (A + B)v = v T Av + v T Bv Theorem 1 (Fact at the root of dualty) A matrx Y s postve semdefnte ff for every postve semdefnte matrx A, A Y 0. Proof: Frst, assume Y 0. Let A be an arbtrary semdefnte matrx. Consder Cholesky decompostons of matrces A and Y : A = V T V, and Y = Z T Z. Then A Y = tr(ay ) = tr(v T V Z T Z) = tr(zv T V Z T ) = tr(zv T (ZV T ) T ) 0. Note that f A s a postve defnte matrx and Y 0, then A Y > 0. Indeed, snce A s nonsngular, V s also nonsngular, therefore ZV T 0, whch mples A Y > 0. The other drecton s trval: v T Y v = Y (vv T ) 0. }{{} 0 Therefore Y s postve semdefnte. Theorem 2 The followng system of equatons s nfeasble ff x 1,...,x m R s.t. x A 0. A 1 X = 0 A 2 X = 0 A m X = 0 X 0 X 0 Proof: Suppose there are no x 1,...,x m R s.t. x A 0. Ths means that the lnear subspace { x A } s dsjont from the cone of postve defnte matrces (whch equals the nteror of the cone of postve semdefnte matrces). By Farkas lemma, there exsts a separatng hyperplane that contans ths lnear space s.t. the semdefnte cone les on one sde of the hyperplane. Let Y be the normal to ths hyperplane. Then Y A = 0, and, by Theorem 1, Y 0. In other words, Y s a feasble soluton of the system of the equatons. We get a contradcton. The proof of the other drecton s straght forward. Assume that there exst x 1,...,x n for whch x A 0; but the system of equatons has a feasble soluton X. We have x A X = x (A X) = x 0 = 0;

7 7 cone of SDP matrces 0 x A Fgure 2: The SDP cone s dsjont from the set x A. on the other hand x A X = ( x A ) X > 0. }{{} 0 We get a contradcton. Ths concludes the proof.