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Tight integrality gaps for vertex-cover semidefinite relaxations in the Lovász-Schrijver Hierarchy Avner Magen Joint work with Costis Georgiou, Toni Pitassi and Iannis Tourlakis University of Toronto

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Minimum Vertex Cover Finding minimum size VC is NP-hard Exist simple 2-approximations All known algs are 2 o(1) approximations! Probabilistically checkable proofs (PCPs) No poly-time 1.36 approximation [Dinur-Safra02] Unique Games Conjecture [Khot02] No poly-time 2 approximation [Khot-Regev03] Alternative (concrete) approach [ABL02, ABLT06]: Rule out approximations by large subfamilies of algorithms

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Linear Programming approach min i V v i v i + v j 1, ij E v i {0,1} 0 v i 1 True Optimum Optimal Fractional Solution Integrality Gap:max Easy to see IG 2 for K n : IG = 2 1/n

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SDP: the ultimate remedy? Vertex Cover on G = (V,E) Tighter relaxation? Smaller integrality gap? min i V (1 + v 0 · v i )/2 (v 0 v i ) · (v 0 v j ) = 0, ij E || v i || 2 = 1, v i R n+1 min i V (1 + x 0 x i )/2 (x 0 x i )(x 0 x j ) = 0, ij E |x i | = 1 Hatami-M-Markakis06: Integrality gap still 2 o(1), even with pentagonal inequalities Semidefinite Programming Relaxations Kleinberg-Goemans98: Integrality gap 2 o(1) Clearly holds in integral case v i { 1,1} (v 0 v i ) · (v 0 v j ) 0, i,j (v i v j ) · (v i v k ) 0, i,j Charikar02: Gap still 2 o(1)

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Systematic Approach: Lovász-Schrijver Liftings [LS91] Procedures LS 0, LS, LS + for tightening linear relaxations Integral hull in n rounds Optimize over rth round relaxation in n O(r) time Very powerful algorithms obtained through small number of rounds: GW94, KZ97, ARV04 algorithms poly-time in LS + All NP in exponential time May view super-constant rounds lower bounds in LS + models as evidence about inapproximability Initial Linear Relaxation Integral Hull Has PSD constraint Sequence of tighter and tighter SDPs Lift to obtain SDP Relaxation n variables n 2 variables Project back to obtain tighter LP

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Previous Lower Bounds for Vertex Cover – without SDP constraints (LS) [ABLT06]: Int. gap 2 o(1) after (log n) LS rounds [Tourlakis06]: Int. gap 1.5 o(1) after (log 2 n) LS rounds [STT06b]: Int. gap 2 o(1) after (n) LS rounds

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Status of SDP variant LS+ Stronger: one round already Implies clique constraint More generally, gives n-θ(G) lower bound on VC (so sparse graph are generally not good) Gives rise to SDPs in the lift phases.

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Integrality gap of 7/6 for LS+ (STT06a) PCP world: Hastad 0.5-hardness for MAX3XOR and the FGLSS reduction imply 7/6-hardness for VC AAT05 proved matching LB (for int. gap) in LS+ world for MAX3XOR STT06b using further ideas from FO06, extend AAT MAX3XOR LB to prove 7/6 int. gap for linear rounds graph family: FGLSS reduction on random MAX3XOR instances Int. gap 7/6 already after one round

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Vertex Cover in LS: results so far SDP version (LS+)? Int. gap 2-o(1) ? # rounds superconsant? ABLT 02,STT 07 NO YES YES STT 06 YES NO YES Charikar 02 YES YES NO New result YES YES YES

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Main Result Theorem: Int. gap 2 o(1) for SDPs resulting after (log n/log log n) LS + rounds One LS + round tighter than [C02] SDP SDPs ruled out incomparable to SDPs with (generalized) triangle and pentagonal inequalities (e.g., [HMM06]) Theorem: Int. gap 2 O(1/log n/log log n) after O(1) LS + rounds Karakostas [K05] SDP gives 2 (1/log n) approximation Use same graph families as [KG98], [C02], [HMM06] SDP solutions rely on sequence of polynomials applying tensor operations on vectors

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x k (x i + x j x 0 ) 0 ij E (x 0 x i )(x j – x 0 ) 0 ij E (x 0 x i )(x 0 x j ) 0 v k · (v i + v j v 0 ) 0 ij E (v 0 v i ) · (v 0 v j ) = 0 ij E (v 0 v i ) · (v 0 v j ) 0 x k (x i + x j x 0 ) 0 ij E (x 0 x i )(x 0 – x j ) = 0 ij E (x 0 x i )(x 0 x j ) 0 Y ik + Y jk Y 0k 0 ij E Y 00 Y 0i Y 0j + Y ij = 0 ij E Y 00 Y 0i Y 0j + Y ij 0 Convert vertex cover LP into an SDP? Multiply linear inequalities to get valid quadratic constraints. Crucially, add integrality conditions: (x 0 x i )x i = 0 E.g., Linearize: replace products x i x j with linear variables Y ij Lifted SDP in (n + 1) 2 variables Project resulting convex body back onto n + 1 variables Y 0i x k (x i + x j x 0 ) 0 ij E (x 0 x i )(x i + x j x 0 ) 0 ij E (x 0 x i )(x 0 x j ) 0 LS + lift-and-project: the quick guide min i V x i x i + x j 1 (i,j) E 0 x i 1 i V (x 0 = 1) x i + x j x 0 0 (i,j) E x i 0 i V x 0 x i 0 i V Ye i,Y(e 0 e i ) K Y 0i = Y ii (x 0 x i )x i = 0 (x 0 = 1) Y is PSD Homogenization: cone K = x i

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How LS and LS+ tighten VC Relaxation One round of LS precisely adds odd-cycle constraints: For all cycles C in G of odd length, i C x i (|C|+1)/2 x 1 + x 2 + x 3 2 One round of LS+ adds more: Clique constraints: For all cliques K in G, i K x i |K| – 1 min i V x i x i + x j 1, ij E 0 x i 1 vs. x 1 + x 2 + x 3 3/2

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Deriving the clique constraints in LS+ 0 x 0 – x i ) (x i + x j – x 0 ) +( (k – x 0 – x i ) i 2 Edge constraint ij Let K be a clique of size k in G SDP condition x i 2 – (k – 1) x 0 2 x i k – 1 After projecting i

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x K (r) if matrix Y s.t. diagonal is x Y is PSD columns K (r 1) Proving Lower Bounds in LS + Hierarchies I.H. LP relaxation K for G with min VC ~ n: x i + x j 1 ij E (½, ½,…) K (1) K (3) K (2) Int. gap of K is 2 – o(1) (½+, ½+ …) Use inductive proof: find appropriate Ys Protection matrix for x Lemma (LS91):

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Frankl-Rödl graphs m-dimensional Hamming cube: n = 2 m points V = { 1,1} m (i, j) E iff (i, j) = (1 )m } parameter Theorem: [Frankl-Rödl87] Max Ind.Set size |B(v,n/2(1- ))| m2 m (1 2 /64) m Cor: If = (log m/m) then max IS is o(2 m ) = o(n) Graphs used for int.gaps in [KK91, AK94, KG95, C02, HMM06] (i, j) = (1 )m

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o(n) Whats so wonderful about them?... Start with a perfect matching Perturb : edges connect vertices of Ham. Dist. (1- )n Vertex Cover = n/2 ``Geometric vertex cover =n/2 +O( )

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Proof Outline In induction: need vectors v i to define matrix Y ij = v i v j Show v i exist whenever x {0, 1, ½ + } n and > 6 Ensure S {0, 1, ½ + } n where O( ) ( / ) round lower bound for x = (½ + )1 Constant and = (log m/m) Int. gap 2 o(1) after (log n/log log n) rounds x K (r) if PSD matrix Y s.t. 1. diagonal is x 2. columns K (r 1) 2. Show some set S K (r 1) where columns conv(S) (i, j) = (1 )m VC 1 o(n) x = (½ + )1

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Back to Frankl-Rödl graphs Natural set {u i } of unit vectors: { 1,1} m (v 0 v i ) · (v 0 v j ) = 0, (i, j) E m 1 Note: u i · u j = 1 2 (i, j)/m Hence (i, j) E u i and u j nearly antipodal Nearly true for v i = u i 2 1 for (i, j) E linear function F of v i · v j (i, j) = (1 )m VC 1 o(n) u i · u j 1 2 1 1 F 1 0 1 v i · v j 1 1 Kleinberg-Goemans: Affine translation on u i to obtain v i F V = { 1,1} m

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Use Kleinberg-Goemans v i for LS + ? Fact: One round of LS + also requires following ineq: Idea (Charikar): Map u i to w i s.t. F(w i · w j ) 0 F(w i · w j ) = 0 if ij E I.e, when u i · u j = 2 1 How? Use tensoring (v 0 v i ) · (v 0 v j ) 0 i,j equality whenever ij E (i, j) = (1 )m VC 1 o(n) u i · u j 1 2 1 1 F(v i · v j ) 1 0 1 v i · v j 1 1 [KG] affine map on u i linear map F(v i · v j ) F(w i · w j ) 1 0 1 Desired mapping on dot-products

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Tensoring u, v R n Tensor product: u v R n 2 Value u i v j at coordinate (i, j) [n] 2 Easy fact: (u v) · (u v) = (u · v) 2 Let P(x) = c 1 x t 1 + … + c q x t q Consider map T P (u) = (c 1 u t 1,…, c q u t q ) Example: P(x) = x 2 + 4x T P (u) = (u u, 2u) R n 2 +2n T P (u) · T P (v) = (u · v) 2 + 4(u · v) 2 = P(u · v) Fact: T P (u) · T P (v) = P(u · v) 2 2 P determines dot-product of resulting vectors Positive coefficients

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Back to finding solution for stronger SDP: Use T P Charikar exhibits appropriate P (i, j) = (1 )m VC 1 o(n) I.e, when u i · u j = 2 1 (v 0 v i ) · (v 0 v j ) 0 i,j equality whenever (i, j) E F(v i · v j ) u i · u j 1 2 1 1 F 1 0 1 Want w i = T P (u i ) s.t. F(w i · w j ) min at (i, j) E u i · u j 1 1 2 1 0 KG C

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Charikar soln gives one round LS + lower bound Charikar vectors define Y ij = v i · v j that: 1. Diagonal is x = (½ + )1 2. Columns K x K (r) if PSD matrix Y s.t. 1. diagonal is x 2. columns K (r 1) I.H. x = (½ + )1 Can Charikar vectors show columns K (1) ? VC = 1 o(n) Must have seq of polynomials Problems: (1) Columns not of form (½ + )1 (2) Charikars vectors work only for one value Values distributed like polynomial of Gaussian

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Making non-uniform columns uniform Columns we want to continue from not of form (½ + )1 Def [STT]: x K is -saturated if for all ij E so that x i, x j < 1 there is surplus: x i + x j 1 + 2 Lemma [STT]: x is -saturated there exists set of vectors x (i) {0, 1, ½ + } n in K s.t. x conv({x (i) }). Can convert columns to (essentially) (½ + )1 IF columns are -saturated Will be safe to ignore 0/1 values distributed like polynomial of Gaussian

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Goal: matrix Y for x with column saturation ( ) Recall P(x) defines T P (u) such that T P (u) · T P (v) = P(u · v) deg(P C ) = O(1/ ) Fact: Y has columns s.t. some edges never have surplus Problem: saturation of close by edges? Saturation Normal. Ham. Dist. from blue edge Necessary: deg(P) · m ! For all P P Bad saturation zone The blue edge ~ P(1) P(1-1/m) P(1)/m Is saturation good enough? = o(m)

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Want column saturation O( ) Precise technical property needed for P: | P(u i · u k ) + P(u j · u k ) | O( ) For all vertices k and all edges ij : [ 1, 1] But u i · u j = 2 1 for all edges ij, so Need | P(x) + P(y) | O( ) over R Red points correspond to 0-1 edges Ignored in saturation calculation 1 1 1 1 1 2 2 1 R 1 1/m x y Domain of P(x) + P(y) |u i ·u k +u j ·u k | 2 |u i ·u k u j ·u k | 2(1- )

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So far: There must be a seq of polys dep. on m. Polynomials must have large degree. Let x {0, 1, ½ + } n Take P(x) = ( x x 1) m/ + x 1/ + (1- x Properties: Minimum at u i · u j, ij E P(1) > m Works as long as > 6 The Columns of Y that is produced by using T P,m (u i ) have saturation O( ) u i · u j 1 12 1 0 KG C P arbitrary > 6 Defining the sequence of tensoring polynomials

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Putting everything together Induction: Have x {0, 1, ½ + } n where > 6 Define Y using T P,m (u i ) Columns have saturation O( ) [STT] Exists S K {0, 1, ½ + } n s.t. columns conv(S) Induction Hypothesis S K (r 1) Take constant and = (log m/m) x K (r) if PSD matrix Y s.t. 1. diagonal is x 2. columns K (r 1) 2. Show some set S K (r 1) where columns conv(S) x = (½ + )1 r = ( / ) (i, j) = (1 )m VC 1 o(n) x K (r)

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Requiring that ||v i -v j || 2 is l1? As is, no l1 inequalities are not implied. The results of [HMM] (showing that metric-cut ineqaities and pentagonal inequalities hold) suggest the examples are still good. Need to Give Sherali Adams LB introduce d ij = ||v i -v j || 2 Add more reqs the LS+ proof need to satisfy.

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Sherali-Adams [SA90] Lift-and-Project Idea: Keep lifting but never project! Simulate third, fourth, etc, degree products with linear vars Only known integrality gap [FK06]: (log n) SA rounds int. gap 2 for MAX-CUT SA + lower bound would inequalities for lifted variables Triangle, pentagonal, etc., inequalities derivable E.g., x 1 x 2 x 3 Y 123 LP not SDP version

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Relations to Unique Games Conjecture (UGC) LS + lower bounds may provide evidence of inapproximability UGC [Khot02] implies optimal inapproximability results for Vertex Cover, MAX-CUT, etc Strong LS +, SA + lower bounds for VC, MAX-CUT

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