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How to Round Any CSP
Prasad RaghavendraUniversity of Washington, Seattle
David Steurer,Princeton University
(In Principle)
Constraint Satisfaction ProblemA Classic Example : Max-3-SAT
Given a 3-SAT formula,Find an assignment to the variables that satisfies the maximum number of clauses.
))()()(( 145532532321 xxxxxxxxxxxx Equivalently the
largest fraction of clauses
Variables : {x1 , x2 , x3 ,x4 , x5} Constraints : 4 clauses
Constraint Satisfaction Problem
Instance :• Set of variables.• Predicates Pi applied on variables
Find an assignment that satisfies the largest fraction of constraints.
Problem :
Domain : {0,1,.. q-1}Predicates : {P1, P2 , P3 … Pr}
Pi : [q]k -> {0,1}
Max-3-SAT
Domain : {0,1}Predicates :
P1(x,y,z) = x ѵ y ѵ z
))()()(( 145532532321 xxxxxxxxxxxx
Theorem: [Raghavendra 08]Assuming Unique Games Conjecture, For every CSP, “a simple semidefinite program (SDP1) gives the best approximation computable efficiently.”
[Raghavendra08]A generic rounding scheme for (SDP1) that is optimal for every CSP under UGC.
Independent of UGC, for 2CSPs, the generic rounding scheme for (SDP1) achieves an
Approximation Ratio ≥ (1-²) Integrality Gap of SDP.
Rounding Algorithm
minimumover all instances
=
value of rounded solution
value of SDP solution
rounding – ratioA ( ¦ )(approximation ratio)
≥ (1-²) integrality gap ( ¦ )
=
value of optimal solution
value of SDP solution
minimumover all instances
For any CSP ¦ and any ²>0, there exists an efficient algorithm A,
Unconditionally, the algorithm A as good as all known algorithms for CSPs
Very Simple : No Invariance Principle, Dictatorship Tests, Unique Games.
Drawbacks•Running Time(A) On CSP over alphabet size q, arity k
•No explicit approximation ratio)(2
)/1,,(2 npolyqkpoly
Computing Integrality Gaps
Theorem:
For any CSP ¦ and any ²>0, there exists an algorithm A to compute integrality gap (¦) within an accuracy ²
Running Time(A) On CSP over alphabet size q, arity k
)/1,,(22qkpoly
Previous Work SDP ALGORITHMS[Charikar-Makarychev-Makarychev 06]
MaxCut [Goemans-Williamson] [Charikar-Wirth]
[Lewin-Livnat-Zwick][Charikar-Makarychev-Makarychev 07]
[Hast] [Charikar-Makarychev-Makarychev 07]
[Frieze-Jerrum][Karloff-Zwick]
[Zwick SODA 98][Zwick STOC 98]
[Zwick 99][Halperin-Zwick 01]
[Goemans-Williamson 01][Goemans 01]
[Feige-Goemans][Matuura-Matsui]
[Trevisan-Sudan-Sorkin-Williamson]
[O’Donnell-Wu] Optimal rounding schemes for MaxCut
ALGORITHM OUTLINERounding Any Constraint Satisfaction Problem
Max Cut
10
15
3
7
11
Max CUTInput : A weighted graph G
Find :A cut with maximum fraction of crossing edges
Eji
jiij vvw),(
2||4
1
Semidefinite Program
Variables : v1 , v2 … vn
| vi |2 = 1
Maximize
Max Cut SDP
10
15
3
7
11
1
1
1
-1
-1
-1
-1-1
-1
v1
v2
v3
v4
v5
MaxCut Rounding Problem
Given a graph on the n - dimensional unit ball,Find the maximum cut of the graph.
Approximation using Finite Models
¦-CSP Instance =
¦-CSP Instance =finite
variablefolding
(identifyingvariables)
optimal solution for
=finite
approximate solution
for =
unfolding ofthe assignment
constant time
Challenge: ensure = finite has a good solution
10
15
3
7
11
1
1
-1
-1
-1
-1-1
-1
-1
1
1
11
Approximation using Finite Models
[Frieze-Kannan]For a dense instance =, it is possible to construct finite model
=finite
OPT(=finite) ≥ (1-ε) OPT(=)
General Method for CSPs
What we will do :
SDP value (=finite) > (1-ε)SDP value (=)
PTAS for dense instances
Analysis of Rounding Scheme
¦-CSP Instance =
¦-CSP Instance =finite
SDP value ®
SDP value > ® - ²
OPT value¯
rounded value¯
010001001010001001
Hence: rounding-ratio for = < (1+²) integrality-ratio for = finite
unfolding
CONSTRUCTING FINITE MODELS (MAXCUT)
Rounding Any Constraint Satisfaction Problem
v1
v2
v3
v4
v5
STEP 1 : Dimension Reduction
• Pick d = 1/ Є4 random Gaussian vectors {G1 , G2 , .. Gd} • Project the SDP solution along these directions.Map vector V
V → V’ = (V G∙ 1 , V G∙ 2 , … V G∙ d)v
1
v3v
4 v5
Constant dimensions
STEP 2 : SurgeryScale every vector V’ to unit length
STEP 3 : Discretization•Pick an Є –net for the
d dimensional sphere• Move every vertex to the nearest point in the Є –net
v2v
2
FINITE MODEL Graph on Є –net points
To Show:
SDP value (=finite) > (1-ε)SDP value (=)
Lemma : “Inner Products are almost preserved under random
projections”
If V’,U’ are random projections of U, V on 1/ ε4 directions,
Pr [ |V U – V’ U’| > ∙ ∙ ε] < ε2
STEP 1 : Dimension Reduction•Project the SDP solution along 1/ Є4 random directions.
STEP 2 : SurgeryScale every vector V’ to unit length
STEP 3 : Discretization•Pick an Є –net for the
d dimensional sphere• Move every vertex to the nearest point in the Є –net
For SDP value (=)Contribution of an edge e = (U,V)
|U-V|2 = 2-2 V U ∙
To Show:
SDP value (=finite) > (1-ε)SDP value (=)SDP Vectors for =finite = Corresponding vectors in Є –net
STEP 1With probability > 1- Є2 ,
| |U-V|2 - |U’-V’|2 | < 2Є
STEP 2With probability > 1- 2Є2 ,
1+ Є < |V’|2 ,|U’|2 < 1- Є, Normalization changes distance by at most 2Є
STEP 3Changes edge length by at most 2Є
For SDP value (=)Contribution of an edge e = (U,V)
|U-V|2 = 2-2 V U ∙
To Show:
SDP value (=finite) > (1-ε)SDP value (=)SDP Vectors for =finite = Corresponding vectors in Є –net
STEP 1With probability > 1- Є2 ,
| |U-V|2 - |U’-V’|2 | < 2Є
STEP 2With probability > 1- 2Є2 ,
1+ Є < |V’|2 ,|U’|2 < 1- Є, Normalization changes distance by at most 2Є
STEP 3Changes edge length by at most 2Є
ANALYSISWith probability 1-3Є2,The contribution of edge e changes by < 6Є
In expectation,For (1-3Є2) edges, the contribution of edge e changes by < 6Є
SDP value (=finite) > SDP value (=) - 6Є – 3Є2
FINITE MODELS FOR GENERAL CSPRounding Any Constraint Satisfaction Problem
Semidefinite Program for CSPs
Variables :For each variable Xa
Vectors {V(a,0) , V(a,1)}
For each clause P = (xa ν xb ν xc),Scalar variables
μ(P,000) , μ(P,001) , μ(P,010) , μ(P,100) , μ(P,011) , μ(P,110) , μ(P,101) , μ(P,111)
))()()(( 145532532321 xxxxxxxxxxxx
Xa = 1 V(a,0) = 0 V(a,1) = 1
Xa = 0 V(a,0) = 1 V(a,1) = 0
If Xa = 0, Xb = 1, Xc = 1
μ(P,000) = 0 μ(P,011) = 1μ(P,001) = 0 μ(P,110) = 0μ(P,010) = 0 μ(P,101) = 0μ(P,100) = 0 μ(P,111) = 0
Objective Function :
PClauses sassignment
PP
3}1,0{
),()(
Constraints : For each clause P,
0 ≤μ(P,α) ≤ 1
For each clause P (xa ν xb ν xc), For each pair Xa , Xb in P,
consitency between vector and LP variables.
V(a,0) V∙ (b,0) = μ(P,000) + μ(P,001) V(a,0) V∙ (b,1) = μ(P,010) + μ(P,011) V(a,1) V∙ (b,0) = μ(P,100) + μ(P,101) V(a,1) V∙ (b,1) = μ(P,100) + μ(P,101)
1),(
P
Semidefinite Relaxation for CSPSDP solution for =:
SDP objective:
for every constraint Á in =- local distributions ¹Á over
assignments to the variables of Á
Example of local distr.: Á = 3XOR(x3, x4, x7)
x3 x4 x7 ¹Á0 0 0 0.10 0 1 0.010 1 0 0 …1 1 1 0.6for every variable xi in =
- vectors vi,1 , … , vi,q
constraints
(also for first moments)
Explanation of constraints:first and second moments of distributions are consistent and form PSD matrix
maximize
Strong and WeakSTRENGTHFor every clause Á in =- local distributions ¹Á over assignments to the variables of Á
Vector variables vi,a within a clause Á satisfy all valid constraints (like triangle inequality)
– the inner products are in the integral hull.WEAKNESS
The above hard constraint is only for variables that participate together in a clause
Throwing away constraints
{vi,a } { μ …}
-Infeasible SDP solution for a instance = , it does not satisfy the consistency for a clause P.
Consider instance =‘ = = - {P}
Now {vi,a } { μ … } is a good SDP solution for =‘
Throw away clauses from CSP
Throw away constraints from SDP relaxation
v1
v2
v3
v4
v5v
1
v3v
4 v5
Constant dimensions
v2v
2
FINITE MODEL CSP on Є –net points
STEP 1 : Dimension Reduction•Project the SDP solution along d =1/ Є4 random directions.
STEP 3 : Discretization•Pick an Є –net for the
d dimensional sphere• Move every variable to the nearest point in the Є –net =finite = discretized instance
STEP 2 : Throw away Discard clauses for which the corresponding inner products are not preserved within Є.
=‘ = New instance
To Show:
SDP value (=finite) > (1-ε)SDP value (=)
SDP Vectors for =finite = Corresponding vectors in Є –net
LP variables { μ …}?
Problem :
The inner products of vectors corresponding to a clause P might not be in the integral hull.( For example : 3 arbitrary vectors in a Є –net are not guaranteed to satisfy triangle inequality )
The initial SDP solution satisfied all the constraints
STEP 1 : Dimension Reduction•Project the SDP solution along d =1/ Є4 random directions.
STEP 3 : Discretization•Pick an Є –net for the
d dimensional sphere• Move every variable to the nearest point in the Є –net =finite = discretized instance
STEP 2 : Throw away Discard clauses for which the corresponding inner products are not preserved within Є.
=‘ = New instance
From STEP 2,
We have discarded clauses for which inner products are not preserved within Є
Discarding a clause P
Forget about constraints corresponding to P
Discretization changes inner product by Є
For every remaining clause, all inner products are within 2Є of what it was.
Smoothing Operation
Canonical SDP SolutionUniform Distribution over all Integral solutions.
Example:Va,0 V∙ a,0 = Va,1 V∙ a,1 = ½Va,0 V∙ b,0 = Va,0 V∙ b,1 = Va,1 V∙ b,0 = Va,1 V∙ b,1 = 1/4
Є –net SolutionSDP Vectors for =finite =
Corresponding vectors in Є –net (1-Є) X
+
=Final SDP solution
IntegralHull
Є X
Є
Consider the inner products corresponding to a single clause P
SDP Objective value remains roughly the same.
Conclusions
• Rounding stronger SDPs.
• More efficient rounding? Can this SDP be solved in constant dimensional space directly?
• Integrality gaps for stronger SDP relaxation of Unique Games
Thank You
Good finite Models from SDP solutions – Dimension Reduction & Discretization
¦-CSP Instance =
¦-CSP Instance =finite
SDP solution for =
compute
Dimension Reduction
Project on randomlow dimensional
subspace
almostSDP solution
for =
Discretize
Move vectors to closest point
on ²-net
almostSDP solution
for =
Rn Rd
identify variableswith same vectors
Theorem: SDP value (=finite) > SDP value (=)
Idea: use almostSDP solution and
do surgery
finite number ofdifferent vectrs
Constraint Satisfaction Problems (CSP)CSP ¦
finite set of allowed types of constraints Á : [q]k {0,1} (alphabet [q], arity k)e.g. ¦ = { 3XOR, 3SAT, 3NAE}
¦-CSP Instance =- variables x1,…,xn
- list of constraints Á of type ¦ on subsets of variables
Goal: Find assignment x 2 [q]n so as to maximize fraction of satisfied constraints opt(=)
Examples: Max-Cut, Max-3SAT,…
PCP Theorem: NP-hard to distinguish opt(=)=1 and opt(=)<0.9 (even for constant k and q)
Approximation Algorithms: Goemans-Williamson, Zwick, CMM, …