Unique Games Approximation Amit Weinstein Complexity Seminar, Fall 2006 Based on: “Near Optimal...

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3 What is Unique Game? A Constraints Graph k – Domain size Objective: Satisfy as many edges as possible

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Unique Games

ApproximationAmit Weinstein

Complexity Seminar, Fall 2006

Based on: “Near Optimal Algorithms for Unique Games" by M. Charikar, K.

Makarychev, Y. Makarychev

2

Outline What is Unique Game?

Definition Solving a Satisfiable Game Generalization: d-to-d games

Known Hardness and Approximation results Integer Programming and SDP

representation Rounding Algorithm

How is it done What does it guarantee

3

What is Unique Game? A Constraints Graph

k – Domain size Objective: Satisfy as many edges as

possible

, ,, |u v u vc x y x y

u

v

w , ,, |u w u wc x y x y

kS

4

MaxCut as a Unique Game

1 0,1 , 1,0 2k

1 11

1

1

1

5

Can we solve a satisfiable game? Greedy !

Go over all possible x’s Complete the assignment Check Solution

x

,u v x

, ,w u u v x

, , ,z w w u u v x

v

u

w

z

6

Generalization A game is called d-to-d if:

For each edge (u,v) Given an assignment to v Only d possible assignments to u will satisfy

this edge So what is a Unique Game?

A 1-to-1 game Can you think of a simple 2-to-2 game?

3-Coloring Can we solve a 2-to-2 satisfiable game?

7

Known Approximations (and bounds) General Unique Game

Approx. 1/k (Random Assignment)

MaxCut:

Approx. of 0.878… using SDP relaxation NP-hard to approx Hastad 02

2LinEqGF2 Approx. 1/2 (Random Assignment) NP-hard to approx Hastad 02

1617

0 1 . , . ,kc k GapUG c NP hard

1112

1 7

2 3

1 4

101

x xx xx x

can be very small

Geomans, Williamson

95

8

Unique Games Conjecture (UGC) This is the main Conjecture of Unique

Games Still haven’t been proven Most people assume it is true

, 0., .

, .

,1k

k

k k

GapUG NP hard

YES INSTANCE

At least of the edges

can be satisfied

1 NO INSTANCEAt most of the edges

can be satisfied

9

Assuming the UGC is true MaxCut

We know approximation 0.878… It is NP-hard to approx. within any factor

Khot, Kindler, Mossel, O’Donnel 04 Again, this means 0.878… is optimal

Vertex Cover We know approximation 2 It is NP-hard to approx. within any factor

Khot, Regev 03 Meaning 2 is optimal

2

0.878...

10

Known Unique Game Approx.

Results:

This Article:

152 10

3

/ 2

1

1 log 1/ 1/

1 log 1/

1 log 1/

1

1 log 1/ log

for OPT for

O k O k

O n O k

O n O k

k const

O k O k

Meaning 1 log ,1kGapUG O n P

logk n

11

Unique Game as Integer Programming We define:

Claim:

And therefore

1. 1 .

0i

f u iu V i k u

f u i

,

2 ,12

1 ,

0

1u v

ku v

i ii u v

f u f vu v

f u f v

,

212

, 1

#u v

k

i iu v E i

u v unsatisfied edges

12

Integer Programming – Edges weight Proof for:

,u v

1v 2v 3v 4v kv

1u 2u 3u 4u ku

0 0 1 0 0

1 0 0 0 00 1 0 0 0

,

2 2 2 2 21 12 2

1

0 0 0 0 1 1 0 0 0u v

k

i ii

u v

,

2 2 2 2 21 12 2

1

0 1 0 0 1 0 0 0 1u v

k

i ii

u v

,

2 ,12

1 ,

0

1u v

ku v

i ii u v

f u f vu v

f u f v

13

Unique Game as Integer Programming Remember:

The program:

1. 1 .

0i

f u iu V i k u

f u i

,

2

2

1

12

, 1

minimize

subject to . 0

1

. 0,1

u v

k

i iu v

i jk

ii

E i

i

u V i j k u u

u V u

u V i k u

u v

14

From Integer Programming to SDP Discrete variables to vectors We also add a few constraints

,

,

212

221

1

2

, 1

, . , , 0 (4)

minimize

0 . , 0 (2)

1 1 (3)

0,1 .

, . 0 , (5)u v

u v

i j

k

i iu v E

i j i j

kk

i iii

i

i i

i

i

u u u V i j k u u

u u V u

u u V i k

u v E i k

u v E i j k u v

u v

u v u

We don’t need

,

212

2

1

, 1

minimize

0 .

1

0,1 .

u v

i jk

ii

k

iu v E i

i

i

u u u V i j k

u u V

u u V i

u

k

v

Triangle Inequalities

on the norms

From now on, all

variables ui are vectors !

15

SDP – Some Intuition

Size = probability Direction = correlation

Small angle – correlated Large angle – uncorrelated Reminder:

2 Pr

, Pr

i

i j

u f u i

u v f u i f v j

, cosu v u v

16

SDP – Solution Illustration

x

y

z

1u

2u

3u1v

3v

2v

1w

3w

2w

1

3

3

f

w

v

f u

f

3

, ,kV u v w

17

Rounding Algorithm – The Idea

,

1, . ,

2u v

i j k

i i

u v V i j u v

u v

~ 0,1N

For simplicity, we assume

We pick a random Gaussian vector g Each coordinate of g

18

Rounding Algorithm – The Idea | ,u iS i g u

1uS

,Pr ?u vf u f v

1Pr , i kg u

Define the Sets: Possible Values Choose a threshold s.t.

Randomly choose from these Sets

What is

19

Rounding Alg. – The Idea Calculation

,

,1

,1

, ,1

,

Pr

Pr

1 Pr

1 Pr

u v

k

u vi

k

v u v uiu v

k

u v u u v viu v

u u v v

u v

f u f v

f u i f v i

i S i SS S

i S SS S

S SS S

Chosen Independ

ent

,, ,Pr u u v vu v u u v v

u v

S Sf u f v S S

S S

Sum over all

possible i

By Definition

1uS

20

Rounding Alg. – The Idea Calc. cont. .

By our choice:

Since there are k such possibilities:

, ,Pr ?u v u u v vf u f v S S

,, , , ~ 0,1

cov , 1u vi iX g ku Y g kv N

X Y

1Pr kX t k

/ 2 1Pr kX t k Y t k k

/ 2, ,Pr u v u u v vf u f v S S k

From the Promise and assumptions

For intuition,

Not accurate

21

What about our assumptions? Lengths assumption

Distance assumption

We repeat the procedure #times ~ vector’s length For vector ui we repeat times Using different random vectors

We choose k random Gaussian vectors 1,..., kg g

2

iu is u k

1

?i

k

ui

s

Starting here

22

The Rounding Algorithm Define Recall:

Define

Define

The Assignment:

We now need to analyze it

0

0 0

ii

uu i

ii

uu

u

2

iu is u k

, , , 1i iu s s i ug u s s

,, |iu u sS i s t

. R uu V f u S

Ignore empty Sets

23

SDP – Rounding Illustration

x

y

z

1u

2u

3u1v

3v

2v

1w

3w

2w

3

, ,kV u v w

1 2 3

3 , 1u u us s s

1 1 1 2 3,1 ,2 ,3 ,1 ,1, , , ,u u u u u

1 3 . 1,ua S a

1f u

24

Rounding Algorithm – Definitions The distance between two vertices:

Also,

Which basically holds: When is the angle between them If one of the vectors is 0, we set

,

212

1u v

k

uv i ii

u v

,

212 u v

iuv i iu v

1 cosiuv i

i

21 ,iuv i

21 1

2 2

2 212

,

2 ,

1 cos

u v u v u v

u u v v

25

Rounding Algorithm – Definitions We define a measure

Notice:

22

,

2 ,i iu vu v

uvi T

T T k

1uv k 2

1(3) : 1k

iiSDP u

26

Rounding Algorithm – Proof Sketch 3 steps, similar to the easy case

Lemma 3.3: Bound

Lemma 3.7: Bound

Averaging

,

,min , . Pr , ,i iu vu v u vs s s f u i s f v i s

Pr ,uvP u v is satisfied

27

Rounding Algorithm – Lemma 3.3

We define:

,

,

2 / 2log1 1

loglog

1 1loglog

, . . min , .

Pr , ,

min 1,

min 1,

i iu v

iuv

iuv

iuv

u v

u v

kkkk

ik uvkk

u v E i k s s s

f u i s f v i s

f

2

2log xkk kf x

28

Rounding Alg. – Lemma 3.3 Proof .

,, ,, ~ 0,1 , cov .,. cos 1i iu v

iu s v s i uvN

, ,

1 1loglog

Pr

min 1,

i iuv

iuv

u s v s

ik uvkk

t t

f

1 2

1 2ku u u ukS s s s

,, ,|i iu v

u u s v sS t t const

, ,iu s s ig u

2

iu is u k

Appendix Lemma

B.3

Appendix Lemma

B.1

29

Rounding Alg. – Lemma 3.3 Proof We get

,

,

, ,

1 1loglog

Pr , ,

1 Pr , ,

1 Pr

min 1,

i iuv

iuv

u v

u u v vu v

u s v su v

ik uvkk

f u i s f v i s

i s S i s SS S

t tS S

f

,u vS S const

By Definiti

on

30

Rounding Algorithm – Proof Sketch 3 steps, similar to the easy case

Lemma 3.3: Bound

Lemma 3.7: Bound

Averaging

,

,min , . Pr , ,i iu vu v u vs s s f u i s f v i s

Pr ,uvP u v is satisfied

31

Rounding Algorithm – Lemma 3.7 .

Proof:

1log log

, . Pr ,

min 1,uv

uv

kk uvk k

u v E P u v is satisfied

f

,

,

min ,

,1 1

1 1loglog

1

21

log log1

Pr , ,

min , min 1,

min 1, 1

u vi iu v

ii iu v uv

iuv

s sk

uv u vi s

ki

u v k uvkki

ki ik

uv uv k uvk ki

P f u i s f v i s

s s f

i f

Our measure properties:

,

2min , 1

i iu v

iu v uv uvs s i k

Lemma 3.3

32

21

log logmin 1, 1

iuv

i ikuv uv uv k uvk k

i M

P i f

21

log logmin 1, 1

uv

i ikuv uv uv k uvk k

i M

P i f

Rounding Alg. – Lemma 3.7 Proof Consider

For any

We know:

So by Markov inequality:

| 2iuv uvM i k

. log 2 logiuv uvi M k k

1

ki

uvi

uv uvi M

uv uvi

i i

Out measure

properties 12uv M

33

Rounding Alg. – Lemma 3.7 Proof The function is convex at [0,1] By Jensen’s inequality:

Allows us to insert the Sum into the function

21 kx f x

i ii i

i i

a xa

a xa

1

log0k

kf

k

x

y

34

Rounding Alg. – Lemma 3.7 Proof The function is convex at

[0,1] By Jensen’s inequality:

Allows us to insert the Sum into the function

21

log log

21log log

21log log

min 1, 1

min 1, 1

min 1, 1

uv

uv

uv

i ikuv uv uv k uvk k

i M

kuv uv k uvk k

kuv k uvk k

P i f

M f

f

21 kx f x

35

Rounding Algorithm – Proof Sketch 3 steps, similar to the easy case

Lemma 3.3: Bound

Lemma 3.7: Bound

Averaging

,

,min , . Pr , ,i iu vu v u vs s s f u i s f v i s

Pr ,uvP u v is satisfied

36

Rounding Algorithm – The Result There is a polynomial time algorithm

(which we saw), that find an assignment which satisfies

given the optimal assignment satisfies at least of the constraints.

1

/ 221log log

min 1, 1 kk k

37

21log log

min 1, 1uv

kuv uv k uvk kP f

21log log

min 1, 1kuv uv k uvk kP f

Rounding Alg. – The Result’s Proof We consider only For So

So averaging over all , using Jensen and the convexity of we get:

Again, we insert the average sum inside.

' , | 2uvE u v E

, ' log 2 loguvu v E k k

, 'u v E

21 kx f x

/ 221log log

min 1, 1 kk k

38

Rounding Algorithm – Proof Sketch 3 steps, similar to the easy case

Lemma 3.3: Bound

Lemma 3.7: Bound

Averaging

,

,min , . Pr , ,i iu vu v u vs s s f u i s f v i s

Pr ,uvP u v is satisfied

39

Proof Meaning Given SDP solution better than We found an assignment We proved it satisfies

This is what we wanted

1

2

logkk

40

Summary Given a Unique Game Input Defined Integer Programming Translated into SDP Used a rounding Algorithm We showed that if at least could

be satisfied Our solution will give:

1

2

logkk

41

Questions? ? ?? ?? ?? ?? ?

? ? ?

? ? ?? ?

42

Rounding Algorithm – Filling Holes Lemma:

We use:

1

ki

uv uv uvi

i

, ,

, ,

,

1

2 22 212

1

2212

1

212

1

cos

2 cos

u v u v

u v u v

u v

ki

uv uvi

k

i i ii ii

k

i i ii ii

k

i uvii

i

u v u v

u v u v

u v

,, 0

u vi iu v

2 2 2

2 2

0 2

2

a b a b ab

a b ab

,

2(5) : 0 ,u vi iiSDP u v u

1 cosiuv i

,

2212 u vuv i ii u v

43

Rounding Algorithm – Filling Holes Lemma:

W.l.o.g. assume

,

2min , 1

i iu v

iu v uv uvs s i k

22,

,

2 22

2 22 1

1 cos

cos

i iu v

iu v

u viuv uv i

i i uki

i

v u s

, ,min ,

i iiu v u vi u v uiu v s s s

,cos

u v i iiv u ,

2(5) : 0 ,u vi iiSDP u v u

1 cosiuv i

2

iu is u k

44

Normal Gaussian vectors properties

, , , , 1 , ~ 0,1iX g u Y g v u v g N

cov ,X Y XY X Y XY

back

2 20, 1 , ~ 0,i i i i i iX g u G N u G N u

,

,

, cos cos

i i j j i j i ji j i j

i j i j i ii j i

XY g u g u u v g g

u v g g u v

u v u v

cos