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 possibleTRANSCRIPT
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