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Fine-Grained Complexity and Algorithm Design Boot Camp Fine-Grained Complexity and Algorithm Design Boot Camp

Fine-Grained Complexity and Algorithm Design Boot Camp - PowerPoint Presentation

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Fine-Grained Complexity and Algorithm Design Boot Camp - PPT Presentation

Lower Bounds Based on SETH D ániel Marx slides by Daniel Lokshtanov Simons Institute Berkeley CA September 4 2015 Insert Academic unit on every page 1 Go to the menu Insert ID: 399513

sat set variables algorithm set sat algorithm variables hitting independent group seth assignment size assignments dominating sets elements time

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Slide1

Fine-Grained Complexity and Algorithm Design Boot CampLower Bounds Based on SETH

Dániel Marx(slides by Daniel Lokshtanov)Simons Institute, Berkeley, CASeptember 4, 2015

Insert

«

Academic

unit»

on every page:

1 Go to the menu «Insert»

2 Choose: Date and time

3 Write the name of your faculty or department in the field «Footer»

4 Choose «Apply to all"Slide2

Tight lower boundsHave seen that

ETH can give tight lower boundsHow tight? ETH «ignores»

constants

in

exponent

How to

distinguish

1.85

n

from

1.0001

n

?Slide3

SATInput: Formula

with m clauses over n boolean variables.Question: Does there exist an

assignment

to

the

variables

that satisfies all clauses?Note: Input can have size superpolynomial in n!

 

Fastest

algorithm

for

SAT

:

2

n

poly(m)Slide4

d-SATHere all clauses have size

dInput size nd

 

Fastest

algorithm

for

2-SAT: n+mFastest algorithm for 3-

SAT: 1.31n

Fastest

algorithm

for

4

-SAT: 1.47n…Fastest algorithm for d-SAT: Fastest algorithm for SAT:

 Slide5

Strong ETHLet

d-SAT has a

algorithm

 

Know:

0

d

1 ETH: s3 0

 

SETH:

1

 

Let

 Slide6

Showing Lower Bounds under SETH

Your ProblemToo fast algorithm?d-SAT

 

The

number

of

9

’s

MUST

be independent of dSlide7

Dominating SetInput: n vertices

, integer kQuestion: Is there a set S of at most k vertices such that

N[S] = V(G)

?

Naive:

n

k+1Smarter: nk+o(1)Assuming ETH: no f(k)no

(k)n

k

/10

?

n

k-1?Slide8

SAT

k

-Dominating Set

Variables

SAT-formula

k

groups, each on

n/k

variables.

One vertex for each of the

2

n/k

assignments to the variables

in the group.Slide9

x

y

x

y

x

y

x

y

x

y

Variables

SAT-formula

k

groups, each on

n/k

variables.

Selecting

one

vertex

from each

cloud corrsponds to selecting

an

assignment

to the

variables

.

CliquesSlide10

x

y

x

y

x

y

x

y

x

y

Variables

SAT-formula

k

groups, each on

n/k

variables.

One vertex per clause in the formula

Edge if the

partial assignment

satisfies the clauseSlide11

SAT  k-Dominating Set

analysisToo fast algorithm for k-Dominating Set: nk-0.01

For

any

fixed

k (like k=3)The output graph has k2

n/k + m 2k

2

n/k

vertices

 If m 2n/k then 2n is at most mk, which is polynomial!So m

 

 

 

 Slide12

Dominating Set, wrapping upA O(n2.99

) algorithm for 3-Dominating Set, or a O(n3.99) algorithm for 4-Dominating Set, or aa

O(n

4.99

)

algorithm for 5-Dominating Set, or a …… would violate SETH.Slide13

TreewidthWe have seen: 2tnO(1)

, 3tnO(1), etc. algorithms and no 2o(t)nO(1) algorithms assuming ETH.Slide14

Independent Set / TreewidthInput: Graph

G, integer k, tree-decomposition of G of width t.Question:

Does

G

have an independent set of size at least k? 

DP: O(2

t

n)

time

algorithm

Can we do it in 1.99t poly(n) time?Next: If yes, then SETH fails!Slide15

Independent Set / TreewidthWill reduce

n-variable d-SAT to Independent Set in graphs of treewidth t, where t

n+d

.

So a

1.99

tpoly(N) algorithm for Independent Set gives a 1.99n+dpoly(n)

O(1.999n) time

algorithm

for

d

-

SAT. Slide16

Independent Sets on an Even Path

tft

f

t

f

t

f

t

f

True

False

In

independent

set

:

Not in

solution

:

first

True

then

FalseSlide17

d

-SAT

Independent

Set

proof

by

example

 

= (a

b

c)

(a

c

d

)

(b

c

d)

 

tftftf

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

a

b

c

d

a

c

b

a

d

c

b

d

cSlide18

Independent Sets Assignments

 

= (a

b

c)

(a

c

d

)

(b

c

d

)

 

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

a

b

c

d

a

c

b

a

d

c

b

d

c

True

True

False

False

But

what

about

the

first

true

then

false

independent

sets

? Slide19

Dealing with truefalse

a

b

c

d

Clause

gadgets

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

Every

variable

flips

true

false

at most

once

!Slide20

Treewidth Boundby picture

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

t

f

a

b

c

d

a

c

b

a

d

c

b

d

c

n

d

Formal

proof

-

exerciseSlide21

Independent Set / Treewidthwrap up

Reduced n-variable d-SAT to Independent Set in graphs of treewidth t, where

t

n+d

.

A

1.99tpoly(N) algorithm for Independent Set gives a 1.99

n+dpoly(n) O(1.999

n

)

time

algorithm

for d-SAT. Thus, no 1.99t algorithm for Independent Set assuming SETHSlide22

Assuming SETH, the following algorithms are optimal: 2

t poly(n) for Independent Set 3t poly(n) for Dominating Set

c

t

poly(n) for c-Coloring 3t

poly(n) for Odd Cycle Transversal 2t

poly

(n)

for Partition Into Triangles 2t poly(n) for Max Cut 2t poly(n) for #Perfect Matching… Slide23

3t lower bound for Dominating Set?

Need to reduce k-SAT formulas on n-variables to Dominating Set in graphs of treewidth t

,

where

 

So

0.58

 Slide24

Hitting Set / nInput: Family F = {S

1,…,Sm} of sets over universe U = {v1, …, vn},

integer

k

.

Question: Does there exist a set X U of

size at most k such that

for

every

S

i F, Si X ? Naive algorithm runs in

time.

 

Next:

implies

that

SETH

fails  Slide25

d

-SAT

Hitting

Set

 

 

 

= (a

b

c)

(a

c

d

)

(b

c

d

)

 

 

 

 

 

 

 

Budget = 4Slide26

d-SAT vs Hitting SetA c

n algorithm for Hitting Set makes a c2n algorithm for d-SAT.Since 1.412n

< 1.9999

n

, a

1.41

n algorithm for Hitting Set violates the SETH.Have a

2n algorithm and a

1.41

n

lower

bound.Next: 2n lower bound Slide27

Hitting SetFor any fixed

, will reduce k-SAT with n variables to Hitting Set with

universe

with

at most

(1+)n elements. 

So a 1.99n algorithm for

Hitting

Set

gives

a

1.99n(1+) 1.999n time algorithm for k-SAT  Slide28

Some deep mathFor

every there exists a natural number g such that, for t =

we

have:

 

 

Why

is

this

relevant?Slide29

d-SAT Hitting Set

 Group the variables into groups of size g, and set

t

=

.

 

Variables:

g

g

g

g

g

t

t

t

t

t

Elements

(1 +

)

variables

 

Elements:

Solution

budget

from

each

group

Will force

from

each

group

Exactly

from

each

group

 Slide30

Analyzing a group

Group

of

g

variables

Group

of

t

elements

2

g assignments to variables

subsets

of

elements

of

size

exactly

.

 InjectionSlide31

Forcing solution

vertices in a group?  

Add

all

subsets

of the group of size

to the family F.

 

Any

set

that picks less than elements the group misses a guard. 

Any

set

that

picks

at

least elements from each group

hits all the guards

 Lets call these sets guardsSlide32

Analyzing

a

group

Group

of

g

variables

Group

of

t

elements

assignments

to variables

subsets

of

elements

of

size

exactly . Injection

What

about

the

element

subsets

of

size

that

do not

correspond

to

assignments

?

 Slide33

Sets of size

 Adding a set of

size

to

the

family F ensures that the «

group complement» set is not picked

.

All

other

sets of size in the group may still be picked in solution. Forbid sets of size that do not correspond

to

assignments

.

 Slide34

d-SAT Hitting Set

 

Variables:

g

g

g

g

g

t

t

t

t

t

Elements:

potential

solutions

assignments

Want

:

Solutions

Satisfying

assignments

 Slide35

Forbidding partial assignmentsPick any

d groups of variables, and consider some assignment to these variables.If this assignment falsifies

we

want

to forbid the corresponding set in the Hitting Set instance from being

selected.

 Slide36

Forbidding partial assignments

Variables

Bad

assignment

Set

added

to

F

to

forbid

the

bad

assignmentSlide37

Forbidding partial assignmentsFor each bad assignment

to at most d groups, forbid it by adding a «bad assignment guard»This adds O(nd

2

gd

) = O(

n

d) sets to F.Slide38

Satisfying Assignments Hitting Sets

 A satisfying assignment has no bad sub-assignments 

corresponds

to a

hitting

set.A hitting set corresponds to an assignment.

If this assignment falsified a

clause

C

,

the assignment would be bad for the d groups C lives in, and miss a bad assignment guard.  Slide39

Hitting Set wrap upCan reduce

n variable d-SAT to n(1+) element Hitting Set. So a cn algorithm for Hitting Set yields a (c+

)

n

algorithm

for d-SAT.A 1.99n algorithm for Hitting Set would violate

SETH. Slide40

ConclusionsSETH can be used to give very

tight running time bounds.SETH recently has been used to give lower bounds for polynomial time solvable problems, and for running time of approximation algorithms.Slide41

Important Open ProblemsCan we show a 2

n lower bound for Set Cover assuming SETH?

Can

we

show a

1.00001

lower

bound

for

3-SAT assuming SETH?