[PPT] - Test Instance Generation Test Instance Generation for MAX 2SAT for PowerPoint Presentation

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Test Instance Generation Test Instance Generation for MAX 2SAT for MAX 2SAT

Mitsuo Motoki Mitsuo Motoki Japan Advanced Inst. of Japan Advanced Inst. of Sci Sci. . and Tech. (JAIST) and Tech. (JAIST)

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Test Instance Generation Test Instance Generation problem problem

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Empirical study of solvers for Empirical study of solvers for combinatorial opt. problem combinatorial opt. problem

Output Solution Output Solution Output Solution Solver Solver Solver Test Instance Test Instance Test Instance Optimal Solution Optimal Solution Optimal Solution Compare Compare Compare

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Benchmark sets / Test Instance Generator Benchmark sets / Benchmark sets / Test Instance Generator Test Instance Generator

Empirical study of solvers for Empirical study of solvers for combinatorial opt. problem combinatorial opt. problem

Output Solution Output Solution Output Solution Solver Solver Solver Test Instance Test Instance Test Instance Optimal Solution Optimal Solution Optimal Solution Compare Compare Compare

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Test instance generator Test instance generator

Ideally

Ideally… …

Generate

Generate all all instances with the opt. solution instances with the opt. solution

Running time is

Running time is polynomial polynomial in the length of in the length of

utput instance
utput instance.

. Test Instance Generator Test Instance Test Instance Generator Generator Parameters Parameters about instance about instance (Random bits) (Random bits) Test Instance Test Instance

Opt. Solution
Opt. Solution

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for NP hard optimization problem for NP hard optimization problem

Unless NP=co

Unless NP=co-

NP

NP, there is , there is no ideal instance generator. no ideal instance generator.

Why?

Why?

Consider the decision

Consider the decision version (NP hard) version (NP hard)

(I, k) (I, k)

Yes Yes No No

? ?

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for NP hard optimization problem for NP hard optimization problem

Unless NP=co

Unless NP=co-

NP

NP, there is , there is no ideal instance generator. no ideal instance generator.

Why?

Why?

Consider the decision

Consider the decision version (NP hard) version (NP hard)

The

The random bits random bits used in the used in the instance generator become a instance generator become a witness for each witness for each “ “yes yes” ” instance, instance,

and also for each

and also for each “ “no no” ” instances. instances.

(I, k) (I, k)

Yes Yes No No

Instance Instance Generator Generator

(I, Opt) (I, Opt)

parameters parameters random bits random bits

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What can we do? What can we do?

Relax some requirements for instance

Relax some requirements for instance generator generator

Can generate instances from

Can generate instances from some some subset subset

f whole instance set
f whole instance set
Outputs a

Outputs a feasible solution feasible solution instead of the instead of the

ptimal solution
ptimal solution

(Outputs optimal solution with high prob.) (Outputs optimal solution with high prob.)

Running time is

Running time is “ “exponential exponential” ” instead of instead of “ “polynomial polynomial” ”

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What can we do? What can we do?

Relax some requirements for instance

Relax some requirements for instance generator generator

Can generate instances from

Can generate instances from some some subset subset

f whole instance set
f whole instance set
Outputs a feasible solution instead of the

Outputs a feasible solution instead of the

ptimal solution
ptimal solution

(Outputs optimal solution with high prob.) (Outputs optimal solution with high prob.)

Running time is

Running time is “ “exponential exponential” ” instead of instead of “ “polynomial polynomial” ”

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Our approach Our approach

Poly. time exact instance generator
Poly. time exact instance generator
The set of instance generated is a

The set of instance generated is a subset subset

f the whole instance set
f the whole instance set
The generator

The generator always always outputs a test

utputs a test

instance with the instance with the optimal

ptimal solution

solution

The running time is

The running time is polynomial polynomial in the length in the length

f the output instance
f the output instance

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New requirements New requirements

The instances generated should be

The instances generated should be hard hard. .

How to guarantee the hardness?

How to guarantee the hardness?

Theoretical way

Theoretical way

For any poly. time exact instance generator, the

For any poly. time exact instance generator, the decision problem over the set of instance decision problem over the set of instance generated is generated is NP NP ∩ ∩ co co-

NP

NP (no more NP complete) (no more NP complete)

How hard to distinguish the instances generated?

How hard to distinguish the instances generated?

(Empirical study)

(Empirical study)

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Poly. time exact instance
Poly. time exact instance

generator for MAX 2SAT generator for MAX 2SAT

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MAX 2SAT MAX 2SAT

Input: 2CNF formula

Input: 2CNF formula

Each clause consists of exactly 2 literals

Each clause consists of exactly 2 literals

Each variable appears at most once in a

Each variable appears at most once in a clause clause

Any clause can appear more than once

Any clause can appear more than once

Question: find a truth assignment

Question: find a truth assignment s.t s.t. .

maximizes

maximizes # of # of satisfied satisfied clauses, clauses,

i.e

i.e., ., minimizes minimizes # of # of unsatisfied unsatisfied clauses. clauses.

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How hard? How hard?

MAX SNP complete

MAX SNP complete

Decision version (Is there an assignment that

Decision version (Is there an assignment that satisfies at least satisfies at least k k clauses?) is clauses?) is NP complete NP complete

For satisfiable 2CNF formulas, poly. time

For satisfiable 2CNF formulas, poly. time solvable (2SAT is in solvable (2SAT is in P P) )

Inapproximability upper bound:

Inapproximability upper bound:

21/22

21/22 ≈ ≈ 0.955 0.955 [H [Hå åstad STOC stad STOC‘ ‘97] 97]

0.945

0.945 (under some unproven conjectures) (under some unproven conjectures) [Khot, Kindler, Mossel, and O [Khot, Kindler, Mossel, and O’ ’Donnell FOCS Donnell FOCS‘ ‘04] 04]

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Related works Related works

Probabilistic generator for MAX

Probabilistic generator for MAX k kSAT SAT [ [Dimitriou Dimitriou CP CP’ ’03] 03]

Unique optimal solution

Unique optimal solution w.h.p w.h.p., ., O( O(n nk

k) clauses

) clauses

Exact/probabilistic generator for MAX

Exact/probabilistic generator for MAX 2SAT [Yamamoto 2SAT [Yamamoto ‘ ‘04] 04]

To characterize opt. solution, requires an

To characterize opt. solution, requires an expander graph expander graph

They use an explicit expander graph

They use an explicit expander graph construction algorithm / a random graph construction algorithm / a random graph

Probabilistic generator for MAX 3SAT

Probabilistic generator for MAX 3SAT [MM COCOON [MM COCOON‘ ‘01] 01]

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Strategy of instance generator Strategy of instance generator

i. i.

Choose Choose t t ∈ ∈{0,1} {0,1}n

n at random as the

at random as the

ptimal solution
ptimal solution

ii. ii.

Combine appropriate number of Combine appropriate number of minimal minimal unsat unsat. . 2CNFs that contains 2CNFs that contains exactly 1 exactly 1 clause falsified by clause falsified by t t

iii. iii.

Add several clauses satisfied by Add several clauses satisfied by t t

There is no assignment that falsifies less #

There is no assignment that falsifies less #

f clauses than # of 2CNFs in ii.
f clauses than # of 2CNFs in ii.
Thus

Thus t t is an optimal solution is an optimal solution

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Implication graph Implication graph

[ [Aspvall Aspvall, , Plass Plass, , Tarjan Tarjan ‘ ‘79] 79]

Transform 2CNF

Transform 2CNF F F into a into a digraph digraph G GF

F

Each vertex corresponds to a

Each vertex corresponds to a literal literal

F

F contains a clause ( contains a clause (a a ∨ ∨b b) ) ⇔ ⇔ G GF

F has edge and

has edge and

Contradictory bicycle

Contradictory bicycle

F

F is is unsat unsat. . ⇔ ⇔ G GF

F has a cycle

has a cycle (and its complement cycle) (and its complement cycle) containing x and containing x and a b → a b → b a → b a →

1 2 2 4 2 3 1 3 4 2 (

) ) ) ( ( ) ( ( ) x x x x x x x x x x ∨ ∨ ∨ ∨ ∨

1

x1 x

1

x1 x

2

x2 x

2

x2 x

3

x3 x

3

x3 x

4

x4 x

4

x4 x

x x

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Minimal Minimal unsat

unsat. 2CNF

. 2CNF F F containing containing exactly 1 clause falsified by exactly 1 clause falsified by t t

2

j

x

2

j

x

1 q

j

x

−1 q

j

x

−

W.l.o.g

W.l.o.g. assume . assume t t =1 =1n

n

G

GF

F is a (simple) contradictory bicycle

is a (simple) contradictory bicycle

F

F has just 1 clause consisting of only has just 1 clause consisting of only negative literals negative literals

⇒

⇒ Cont. cycle

Cont. cycle has just 1 edge from

has just 1 edge from positive literal to negative literal positive literal to negative literal

Cont. cycle contains some variables
Cont. cycle contains some variables

as positive and negative literal as positive and negative literal

⇒

⇒ There is exactly 1 edge from There is exactly 1 edge from negative literal to positive literal negative literal to positive literal

unsat. 2CNF

. 2CNF F F containing containing exactly 1 clause falsified by exactly 1 clause falsified by t t

2

j

x

2

j

x

1 q

j

x

−1 q

j

x

−

W.l.o.g

W.l.o.g. assume . assume t t =1 =1n

n

G

GF

F is a (simple) contradictory bicycle

is a (simple) contradictory bicycle

F

F has just 1 clause consisting of only has just 1 clause consisting of only negative literals negative literals

⇒

⇒ Cont. cycle

Cont. cycle has just 1 edge from

has just 1 edge from positive literal to negative literal positive literal to negative literal

Cont. cycle contains some variables
Cont. cycle contains some variables

as positive and negative literal as positive and negative literal

⇒

⇒ There is exactly 1 edge from There is exactly 1 edge from negative literal to positive literal negative literal to positive literal

We can divide a cont. cycle into a

x

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Instance generator algorithm Instance generator algorithm

Input: # Input: # n n of vars.

f vars.

i. i.

Let Let F F be an empty formula be an empty formula

ii. ii.

Choose Choose t t ∈ ∈ {0,1} {0,1}n

n at random

at random

iii. iii.

Choose min. # Choose min. # k k ( (≥ ≥0) of 0) of unsat

unsat. clauses

. clauses

iv. iv.

for for i i =1 =1 to to k k do do

Generate a 2CNF in Generate a 2CNF in B Bt

t at random and add

at random and add to to F F

v. v.

Add clauses in Add clauses in C Ct

t to

to F F at random at random

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The set The set I I of instances generated

f instances generated
B

Bt

t : the set of minimal

: the set of minimal unsat

unsat. 2CNFs

. 2CNFs containing exactly 1 clause falsified by containing exactly 1 clause falsified by t t

C

Ct

t : the set of clauses satisfied by

: the set of clauses satisfied by t t

I

I = { = {F F ∈ ∈2CNF | 2CNF |F F consists of elements consists of elements

f
f B

Bt

t and

and C Ct

t for some

for some t t } }

F

F ∈ ∈I I ⇔ ⇔ min. # of

min. # of unsat
unsat. clauses = max. # of

. clauses = max. # of

cont. bicycles
cont. bicycles
Both

Both t t and a partition into elements of and a partition into elements of B Bt

t and

and C Ct

t become a witness of

become a witness of F F ∈ ∈I I

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Hardness results Hardness results

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Hardness of Hardness of the instances generated the instances generated

To distinguish the set

To distinguish the set I I of instances

f instances

generated is NP complete generated is NP complete

I.e., finding an opt. solution is at least as hard

I.e., finding an opt. solution is at least as hard as finding a sat. assign. of satisfiable 3CNFs as finding a sat. assign. of satisfiable 3CNFs

To approximately distinguish the set

To approximately distinguish the set I I of

f

instances generated is also NP hard instances generated is also NP hard

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I I is NP complete (1) is NP complete (1)

Reduction from 3SAT

Reduction from 3SAT

For any 3CNF

For any 3CNF F F3CNF

3CNF =

=c c1

1∧

∧c c2

2∧

∧… …∧ ∧c cm

m,

, transform each clause transform each clause c ci

i=(

=(l li

i,1 ,1∨

∨l li

i,2 ,2∨

∨l li

i,3 ,3) into

) into and let and let F F2CNF

2CNF=

=∧

∧i

i b

bi

i .

.

Note that new variables

Note that new variables y yi

i,1 ,1 and

and y yi

i,2 ,2 appear

appear only

nly in

in b bi

i

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

( ) ( ) ( ) ( ( ) ( ) ( ( ) ) )

i i i i i i i i i i i i i i i i i

y b y l y y l y l y l y l l y l y = ∨ ∧ ∧ ∧ ∧ ∧ ∧ ∨ ∨ ∨ ∨ ∧ ∨ ∨ ∨

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

( ) ( ) ( ) ( ( ) ( ) ( ( ) ) )

i i i i i i i i i i i i i i i i i

y b y l y y l y l y l y l l y l y = ∨ ∧ ∧ ∧ ∧ ∧ ∧ ∨ ∨ ∨ ∨ ∧ ∨ ∨ ∨

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,1 i

y

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I I is NP complete (3) is NP complete (3)

If

If F F3CNF

3CNF =

=c c1

1∧

∧c c2

2∧

∧… …∧ ∧c cm

m is satisfiable

is satisfiable

F

F2CNF

2CNF has

has m m contradictory bicycle contradictory bicycle

Min. # of
Min. # of unsat
unsat. clauses in

. clauses in F F2CNF

2CNF is

is m m

F

F2CNF

2CNF ∈

∈ I I

If

If F F3CNF

3CNF is unsatisfiable

is unsatisfiable

F

F2CNF

2CNF has

has m m contradictory bicycle contradictory bicycle

Min. # of
Min. # of unsat
unsat. clauses in

. clauses in F F2CNF

2CNF is at least

is at least m m +1 +1

F

F2CNF

2CNF ∉

∉ I I

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Hardness for approximation Hardness for approximation

If

If c ci

i ∈

∈ F F3CNF

3CNF is falsified

is falsified by the opt. solution, by the opt. solution, 2 clauses in 2 clauses in b bi

i are

are falsified by corresponding falsified by corresponding assignment assignment

If min. # of

If min. # of unsat

unsat. clauses

. clauses in in F F3CNF

3CNF is

is k k, min. # of , min. # of unsat unsat. . clauses in clauses in F F2CNF

2CNF is

is m m + +k k

l li,

i,1 1 l

li,

i,2 2 l

li,

i,3 3 y

yi,1

i,1 y

yi,2

i,2 # of

# of unsat unsat. . 2 2 1 1 2 2 1 1 3 3 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ∗ ∗ 1 1 1 1 1 1 ∗ ∗ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ∗ ∗ 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Hardness for approximation Hardness for approximation

The reduction from 3SAT is a gap

The reduction from 3SAT is a gap reserving reduction reserving reduction

If

If F F3CNF

3CNF is satisfiable, min. # of unsatisfiable

is satisfiable, min. # of unsatisfiable clauses in clauses in F F2CNF

2CNF is

is m m

If 1/8 fraction of F

If 1/8 fraction of F3CNF

3CNF is unsatisfiable, min.

is unsatisfiable, min. # of # of unsat

unsat. clauses in

. clauses in F F2CNF

2CNF is m+m/8 = 9m/8

is m+m/8 = 9m/8

If we can approximate any member of

If we can approximate any member of I I within , we can distinguish within , we can distinguish satisfiable 3CNFs and satisfiable 3CNFs and unsat

unsat. 3CNFs

. 3CNFs

8 9 / 55 6 8 8 5 m m m m − = − 8 9 / 55 6 8 8 5 m m m m − = −

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Future works Future works

Improve the hardness for

Improve the hardness for approximation (55/56 approximation (55/56 ≈ ≈ 0.982). 0.982).

Cf.
Cf. imapproximability

imapproximability: : 21/22 21/22≈ ≈0.955 0.955, , 0.945 0.945

Estimate appropriate values for

Estimate appropriate values for parameters parameters

Analyze the distribution/expectation of #

Analyze the distribution/expectation of #

f contradictory bicycles in random 2CNFs
f contradictory bicycles in random 2CNFs
Instance generator for other problems