Algorithms for SAT Based on Search in Hamming Balls

Author : Evgeny Dantsin, Edward A. Hirsch,

and Alexander Wolpert

Speaker : 羅正偉

Outline

Definition of SAT problem Introduction of this paper Definitions and notation Randomized Algorithm

Preliminary: K-SAT problem

m

j

n

jm

CCCm

C

xxxn

KCCCCm

,...,, clauses

thesatisfies that assignmentA truth :Output

form. normal econjunctiv is

each For .,...,, riablesboolean va on

|| with ,...,, clauses :Input

21

21

21

Preliminary-A fact

2-SAT is in P, but 3-SAT is NP-complete.

Preliminary-Hamming Ball

Definition: The Hamming ball of radius R around a truth assignment A is the set of all assignments whose Hamming distance to A is less than or equal to R.

The assignment A is called the center of the ball.

The volume of a ball is the number of assignments that belong to the ball.

Introduction

In this paper a randomized algorithm for SAT is given, and its derandomized version is the first non-trivial bound for a deterministic SAT algorithm with no restriction on clause length.

For k-SAT, Schuler’s algorithm is better than this one.

Notation

(0,1)on defined x)-(1x)log-(1-x-xlogH(x)

T to settingby obtained formula the:|F

),..., riableboolean va (i.e. literal :

R. radius of {0,1}in ball Hamming a of volume:R)V(n,

Aat centered R, radius , {0,1}in ball Hamming a :R)B(A,

ball Hamming a of radius :R

formula CNF a :F

22

i

1i

n

n

l

xxl

Tl

n

i

Fact about H(x)

The graph of H(x) is

x

y

A bound of V(n,R)

formula. sStirling’by proceed then and ,

R)C(n,C(n,1)C(n,0)R)V(n, Proof. 2.

2),(2)/1(8

1

have wen/2,RFor 1.

)/()/(

nnRHnnRH RnVnRR

Randomized Algorithm

The randomized algorithm is called Random-Balls, and it invokes procedures called Ball-Checking and Full-Ball-Checking.

Ball-Checking algorithms

no""5.return

Sreturn S, assignment satisfying a returns thisif

1),-RT,|A T,|(F Invoke

k to1 i4.For

Aby falsified is that Fin ... clauseshortest a 3.Choose

no""“"return clauseempty an contains For 0R 2.If

Areturn A,under trueare Fin clauses all 1.If

no""“"or assignment satisfying :

Rnumber A, assignment F, formula :

1

ii

k

llingBall-Check

ll

ing(F,A,R)Ball-Check

Output

Input

Procedure

Observations

The recursion depth is at most R Any literal is altered at most once during

execution of Ball-Checking, and at any time, those remaining variables are assigned as in the original assignment

Lemma 1

There is a satisfying assignment in B(A,R) iff Ball-Checking(F,A,R) returns a satisfying assignment in B(A,R)

Proof. By induction of R (see reference)

Lemma2

The running time of Ball-Checking(F,A,R) is at most , where k is the maximum length of clauses occurring at step 3 in all recursive calls.

Proof. The recursion depth is at most R and the maximum degree of branching is at most k.

Rpoly(n)mk

Full Ball Checking

Procedure Full-Ball-Checking(F,A,R)Input: formula F over variables

, assignment A, number ROutput: satisfying assignment or “no”

1.Try each assignment A’ in B(A,R), if it satisfies F, return it.

2.Return “no”

nxx ,...,1

Observation

Full-Ball-Checking runs in time poly(n)mV(n,R)

Randomized algorithm

no""3.return

found. assignmentany return R),A,,Checking(F-Ball

invoke else R),A,,Checking(F-Ball-Full invoke A,by satisfied

literals R andA by falsified literals with clauses a contains F (b)If

randomat uniformly A assignmentan (a)Choose

:following the timesN2.Repeat

2R/n)-8R(1 1.N

no""or assignment satisfying :

nR0 s.t. R, numbers s,n variableover F formula :

)R,(F,

))/(1(

l

ll

llsRandom-Bal

nRHn

Output

Input

Algorithm

Lemma 3

Correctness of Random-Balls.

For any R, l, the following holds:(a)If F is unsatisfiable, then Random-Balls

returns “no”, (b)else Random-Balls finds a satisfying

assignment with probabability 1/2

Proofs of lemma 3

e

enRR

NnnnRH

1/1/n)-(1 since follows inequalitylast The

/1))/1(8

21()R)/2V(n,-(1

A]chosen any contain t doesn' R),Pr[B(A'

chosen] R)B(A,any in coverednot is Pr[A'

any), (if A' assignment satisfying aFor (b)

trivial.is (a)

Proof

n

)/(Nn

A bound of V(n,R)

formula. sStirling’by proceed then and ,

R)C(n,C(n,1)C(n,0)R)V(n, Proof. 2.

2),(2)/1(8

1

have wen/2,RFor 1.

)/()/(

nnRHnnRH RnVnRR

REVIEW!!

Probability of correctness

So the prob. of correctness = 1 – 1/e > 1/2 Choosing N to be k times larger will reduce

the probability of error to less than . Note this doesn’t ruin the time complexity

we need.

ke

Lemma 4

Consider the execution of Random-Balls(F,R,l) that invokes Ball-Checking. For any input R, l, the maximum length of clauses chosen at step 3 of Procedure Ball-Checking is less than l.

Ball-Checking algorithms

no""5.return

Sreturn S, assignment satisfying a returns thisif

1),-RT,|A T,|(F Invoke

k to1 i4.For

Aby falsified is that Fin ... clauseshortest a 3.Choose

no""“"return clauseempty an contains For 0R 2.If

Areturn A,under trueare Fin clauses all 1.If

no""“"or assignment satisfying :

Rnumber A, assignment F, formula :

1

ii

k

llingBall-Check

ll

ing(F,A,R)Ball-Check

Output

Input

Procedure

REVIEW!!

Proof of lemma 4

ion.contradict a A,by satisfied literals Rmost at has

D R,|C|-|D| R, isdepth recursion theSince A.by falsified

literals k has D Thus A.by falsified re’ that theyso

A, original in the as assigned are ,..., of variables

The D. clause some ofdecendent a is clause Then this

3. stepat chosen be k),...(C clause aLet

1

1

l

ll

lll

k

k

Lemma 5

For any R, l, let p be the probability (taken over random assignment A) that Random-Balls invokes Full-Ball-Checking. Then

)1)((m2p

Rl

RHl

Proof of lemma 5

||

||)||/),|(min(|

D||

D|D|)||D|-t)/H((|

D||

D||

D|-|

2

2

2/||),|min(|

-t|D| since 2

2

2

-t)|D||,DV(|

2

-t)|D||,DC(||,1)DC(||,0)DC(|2

2t))|,DC(|-1)|D||,DC(||)D||,D(C(|

A]by falsified literals -R)|D|,max(least tat has Pr[D

2(b)] ofcondition thesatisfies Pr[D

F,in D clause aFix

D

DDRlDH

n

n

DRlD

l

Proof of lemma 5(conti.)

required as

,|D|such since m2

2(b)]condition satisfies D Pr[some Thus

2

|D|))H(R/(R|D||)D|H(R/

exponentR,-|D| If

|D|))H(R/(R

|D|R))/(-H(1|D||)D|/-H(1

exponentR,-|D| If

1) - )))(H(R/(R(

1) - )))(H(R/(R(D||

l

l

l

l

lll

l

ll

l

因此上式

上式分子的

上式分子的

Fact about H(x)

The graph of H(x) is

x

y

REVIEW!!

Theorem 1

n0.712-n2 2 m poly(n)

mostat is )R,(F,

of timerunning expected

the,n1.87 and n0.339RFor

llsRandom-Bal

l

Proof of theorem 1

)2(m m poly(n)

)R)V(n,(p m poly(n)

)*p)-(1R)V(n,(p m poly(n) T

).algorithms checking Ball theofeither

(by Checking-Ball of timerunning expected

theis T whereT,*N estimate toneed We

)()1)((R

nn

RH

lR

RHl

R

R

l

l

l

Proof of Theorem 1(conti.)

--

2

--n2

)()1)(()nn

RH( -n

2 2 minimize toneed weThus

,Rlog-H(R/n)n and ))R

RH(-(1

where

)2(22poly(n)m

)2(m2poly(n)m

TN Thus

ll

l

l Rn

n

RH

lR

RHl

Proof of Theorem 1 (conti.)

Assign R=a ,l=b ,where a < b constants. We use the fact ln(1+x)=x+o(x).

n n

Proof of theorem1(conti.)

Proof of Theorem 1 (conti.)

Taking a=0.339, b=1.87, we have Φ ,ψ>0.712 , proving the theorem.n

Reference

E. Dantsin, A. Goerdt, E. A. Hirsch, R. Kannan, J. Kleinberg, C. Papadimitriou, P. Raghavan, and U. Schoning. A deterministic

algorithm for K-SAT based on local search. Theoretical Computer Science, 289(1):69-83, October 2002.

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k)

1

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