CHF 4 SAT - cs.sfu.ca
Transcript of CHF 4 SAT - cs.sfu.ca
We now allow large conjunctions to disjunctions-
Observe : ¢pnQ)nR) = LPNCQNRD. can write (pnQiR) as also equivalent .
More generally:&t (Air Azn . . - n Ar) LEA,and Lt Az and . . . dFAn
& f- (A , v Azv . . - v An)⇒ a FA, or at Azor . - - at An.
Eg: (A , n Az n As n - - - n An) (P u Q u R vs v T)
We also use notation : A {A. , Az. . . . 3 = (Air Az n - - - )
¥sAi = VLAI lies}etc.
TerminologyLiteral : atom a negated atom P
,'P.Q.IQ . . .
clause : disjunction of literals (Pr Q virus)cube : conjunction of literals ( Pn Qn > Res)
CHF Formula : Conjunction of clauses .
(PrQuip) n(RriQ)n(Put R)
DNF Formula : Disjunction of cubes
(pnQnR) v GPnnR)v(QniR)
We may use setnotation for CNF :
{{ P.QiRHP.ws} I CLPRQVIR) n (Pri Q))
Eat : Every formula has a logically equivalent CHF formulaand a logically equivalent DNF formula .
These can be constructed from the truth table :
= (p n ( nprr))
p r 9 clauses cubes- --- -
T T T - ( p n r)T F F Gpvr) -
F T F ( pur) -
FFF(pi
CNF (Pvr)n( prir)n Cprr)DNFI Cpnr)
tact : For every r , there are formulas In and 4h st .- the smallest CHF = to 4in has 2
"
clauses- the smallest DNF = to win has 2
"
cubes
4n= ( p , nQ.) v ( pzn Qe) r (Psn QD . . . ( pun Qu).
Problem : SAT Solvers take input in CNF,
so we need an effiewt way to
transform to CNF
SAIInstance : A propositional CHF formula
Q
Question : Is 9 satisfiable?
E ✓ ( Npr :o) n (purrs)n Grip) n Gsu :p) Lp ,girl
X ↳ of n↳ v±)n # ±) lip ,
Site of a clause is its # of literals
4 is K- CHF if : . it is in CHF , and• has no clause largerthan K .
k¥4or anyke H ) : ,
Instance: K - CHF formula 9
Question : Is 4 satisfiabhe .
Factsk - SAT is NP-complete , for each K73
2- SAT 4 1- SAT have linear - time algorithms.
.
1-SAT: ( p ) . (g) n Gr) n (s) n (g) n (t) nfs)
Deciding if aCHF formula is a tautology has a
linear - time algorithm .
( prog) n . . .
( prgrr rig)
Efficieuttraustormatimsonformi.la#CNF-o3C-HF- Can eliminate a long clause by introducing a new atom
:
④{(d ' ✓Elf " II r
- - " ) i l ) el ) . . . ①
Career -tip ne-iXL InC)②( where t, is
"new
' )•
• ① is satisfie ble iff ② is.
. As long as there is a clause of length > 3 :
- eliminate a long clause as in (* )
- linear time,linear increase in site ,
• # new variables linear in site .
Negation .
is NNF = negations only on atoms> (price) tip nQ)x ✓
Linear - time algorithm to transform apropositional formula to NNF.
Repeat :replace a sub formula ' far B) with GAMB)
"" ' ( AnB) with GAnB)
" " 7nA with A
Tseitin's Polytime transformation to CNE-
F-for four))
i.
Claim '.
⇒ at t ⇒ at 4
2) PE9 ⇒ some extension of Bsatisfies M .
Resolutionproofspes. Rule: (LrA)GLvB)_ A. Bare
(Ar B) arbitrarydisjunction,
• Observe : { Ur A) ,Glr B)3 FCARB) oftitera.lyResolution Derivation-Sequence IT -_ (9,4 , . . . ,
Cm) of clauses is
a resolution derivation of Cfrom 17
if :. Cm=C
• for each Ci , either CiEP
or Ci is the resolvautflj.lkwith jik hi .
F5E read - dem . of Cp) from{(pvQuR7 , GRUQ) ,(PVIQI}(Pr Qv R) GRVQ) .
(D)-
.
Prop. 2 . If there is a resolution derivation-
of C fromT then TEC .
Consider : (p)(. \ is unsatisfiede.
C) . sometimes write D fall .
Beth: A resolutionrefutati of IT is a resolutionderivation of D from 17 .
( if D)
lonstructingaRefutationonasemant.IT#Semantic Tree faT={ CPRQ) , (Pvr) ,R,GRvs)
.GS#)i...o/iPDLp70FTl*p)tip) I
{RQ7&↳piQp.im/7YiQ)GRQ) - n.LPIR) (RR)
( RQ1R)o/\oLRQiRY 4RRQ#¥RiQYbag) A LR11
4BR TIPIRQ.is)to;D tR
Prof. If Tisunsatisfiable , then These resolution refutation .
lonstructingaRefutationonasemant.IT#Semantic Tree faT={ CPRQ) , ✓ R)
, GQVR) .GR vs) ,GsriQ) ,- . -
Yip•
-
↳ aa
{RQ7 ftp.QQ#RX4piN\ "
.
{ RQ1M 's @LRQIRYKRR.QA RRiQY
^tip,RiQ,s) TIRRQ.is)
Prof : If Tisunsatisfiable , then These resolution refutation .
B-acktrackingAIgorith-mfaSATBTIT.NLif 14) = false fou C threturn
'
UN SAT'
else if I is total
return'SAT
'
elsed- a literal from 4 sit . I (e) undefined
if BT6P ,I used}) returns ' SAT '
return ' SAT'
elsereturn BTLR , tubes)
}
start with BT1R,0 ) .
[we are representing a partial t.cn .as a set of literals .
Observation :
• For unsatisfiedhe F :
• the tree of recursive calls induced
by BTCR ,0) is a semantic tree ta T
• At each call BTCT ,a) , d is the p.to .
labelling the corresponding tree node .
• So : BTCT ,0) implicitly constructs a
resolution refutation of P .
→-