Solving the Resolution-Free SAT Problem by Hyper-Unit Propagation in Linear Time Gábor Kusper gkusper@risc.uni-linz.ac.at  Research Institute for Symbolic Computation (RISC-Linz) Johannes Kepler University, Linz, Austria http://www.risc.uni-linz.ac.at/

Outline Motivation Hyper-Unit Propagation Resolution-Free SAT Problem Unicorn-SAT Algorithm Example General-Unicorn-SAT Algorithm Conclusion

Motivation Theorema, Prof. Buchberger http://www.theorema.org/ A1, ..., An G A1, ..., An, G is unsatisfiable

Unit Propagation a  b  c a  b a  c a  b  c a  b  c by a

Hyper-Unit Propagation a  b  c a  b a  c a  b  c a  b  c a  b a  c a  b  c HUP by a  b Hyper-Unit Prop. (HUP) by assignment A on S is denoted by S[A].

Resolution Resolution (tautologies are not allowed): a  b a  b a  c Examples Resolution OK: b  c a b  b  c a  b a  c a  b  c a  b  c Examples Resolution-Free: a  a  c a  b  c a  b  c a  b  c a  b  c Tautologies! Example Resolution-Free Clause Set:

Definition A clause set is resolution-free iff resolution cannot be performed on any two clauses of the clause set.

L(C): Negation of a Resolution-Mate a  b  c Let C be: negate a literal A Resolution-Mate of C: a  b  c a  b  c a  b  c These are clauses! L(C): negate all but one literal negate A Sub-Model generated from C: a  b  c a  b  c a  b  c These are assignments!

Definition We obtain a resolution-mate of a clause by negating one literal of it. A sub-model generated from clause C is the negation of a resolution-mate of C. It is denoted by L(C). By definition L() = . L(C) is an assignment, L(C) is a clause. L(C) is a resolution-mate of clause C.

Res.-Mate is not in S If S is resolution-free, S, CS and B is a resolution-mate of C. Then BS. Proof: C resolves with B! a  b  c a  b  c a  b  c a  b  c a  b  c a  b  c

 is not in S after Sub-Model Propagation If S is resolution-free, S, CS and C is minimal (B = L(C) ,i.e., B is a resolution-mate of C). Then S[L(C)]. Proof: Since C is minimal it suffices to show that B  S. It is known from the previous slide. a resolution- mate: S is resolution-free: a sub-model: a  b a  b  c a  b  c minimal: subsumed: not subsumed: a  b a  b a  b  c a  b  c a  b  c a  b  c

In Picture START input: S S[A] S[A] HUP S is L[C] = A := a  b by A resolution-free and S HUP by A L[C] = A := a  b Let A negation of B B := a  b a  b if S[A] =  then M := A Let B a res.-mate of C Let C be a min. clause from S C:= a  b END

HUP Preserves Resolution-Free-ness If S is resolution-free and A is an assignment. Then S[A] is resolution-free. Proof: Assume S[A] is not resolution-free. We show that this contradict with S is resolution-free. S: S[A]: ... y ... x ... ... ... y ... x ...

Unicorn-SAT Algorithm START input: S i := 1 S[Ai] is res.-free S[Ai] S := S[Ai], i := i+1 S is resolution-free and S HUP by Ai Ai := a  b Let Ai negation of B B := a  b a  b if S[Ai] =  then M := A1A2...Ai Let B a res.-mate of C Let C be a min. clause from S C:= a  b END

Termination If S  , S and CS. Then S[L(C)] has fewer clauses than S and contains fewer variables than S. Proof: Hyper-Unit propagation inherits this property from unit propagation.

Example sub-model a  b  c a  b  c a  b  d  e a  b  c  d a  d  e a  b  c  d  e a  b  c  d  e a  b  c  d  e b  c  d  e sub-model a  b  c HUP sub-mo. d  e d  e d  e HUP a  b  c  d  e

Time Complexity # variables : n # clauses : m O(nm) Argument: HUP by Ai can be simulated by length(Ai) UP. If we need k HUP then Summa(length(Ai), i:=1..k) = n. So we can simulate k HUP by n UP. Since UP is O(m) time method Unicorn-SAT is O(nm).

General-Unicorn-SAT Algorithm if i = 0 then unsatisfiable satisfiable if Si i := i-1 i := i+1 if Si+1 =  if Si+1 Si := Sires(C,L(C)) Correct the guess! Si Si+1 if Si then Ai := L(C), where C is a min. clause form Si. Guess a solution! if Si+1 i := i+1 i := i-1 START: i := 1, S0 := S, A0 := , Si := Si-1[Ai-1]

Conclusion DPLL like algorithms General-Unicorn-SAT use unit propagation incrementally construct a solution General-Unicorn-SAT uses hyper-unit propagation guesses a sub-model & correct it if necessary

Thank you for your attention!