Easily refutable subformulas of large random 3CNF formulas

1
Easily refutable subformulas of large random 3CNF
formulas

• Uriel Feige and Eran Ofek

The Weizmann Institute of Science
2
Heuristic for 3SAT

• A 3CNF formula
• 3SAT is NP-hard.
• A heuristic may try to
• Find a satisfying assignment (if exists).
• Refute the input formula F.
• What is a good refutation Heuristic? Refutes
  most unsatisfiable formulas.

3
Generating unsatisfiable formulas

• A refutation algorithm is ?-good if it refutes
  most random formulas with density ?.
• If refutation is hard in this model then
  approximation of some NP-hard problems is also
  hard.

Pr(SAT)
m clauses n variables
1
0
m/n density
3.52 KKL,HS
4.596 JSV
4
Certification Algorithm

• An algorithm is sound with respect to a property
  P if
• The algorithm is complete with respect to some
  probability measure on the instances if in
  addition

Algorithm dont know
not P
P
Algorithm accepts
5
Examples for certification algorithms

• The Lovasz Theta function upper bounds the size
  of the maximum independent set of a given graph.
  Juhasz82 For almost all graphs its value is of
  order n½.
• Zwick99 SDP Max-Cut relaxation such that
• If MaxCut(G) lt (½ e) E(G), then SDP(G) lt
  (½ d) E(G) (where d ? 0 as e? 0).
• Corollary we can certify that a random graph has
  no (½ d)-cut.We will use this corollary
  extensively.

6
More examples

• Random kCNF formula with ckn clauses is
  unsatisfiable.GK01 gives an algorithm which
  refutes most random kCNF formulas with poly(log
  n)nk/2 clauses but only for even k.(improved to
  O(nk/2) clauses in CGLS03,FO03).
• FGK01 gives a refutation algorithm for random
  3CNF formulas with cn3/2e clauses (improved to
  poly(log(n)) n3/2 clauses in GL03).

7
Our result

• Theorem There is an efficient algorithm that
  refutes most 3CNF formulas with cn3/2 clauses
  (for some constant c).Builds upon earlier work.
  Our main tool is a reduction to refutation of
  random 2XOR formulas.
• Experimental results on large formulas.

8
Refuting random 2XOR

• Random 2XOR F with cn clauses is at most
  (½d)-satisfiable.
• We can efficiently certify it
• Reduction to Max-Cut
• If F is (½ e)-satisfiable, then G has a (½
  e)-cut(every satisfied clause/edge crosses the
  assignment cut).
• GF is random, has no (½ e)-cut and we can
  certify that it doesnt have (½ d)-cut.

True
False
9
The 3XOR Lemma Feige 02

• For a random F we can (almost surely) certifyif
  F is satisfiable as 3CNF ? it is (1 -
  e)-satisfiable as 3XOR.
• Proof Assume A is a satisfying assignment. We
  can certify

The number of satisfied literalsis bounded by
(3/2 e)m
The fraction of clauses satisfiedas 3AND is at
least ¼-e
ai fraction of clauses in F satisfied i times
(i1,2,3)
F F2XOR (x1,x4,x7) ? (x1,x4),(x1,x7),(x4
,x7)

• a1 a2 a3 1
• a1 2a2 3a3 (3/2 e)
• a3 ¼ - e

3AND ? 0/3 satisfied clauses 3NAE ? 2/3
satisfied clauses
If (¾ e)m clauses are satisfied as 3NAE, then
F2XOR is (½ 2e)-satisfiable (but we refute this
case).
Using A, almost all the clauses are satisfied
once or 3 times
10
Refuting random 3CNF
Certify
F1, F2 have the 3XOR property.
F2lin is at most (½ e)-satisfiable as 2LIN.
F2lin is (1- 2e)-sat as 2LIN
? Fi is (1- e)-sat as 3XOR
Assume F is sat
11
Implementation

• A variant of the algorithm refuted random
  formulas with 50,000 variables and 2.43 n3/2
  27,335,932 clauses (73 minutes on a machine with
  1700MHz CPU, 2GB memory and 256K cache).

12
Open problems

• Break the cn3/2 barrier for random 3CNF
  refutation.
• BsW this can not be done using resolution
  random 3CNF formulas with m lt n3/2-e require
  exponential length resolution refutation.
• Give a running-time/clause-density tradeoff a
  refutation algorithm that runs in time
  O(poly(n)expn/c2) and almost surely refutes a
  random 3CNF formula with cn clauses.