Phase Transitions of PP-Complete

Satisfiability Problems

D. Bailey, V. Dalmau, Ph.G. KolaitisD. Bailey, V. Dalmau, Ph.G. KolaitisComputer Science DepartmentComputer Science Department

UC Santa CruzUC Santa Cruz

Phase Transition Phenomena

First observed in NP-complete problems:First observed in NP-complete problems:SAT, CSP, Number Partitioning, …SAT, CSP, Number Partitioning, …

Recently also observed in problems that are Recently also observed in problems that are complete for higher complexity classescomplete for higher complexity classes

Observed in PSPACE-complete problems:Observed in PSPACE-complete problems: QBF: Cadoli et al. ; Gent & WalshQBF: Cadoli et al. ; Gent & Walsh SSAT: Littman; Littman-Majercik-PitassiSSAT: Littman; Littman-Majercik-Pitassi

PP: Probabilistic NP

PP: there is a polynomial time NTM such thatPP: there is a polynomial time NTM such thatan input is accepted iff at least half of the computations an input is accepted iff at least half of the computations are accepting.are accepting.Simon, 1975; Gill, 1977Simon, 1975; Gill, 1977

Prototypical PP-complete problem:Prototypical PP-complete problem:MAJSAT: given a CNF-formula MAJSAT: given a CNF-formula , is it satisfied by at , is it satisfied by at least half of the possible truth assignments?least half of the possible truth assignments?

PP-complete problem in AI:PP-complete problem in AI:Orponen, 1990; Roth, 1996Orponen, 1990; Roth, 1996

PP vs. Other Complexity Classes

PP contains both NP and coNP.PP contains both NP and coNP.

PP is contained in PSPACE.PP is contained in PSPACE.

P P PP PP = P = P # P # P (Angluin, 1980). Hence, (Angluin, 1980). Hence, PP captures the complexity of counting.PP captures the complexity of counting. PP is PP is highly intractable, highly intractable, since P since P # P # P contains the contains the

polynomial hierarchy PHpolynomial hierarchy PH(Toda, 1990). (Toda, 1990).

Phase Transitions in PP Problems

Study PP-complete satisfiability problems Study PP-complete satisfiability problems under the fixed clauses-to-variables model.under the fixed clauses-to-variables model.

First natural choice to study: MAJ 3SATFirst natural choice to study: MAJ 3SAT

Phase Transitions in PP Problems

Study PP-complete satisfiability problems under the Study PP-complete satisfiability problems under the fixed clauses-to-variables model.fixed clauses-to-variables model.

First natural choice to study: MAJ 3SAT.First natural choice to study: MAJ 3SAT.

However,However, MAJ 3SAT is not known to be PP-complete.MAJ 3SAT is not known to be PP-complete.

MAJ 3SAT has no phase transition: for every r, almost MAJ 3SAT has no phase transition: for every r, almost all random 3CNF formulas are satisfied by less than all random 3CNF formulas are satisfied by less than half of all possible truth assignments.half of all possible truth assignments.

Square Root 3SAT

Square Root 3SAT - #3SAT( Square Root 3SAT - #3SAT( 2 2n/2 n/2 ): ): given a 3CNF-formula given a 3CNF-formula , is it satisfied by , is it satisfied by at least at least 22n/2 n/2 truth assignments?truth assignments?

Intuitively, #3SAT( Intuitively, #3SAT( 2 2n/2 n/2 ) asks whether at ) asks whether at least one of the first n/2 bits of the number least one of the first n/2 bits of the number of satisfying truth assignments is equal to 1.of satisfying truth assignments is equal to 1.

Square Root 3SAT

Square Root 3SAT - #3SAT( Square Root 3SAT - #3SAT( 2 2n/2 n/2 ):):given a 3CNF formula given a 3CNF formula , is it satisfied by at least , is it satisfied by at least 22n/2 n/2 truth assignments?truth assignments?

Intuitively, #3SAT( Intuitively, #3SAT( 2 2n/2 n/2 ) asks whether at least ) asks whether at least one of the first n/2 bits of the number of satisfying one of the first n/2 bits of the number of satisfying truth assignments is equal to 1.truth assignments is equal to 1.

Theorem: Theorem: #3SAT( #3SAT( 2 2n/2 n/2 ) is PP-complete.) is PP-complete.

Phase Transition Definition F(n,r): space of all 3CNF-formulas F(n,r): space of all 3CNF-formulas with with

n variables and rn clauses.n variables and rn clauses. X: random variable on F(n,r), such that XX: random variable on F(n,r), such that X = number of satisfying assignments of = number of satisfying assignments of .. S: an assertion about X.S: an assertion about X. X has phase transition at r* iff for all r X has phase transition at r* iff for all r

If r < r*, then Pr[ S is trueIf r < r*, then Pr[ S is true ] ] 1; 1; If r > r*, then Pr[ S is true ] If r > r*, then Pr[ S is true ] 0. 0.

Phase Transition Conjecture F(n,r): space of all 3CNF-formulas F(n,r): space of all 3CNF-formulas with with

n variables and rn clauses.n variables and rn clauses.

X: random variable on F(n,r) such thatX: random variable on F(n,r) such thatXX = number of satisfying assignments of = number of satisfying assignments of ..

Conjecture:Conjecture: There is a ratio r* such that There is a ratio r* such that If r < r*, then Pr[ X If r < r*, then Pr[ X 2 2n/2 n/2 ] ] 1; 1; If r > r*, then Pr[ X If r > r*, then Pr[ X 2 2n/2 n/2 ] ] 0. 0.

Evidence for the Phase Transition

Analytical results that yield upper and lower Analytical results that yield upper and lower bounds for r*.bounds for r*.

Experimental results suggesting thatExperimental results suggesting that r* r* 2.5 2.5

Upper and Lower Bounds for r*

Theorem:Theorem: 0.9227 0.9227 r* r* 2.595 2.595

Hint of Proof:Hint of Proof: Upper Bound: Upper Bound: Markov’s inequality. Markov’s inequality. Lower Bound:Lower Bound: Covering partial Covering partial

assignments. assignments.

Upper Bound for r*

From Markov’s inequality, From Markov’s inequality, Pr[ X Pr[ X 22n/2 n/2 ] ] E(X)/ 2E(X)/ 2n/2 n/2

E(X) = 2E(X) = 2nn(7/8)(7/8)rnrn

Pr[ X Pr[ X 22n/2 n/2 ] ] 22n/2n/2(7/8)(7/8)rnrn

If , 2If , 21/21/2(7/8)(7/8)r r 1, then Pr [ X 1, then Pr [ X 22n/2 n/2 ] ] 0 as 0 as n n

So if, r So if, r 2.595, then Pr [ X 2.595, then Pr [ X 22n/2 n/2 ] ] 0 as 0 as n n

Lower Bound General Approach

Show that a randomly selected formula Show that a randomly selected formula (x(x11,…,x,…,xnn) has at least 2) has at least 2n/2n/2 satisfying satisfying assignments assignments by finding a partial by finding a partial assignment with assignment with 22n/2n/2 variables that variables that covers covers ..

Covers means: the partial assignment makes Covers means: the partial assignment makes the formula true, regardless of the truth the formula true, regardless of the truth value of the un-assigned variables.value of the un-assigned variables.

Finding Covering Partial Assignments To show r* To show r* , derive from the existence , derive from the existence

of at least one satisfying assignment fromof at least one satisfying assignment from a 2CNF theorem or froma 2CNF theorem or from a 3SAT lower bound theorem.a 3SAT lower bound theorem.

To show r* To show r* use Achlioptas’ use Achlioptas’ technique to analyze a simple randomized technique to analyze a simple randomized algorithm, called the Extended Unit Clause.algorithm, called the Extended Unit Clause.

1/2 Lower Bound for r*

Random n variable 3CNF Random n variable 3CNF with m clauses, with m clauses, such that m/n such that m/n 1/2.1/2.

has at least one satisfying assignment, has at least one satisfying assignment, by by lower bound theorem for 3SAT.lower bound theorem for 3SAT.

Build partial assignment Build partial assignment with with n/2 variables n/2 variables that covers that covers for each of the for each of the n/2 clauses in n/2 clauses in , , choose some literal in the clause that is true choose some literal in the clause that is true under under and set and set to also make that literal true. to also make that literal true.

Experimental Results

Implemented a Implemented a thresholdthreshold version of version of Birnbaum & Lozinskii’s CDP algorithm.Birnbaum & Lozinskii’s CDP algorithm.

Experiments on random 3CNF-formulas Experiments on random 3CNF-formulas with n = 10, 20, 30, 40, and 50 variables.with n = 10, 20, 30, 40, and 50 variables.

Probability curves cross at r Probability curves cross at r 2.5 2.5 The average number of recursive calls The average number of recursive calls

peaks near 2.5peaks near 2.5

Birnbaum & Lozinskii’s CDP

recursive function CDP(recursive function CDP(n)n) if if is empty, return 2 is empty, return 2nn

if if contains an empty clause, return 0 contains an empty clause, return 0 if if contains unit clause {t}, return contains unit clause {t}, return

CDP(CDP(,n-1), where,n-1), where

contains all clauses in contains all clauses in that do not that do not

contain t ;contain t ; the literal the literal t is removed if present.t is removed if present.

CDP (cont.)

otherwise choose any variable x in otherwise choose any variable x in return return CDP(CDP( ,n-1) + CDP( ,n-1) + CDP( ,n-1), where ,n-1), where

contains all clauses in contains all clauses in that do not that do not

contain x, with the literal contain x, with the literal x removed if x removed if present. present.

contains all clauses in contains all clauses in that do not that do not

contain contain x, with the literal x removed if x, with the literal x removed if present.present.

Threshold CDP

Accumulate partial counts in recursive calls Accumulate partial counts in recursive calls of CDP.of CDP.

Return yes when accumulated count equals Return yes when accumulated count equals or exceeds threshold.or exceeds threshold.

Return no otherwise.Return no otherwise. Can also use upper bound tracking to Can also use upper bound tracking to

terminate and return no earlier.terminate and return no earlier.

Concluding Remarks

Evidence for a phase transition in a natural Evidence for a phase transition in a natural PP-complete satisfiability problem.PP-complete satisfiability problem.

Analytical upper bound obtained via Markov’s Analytical upper bound obtained via Markov’s inequality is quite close to the value of the critical inequality is quite close to the value of the critical ratio suggested by the experiments.ratio suggested by the experiments.

Next steps:Next steps: Obtain tighter upper and lower bounds.Obtain tighter upper and lower bounds. Characterize the rate of growth of the average Characterize the rate of growth of the average

number of recursive calls as r and n vary.number of recursive calls as r and n vary.