Generating Hard Satisfiability Problems1 Bart Selman, David Mitchell, Hector J. Levesque Presented by Xiaoxin Yin

Generating Hard Satisfiability Problems

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Bart Selman, David Mitchell, Hector J. Levesque

Presented by Xiaoxin Yin


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Can SAT be solved in polynomial time in average case? In Goldberg’s paper [Goldberg ‘79] it is

claimed that SAT can be solved “on average” in polynomial time.

Goldberg’s model of generating formulas m clauses, n variables, each literal has probability

p to be in each clause Is SAT really easy?


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Can SAT be solved in polynomial time in average case? Goldberg’s SAT formulas are easy to solve

[Franco & Paull ‘83] Theorem 1: Number of truth assignments for

a formula is greater than 2n(1-ε) with probability 1

Theorem 2: By randomly guessing truth assignments w times,

Pr(success)=1 – m/(2αnw)assume each clause has at least αr literals


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K-SAT is a harder problem K-SAT: each clause has K literals


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Problem Definition of K-SAT Generate formulas of

N variables K literals per clause (K = 3 in this paper) M clauses

To generate a clause Randomly choose K distinct variables Negate each with probability 0.5

Generate formulas with certain ratio of clauses-to-variables


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DP Procedure DP: a backtracking depth-first search in the

space of all truth assignments Procedure DP

Given a set of clauses Σ defined over a set of variables V

Set the value of a variable v and call DP on the simplified formula If this call returns “satisfiable”, then return

“satisfiable” Set v to the opposite value, return the result of

calling DP on the re-simplified formula


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Three Common Rules for DP The unit clause rule

If a clause contains only one literal, set it to true The pure literal rule

If a formula contains a literal but not its complement, set it to true

The smallest clause rule If none of the above rules applies, set a variable in

a smallest clause The last two rules are not used in this paper


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DP’s Performances on K-SAT Ratio of clauses-to-variables significantly affects the

hardness of formulas


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Ratio of Clauses-to-variables vs. Computational Cost Another example in [Mitchell & Levesque ‘96]


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50% Point

50% point: given a certain N (# variables), the point that 50% of generated formulas are satisfiable

50% point is stable w.r.t. ratio of clauses-to-variables

N 20 50 100 150

50% point 4.55 4.36 4.31 4.3


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50% Point – The Hardest Point (cont.) 50% point is close to the location of peak hardness


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Satisfiable vs. Unsatisfiable Formulas Short formulas – under-constrained, have many satisfying

assignments Long formulas – over-constrained, contradictions can often

be easily found


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Finding All Satisfying Assignments or Contradiction Let DP search the full space, until finding a contradiction Given a set of variables

Ratio of clauses to variables increases → Search space decreases


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Finding All Satisfying Assignments or Contradiction (cont.) Computational cost of DP – monotonically

decreases for increasing ratios of clauses to variables


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Satisfiablity when ratio of clause-to-variable is small Pure literal rule: if a formula contains a literal

but not its complement Set this literal to 1 Remove all clauses containing this literal

For 3-CNF with up to 1.63n clauses, pure literal rule by itself finds satisfying assignments with high probability [Broder, Frieze & Upfal ‘93]


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Satisfiablity when ratio of clause-to-variable is small Smallest clause rule:

Choose a (random) literal in a (random) smallest clause

For 3-CNF with less than 3.003n clauses, by smallest clause rule one can find satisfying assignments with high probability [Frieze & Suen ‘92]

By “high probability” we mean Pr → 1 as n → ∞


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Satisfiablity when ratio of clause-to-variable is large When c>4.762, a random 3-SAT formula is

unsatisfiable with high probability [Kamath et al ‘94]. Some intuitions of proof:

Consider a certain assignment Z1, each clause is true with probability 7/8.

Let #F denote number of satisfying assignments on F. E[#F] = 2n(7/8)cn.

By Markov inequality, P[#F>0]≤E[#F]=(2∙(7/8)c)n

This probability is exponentially small when c > 5.191


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DP vs. Resolution DP searches for satisfying assignments as

well as contradictions Resolution searches for contradictions Some result about resolution

In k-SAT problems of cn clauses, when k ≥ 3 and c2–k ≥ 0.7, with probability tending to 1 as n goes to infinity, a randomly chosen formula of cn clauses is unsatisfiable, but there exists ε>0 such that every resolution proof must generate at least (1+ε)n clauses [Chvatal & Reed ‘92]

c=5.6 when k=3


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Satisfiability for Different K For a random formula Fk(n, cn)

ck = sup{ c : Fk(n, cn) is satisfiable with high prob} ck

*= inf{ c : Fk(n, cn) is unsatisfiable with high prob}

[Kirousis et al ‘98]

ck* ≤ 2k ln2 – (1+ln2)/2

[Achlioptas and Peres ‘03]

ck = ck* (1 – o(1))

ck ≥ 2k ln2 – (k+1)ln2/2 – 1 – δk

(for a certain sequence δk→0)

ck ck*

satisfiable unsatisfiableundetermined


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Satisfiability for Different K (cont.) Lower and upper bounds for different K

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rati

o Lower

Upper


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K-SAT with different K’s When K is larger → Higher satisfiability, larger

search space when n is large


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P-SAT with different K’s Each variable has certain probability to appear in

each clause Each clause has K variables on average


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Solving SAT by Local Search [B. Selman et al ‘92] “A new method for

solving hard satisfiability problems” GSAT – greedily search in the space of

assignments GSAT algorithm

Repeat for MAX-TRIES timesrandomly generate an assignmentrepeat for MAX-FLIPS times

flip a variable to get largest increase in number of satisfied clauses


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Solving SAT by Local Search (cont.) Performance of GSAT

Drawback – cannot prove unsatisfiability

0.1

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num of var

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GSAT

DP


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Thank you!


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Additional Contents

Performances of other approaches for solving K-SAT are presented in [Larrabee & Tsuji ‘93], which shows similar results (easy-hard-easy pattern)


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Solving 2-SAT in Linear Time

Choose a variable x and assign a value (e.g. x=1) Remove all clauses that are true Set values to all variables whose values are decided Propagate in this way until nothing can be done If contradiction happens, return false

A set of clauses are left that are independent with the removed ones If these clauses are satisfiable, return true Else return false

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Generating Hard Satisfiability Problems1 Bart Selman, David Mitchell, Hector J. Levesque Presented by Xiaoxin Yin