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Conditional Lower Bounds for Failed Literals and Related Techniques - - PowerPoint PPT Presentation

Conditional Lower Bounds for Failed Literals and Related Techniques Matti Jrvisalo Janne H. Korhonen* University of Helsinki, Department of Computer Science, and Helsinki Institute for Information Technology HIIT 1.


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Conditional Lower Bounds for Failed Literals and Related Techniques

Matti Järvisalo Janne H. Korhonen*

  • University of Helsinki, Department of Computer Science, and

Helsinki Institute for Information Technology HIIT

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Background: Formula Simplification

1.

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formula simplification CNF formula “simpler” formula

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CNF formula SAT/ UNSAT SAT solver formula simplification

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CNF formula SAT/ UNSAT SAT solver formula simplification

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CNF formula SAT/ UNSAT SAT solver formula simplification simplification techniques faster stronger

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Main Result: Lower Bound for Failed Literal Existence

2.

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  • Unit propagation
  • apply until fixpoint
  • we write F ⊢up (ℓ) if (ℓ) can be derived from F by

repeated application of unit resolution rule (¬ℓ1), (¬ℓ2), …, (¬ℓk) (ℓ)

}

(ℓ1∨…∨ℓk∨ℓ)

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  • Unit propagation
  • apply until fixpoint
  • we write F ⊢up (ℓ) if (ℓ) can be derived from F by

repeated application of unit resolution rule

  • Failed literals
  • a literal ℓ ∈ F is a failed literal if F∧(ℓ) ⊢up (ℓ’), (¬ℓ’)


for some ℓ’ ∈ F

  • replace F with F∧(¬ℓ) if ℓ is a failed literal

(ℓ)

}

(¬ℓ1), (¬ℓ2), …, (¬ℓk) (ℓ1∨…∨ℓk∨ℓ)

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Failed literal existence problem Input: CNF formula F Problem: Decide whether F has a failed literal

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  • Upper bounds (assuming bounded clause width)
  • unit propagation O(n+m)
  • failed literal existence O(n(n+m))
  • failed literal elimination fixpoint O(n2(n+m))
  • Can we do any better?

Failed literal existence problem Input: CNF formula F Problem: Decide whether F has a failed literal

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  • Theorem. If failed literal existence can solved in

O((n+m)2-ε) time on Horn-3-CNFs for some ε > 0, then CNF-SAT can be solved in time 2(1-ε/2)n poly(n,m)

  • n formulas of unrestricted clause length.
  • Recall that CNF formula is Horn if each clause has at

most one unnegated variable

  • Horn-SAT is solvable in linear time
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  • Theorem. If failed literal existence can solved in

O((n+m)2-ε) time on Horn-3-CNFs for some ε > 0, then CNF-SAT can be solved in time 2(1-ε/2)n poly(n,m)

  • n formulas of unrestricted clause length.
  • We do not know how to solve CNF-SAT in time


2(1-ε)n poly(n,m) for any ε > 0

  • This would give exponential speed-up for CNF-SAT!
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lim inf {δ : k-SAT can be solved in time O(2δn)} = 1

n→∞

[Calabro, Impagliazzo, and Paturi 2009]

CNF-SAT with unrestricted clause length cannot be solved in time 2(1-ε)n poly(n,m) for any ε > 0

The Strong Exponential Time Hypothesis

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  • Corollary. Failed literal existence restricted to Horn-3-

CNFs cannot be solved in time O((n+m)2-ε) for any ε > 0 unless SETH fails.

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  • Corollary. Failed literal existence restricted to Horn-3-

CNFs cannot be solved in time O((n+m)2-ε) for any ε > 0 unless SETH fails.

  • Compare with other similar results: for any ε > 0 we

cannot solve

  • k-dominating set for k ≥ 3 in time O((n+m)k-ε)
  • 2-SAT with O(n) clauses and two unrestricted

length clauses in time O(n2-ε)

  • Local alignment of two binary strings in time O(n2-ε)

[Pătraşcu and Williams 2010] [Abboud, Vassilevska Williams, and Weimann 2014]

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Proof of the Failed Literal Existence Lower Bound

3.

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Proof Overview

3a.

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  • Theorem. If failed literal existence can solved in

O((n+m)2-ε) time on Horn-3-CNFs for some ε > 0, then CNF-SAT can be solved in time 2(1-ε/2)n poly(n,m)

  • n formulas of unrestricted clause length.
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Horn-3-CNF O((n+m)2-ε) failed literal / no failed literals n variables m clauses F

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CNF formula 2(1-ε/2)n poly(n,m) F n variables m clauses SAT / UNSAT

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CNF formula n variables m clauses F

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CNF formula Reduction 2n/2 poly(n,m) N = 2n/2 poly(n,m) M = 2n/2 poly(n,m) n variables m clauses Horn-3-CNF F F’

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CNF formula Reduction 2n/2 poly(n,m) O((N+M)2-ε) = 2(1-ε/2)n poly(n,m) N = 2n/2 poly(n,m) M = 2n/2 poly(n,m) n variables m clauses Horn-3-CNF failed literal / no failed literals F F’

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CNF formula Reduction 2n/2 poly(n,m) O((N+M)2-ε) = 2(1-ε/2)n poly(n,m) SAT / UNSAT F N = 2n/2 poly(n,m) M = 2n/2 poly(n,m) n variables m clauses F’ Horn-3-CNF failed literal / no failed literals

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Reduction from CNF-SAT to Failed Literal Elimination

3b.

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CNF formula Reduction 2n/2 poly(n,m) O((N+M)2-ε) = 2(1-ε/2)n poly(n,m) SAT / UNSAT F N = 2n/2 poly(n,m) M = 2n/2 poly(n,m) n variables m clauses F’ Horn-3-CNF failed literal / no failed literals

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n variables input formula F

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n variables n/2 “upper” variables n/2 “lower” variables input formula F

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n variables n/2 “upper” variables n/2 “lower” variables 2n/2 partial truth assignments P 2n/2 partial truth assignments Q input formula F

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constructing the output formula F’ P Q ~ m clauses in F c1 c2 c3 … cm ~ 2n/2 partial truth assignments in Q ~ 2n/2 partial truth assignments in P yp yq

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constructing the output formula F’ P Q c1 c2 c3 … cm yp yq add clause (yp → ci) if p(Ci) = 1

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constructing the output formula F’ P Q yp yq add clause ((⋀i∈S ci)→ yq) where S = { i : q(Ci) ≠ 1 }

add clause (yp → ci) if p(Ci) = 1

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constructing the output formula F’ P Q yp yq add clause ((⋀i∈S ci)→ yq) where S = { i : q(Ci) ≠ 1 }

add clause (yp → ci) if p(Ci) = 1 F∧(yp) ⊢up (yq) pq(F) = 1 if and only if

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Extensions and Open Questions

4.

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  • Assuming SETH, for any ε > 0 we cannot solve
  • asymmetric tautology existence on Horn-3-CNFs in

time O((n+m)2-ε)

  • asymmetric literal existence on Horn-3-CNFs in time

O((n+m)2-ε)

  • singleton arc consistency on (3,2)-CSPs in time


O((n+m)2-ε)

  • k-step lookahead lower bound?
  • Fix values for k variables, do unit propagation
  • We can probably show lower bound vs. O((n+m)k+1-ε) time

Extensions of the Main Result

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  • Failed literal existence on 2-CNFs?
  • CNF version requires clause length 3
  • CSP version requires domain size 3
  • Maybe one can do better on 2-CNFs?
  • Failed literal elimination fixpoint?
  • Lower bound O((n+m)3-ε)?

Open Questions

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Questions, comments?

Thank you!

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  • clause C = (ℓ1∨…∨ℓk) ∈ F is an asymmetric tautology

if (F \ C)∧(¬ℓ1)∧…∧(¬ℓk) ⊢up (ℓ’), (¬ℓ’) for some ℓ’ ∈ F

  • replace F with F \ C
  • literal ℓ in a clause C ∈ F is an asymmetric literal if

F∧(ℓ) ⊢up (ℓ’) for some ℓ’ ∈ C \ {ℓ}

  • replace F with (F \ C) ∧ (C \ ℓ)

Definitions: Asymmetric Tautologies and Literals