Space Complexity of Polynomial Calculus Massimo Lauria Sapienza - - PowerPoint PPT Presentation

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Space Complexity of Polynomial Calculus Massimo Lauria Sapienza - - PowerPoint PPT Presentation

Space Complexity of Polynomial Calculus Massimo Lauria Sapienza Universit` a di Roma L OGICAL A PPROACHES TO B ARRIERS IN C OMPLEXITY II C AMBRIDGE 2012 (joint work with Y. Filmus, J. Nordstr om, N.Thapen and N. Zewi) SAT has been


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Space Complexity of Polynomial Calculus

Massimo Lauria

Sapienza – Universit` a di Roma

LOGICAL APPROACHES TO BARRIERS IN COMPLEXITY II CAMBRIDGE 2012 (joint work with Y. Filmus, J. Nordstr¨

  • m, N.Thapen and N. Zewi)
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“SAT has been solved”

—the daring practitioner—

“SAT is clearly hard”

—the savvy theorist—

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“SAT solving can get better. . . ”

—the relentless coder—

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Next step is to study memory requirements

  • f algebraic sat-solvers/theorem provers.
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Modern SAT solvers are based on Resolution

but

algebraic reasoning can be beneficial.

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shorter proofs

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shorter proofs simple proof structure

(e.g. Polynomial Calculus)

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shorter proofs simple proof structure

(e.g. Polynomial Calculus)

proof search

(e.g. Buchberger)

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shorter proofs simple proof structure

(e.g. Polynomial Calculus)

proof search

(e.g. Buchberger)

implementations

(e.g POLYBORI)

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. . . NOT ENOUGH TO SWITCH

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. . . NOT ENOUGH TO SWITCH

modern solvers are very optimized

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. . . NOT ENOUGH TO SWITCH

modern solvers are very optimized to change paradigm is bad for SAT races

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. . . NOT ENOUGH TO SWITCH

modern solvers are very optimized to change paradigm is bad for SAT races their main issue is space, not proof length

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SPACE IN MODERN SAT-SOLVERS

during a running, solvers learn clauses memory fills very quickly retaining the right clauses if fundamental

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DOES ALGEBRA HELP?

memory requirements relation between space and time

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OUR RESULTS

width variables space PCR

n n2 Ω(n)

[Alekhnovich et al. 02] PC

3 n2 Θ(n)

pigeonhole principle PC

O(1) n O(n)

any formula PCR

2 log n n log n Ω(n)

bit-pigeonhole principle PCR

4 n2 Ω(n)

xor-pigeonhole principle

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OUR RESULTS

width variables space PCR

n n2 Ω(n)

[Alekhnovich et al. 02] PC

3 n2 Θ(n)

pigeonhole principle PC

O(1) n O(n)

any formula PCR

2 log n n log n Ω(n)

bit-pigeonhole principle PCR

4 n2 Ω(n)

xor-pigeonhole principle

IN THIS TALK WE SKETCH THIS PROOF!

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OUTLINE

1

Algebraic proof system PCR

2

Model space complexity in PCR refutations

3

We sketch the proof of a space lower bound for PCR

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ALGEBRAIC PROOF SYSTEM

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PROOF SYSTEMS

Deterministic polynomial time P(·, ·) if F ∈ UNSAT then P(F, π) = 1 for some π ∈ {0, 1}∗ if F ∈ UNSAT then P(F, π) = 0 for all π ∈ {0, 1}∗

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PROOF SYSTEMS IN COMPLEXITY THEORY

There is P where any unsat formula has a “short” refutation in P ⇐ ⇒

NP=coNP

Cook-Reckhow program (1979)

Prove proof length lower bound for stronger and stronger system in order to prove NP=CONP

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PROOF SYSTEMS AND SAT SOLVERS

The trace of “SAT-solver(F)=unsat” is a refutation for F. DLL − → tree-like resolution Clause Learning − → regular WRTL [BHJ ’08] CL + Restarts − → resolution CRYPTOMINISAT − → fragments of PCR on GF(2) POLYBORI − → PC on GF(2)

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POLYNOMIAL CALCULUS (PCR)

CNF formula − → set of polynomials SAT assignments − → common roots true − → false − → 1 variable x − → x,¯ x x ∈ {true, false} − → x2 − x x + ¯ x − 1 x ∨ ¬y ∨ ¬z ∨ s ∨ t − → x · ¯ y · ¯ z · s · t

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POLYNOMIAL CALCULUS (PCR)

LINEAR COMBINATION

p q αp + βq

MULTIPLICATION

p xp

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F ⊢ 1 iff F ∈ UNSAT (SOUNDNESS) INFERENCE PRESERVES COMMON ROOTS (COMPLETENESS) SIMULATES DECISION TREES

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PC defined in [CEI96] and PCR in [ABRW02]; PCR strictly better than resolution in proof length; Size-Degree Trade-off [IPS99,GL10a]; Exponential lower bounds on length are known [Raz98,AR03, BGIP01, BI10, IPS99, Raz98]; Proof search is hard [GL10b] based on [AR08].

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SPACE COMPLEXITY OF PCR

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· · · →   xz + yz xz − 1  

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· · · →   xz + yz xz − 1   →   xz + yz xz − 1 1 − yz   inference step from polynomials in memory

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· · · →   xz + yz xz − 1   →   xz + yz xz − 1 1 − yz   →   xz + yz — 1 − yz   inference step from polynomials in memory erasure of a polynomial

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· · · →   xz + yz xz − 1   →   xz + yz xz − 1 1 − yz   →   xz + yz 1 − yz   →   xz + yz x2 − x 1 − yz   · · · inference step from polynomials in memory erasure of a polynomial logical axiom/initial polynomial download

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Space measure: #monomials in a configuration

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Space measure: #monomials in a configuration

    xz + yz xz − 1 1 − yz    

(this configuration counts as space six)

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Space measure: #monomials in a configuration

    xz + yz xz − 1 1 − yz    

(this configuration counts as space six)

Roads not taken

O(1) polynomials are always sufficient (#polynomials) too expensive compared to implementations (#symbols)

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LITTLE IS KNOWN FOR PCR SPACE

Lower bounds for wide CNFs [Alekhnovich et al. 2002] Length-Space trade-offs [Huynh, Nordstr¨

  • m, 2012]

Lower bounds for narrow CNFs [FLNTZ 2012]

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LOWER BOUND FOR

NARROW CNFS

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BIT-PIGEONHOLE PRINCIPLE

Fix n = 2m: there is no injective function F : [n + 1] → {0, 1}m.

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BIT-PIGEONHOLE PRINCIPLE

Fix n = 2m: there is no injective function F : [n + 1] → {0, 1}m. For each: two pigeons a and b hole s ∈ {0, 1}m (fa,1 = s1) ∨ · · · ∨ (fa,m = sm)

  • F(a)=s
  • (fb,1 = s1) ∨ · · · ∨ (fb,m = sm)
  • F(b)=s
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EXAMPLE

F(1) = 1101 or F(3) = 1101 translates to (¬f1,1 ∨ ¬f1,2 ∨ f1,3 ∨ ¬f1,4)

  • (¬f3,1 ∨ ¬f3,2 ∨ f3,3 ∨ ¬f3,4)
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THEOREM

Any PCR refutation of the “Bit” pigeohole principle has a configuration with at least n/8 monomials.

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Proof Sketch

. . .    . . .       . . .       . . .       . . .    1 = 0

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Proof Sketch

. . .    . . .       . . .       . . .       . . .       . . .       . . .       . . .       . . .    1 = 0

1 a parallel sequence of (. . .) such that (. . .) [. . .] ;

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Proof Sketch

. . .    . . .       . . .       . . .       . . .       . . .       . . .       . . .       . . .    1 = 0

1 a parallel sequence of (. . .) such that (. . .) [. . .] ; 2 size of (. . .) if at most twice the size of [. . .]; (assuming monomial space ≤ n/8: )

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Proof Sketch

. . .    . . .       . . .       . . .       . . .       . . .       . . .       . . .       . . .    1 = 0

1 a parallel sequence of (. . .) such that (. . .) [. . .] ; 2 size of (. . .) if at most twice the size of [. . .]; (assuming monomial space ≤ n/8: ) 3 (. . .) of size ≤ n/4 are all satisfiable;

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Proof Sketch

. . .    . . .       . . .       . . .       . . .       . . .       . . .       . . .       . . .    1 = 0

1 a parallel sequence of (. . .) such that (. . .) [. . .] ; 2 size of (. . .) if at most twice the size of [. . .]; (assuming monomial space ≤ n/8: ) 3 (. . .) of size ≤ n/4 are all satisfiable; 4 contradiction since (. . .) [1 = 0]

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SPECIAL CONFIGURATIONS

(2-CNFs where no pigeon is mentioned twice)

     ¬f1,3 ∨ f4,2 f5,1 ∨ f7,3 . . . ¬f6,5 ∨ ¬f2,3      A satisfying assignment: satisfies the 2-CNFs no collision on the occurring pigeons

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Observation

Any special configuration with at most n/4 clauses is satisfiable.

Proof.

     ¬f1,3 ∨ f4,2 f5,1 ∨ f7,3 . . . ¬f6,5 ∨ ¬f2,3     

1

at most n/2 pigeons;

2

at least n/2 + 1 free holes per pigeon;

3

at least one free hole per satisfied literal.

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Input: (M0, M1, . . . , Ml)

with |Mi| ≤ n/8

Output: (S0, S1, . . . , Sl)

such that |Si| ≤ 2|Mi| and Si implies Mi

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Input: (M0, M1, . . . , Ml)

with |Mi| ≤ n/8

Output: (S0, S1, . . . , Sl)

such that |Si| ≤ 2|Mi| and Si implies Mi

S0 := ∅ [Initial configuration] Si+1 := Si [Inference] Si+1 := Si [Logical axioms]

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Input: (M0, M1, . . . , Ml)

with |Mi| ≤ n/8

Output: (S0, S1, . . . , Sl)

such that |Si| ≤ 2|Mi| and Si implies Mi

S0 := ∅ [Initial configuration] Si+1 := Si [Inference] Si+1 := Si [Logical axioms] [Download of F(a) = s ∨ F(b) = s] Si+1 := Si [a, b ∈ Si] Si+1 := Si ∪ {fa,1 ∨ fb,1} [a, b ∈ Si] Si+1 := Si ∪ {fa,1 ∨ fc,1} for some c ∈ Si [a ∈ Si, b ∈ Si]

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[Erasure step] is the hard case How may clauses in 2-CNF influence the value of a monomial? ≤ 2 space complexity preserved; > 2 weak influence, most clauses can be removed.

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PROOF RECAP

Assuming monomial space ≤ n/8: a corresponding sequence of small 2-CNFs; all such small 2-CNFs are “satisfiable”; last memory configuration is satisfiable. (contradiction)

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WRAPPING UP

PCR is a candidate model of future SAT solvers; we need to study space to discover if it is the case; we have sketched a space lower bounds for 2 log n-CNFs.

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THINGS I WANT YOU TO WORK ON

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THINGS I WANT YOU TO WORK ON

A linear PCR space lower bounds for (random) 3-CNFs space bounds for other proof systems (e.g. cutting planes) trading space for time

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THINGS I WANT YOU TO WORK ON

A linear PCR space lower bounds for (random) 3-CNFs space bounds for other proof systems (e.g. cutting planes) trading space for time theoretical space bounds vs memory usage improve PCR implementations

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Thank you

  • Y. Filmus

J.Nordstr¨

  • m
  • N. Taphen
  • N. Zewi