Space complexity of cutting planes refutations . . . . . Nicola - PowerPoint PPT Presentation

. . Space complexity of cutting planes refutations . . . . . Nicola Galesi, Pavel Pudl´ ak, Neil Thapen Nicola Galesi Sapienza - University of Rome June 19, 2015 () June 19, 2015 1 / 32

Cutting planes proofs A refutational system for CNFs as linear inequalities ∨ ∨ ∑ ∑ x i ∨ ¬ x i �− → x i + ( 1 − x i ) ≥ 1 . i ∈ P i ∈ N i ∈ P i ∈ N Rules : Axioms, Linear Combination, Cut Rule x ≥ 0, − x ≥ − 1 ∑ λ 1 ∑ λ k ∑ s λ i x i ≥ t i x i ≥ t 1 · · · i x i ≥ t k ∑ λ i x i ≥ ⌈ t / s ⌉ and ∑ (∑ ) j s j λ j x i ≥ ∑ j s j t j i where s 1 , . . . , s k and s must be strictly positive integers, and the linear combination rule can take any number of premises. () June 19, 2015 2 / 32

Refuting CNFs in CP An example of CP derivation x + y ≥ 1 x + ( 1 − y ) ≥ 1 ( 1 − x ) + y ≥ 1 ( 1 − x ) + ( 1 − y ) ≥ 1 2 x ≥ 1 − 2 x ≥ − 1 x ≥ 1 − x ≥ 0 0 ≥ 1 () June 19, 2015 3 / 32

Memory configurations A memory configuration M is a set of linear inequalities. A CP derivation of I from F is given by a sequence M 0 , . . . , M ℓ of memory configurations The sequence must satisfy that : M 0 is empty, that I ∈ M ℓ , and that for each i < ℓ , M i + 1 is obtained from M i in one of three ways: Axiom download : M i + 1 = M i ∪ { J } for some J ∈ F Inference : M i + 1 = M i ∪ { J } where J follows from M i by an inference rule, or is a Boolean axiom Erasure : M i + 1 ⊂ M i . A CP refutation of F is a CP derivation of 0 ≥ 1 from F . () June 19, 2015 4 / 32

Space measures Three measures of the space taken by a memory configuration M . The inequality space is the number of inequalities in M . The variable space is the sum, over all inequalities J in M , of the number of distinct variables appearing in J with a non-zero coefficient. the total space , the sum, over all inequalities J in M , of the length in binary of all non-zero coefficients in J and of the constant term of J (ignoring signs). For each measure, the corresponding space of a refutation Π is the maximum space of any configuration M i in Π . The corresponding space needed to refute a set of inequalities F is the minimum space of any refutation of F . () June 19, 2015 5 / 32

Space Complexity for Resolution A memory configuration M contains clauses instead of linear inequalities. The clause space is the number of clauses in M . The total space is the overall number of variables (with repetitions) appearing in M Many works studying clause space for Resolution. n number of vars. several Ω( n ) lower bounds (PHP , random 3CNFs,.....) [ET99,ABRW00,....] straightforward n + 1 upper bound. Few works exploring total space for Resolution. Ω( n 2 ) lower bounds (PHP [ABRW00], random k-CNFs [BGT14, BBGHMW15])) straightforward O ( n 2 ) upper bound. () June 19, 2015 6 / 32

Res vs CP Focus on proof size to compare proof strength of CP and Res. CP efficiently simulates Resolution (see example). CP is exponentially stronger than Resolution: PHP ha poly size CP proofs, but requires exp size proofs in Resolution () June 19, 2015 7 / 32

Almost no result for space in CP G¨ o¨ os and Pitassi [GP14] give a family of CNFs of size m which cannot simultaneously be refuted with small inequality space and small length — the space s and length ℓ of every CP refutation must satisfy s log ℓ ≥ m 1 / 4 − o ( 1 ) . No explicit lower bound, no explicit upper bound was known so far. () June 19, 2015 8 / 32

Contradictions The complete tree contradiction CT n is a CNF in n variables x 0 , . . . , x n − 1 , with 2 n clauses. For each assignment α , it contains the clause ∨ ∨ x i ∨ ¬ x i i ∈ Z i ∈ A where A = { i : α ( x i ) = 1 } and Z = { i : α ( x i ) = 0 } . This clause is falsified by α and by no other assignment. The pigeonhole principle PHP n x ij + x i ′ j ≤ 1, i ̸ = i ′ < n + 1 , j < n . (injectivity) ∑ j < n x ij ≥ 1, i < n + 1. (totality) () June 19, 2015 9 / 32

Results for cutting planes . Theorem . . . CT n has a CP refutation with inequality space 5. . . . . . Observations In Res and PCR , CT n requires (clause and monomial space) Ω( n ) proof uses coefficients of value O ( 2 n ) () June 19, 2015 10 / 32

Consequences . . . Any set of linear unsatisfiable linear inequalities (in particular 1 UNSAT CNFs) has proof of inequality space 5. . . . O ( n 2 ) total space is sufficient to refute any unsatisfiable set of 2 linear inequalities 1 . . . any UNSAT set L of linear inequalities over n variables, with max 3 coefficient M , can refuted using coefficients of value bounded by max { L , 2 n } . . . any UNSAT set L of linear inequalities over n variables can be 4 refuted in variable space O ( n ) 1 Notice that, restricted to CNFs, the upper bound follows from the O ( n 2 ) upper bound for total space in resolution () June 19, 2015 11 / 32

Consequences . Proposition . . . Let F be an unsatisfiable CNF. The minimal width of refuting F in resolution is at most the variable space of refuting F in CP. . . . . . Use Res width lower bounds to get optimality for variable space. . Theorem . . . With high probability the variable space of refuting a random k-CNF in CP is Θ( n ) . . . . . . () June 19, 2015 12 / 32

Cutting planes with bounded coefficients CP k : coefficients bounded in absolute value by k . . Theorem (CCT) . . . CP 2 is exponentially stronger than Res. . . . . . Proof PHP n has poly size CP 2 proofs. CP 2 p -simulates Res Known proofs of PHP n either use constant space and linear coefficients, or O ( log n ) spaceandconstantcoefficients () June 19, 2015 13 / 32

Results for cutting planes with bounded coefficients . Theorem . . . PHP n has polynomial size CP 2 refutations with inequality space 5. . . . . . . Theorem . . . For any constant k ∈ N , the complete tree contradiction CT n requires inequality space Ω( log log log n ) to refute in CP k . . . . . . () June 19, 2015 14 / 32

Proof sketch for CT n upper bound . Theorem . . . CT n has a CP refutation with inequality space 5. . . . . . Obs : let b ≥ 1. In space 3, ∑ i ∈ S λ i x i + ∑ i ∈ T λ i ( 1 − x i ) ≥ b . . . . ∑ i ∈ S x i + ∑ i ∈ T ( 1 − x i ) ≥ 1 Proof c ≥ max { b , λ i } . Add ( c − λ i ) x i ≥ 0 to the inequality for each i ∈ S Add ( c − λ i )( 1 − x i ) ≥ 0 for each i ∈ T . Divide by c and round. () June 19, 2015 15 / 32

Proof sketch for CT n upper bound Let a < 2 n . Then ( a ) 0 , . . . , ( a ) n − 1 for the bits of the binary expansion of a , so that a = ∑ 2 i ( a ) i . I b for the clause of CT n which is falsified exactly by the assignment x i �→ ( b ) i . For a ∈ N , define the inequality T a as ∑ 2 i x i ≥ a . T a : The assignments falsifying T a are exactly those lexicographically strictly less than a . In other words, T a is equivalent to the conjunction of the inequalities I b over all b < a , () June 19, 2015 16 / 32

Proof sketch for CT n upper bound . Claim . . . For a < 2 n , in space 5 T a , I a ⊢ T a + 1 . . . . . . Then we proceed deriving T 0 , T 1 , T 2 , . . . , T 2 n − 1 and finally deriving a contradiction from T 2 n − 1 and I 2 n − 1 . () June 19, 2015 17 / 32

Proof sketch for CT n upper bound For the inductive step, fix a < 2 n . Let A = { i < n : ( a ) i = 1 } Z = { i < n : ( a ) i = 0 } . Define two inequalities ∑ L k ∑ M a : x i ≥ 1 a : x k + x i ≥ 1 , k ∈ A . i > k i ∈ Z i ∈ Z Obs if β ≥ a , then β satisfies L k a for each k ∈ A . If β > a , then β also satisfies M a . () June 19, 2015 18 / 32

Proof sketch for CT n upper bound In space at most 5: Claim 1 T a , I a ⊢ M a Claim 2 T a ⊢ L k a , for any k ∈ A . Using these two claims, we can then show Claim 3 We can derive T a + 1 from T a and I a in space 4. () June 19, 2015 19 / 32

Claim 1 I a = ∑ i ∈ Z x i + ∑ i ∈ A ( 1 − x i ) ≥ 1 . From Axioms get: ( 2 i − 1 ) x i ≥ 0 ( 2 i − 1 )( 1 − x i ) ≥ 0 . ∑ ∑ and i ∈ Z i ∈ A Sum axioms with T a , and I a and get ∑ 2 i x i ≥ 1 . 2 i ∈ Z Use Obs to get M a () June 19, 2015 20 / 32

Claim 2 Rearrangements of T a , plus axioms multiplied by the right coefficients plus Obs () June 19, 2015 21 / 32

Claim 3 Write M a in the REG1 For each k ∈ A , we use Claim 2 to write L k a in REG2, and then multiply it by 2 k , giving 2 k x k + 2 k ∑ x i ≥ 2 k . i > k i ∈ Z Repeat for each k ∈ A in turn, each time adding the result to REG1. At the end of this process, REG 1 contains the inequality ( ∑ 2 k ) ∑ 2 k x k + ∑ ∑ ∑ 2 k . x i + x i ≥ 1 + k < i k ∈ A i ∈ Z i ∈ Z k ∈ A k ∈ A Right hand side is a + 1. Use axioms multiplied by the right coefficients to get T a + 1 () June 19, 2015 22 / 32

PHP n using space 5 in CP 2 Standard proofs of the PHP n . . . from axioms x ij + x i ′ j ≤ 1 derive for all j < n 1 ∑ x ij ≤ 1 i < n + 1 . . . sum over all j < n getting: ∑ ∑ i < n + 1 x ij < n 2 j < n . . . sum for all i < n + 1 the axioms ∑ j < n x ij ≥ 1 getting: 3 ∑ ∑ i < n + 1 x ij ≥ n j < n () June 19, 2015 23 / 32