Resolution over Linear Equations: (Partial) Survey & Open Problems Iddo Tzameret Royal Holloway, University of London (Based mainly on Raz-T. 2008, Itsykson-Sokolov 2014, and ongoing joint work with Fedor Part)

Resolution over Linear Equations R(linℜ) Proof-lines are disjunction of linear equations over ring ℜ: 𝐿 1 = 𝑎 1 ∨…∨ 𝐿 𝑚 = 𝑎 𝑚 Rules Resolution: 𝐷∨𝐿=𝑎 𝐸∨𝐿′=𝑏 𝐷∨𝐸∨(𝐿− 𝐿 ′ =𝑎−𝑏) Simplification: 𝐷∨ 𝑏=𝑎 𝐷 An R(linℜ) refutation of a collection of disjunctions of linear equations K is a proof of the empty disjunction from K. Introduced by Raz, T. 2008 (over ℤ; in unary representation). See also, R(CP*) in Krajicek 1998 Weakening: 𝐷 𝐷 ∨𝐸 if b≠a Boolean Axiom: ( 𝑥 𝑖 =0)∨ (𝑥 𝑖 =1)

Example: R(linℤ) Refuting CNFs: Replace positive literals by 𝑥 𝑖 =1 and negative literals by 𝑥 𝑖 =0 Size is number of symbols with integers in unary Example: R(linℤ) Refute: ( 𝑥 1 + 𝑥 2 =3) 𝑥 1 =0 ∨ 𝑥 1 =1 𝑥 2 =0 ∨ 𝑥 2 =1 𝑥 2 =0 ∨ 𝑥 2 =1 𝑥 1 +𝑥 2 =0 ∨ 𝑥 1 =1 ∨ 𝑥 2 =1 𝑥 1 +𝑥 2 =0 ∨ 𝑥 1 +𝑥 2 =1 ∨ 𝑥 2 =1 If x2=0 𝑥 1 =0 ∨ 𝑥 1 =1 𝑥 2 =0 ∨ 𝑥 2 =1 𝑥 2 =0 ∨ 𝑥 2 =1 𝑥 1 +𝑥 2 =1 ∨ 𝑥 1 =1 ∨ 𝑥 2 =0 𝑥 1 +𝑥 2 =1 ∨ 𝑥 1 +𝑥 2 =2 ∨ 𝑥 2 =0 If x2=1 𝑥 1 +𝑥 2 =0 ∨ 𝑥 1 +𝑥 2 =1 ∨ 𝑥 1 +𝑥 2 =2 𝑥 1 +𝑥 2 =0 ∨ 𝑥 1 +𝑥 2 =1 ∨ 𝑥 1 +𝑥 2 =2 ∨ 0=1 ( 𝑥 1 + 𝑥 2 =3) ⎕

Natural (“minimal”) extension of resolution that can “count”. Motivation Natural (“minimal”) extension of resolution that can “count”. First step towards Frege+Counting Gates lower bounds: R(lin 𝔽 2 ): “weakest” subsystem of AC0[2]-Frege for which we don’t know lower bounds.

Some Upper Bounds R(linℤ) ⊢* PHP (in CNF) (Raz-T. 2008) R(linℤ) ⊢* Tseitin (mod q) (in CNF) (Raz-T. 2008) Simulations R(linℤ) simulates CP with small coefficients (Raz-T. 2008) R(linℤ) simulates R(lin 𝔽 2 ) (Itsykson-Sokolov 2014)

Lower Bounds R0(linℜ): restrict R(linℜ) to operate with Constant many distinct linear forms in a clause (excluding single variables, that can occur freely); Coefficients of variables are constants. Exponential lower bounds on R0(linℜ) via monotone interpolation: clique/coloring tautologies (Raz-T. 2008)

R(lin 𝔽 2 ) Most results from Itsykson-Sokolov 2014 Focused on tree-like refutations Over 𝔽2, so don’t need Boolean axioms

Some Upper Bounds Tree-like R(lin 𝔽 2 ) ⊢* unsatisfiable 𝐴 𝒙 = 𝒃 (Itsykson-Sokolov 2014) Note: no disjunctions in initial clauses Tree-like R(lin 𝔽 2 ) ⊢* Graph Matching Principle “no perfect matching in graphs with odd number of nodes”

Linear Decision Trees A linear decision tree for unsat CNF C is a tree with: Linear forms f on nodes; f=0 go to left child; f=1 go to right child; Clause 𝐶 𝑖 on leaf, if system of equations on its path imply ¬ 𝐶 𝑖 . 𝑥 1 +𝑥 3 +𝑥 5 +𝑥 6 =0 𝑥 1 +𝑥 3 +𝑥 5 +𝑥 6 =1 𝑥 1 +𝑥 2 =1 𝑥 1 +𝑥 3 =1 𝑥 2 +𝑥 4 =1 𝑥 2 +𝑥 4 =0 𝒕𝒉𝒆 𝒑𝒂𝒕𝒉⊨¬𝐶 1 𝒕𝒉𝒆 𝒑𝒂𝒕𝒉⊨¬𝐶 3

Linear Decision Trees Linear decision tree for unsat CNF C ≈ Tree-like R(lin 𝔽 2 ). 𝑥 1 +𝑥 3 +𝑥 5 +𝑥 6 =0 𝑥 1 +𝑥 3 +𝑥 5 +𝑥 6 =1 𝑥 1 +𝑥 2 =1 𝑥 1 +𝑥 3 =1 𝑥 2 +𝑥 4 =1 𝑥 2 +𝑥 4 =0 𝒕𝒉𝒆 𝒑𝒂𝒕𝒉⊨¬𝐶 1 𝒕𝒉𝒆 𝒑𝒂𝒕𝒉⊨¬𝐶 3

Tree-like R(lin 𝔽 2 ) ⊢* unsat 𝐴 𝒙 = 𝒃 𝐴 𝒙 = 𝒃 is written as a CNF (assume number of rows=4) Just build the linear decision tree 𝐴 1 𝒙 =0 𝐴 1 𝒙 = 𝑏 1 𝐴 2 𝒙 = 𝑏 2 Full decision tree for variables in 𝐴 1 𝒙 is proportional to CNF refpresentation of 𝐴 1 𝒙 𝐴 3 𝒙 = 𝑏 3 𝐴 4 𝒙 = 𝑏 4 𝐴 4 𝒙 =1 Full decision tree for variables in 𝐴 4 𝒙 is proportional to CNF refpresentation of 𝐴 4 𝒙 this node is not reached, by assumption on unsatisfiability of the system. so we put the dt on the parent node

Lower Bounds Exponential lower bounds on tree-like R(lin 𝔽 2 ) for the n+1 to n PHP (IS 2014) More tree-like R(lin 𝔽 2 ) lower bounds

Tree-like R(lin 𝔽 2 ) PHP Lower Bounds Use Impagliazzo-Pudlak game technique: Given an unsatisfiable CNF, Prover and Delayer play in turns: Prover: Asks Delayer the value of some linear form Delayer: Answers 0/1. Answers “you choose!”, earning 1 point. Game ends when Prover exposes a contradiction between equations accumulated to an initial clause (or if equations accumulated are unsatisfiable). Thm (IS14): If Delayer has a strategy to always earn k points, then linear decision tree is ≥ 2k. Delayer can earn at least (n-1)/2 points. And so the lower bound on decision trees is 2(n-1)/2.

Open Problems

Open Problems R(lin 𝔽 2 ) lower bounds Good candidate: PHP

Depth-3 IPS over 𝔽2 simulates tree-like R(lin 𝔽 2 ) Algebraic Approach Depth-3 IPS over 𝔽2 simulates tree-like R(lin 𝔽 2 ) Probably: Depth-4 IPS over 𝔽2 simulates dag-like R(lin 𝔽 2 ) Use Grigoriev-Razborov 2000 lower bound (cf. Kumar-Sapsharishi 2017)? Feasible Monotone Interpolation Krajicek 2017, Krajicek-Oliviera 2017 R(lin 𝔽 2 ) lower bound is reduced to a monotone circuit lower bound with oracle access

Communication Complexity Approaches Sokolov 2017 Communication complexity protocol (tree) generalized into a DAG; lower bounds attempt on this Further Related Results Garlik and Kolodziejczyky, 2017: “Some subsystems of constant-depth Frege with parity” If R(linℤ) is (weakly) automatizable then random 3CNFs with O(n1.4) clauses can be refuted in polynomial-deterministic time (T. 2014) ‘’Simulation’’ of Feige-Kim-Ofek (2006) witnesses.

Other Open Problems Random 3CNF lower bounds for tree-like R(lin 𝔽 2 ) Weak automatizability of R(lin 𝔽 2 ) implies PTIME refutation algorithm for random 3CNFs with O(n1.4) clauses? (Known for R(linℤ))

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Appendix

Some Upper Bounds Proof of m to n PHP (for any m) [DAG like proof]: Pigeon Axioms: Hole Axioms: For every pigeon i: For every hole j: Summing by pigeons (1), all variables sum up to a value from m,m+1,…,nm Summing by holes (2), all variables sum up to a value from 0,1,…,n

Tree-like R(lin 𝔽 2 ) PHP Lower Bounds Lemma: Let 𝐴 𝒙 = 𝒃 (over all variables 𝑥𝑖𝑗). Assume number of equations ≤ (𝑛−1)/2. For each pigeon 𝑖: if there is a proper solution, then there is a proper solution that satisfies pigeon 𝑖 axiom. Proof. If we have a proper solution to 𝐴 𝒙 = 𝒃 , then since number of linear equations is ≤ (𝑛−1)/2, we have enough “slack” to sufficiently modify the assignment while forcing pigeon 𝑖 to some hole.

Tree-like R(lin 𝔽 2 ) PHP Lower Bounds Lemma: Let 𝐴 𝒙 = 𝒃 (over all variables 𝑥𝑖𝑗). Assume number of equations ≤ (𝑛−1)/2. For each pigeon 𝑖: if there is a proper solution, then there is a proper solution that satisfies pigeon 𝑖 axiom. Concluding the lower bound: Prover strategy: Delayer asks value of f: If accumulated equations T properly imply f=a, answer a; If T has proper solution then T ⋃ {f=a} has proper solution Otherwise, answer “You choose!”.

Tree-like R(lin 𝔽 2 ) PHP Lower Bounds Concluding the lower bound: Prover strategy: Delayer asks value of f: If accumulated equations T properly imply f=a, answer a; If T has proper solution then T ⋃ {f=a} has proper solution Otherwise, answer “You choose!”. Delayer earns > (𝑛−1)/2 points: While Delayer points ≤ (𝑛−1)/2, we have: Consider only chosen accumulated equations T (other equations are propery implied by it). Since T has ≤ (𝑛−1)/2 equations, for every pigeon i there’s a proper solution that maps it somewhere, so no pigeon axiom is falsified. Since T has proper solution, hole axioms are not falsified.