Are Short Proofs Narrow? QBF Resolution is not so Simple. Meena - - PowerPoint PPT Presentation

Are Short Proofs Narrow? QBF Resolution is not so Simple.

Meena Mahajan

The Institute of Mathematical Sciences, HBNI, Chennai.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Credits

partially supported by the EU Marie Curie IRSES grant CORCON. joint work with Olaf Beyersdorff Univ of Leeds, UK Leroy Chew Univ of Leeds, UK Anil Shukla formerly at IMSc, Chennai now at IIT Jodhpur In ACM Transactions on Computational Logic 19(1) 2018, 1:1-1:26. (preliminary version in STACS 2016)

Dagstuhl 18051 Proof Complexity Meena Mahajan

The Resolution Proof System

C: bag of clauses. ˜ a: assignment to the variables. If ˜ a satisfies Then ˜ a satisfies C = . . . A ∨ x B ∨ ¬ x . . . C′ = . . . A ∨ x B ∨ ¬ x . . . A ∨ B

Dagstuhl 18051 Proof Complexity Meena Mahajan

The Resolution Proof System

C: bag of clauses. ˜ a: assignment to the variables. If ˜ a satisfies Then ˜ a satisfies C = . . . A ∨ x B ∨ ¬ x . . . C′ = . . . A ∨ x B ∨ ¬ x . . . A ∨ B C0 ∈ SAT = ⇒ C1 ∈ SAT = ⇒ . . . = ⇒ Ct−1 ∈ SAT = ⇒ Ct ∈ SAT C0 ∈ SAT ⇐ . . . ⇐ Ci ∈ SAT ⇐ . . . ⇐ Ct ∈ SAT ⇐ ∈ Ct

Dagstuhl 18051 Proof Complexity Meena Mahajan

Extending Resolution to QBFs

QBFs: Quantified Boolean Formulas W.l.o.g., QBF in prenex CNF: Q x · F( x); F a set of clauses. Resolution is sound: If Q x · F(x) is true, and we add a clause C to F through resolution to get F ′, then Q x · F ′(x) is also true.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Extending Resolution to QBFs

QBFs: Quantified Boolean Formulas W.l.o.g., QBF in prenex CNF: Q x · F( x); F a set of clauses. Resolution is sound: If Q x · F(x) is true, and we add a clause C to F through resolution to get F ′, then Q x · F ′(x) is also true. But Resolution alone is not enough. Consider ∃x ∀u (¬ u ∨ x) (u ∨ ¬ x). Resolution can add (x ∨ ¬ x) or (u ∨ ¬ u). Useless. Universal variable u has to be handled differently. Two ways to proceed, modelling

• CDCL-based solvers
• expansion-based solvers

Dagstuhl 18051 Proof Complexity Meena Mahajan

The Evaluation Game on QBFs

QBF Q x · F( x) Two players, Red and Blue, step through quantifier prefix left-to-right. Red picks values for ∃ variables, Blue for ∀ variables. Assignment constructed: ˜ a. Red wins a run of the game if F(˜ a) true. Otherwise Blue wins.

Dagstuhl 18051 Proof Complexity Meena Mahajan

The Evaluation Game on QBFs

QBF Q x · F( x) Two players, Red and Blue, step through quantifier prefix left-to-right. Red picks values for ∃ variables, Blue for ∀ variables. Assignment constructed: ˜ a. Red wins a run of the game if F(˜ a) true. Otherwise Blue wins. example: ∃x ∀u (¬ u ∨ x) (u ∨ ¬ x).

Dagstuhl 18051 Proof Complexity Meena Mahajan

The Evaluation Game on QBFs

QBF Q x · F( x) Two players, Red and Blue, step through quantifier prefix left-to-right. Red picks values for ∃ variables, Blue for ∀ variables. Assignment constructed: ˜ a. Red wins a run of the game if F(˜ a) true. Otherwise Blue wins. example: ∃x ∀u (¬ u ∨ x) (u ∨ ¬ x). Red: x = 1, Blue: u = 1: Red wins Red: x = 1, Blue: u = 0: Blue wins Red: x = 0, Blue: u = 1: Blue wins Blue can always win: set u = x.

Dagstuhl 18051 Proof Complexity Meena Mahajan

The Evaluation Game on QBFs

QBF Q x · F( x) Two players, Red and Blue, step through quantifier prefix left-to-right. Red picks values for ∃ variables, Blue for ∀ variables. Assignment constructed: ˜ a. Red wins a run of the game if F(˜ a) true. Otherwise Blue wins. example: ∃x ∀u (¬ u ∨ x) (u ∨ ¬ x). Red: x = 1, Blue: u = 1: Red wins Red: x = 1, Blue: u = 0: Blue wins Red: x = 0, Blue: u = 1: Blue wins Blue can always win: set u = x. Q x · F( x) false if and only if Blue has a winning strategy. Use this to extend Resolution.

Dagstuhl 18051 Proof Complexity Meena Mahajan

The ∀ reduction rule

Consider this scenario: Q x · F( x) is true. So Red has a winning strategy. F( x) has a clause C in which the rightmost variable (as per order in Q x) is a universal variable u. i.e. C = A ∨ ℓ; ℓ ∈ {u, ¬ u}; all variables in A are left of u.

Dagstuhl 18051 Proof Complexity Meena Mahajan

The ∀ reduction rule

Consider this scenario: Q x · F( x) is true. So Red has a winning strategy. F( x) has a clause C in which the rightmost variable (as per order in Q x) is a universal variable u. i.e. C = A ∨ ℓ; ℓ ∈ {u, ¬ u}; all variables in A are left of u. Then by the time Blue has to fix u, Red’s strategy must ensure that sub-clause A is already satisfied. That is, Red has a winning strategy on Q x · [F( x) ∧ A]. So Q x · [F( x) ∧ A] is also true. Equivalently: If Blue has a winning strategy on Q x · [F( x) ∧ A], then Blue has a winning strategy on Q x · F( x).

Dagstuhl 18051 Proof Complexity Meena Mahajan

The proof system Q-Res

Q x · C Grow the bag of clauses C using Resolution with existential pivots: If A ∨ x and B ∨ ¬ x are in the bag, can add A ∨ B (provided not a tautology), ∀-Reduction for universal variables: If A ∨ ℓ(u) in the bag and and all variables in A are left of u, can add A, until the empty clause is added.

Dagstuhl 18051 Proof Complexity Meena Mahajan

The proof system Q-Res (cont’d)

Sound: A derivation of reveals a winning strategy for Blue.

[KleineB¨ uningKarpinskiFl¨

• gel 1995]

Complete: Use a winning strategy of Blue to decide which clauses to derive.

Suffices to eliminate variables in right-to-left order of quantification blocks (Level-ordered Q-Res)

Dagstuhl 18051 Proof Complexity Meena Mahajan

A derivation in Q-Res

∃e ∀u ∃c ∃d [(¬ e ∨ c), (¬ u ∨ c), (e ∨ d), (u ∨ d), (¬ c ∨ ¬ d)]

Dagstuhl 18051 Proof Complexity Meena Mahajan

A derivation in Q-Res

∃e ∀u ∃c ∃d [(¬ e ∨ c), (¬ u ∨ c), (e ∨ d), (u ∨ d), (¬ c ∨ ¬ d)] e u c d (¯ u ∨ c) (e ∨ d) (¯ u ∨ c) d (e ∨ d) (¯ c ∨ ¯ d) 1 u c d (¯ e ∨ c) (u ∨ d) (¯ e ∨ c) d (u ∨ d) (¯ c ∨ ¯ d) 1

Dagstuhl 18051 Proof Complexity Meena Mahajan

A derivation in Q-Res

∃e ∀u ∃c ∃d [(¬ e ∨ c), (¬ u ∨ c), (e ∨ d), (u ∨ d), (¬ c ∨ ¬ d)] e u c d (¯ u ∨ c) (e ∨ d) (¯ u ∨ c) d (e ∨ d) (¯ c ∨ ¯ d) 1 u c d (¯ e ∨ c) (u ∨ d) (¯ e ∨ c) d (u ∨ d) (¯ c ∨ ¯ d) 1 ¯ u ∨ c e ∨ ¯ c ¯ e ∨ c u ∨ ¯ c

Dagstuhl 18051 Proof Complexity Meena Mahajan

A derivation in Q-Res

∃e ∀u ∃c ∃d [(¬ e ∨ c), (¬ u ∨ c), (e ∨ d), (u ∨ d), (¬ c ∨ ¬ d)] e u c d (¯ u ∨ c) (e ∨ d) (¯ u ∨ c) d (e ∨ d) (¯ c ∨ ¯ d) 1 u c d (¯ e ∨ c) (u ∨ d) (¯ e ∨ c) d (u ∨ d) (¯ c ∨ ¯ d) 1 ¯ u ∨ c e ∨ ¯ c ¯ e ∨ c u ∨ ¯ c ¯ u ∨ e ¯ e ∨ u

Dagstuhl 18051 Proof Complexity Meena Mahajan

A derivation in Q-Res

∃e ∀u ∃c ∃d [(¬ e ∨ c), (¬ u ∨ c), (e ∨ d), (u ∨ d), (¬ c ∨ ¬ d)] e u c d (¯ u ∨ c) (e ∨ d) (¯ u ∨ c) d (e ∨ d) (¯ c ∨ ¯ d) 1 u c d (¯ e ∨ c) (u ∨ d) (¯ e ∨ c) d (u ∨ d) (¯ c ∨ ¯ d) 1 ¯ u ∨ c e ∨ ¯ c ¯ e ∨ c u ∨ ¯ c ¯ u ∨ e ¯ e ∨ u e ¯ e

Dagstuhl 18051 Proof Complexity Meena Mahajan

A derivation in Q-Res

∃e ∀u ∃c ∃d [(¬ e ∨ c), (¬ u ∨ c), (e ∨ d), (u ∨ d), (¬ c ∨ ¬ d)] e u c d (¯ u ∨ c) (e ∨ d) (¯ u ∨ c) d (e ∨ d) (¯ c ∨ ¯ d) 1 u c d (¯ e ∨ c) (u ∨ d) (¯ e ∨ c) d (u ∨ d) (¯ c ∨ ¯ d) 1 ¯ u ∨ c e ∨ ¯ c ¯ e ∨ c u ∨ ¯ c ¯ u ∨ e ¯ e ∨ u e ¯ e

• Dagstuhl 18051 Proof Complexity

Meena Mahajan

Expansion-Based Systems

∀uQ x · F(u, x) is true

• [Q

x · F(0, x)] ∧ [Q x · F(1, x)] is true

• Q

xu/0Q xu/1 ·

• F(0,

xu/0) ∧ F(1, xu/1)

• is true

Expand the initial formula judiciously, on the fly. Then use standard resolution. Expansion-based systems: ∀Exp+Res [Janota,Marques-Silva 2015], IR [Beyersdorff,Chew,Janota 2014].

Dagstuhl 18051 Proof Complexity Meena Mahajan

Expansion-Based Systems (cont’d)

τ, σ: partial assignments to universal variables. l[τ] denotes lσ, where σ is the sub-assignment of τ setting variables left of l.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Expansion-Based Systems (cont’d)

τ, σ: partial assignments to universal variables. l[τ] denotes lσ, where σ is the sub-assignment of τ setting variables left of l. ∀Exp+Res: ∀ Expansion and Resolution

For initial clause C, use any of the clauses {l[τ] | l ∈ C, l is existential} ∪ {τ(l) | l ∈ C, l is universal}. τ: assignment to all universal variables. Use Resolution.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Expansion-Based Systems (cont’d)

τ, σ: partial assignments to universal variables. l[τ] denotes lσ, where σ is the sub-assignment of τ setting variables left of l. ∀Exp+Res: ∀ Expansion and Resolution

For initial clause C, use any of the clauses {l[τ] | l ∈ C, l is existential} ∪ {τ(l) | l ∈ C, l is universal}. τ: assignment to all universal variables. Use Resolution.

IR: Instantiation, Resolution, Completion

For initial clause C, use any of the clauses {l[τ] | l ∈ C, l is existential} ∪ {τ(l) | l ∈ C, l is universal}. τ: assignment to universal variables in C. Use Resolution. Use Completion to extend annotations.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Part of a derivation in ∀Exp+Res

∃e ∀u ∃c ∃d

• (¯

e ∨ c), (¯ u ∨ c), (e ∨ d), (u ∨ d), (¯ c ∨ ¯ d)

• e

u c d (¯ u ∨ c) (e ∨ d) (¯ u ∨ c) d (e ∨ d) (¯ c ∨ ¯ d) 1 u c d (¯ e ∨ c) (u ∨ d) (¯ e ∨ c) d (u ∨ d) (¯ c ∨ ¯ d) 1 Dagstuhl 18051 Proof Complexity Meena Mahajan

Part of a derivation in ∀Exp+Res

∃e ∀u ∃c ∃d

• (¯

e ∨ c), (¯ u ∨ c), (e ∨ d), (u ∨ d), (¯ c ∨ ¯ d)

• e

u c d (¯ u ∨ c) (e ∨ d) (¯ u ∨ c) d (e ∨ d) (¯ c ∨ ¯ d) 1 u c d (¯ e ∨ c) (u ∨ d) (¯ e ∨ c) d (u ∨ d) (¯ c ∨ ¯ d) 1 cu/1 cu/1 e ∨ du/1 ¯ cu/1 ∨ ¯ du/1 Dagstuhl 18051 Proof Complexity Meena Mahajan

Part of a derivation in ∀Exp+Res

∃e ∀u ∃c ∃d

• (¯

e ∨ c), (¯ u ∨ c), (e ∨ d), (u ∨ d), (¯ c ∨ ¯ d)

• e

u c d (¯ u ∨ c) (e ∨ d) (¯ u ∨ c) d (e ∨ d) (¯ c ∨ ¯ d) 1 u c d (¯ e ∨ c) (u ∨ d) (¯ e ∨ c) d (u ∨ d) (¯ c ∨ ¯ d) 1 cu/1 cu/1 e ∨ du/1 ¯ cu/1 ∨ ¯ du/1 cu/1 e ∨ ¯ cu/1 Dagstuhl 18051 Proof Complexity Meena Mahajan

Part of a derivation in ∀Exp+Res

∃e ∀u ∃c ∃d

• (¯

e ∨ c), (¯ u ∨ c), (e ∨ d), (u ∨ d), (¯ c ∨ ¯ d)

• e

u c d (¯ u ∨ c) (e ∨ d) (¯ u ∨ c) d (e ∨ d) (¯ c ∨ ¯ d) 1 u c d (¯ e ∨ c) (u ∨ d) (¯ e ∨ c) d (u ∨ d) (¯ c ∨ ¯ d) 1 cu/1 cu/1 e ∨ du/1 ¯ cu/1 ∨ ¯ du/1 cu/1 e ∨ ¯ cu/1 e e Dagstuhl 18051 Proof Complexity Meena Mahajan

Part of a derivation in ∀Exp+Res

∃e ∀u ∃c ∃d

• (¯

e ∨ c), (¯ u ∨ c), (e ∨ d), (u ∨ d), (¯ c ∨ ¯ d)

• e

u c d (¯ u ∨ c) (e ∨ d) (¯ u ∨ c) d (e ∨ d) (¯ c ∨ ¯ d) 1 u c d (¯ e ∨ c) (u ∨ d) (¯ e ∨ c) d (u ∨ d) (¯ c ∨ ¯ d) 1 cu/1 cu/1 e ∨ du/1 ¯ cu/1 ∨ ¯ du/1 cu/1 e ∨ ¯ cu/1 e e ¯ e ∨ cu/0 ¯ e ∨ cu/0 du/0 ¯ cu/0 ∨ ¯ du/ ¯ e ∨ cu/0 ¯ cu/0 ¯ e ¯ e Dagstuhl 18051 Proof Complexity Meena Mahajan

Relative power

Q-Res ∀Exp+Res IR

Dagstuhl 18051 Proof Complexity Meena Mahajan

Relative power

Q-Res ∀Exp+Res IR Q-ResT ∀Exp+ResT IRT

Dagstuhl 18051 Proof Complexity Meena Mahajan

Relative power

Q-Res ∀Exp+Res IR Q-ResT ∀Exp+ResT IRT

Dagstuhl 18051 Proof Complexity Meena Mahajan

The Width of Resolution proofs

In Resolution, Short proofs are narrow.

(Size of proof: number of steps. Width of proof: max width of clause in proof.)

Theorem ([Ben-Sasson,Wigderson 2001])

For all unsatisfiable CNFs F: S( ResT F) ≥ 2w

• Res F
• −w(F) .

(tree-like proofs; no reusing clauses)

S( Res F) = exp

• Ω
• w
• Res F
• − w(F)

2 # of variables

• .

(dag-like proofs)

Dagstuhl 18051 Proof Complexity Meena Mahajan

The Width of Resolution proofs

In Resolution, Small-space proofs are narrow.

(Space of proof: pebbling number of underlying graph.)

Theorem ([Atserias,Dalmau 08])

For all unsatisfiable CNFs F the following relation holds: CSpace( Res F) ≥ w( Res F) − w(F) + 1. Do these relations carry over to QBF proof systems?

Dagstuhl 18051 Proof Complexity Meena Mahajan

The Size-Space-Width relations in Q-Res

∀ u1u2 . . . un ∃ e0e1 . . . en   (e0) (¬ ei−1 ∨ ui ∨ ei) for i ∈ [n] (¬ en)  

Dagstuhl 18051 Proof Complexity Meena Mahajan

The Size-Space-Width relations in Q-Res

∀ u1u2 . . . un ∃ e0e1 . . . en   (e0) (¬ ei−1 ∨ ui ∨ ei) for i ∈ [n] (¬ en)   Using Resolution, derive u1 ∨ . . . ∨ un. (n + 1 steps) Then using ∀Red, derive . (n steps) So proof of size O(n). Even tree-like. Space requirement O(1).

Dagstuhl 18051 Proof Complexity Meena Mahajan

The Size-Space-Width relations in Q-Res

∀ u1u2 . . . un ∃ e0e1 . . . en   (e0) (¬ ei−1 ∨ ui ∨ ei) for i ∈ [n] (¬ en)   Using Resolution, derive u1 ∨ . . . ∨ un. (n + 1 steps) Then using ∀Red, derive . (n steps) So proof of size O(n). Even tree-like. Space requirement O(1). However: Any proof must derive u1 ∨ . . . ∨ un. So width of any proof Ω(n).

Dagstuhl 18051 Proof Complexity Meena Mahajan

Re-defining Width For QBFs

Problem: accumulation of universal variables. Possible fix: Redefine Width∃. Count only existential variables. Now does an analogue of the short-proofs-are-narrow hold?

Dagstuhl 18051 Proof Complexity Meena Mahajan

Re-defining Width For QBFs

Problem: accumulation of universal variables. Possible fix: Redefine Width∃. Count only existential variables. Now does an analogue of the short-proofs-are-narrow hold? No! Completion Principle: short clausal encoding of Propositional version: ∃ X ∈ {0, 1}n×n (∃ all-1s row) ∧ (∃all-0s column) QBF version: ∃ X ∈ {0, 1}n×n ∀ z (z ∨ ∃ all-1s row) ∧ (¬ z ∨ ∃all-0s column)

Dagstuhl 18051 Proof Complexity Meena Mahajan

Completion Principle: Propositional

∃ X ∈ {0, 1}n×n ∃ a ∈ {0, 1}n ∃ b ∈ {0, 1}n A : (¬ a1 ∨ . . . ¬ an)

(use aux variables to get 3-clauses)

B : (¬ b1 ∨ . . . ¬ bn)

(use aux variables to get 3-clause)

Rowij : (ai ∨ Xij) i, j ∈ [n].

(ai false forces all-1s row i)

Colij : (bj ∨ ¬ Xij) i, j ∈ [n].

(bj false forces all-0s row j)

Dagstuhl 18051 Proof Complexity Meena Mahajan

Completion Principle: Propositional

∃ X ∈ {0, 1}n×n ∃ a ∈ {0, 1}n ∃ b ∈ {0, 1}n A : (¬ a1 ∨ . . . ¬ an)

(use aux variables to get 3-clauses)

B : (¬ b1 ∨ . . . ¬ bn)

(use aux variables to get 3-clause)

Rowij : (ai ∨ Xij) i, j ∈ [n].

(ai false forces all-1s row i)

Colij : (bj ∨ ¬ Xij) i, j ∈ [n].

(bj false forces all-0s row j)

Resolution proof; Tree-like; size O(n2). space O(1).

Short clauses − → A, B. A, Row∗j − → X (j) = X1j ∨ . . . ∨ Xnj. X (j), Col∗j − → bj. B, b1, . . . , bn − → .

Dagstuhl 18051 Proof Complexity Meena Mahajan

Completion Principle: Propositional

∃ X ∈ {0, 1}n×n ∃ a ∈ {0, 1}n ∃ b ∈ {0, 1}n A : (¬ a1 ∨ . . . ¬ an)

(use aux variables to get 3-clauses)

B : (¬ b1 ∨ . . . ¬ bn)

(use aux variables to get 3-clause)

Rowij : (ai ∨ Xij) i, j ∈ [n].

(ai false forces all-1s row i)

Colij : (bj ∨ ¬ Xij) i, j ∈ [n].

(bj false forces all-0s row j)

Resolution proof; Tree-like; size O(n2). space O(1).

Short clauses − → A, B. A, Row∗j − → X (j) = X1j ∨ . . . ∨ Xnj. X (j), Col∗j − → bj. B, b1, . . . , bn − → .

Hence, there is a tree-like proof of width O(log n).

Dagstuhl 18051 Proof Complexity Meena Mahajan

Completion Principle: QBF

∃ X ∈ {0, 1}n×n ∀z ∃ a ∈ {0, 1}n ∃ b ∈ {0, 1}n A : (¬ a1 ∨ . . . ¬ an)

(use aux variables to get 3-clauses)

B : (¬ b1 ∨ . . . ¬ bn)

(use aux variables to get 3-clause)

Rowij : (z ∨ ai ∨ Xij) i, j ∈ [n].

(z, ai false forces all-1s row i)

Colij : (¬ z ∨ bj ∨ ¬ Xij) i, j ∈ [n].

(¬ z, bj false forces all-0s row j)

Dagstuhl 18051 Proof Complexity Meena Mahajan

Completion Principle: QBF

∃ X ∈ {0, 1}n×n ∀z ∃ a ∈ {0, 1}n ∃ b ∈ {0, 1}n A : (¬ a1 ∨ . . . ¬ an)

(use aux variables to get 3-clauses)

B : (¬ b1 ∨ . . . ¬ bn)

(use aux variables to get 3-clause)

Rowij : (z ∨ ai ∨ Xij) i, j ∈ [n].

(z, ai false forces all-1s row i)

Colij : (¬ z ∨ bj ∨ ¬ Xij) i, j ∈ [n].

(¬ z, bj false forces all-0s row j)

As before, tree-like proof of size O(n2), with space requirement O(1).

Derive A; Derive each X (j) ∨ z; Drop the z to get X (j); Derive B; Derive each ¬ z ∨ bj; Derive ¬ z; Drop ¬ z.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Completion Principle: QBF

∃ X ∈ {0, 1}n×n ∀z ∃ a ∈ {0, 1}n ∃ b ∈ {0, 1}n A : (¬ a1 ∨ . . . ¬ an)

(use aux variables to get 3-clauses)

B : (¬ b1 ∨ . . . ¬ bn)

(use aux variables to get 3-clause)

Rowij : (z ∨ ai ∨ Xij) i, j ∈ [n].

(z, ai false forces all-1s row i)

Colij : (¬ z ∨ bj ∨ ¬ Xij) i, j ∈ [n].

(¬ z, bj false forces all-0s row j)

As before, tree-like proof of size O(n2), with space requirement O(1).

Derive A; Derive each X (j) ∨ z; Drop the z to get X (j); Derive B; Derive each ¬ z ∨ bj; Derive ¬ z; Drop ¬ z.

But, we show: Any proof must have width∃ = Ω(n).

Dagstuhl 18051 Proof Complexity Meena Mahajan

Completion Principle: Lower bound on width

Completion Principle (QBF): Every proof has large width∃. Proof Idea: Associate subsets of [n] with literals and clauses in proof. Singleton sets at Xij; empty set at z, ¬ z. Large sets (size n − 1) at ai, bj. Singletons at ¬ ai, ¬ bj. Initial clauses all have size n. Show: All resolution steps before first ∀Reduction preserve set size. First ∀-reduction: D D′. Dropping z allowed, so only X variables in D. Set size large, so many X variables.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Width∃ in non-tree-like proofs

Completion Principle proofs in Q-Res: Width∃ is Θ(n), but Number of variables also Θ(n2). This does not say anything about DAG-like proofs.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Width∃ in non-tree-like proofs

Completion Principle proofs in Q-Res: Width∃ is Θ(n), but Number of variables also Θ(n2). This does not say anything about DAG-like proofs. A different QBF destroys the size-width and space-width relations.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Width∃ in non-tree-like proofs (cont’d)

∃e1∀u1∃c1∃d1 ∃e2∀u2∃c2∃d2 . . . ∃en∀un∃cn∃dn for i ∈ [n], (¬ ei ∨ ci) (ei ∨ di) (¬ ui ∨ ci) (ui ∨ di) ¬ c1 ∨ ¬ d1 ∨ ¬ c2 ∨ ¬ d2 ∨ . . . ∨ ¬ cn ∨ ¬ dn Winning strategy for universal player: ui = ¬ ei.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Width∃ in non-tree-like proofs (cont’d)

∃e1∀u1∃c1∃d1 ∃e2∀u2∃c2∃d2 . . . ∃en∀un∃cn∃dn for i ∈ [n], (¬ ei ∨ ci) (ei ∨ di) (¬ ui ∨ ci) (ui ∨ di) ¬ c1 ∨ ¬ d1 ∨ ¬ c2 ∨ ¬ d2 ∨ . . . ∨ ¬ cn ∨ ¬ dn Winning strategy for universal player: ui = ¬ ei. (Encode last clause with additional ∃ variables as short clauses.)

Dagstuhl 18051 Proof Complexity Meena Mahajan

Width∃ in non-tree-like proofs (cont’d)

∃e1∀u1∃c1∃d1 ∃e2∀u2∃c2∃d2 . . . ∃en∀un∃cn∃dn for i ∈ [n], (¬ ei ∨ ci) (ei ∨ di) (¬ ui ∨ ci) (ui ∨ di) ¬ c1 ∨ ¬ d1 ∨ ¬ c2 ∨ ¬ d2 ∨ . . . ∨ ¬ cn ∨ ¬ dn Winning strategy for universal player: ui = ¬ ei. (Encode last clause with additional ∃ variables as short clauses.) Short proofs in Q-Res. Size nO(1). Space O(1). We show: Width∃ of any Q-Res proof Ω(n).

Dagstuhl 18051 Proof Complexity Meena Mahajan

Width∃ in non-tree-like proofs (cont’d)

∃e1∀u1∃c1∃d1 ∃e2∀u2∃c2∃d2 . . . ∃en∀un∃cn∃dn for i ∈ [n], (¬ ei ∨ ci) (ei ∨ di) (¬ ui ∨ ci) (ui ∨ di) ¬ c1 ∨ ¬ d1 ∨ ¬ c2 ∨ ¬ d2 ∨ . . . ∨ ¬ cn ∨ ¬ dn Winning strategy for universal player: ui = ¬ ei. (Encode last clause with additional ∃ variables as short clauses.) Short proofs in Q-Res. Size nO(1). Space O(1). We show: Width∃ of any Q-Res proof Ω(n). Large width requirement does not give size lower bound.

Dagstuhl 18051 Proof Complexity Meena Mahajan

What went wrong?

Small but crucial step in the propositional case:

Lemma ([Ben-Sasson,Wigderson 2001])

If a clause A can be derived from F|x=1 in width w, then the clause A ∨ ¬ x can be derived from F in width w + 1 (possibly using a weakening rule at the end).

Dagstuhl 18051 Proof Complexity Meena Mahajan

What went wrong?

Small but crucial step in the propositional case:

Lemma ([Ben-Sasson,Wigderson 2001])

If a clause A can be derived from F|x=1 in width w, then the clause A ∨ ¬ x can be derived from F in width w + 1 (possibly using a weakening rule at the end). Fails for Q-Res. Hard(X): Any n-variable formula needing Ω(n) width in Resolution. ψn = ∃X∃a ∀u ∃b [Hard(X) ∧ (¬ a) ∧ (a ∨ u ∨ ¬ b)] . ψn|b=1 derives in 2 steps, width 2. To derive ¬ b from ψn, must derive from Hard(X) and then weaken.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Expansion-based systems

Only resolution rule used. Hence Size-width, space-width relations do hold for ∀Exp+ResT, IRT.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Expansion-based systems

Only resolution rule used. Hence Size-width, space-width relations do hold for ∀Exp+ResT, IRT. They also hold for ∀Exp+Res, IR. But not useful. Number of variables itself can be exponential. (Recall: different annotations count as different variables.)

Dagstuhl 18051 Proof Complexity Meena Mahajan

Expansion-based systems: trees vs DAGs

Q-Res ∀Exp+Res IR Q-ResT ∀Exp+ResT IRT

Dagstuhl 18051 Proof Complexity Meena Mahajan

Expansion-based systems: trees vs DAGs

Q-Res ∀Exp+Res IR Q-ResT ∀Exp+ResT IRT

Dagstuhl 18051 Proof Complexity Meena Mahajan

Expansion-based systems: trees vs DAGs

Q-Res ∀Exp+Res IR Q-ResT ∀Exp+ResT IRT

Dagstuhl 18051 Proof Complexity Meena Mahajan

Expansion-based systems: trees vs DAGs

Q-Res ∀Exp+Res IR Q-ResT ∀Exp+ResT IRT

IR and ∀Exp+Res DAG proofs: Exponentially separated Tree proofs: Same power

Dagstuhl 18051 Proof Complexity Meena Mahajan

Expansion-based systems: trees vs DAGs

Q-Res ∀Exp+Res IR Q-ResT ∀Exp+ResT IRT

IR and ∀Exp+Res DAG proofs: Exponentially separated Tree proofs: Same power

Dagstuhl 18051 Proof Complexity Meena Mahajan

To summarize

Resolution based proof-systems for QBFs: Width useless in Q-Res. Width∃ useless in Q-Res. Width lower bounds give size/space lower bounds in ∀Exp+ResT, IRT. Width lower bounds not useful by themselves in ∀Exp+Res, IR.

Dagstuhl 18051 Proof Complexity Meena Mahajan

To summarize

Resolution based proof-systems for QBFs: Width useless in Q-Res. Width∃ useless in Q-Res. Width lower bounds give size/space lower bounds in ∀Exp+ResT, IRT. Width lower bounds not useful by themselves in ∀Exp+Res, IR. The picture is murky! Perhaps width is not a central parameter in QBF resolution.

Dagstuhl 18051 Proof Complexity Meena Mahajan

Open Questions

Can we show any useful width-based lower bounds for general ∀Exp+Res, IR? What about other Resolution-based QBF systems? Our width∃ lower bound crucially uses the Q-Res limitation of no universal pivots. Is there some other parameter that plays the same role (as width in propositional Resolution) in understanding size and space?

Dagstuhl 18051 Proof Complexity Meena Mahajan

Thank you

Dagstuhl 18051 Proof Complexity Meena Mahajan