A New Kind of Tradeoffs in Propositional Proof Complexity

Alexander Razborov
2016 Journal of the ACM  
We exhibit an unusually strong tradeoff in propositional proof complexity that significantly deviates from the established pattern of almost all results of this kind. Namely, restrictions on one resource (width in our case) imply an increase in another resource (tree-like size) that is exponential not only with respect to the complexity of the original problem, but also to the whole class of all problems of the same bit size. More specifically, we show that for any parameter k = k(n) there are
more » ... nsatisfiable k-CNFs that possess refutations of width O(k), but such that any tree-like refutation of width n 1−ϵ /k must necessarily have doubly exponential size exp(n Ω(k) ). This means that there exist contradictions that allow narrow refutations, but in order to keep the size of such a refutation even within a single exponent, it must necessarily use a high degree of parallelism. Our construction and proof methods combine, in a non-trivial way, two previously known techniques: the hardness escalation method based on substitution formulas and expansion. This combination results in a hardness compression approach that strives to preserve hardness of a contradiction while significantly decreasing the number of its variables. 1 Recall that we are considering non-uniform models. ⊕ j∈Ji(A) x j ⊕ ϵ ⊕ 1 and, since J i (A) ⊆ J, it has been already decided by α. As E and E are simultaneously falsifiable by our assumption (and α falsifies y ϵ i [A]), α(y i [A]) is actually equal toε. It only remains to show that α ′ can be extended in such a way that it sets all y i [A] for i ∈ Ker ( J ) \ Ker(J) to fixed values predetermined to falsify the formula E[A] . Let A ′ be the matrix obtained from A \ J by additionally removing the column j from it. Since A\J is an (r/2, 3/2)-boundary expander, A ′ is an (r/2, 1/2)-boundary expander. Also, Ker ( J ) \Ker(J) is a set of rows of cardinality ≤ r/2, therefore ∂ A ′ (I) ̸ = ∅ for every non-empty subset I ⊆ Ker ( J ) \ Ker(J). Applying reverse induction, we can write Ker ( J ) \ Ker(J) as an ordered set: Ker ( J ) \ Ker(J) = {i 1 , . . . , i ℓ } in such a way that for every ν ∈ [ℓ] the set of columns . Now, we first extend α ′ to {x j | j ∈ (
doi:10.1145/2858790 fatcat:7r76sjqs5vehvam4fz4zbfavbu