Clifford Smyth

A dual version of Reimer's inequality and a proof of Rudich's conjecture

We prove a dual version of the celebrated inequality of D. Reimer (a.k.a. the van den Berg-Kesten conjecture). We use the dual inequality to prove a combinatorial conjecture of S. Rudich motivated by questions in cryptographic complexity. One consequence of Rudichs Conjecture is that there is an oracle relative to which one-way functions exist but one-way permutations do not. The dual inequality has another combinatorial consequence which allows R. Impagliazzo and S. Rudich to prove that if P=NP then NP/spl cap/coNP/spl sube/i.o.AvgP relative to a random oracle.

The dual bkr inequality and rudich's conjecture

Let be a set of terms over an arbitrary (but finite) number of Boolean variables. Let U( ) be the set of truth assignments that satisfy exactly one term in . Motivated by questions in computational complexity, Rudich conjectured that there exist ∊, δ > 0 such that, if is any set of terms for which U( ) contains at least a (1−∊)-fraction of all truth assignments, then there exists a term t ∈ such that at least a δ-fraction of assignments satisfy some term of sharing a variable with t [8]. We prove a stronger version: for any independent assignment of the variables (not necessarily the uniform one), if the measure of U( ) is at least 1 − ∊, there exists a t ∈ such that the measure of the set of assignments satisfying either t or some term incompatible with t (i.e., having no satisfying assignments in common with t) is at least

Revolutionaries and Spies

\Gd = 1-\Ge-\frac{4\Ge}{1-\Ge}

Restricted Stirling and Lah number matrices and their inverses

. (A key part of the proof is a correlation-like inequality on events in a finite product probability space that is in some sense dual to Reimers inequality [11], a.k.a. the BKR inequality [5], or the van den Berg–Kesten conjecture [3].)

Equilateral Sets in {\mathcal{l}}_{p}^{d}

Abstract Revolutionaries and Spies is a game, G ( G , r , s , k ) , played on a graph G between two teams: one team consists of r revolutionaries, the other consists of s spies. To start, each revolutionary chooses a vertex as its position. The spies then do the same. (Throughout the game, there is no restriction on the number of revolutionaries and spies that may be positioned on any given vertex.) The revolutionaries and spies then alternate moves with the revolutionaries going first. To move, each revolutionary simultaneously chooses to stay put on its vertex or to move to an adjacent vertex. The spies move in the same way. The goal of the revolutionaries is to place k of their team on some vertex v in such a way that the spies cannot place one of their spies at v in their next move; this is a win for the revolutionaries. If the spies can prevent this forever, they win. There is no hidden information; the positions of all revolutionaries and spies is known to both sides at all times. We will present a number of basic results as well as the result that if G ( Z 2 , r , s , 2 ) is a win for the spies, then s ≥ 6 ⌊ r 8 ⌋ . (Here allowable moves in Z 2 consist of one-step horizontal, vertical or diagonal moves.)

A Probabilistic Characterization of the Dominance Order on Partitions

Given

Reimer's Inequality on a Finite Distributive Lattice

R \subseteq \mathbb{N}

A Dual Version of Reimer's Inequality and a Proof of Rudich's Conjecture

let

Enumeration of non-crossing pairings on bit strings

{n \brace k}_R

Revolutionaries and spies on trees and unicyclic graphs

,