Tight Bounds on the Average Sensitivity of k -CNF

: The average sensitivity of a Boolean function is the expectation, given a uniformly random input, of the number of input bits which when ﬂipped change the output of the function. Answering a question by O’Donnell, we show that every Boolean function represented by a k -CNF (or a k -DNF) has average sensitivity at most k . This bound is tight since the parity function on k variables has average sensitivity k .


Introduction and Results
For x ∈ {0, 1} n and i ∈ {1, . . ., n}, let x i denote x with the i-th bit flipped.Let f : {0, 1} n → {0, 1} be a Boolean function on n variables.The sensitivity of f at x, denoted by s( f , x), is the number of bits i for which f (x) = f (x i ).The average sensitivity (also known as total influence) of f , denoted by S( f ), is the expected sensitivity at a random input: The average sensitivity is one of the most studied concepts in the analysis of Boolean functions (see, e. g., [3,4,5,7]).
A literal is a Boolean variable or its negation.Let k be a nonnegative integer.A k-clause is a disjunction of at most k literals, and a k-term is a conjunction of at most k literals.A k-CNF function is a conjunction of k-clauses, and a k-DNF function is a disjunction of k-terms.
Boppana [1] proved that S( f ) ≤ 2k for every k-DNF (as well as k-CNF) function f .Recently, Traxler [9] improved this upper bound to S( f ) ≤ 1.062k.This is nearly optimal, since the parity function on k variables, which is obviously a k-CNF function, has average sensitivity k.
In this note, we close the gap by showing: This solves an open problem posed by O'Donnell in 2007 (see [6]).
2 Proof of Theorem 1.1 Our proof is a small modification of the proof of the 1.062k upper bound by Traxler [9], which is based on a clever use of the Paturi-Pudlák-Zane algorithm (PPZ algorithm, in short) for k-SAT [8].
Let f be a k-CNF function.(The k-DNF case is dual.)Note that Traxler's bound is in fact where z is the probability that f outputs 1.This upper bound is larger than k when 0.25 < z < 0.5.We introduce the distribution D f over {0, 1} n ∪ {⊥}, which is essentially identical to the distribution used in Traxler's proof.
Consider the algorithm eppz( f ) that takes f as input and tries to choose a satisfying assignment for f .The algorithm first chooses uniformly at random some permutation π on the index set {1, . . ., n} of the variables.Then, for j = 1, . . ., n, it does the following: it sets the variable x π( j) to 1 if the single-variable clause (x π( j) ) is in f and to 0 if the single-variable clause (x π( j) ) is in f .We say that in these two cases "x π( j) is forced."Otherwise x π( j) is set to 0 or 1 uniformly at random.Each time, the formula is syntactically simplified, i. e., all clauses which became satisfied are deleted.At the end, the algorithm outputs x.If the algorithm ever produces two contradictory unit clauses, then it just "gives up" and outputs "⊥." Define D f as where the probability is over all the random choices made in eppz .In what follows, we are only interested in the value of D f (x) for x ∈ f −1 (1).Note that a similar algorithm was introduced in [2] for obtaining a lower bound on the success probability of the PPZ algorithm.For x ∈ f −1 (1), let t( f , π, x, i) denote the indicator variable for whether x i is forced or not, given that π was chosen and x output. 1 Note that given that π is chosen and x is output, all of the other random choices of eppz( f ) are fixed; i. e., there is only one outcome that leads to a given π and x.

TIGHT BOUNDS ON THE AVERAGE SENSITIVITY OF k-CNF
The key observation that relates the distribution D f to the sensitivity of f is (essentially from [8]) that, for every x ∈ f −1 (1), if f (x) = f (x i ), i. e., f is sensitive at x for the i-th bit, then We include the proof of Eq. ( 1) for completeness.If 1 = f (x) = f (x i ) = 0, then there must be a clause C such that the only literal in C set to 1 by x is the literal of the i-th variable.The variable x i is forced by eppz if i appears last in π among all variable indices occurring in C.This happens with probability at least 1/k since C has at most k literals.This establishes Eq. ( 1).
In order to show S( f . This is because we may combine The proof is finished by observing This completes the proof of the theorem.