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Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao

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## (Gil Kalai) The entropy/influence conjecture

16 August, 2007 in guest blog, math.CA, math.PR, question | Tags: boolean functions, entropy, Fourier transform, Gil Kalai, influence | by Gil Kalai | 18 comments

[This post is authored by Gil Kalai, who has kindly “guest blogged” this week’s “open problem of the week”. - T.]

The entropy-influence conjecture seeks to relate two somewhat different measures as to how a boolean function has concentrated Fourier coefficients, namely the total influence and the entropy.

We begin by defining the total influence. Let $\{-1,+1\}^n$ be the discrete cube, i.e. the set of $\pm 1$ vectors $(x_1,\ldots,x_n)$ of length n. A boolean function is any function $f: \{-1,+1\}^n \to \{-1,+1\}$ from the discrete cube to {-1,+1}. One can think of such functions as “voting methods”, which take the preferences of n voters (+1 for yes, -1 for no) as input and return a yes/no verdict as output. For instance, if n is odd, the “majority vote” function $\hbox{sgn}(x_1+\ldots+x_n)$ returns +1 if there are more +1 variables than -1, or -1 otherwise, whereas if $1 \leq k \leq n$, the “$k^{th}$ dictator” function returns the value $x_k$ of the $k^{th}$ variable.

We give the cube $\{-1,+1\}^n$ the uniform probability measure $\mu$ (thus we assume that the n voters vote randomly and independently). Given any boolean function f and any variable $1 \leq k \leq n$, define the influence $I_k(f)$ of the $k^{th}$ variable to be the quantity

$I_k(f) := \mu \{ x \in \{-1,+1\}^n: f(\sigma_k(x)) \neq f(x) \}$

where $\sigma_k(x)$ is the element of the cube formed by flipping the sign of the $k^{th}$ variable. Informally, $I_k(f)$ measures the probability that the $k^{th}$ voter could actually determine the outcome of an election; it is sometimes referred to as the Banzhaf power index. The total influence I(f) of f (also known as the average sensitivity and the edge-boundary density) is then defined as

$I(f) := \sum_{k=1}^n I_k(f).$

Thus for instance a dictator function has total influence 1, whereas majority vote has total influence comparable to $\sqrt{n}$. The influence can range between 0 (for constant functions +1, -1) and n (for the parity function $x_1 \ldots x_k$ or its negation). If f has mean zero (i.e. it is equal to +1 half of the time), then the edge-isoperimetric inequality asserts that $I(f) \geq 1$ (with equality if and only if there is a dictatorship), whilst the Kahn-Kalai-Linial (KKL) theorem asserts that $I_k(f) \gg \frac{\log n}{n}$ for some k. There is a result of Friedgut that if $I(f)$ is bounded by A (say) and $\varepsilon > 0$, then f is within a distance $\varepsilon$ (in $L^1$ norm) of another boolean function g which only depends on $O_{A,\varepsilon}(1)$ of the variables (such functions are known as juntas).

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## (Gil Kalai) The weak epsilon-net problem

22 April, 2007 in guest blog, math.MG, question | Tags: convex geometry, entropy, epsilon-net, Gil Kalai | by Gil Kalai | 13 comments

[This post is authored by Gil Kalai, who has kindly “guest blogged” this week’s “open problem of the week”. - T.]

This is a problem in discrete and convex geometry. It seeks to quantify the intuitively obvious fact that large convex bodies are so “fat” that they cannot avoid “detection” by a small number of observation points. More precisely, we fix a dimension d and make the following definition (introduced by Haussler and Welzl):

• Definition: Let $X \subset {\Bbb R}^d$ be a finite set of points, and let $0 < \epsilon < 1$. We say that a finite set $Y \subset {\Bbb R}^d$ is a weak $\epsilon$-net for X (with respect to convex bodies) if, whenever B is a convex body which is large in the sense that $|B \cap X| > \epsilon |X|$, then B contains at least one point of Y. (If Y is contained in X, we say that Y is a strong $\epsilon$-net for X with respect to convex bodies.)

For example, in one dimension, if $X = \{1,\ldots,N\}$, and $Y = \{ \epsilon N, 2 \epsilon N, \ldots, k \epsilon N \}$ where k is the integer part of $1/\epsilon$, then Y is a weak $\epsilon$-net for X with respect to convex bodies. Thus we see that even when the original set X is very large, one can create a $\epsilon$-net of size as small as $O(1/\epsilon)$. Strong $\epsilon$-nets are of importance in computational learning theory, and are fairly well understood via Vapnik-Chervonenkis (or VC) theory; however, the theory of weak $\epsilon$-nets is still not completely satisfactory.

One can ask what happens in higher dimensions, for instance when X is a discrete cube $X = \{1,\ldots,N\}^d$. It is not too hard to cook up $\epsilon$-nets of size $O_d(1/\epsilon^d)$ (by using tools such as Minkowski’s theorem), but in fact one can create $\epsilon$-nets of size as small as $O( \frac{1}{\epsilon} \log \frac{1}{\epsilon} )$ simply by taking a random subset of X of this cardinality and observing that “up to errors of $\epsilon$“, the total number of essentially different ways a convex body can meet X grows at most polynomially in $1/\epsilon$. (This is a very typical application of the probabilistic method.) On the other hand, since X can contain roughly $1/\epsilon$ disjoint convex bodies, each of which contains at least $\epsilon$ of the points in X, we see that no $\epsilon$-net can have size much smaller than $1/\epsilon$.

Now consider the situation in which X is now an arbitrary finite set, rather than a discrete cube. More precisely, let $f(\epsilon,d)$ be the least number such that every finite set X possesses at least one weak $\epsilon$-net for X with respect to convex bodies of cardinality at most $f(\epsilon,d)$. (One can also replace the finite set X with an arbitrary probability measure; the two formulations are equivalent.) Informally, f is the least number of “guards” one needs to place to prevent a convex body from covering more than $\epsilon$ of any given territory.

• Problem 1: For fixed d, what is the correct rate of growth of f as $\epsilon \to 0$?

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