Robustesse de la division d'une junte

On dit qu'une fonction booléenne $f: \{0,1\}^n \to \{0,1\}$ est une $k$ junta si $f$ a au plus $k$ variables d'influence.

Soit $f: \{0,1\}^n \to \{0,1\}$ une junte de $2k$ . Notons les variables de $f$ par $x_1, x_2, \ldots, x_n$ . Fix

$$S_1 = \left\{ x_1, x_2, \ldots, x_{\frac{n}{2}} \right\},\quad S_2 = \left\{ x_{\frac{n}{2} + 1}, x_{\frac{n}{2} + 2}, \ldots, x_n \right\}.$$ Clairement, il existe$S \in \{S_1, S_2\}$tel que$S$contient au moins$k$des variables d'influence de$f$.

Soit maintenant $\epsilon > 0$ , et supposons que $f: \{0,1\}^n \to \{0,1\}$ est $\epsilon$ loin de chaque $2k$ junta (c'est-à-dire qu'il faut changer une fraction d'au moins $\epsilon$ des valeurs de $f$ pour en faire une junte à $2k$ ). Pouvons-nous faire une version "robuste" de la déclaration ci-dessus? Autrement dit, existe-t-il une constante universelle $c$ et un ensemble $S \in \{S_1, S_2\}$tel que $f$ est $\frac{\epsilon}{c}$ loin de chaque fonction qui contient au plus$k$variables d'influence dans$S$?

Remarque: Dans la formulation originale de la question, $c$ était fixé à $2$ . L'exemple de Neal montre qu'une telle valeur de $c$ ne suffit pas. Cependant, comme dans les tests de propriétés, nous ne nous préoccupons généralement pas trop des constantes, j'ai un peu assoupli la condition.

La réponse est oui". La preuve est par contradiction.

Pour plus de commodité, notons les premières n / 2 variables par x et les secondes n / 2 variables par y . Supposons que f ( x , y ) soit δ- proche d'une fonction f 1 ( x , y ) qui ne dépend que des coordonnées k de x . Désignons ses coordonnées influentes par T 1 . De même, supposons que f ( x , y ) soit$n/2$$x$$n/2$$y$$f(x,y)$$\delta$$f_1(x,y)$$k$$x$$T_1$$f(x,y)$$\delta$-close to a function $f_2(x,y)$ which depends only on $k$ coordinates of $y$. Denote its influential coordinates by $T_2$. We need to prove that $f$ is $4\delta$- close to a $2k$-junta $\tilde f(x,y)$.

Disons que ( x 1 , y 1 ) ∼ ( x 2 , y 2 ) si x 1 et x 2 s'accordent sur toutes les coordonnées en T 1 et y 1 et y 2 s'accordent sur toutes les coordonnées en T 2 . Nous choisissons uniformément au hasard un représentant de chaque classe d'équivalence. Soit ( ˉ x , ˉ y ) le représentant de la classe de ( x $(x_1,y_1) \sim (x_2,y_2)$$x_1$$x_2$$T_1$$y_1$$y_2$$T_2$$(\bar x, \bar y)$y ) . Définissez ˜ f comme suit: ˜ f ( x , y ) = f ( ˉ x , ˉ y ) $(x,y)$$\tilde f$

$$\tilde f(x,y) = f(\bar x, \bar y).$$

Il est évident que ˜ f est une junte de 2 k (cela ne dépend que des variables de T 1 ∪ T 2 ) . Nous prouverons qu'il est à la distance 4 δ de f dans l'attente.$\tilde f$$2k$$T_1 \cup T_2)$$4\delta$$f$

Nous voulons prouver que Pr ˜ f ( Pr x , y ( ˜ f ( x , y ) ≠ f ( x , y ) ) ) = Pr ( f ( ˉ x , ˉ y ) ≠ f ( x , y ) ) ≤ 4 δ , où x et y sont choisis uniformément au hasard. Considérons un vecteur aléatoire

$$\Pr_{\tilde f}(\Pr_{x,y}(\tilde f(x,y) \neq f(x,y))) = \Pr(f(\bar x, \bar y) \neq f(x,y)) \leq 4\delta,$$ $x$$y$$\tilde x$ obtained from $x$ by keeping all bits in $T_1$ and randomly flipping all bits not in $T_1$, and a vector $\tilde y$ defined similarly. Note that $$\Pr(\tilde f(x,y) \neq f(x,y)) = \Pr(f(\bar x, \bar y) \neq f(x,y))= \Pr(f(\tilde x, \tilde y) \neq f(x,y)).$$

We have,

$$\Pr(f(x,y) \neq f(\tilde x, y)) \leq \Pr(f(x,y) \neq f_1(x, y)) + \Pr(f_1(x,y) \neq f_1(\tilde x, y)) + \Pr(f_1(\tilde x,y) \neq f(\tilde x, y)) \leq \delta + 0 + \delta = 2\delta.$$

Similarly, $\Pr(f(\tilde x,y) \neq f(\tilde x, \tilde y)) \leq 2\delta$. We have

$$\Pr(f(\bar x, \bar y) \neq f(x,y)) \leq 4\delta.$$ QED

It easy to “derandomize” this proof. For every $(x,y)$, let $\tilde f(x,y) = 1$ if $f(x,y) = 1$ for most $(x',y')$ in the equivalence class of $(x,y)$, and $\tilde f(x,y) = 0$, otherwise.

The smallest $c$ that the bound holds for is $c = \frac{1}{\sqrt 2 - 1} \approx 2.41$.

Lemmas 1 and 2 show that the bound holds for this $c$. Lemma 3 shows that this bound is tight.

(In comparison, Juri's elegant probabilistic argument gives $c=4$.)

Let $c=\frac{1}{\sqrt 2 - 1}$. Lemma 1 gives the upper bound for $k=0$.

Lemma 1: If $f$ is $\epsilon_g$-near a function $g$ that has no influencing variables in $S_2$, and $f$ is $\epsilon_h$-near a function $h$ that has no influencing variables in $S_1$, then $f$ is $\epsilon$-near a constant function, where $\epsilon \le \frac{(\epsilon_g+\epsilon_h)/2}{c}$.

Proof. Let $\epsilon$ be the distance from $f$ to a constant function. Suppose for contradiction that $\epsilon$ does not satisfy the claimed inequality. Let $y=(x_1,x_2,\ldots,x_{n/2})$ and $z=(x_{n/2}+1,\ldots,x_n)$ and write $f$, $g$, and $h$ as $f(y,z)$, $g(y,z)$ and $h(y,z)$, so $g(y,z)$ is independent of $z$ and $h(y,z)$ is independent of $y$.

(I find it helpful to visualize $f$ as the edge-labeling of the complete bipartite graph with vertex sets $\{y\}$ and $\{z\}$, where $g$ gives a vertex-labeling of $\{y\}$, and $h$ gives a vertex-labeling of $\{z\}$.)

Let $g_0$ be the fraction of pairs $(y,z)$ such that $g(y,z) = 0$. Let $g_1=1-g_0$ be the fraction of pairs such that $g(y,z) = 1$. Likewise let $h_0$ be the fraction of pairs such that $h(y,z) = 0$, and let $h_1$ be the fraction of pairs such that $h(y,z) = 1$.

Without loss of generality, assume that, for any pair such that $g(y,z) = h(y,z)$, it also holds that $f(y,z) = g(y,z) = h(y,z)$. (Otherwise, toggling the value of $f(y,z)$ allows us to decrease both $\epsilon_g$ and $\epsilon_h$ by $1/2^n$, while decreasing the $\epsilon$ by at most $1/2^n$, so the resulting function is still a counter-example.) Say any such pair is ``in agreement''.

The distance from $f$ to $g$ plus the distance from $f$ to $h$ is the fraction of $(x,y)$ pairs that are not in agreement. That is, $\epsilon_g + \epsilon_h = g_0 h_1 + g_1 h_0$.

The distance from $f$ to the all-zero function is at most $1 - g_0 h_0$.

The distance from $f$ to the all-ones function is at most $1-g_1 h_1$.

Further, the distance from $f$ to the nearest constant function is at most $1/2$.

Thus, the ratio $\epsilon/(\epsilon_g+\epsilon_h)$ is at most

$$\frac{\min(1/2, 1-g_0 h_0, 1-g_1 h_1)}{g_0 h_1 + g_1 h_0},$$ where $g_0,h_0 \in [0,1]$ and $g_1 = 1-g_0$ and $h_1=1-h_0$.

By calculation, this ratio is at most $\frac{1}{2(\sqrt 2 - 1)} = c/2$. QED

Lemma 2 extends Lemma 1 to general $k$ by arguing pointwise, over every possible setting of the $2k$ influencing variables. Recall that $c=\frac{1}{\sqrt 2 - 1}$.

Lemma 2: Fix any $k$. If $f$ is $\epsilon_g$-near a function $g$ that has $k$ influencing variables in $S_2$, and $f$ is $\epsilon_h$-near a function $h$ that has $k$ influencing variables in $S_1$, then $f$ is $\epsilon$-near a function $\hat f$ that has at most $2k$ influencing variables, where $\epsilon \le \frac{(\epsilon_g+\epsilon_h)/2}{c}$.

Proof. Express $f$ as $f(a,y,b,z)$ where $(a,y)$ contains the variables in $S_1$ with $a$ containing those that influence $h$, while $(b,z)$ contains the variables in $S_2$ with $b$ containing those influencing $g$. So $g(a,y,b,z)$ is independent of $z$, and $h(a,y,b,z)$ is independent of $y$.

For each fixed value of $a$ and $b$, define $F_{ab}(y,z) = f(a,y,b,z)$, and define $G_{ab}$ and $H_{ab}$ similarly from $g$ and $h$ respectively. Let $\epsilon^g_{ab}$ be the distance from $F_{ab}$ to $G_{ab}$ (restricted to $(y,z)$ pairs). Likewise let $\epsilon^h_{ab}$ be the distance from $F_{ab}$ to $H_{ab}$.

By Lemma 1, there exists a constant $c_{ab}$ such that the distance (call it $\epsilon_{ab}$) from $F_{ab}$ to the constant function $c_{ab}$ is at most $(\epsilon^h_{ab} + \epsilon^g_{ab})/(2c)$. Define $\hat f(a,y,b,z) = c_{ab}$.

Clearly $\hat f$ depends only on $a$ and $b$ (and thus at most $k$ variables).

Let $\epsilon_{\hat f}$ be the average, over the $(a,b)$ pairs, of the $\epsilon_{ab}$'s, so that the distance from $f$ to $\hat f$ is $\epsilon_{\hat f}$.

Likewise, the distances from $f$ to $g$ and from $f$ to $h$ (that is, $\epsilon_g$ and $\epsilon_h)$ are the averages, over the $(a,b)$ pairs, of, respectively, $\epsilon^g_{ab}$ and $\epsilon^h_{ab}$.

Since $\epsilon_{ab} \le (\epsilon^h_{ab} + \epsilon^g_{ab})/(2c)$ for all $a, b$, it follows that $\epsilon_{\hat f} \le (\epsilon_g + \epsilon_h)/(2c)$. QED

Lemma 3 shows that the constant $c$ above is the best you can hope for (even for $k=0$ and $\epsilon=0.5$).

Lemma 3: There exists $f$ such that $f$ is $(0.5/c)$-near two functions $g$ and $h$, where $g$ has no influencing variables in $S_2$ and $h$ has no influencing variables in $S_1$, and $f$ is $0.5$-far from every constant function.

Proof. Let $y$ and $z$ be $x$ restricted to, respectively, $S_1$ and $S_2$. That is, $y=(x_1,\ldots,x_{n/2})$ and $z=(x_{n/2+1},\ldots,x_n)$.

Identify each possible $y$ with a unique element of $[N]$, where $N=2^{n/2}$. Likewise, identify each possible $z$ with a unique element of $[N]$. Thus, we think of $f$ as a function from $[N]\times[N]$ to $\{0,1\}$.

Define $f(y,z)$ to be 1 iff $\max(y,z) \ge \frac{1}{\sqrt 2}N$.

By calculation, the fraction of $f$'s values that are zero is $(\frac{1}{\sqrt 2})^2 = \frac{1}{2}$, so both constant functions have distance $\frac{1}{2}$ to $f$.

Define $g(y,z)$ to be 1 iff $y\ge \frac{1}{\sqrt 2}N$. Then $g$ has no influencing variables in $S_2$. The distance from $f$ to $g$ is the fraction of pairs $(y,z)$ such that $y<\frac{1}{\sqrt 2}N$ and $z\ge \frac{1}{\sqrt 2}N$. By calculation, this is at most $\frac{1}{\sqrt 2}(1-\frac{1}{\sqrt2}) = 0.5/c$

Similarly, the distance from $f$ to $h$, where $h(y,z)=1$ iff $z\ge \frac{1}{\sqrt 2}N$, is at most $0.5/c$.

QED