Robustez de dividir una junta

Decimos que una función booleana $f: \{0,1\}^n \to \{0,1\}$ es una $k$ -junta si $f$ tiene como máximo $k$ variables de influencia.

Sea $f: \{0,1\}^n \to \{0,1\}$ ser un $2k$ -junta. Denote las variables de $f$ por $x_1, x_2, \ldots, x_n$ . Arreglo

$$S_1 = \left\{ x_1, x_2, \ldots, x_{\frac{n}{2}} \right\},\quad S_2 = \left\{ x_{\frac{n}{2} + 1}, x_{\frac{n}{2} + 2}, \ldots, x_n \right\}.$$ Claramente, existe$S \in \{S_1, S_2\}$tal que$S$contiene al menos$k$de las variables influyentes de$f$.

Ahora deje $\epsilon > 0$ , y suponga que $f: \{0,1\}^n \to \{0,1\}$ es $\epsilon$ -mucho de cada $2k$ -junta (es decir, uno tiene que cambiar una fracción de al menos $\epsilon$ de los valores de $f$ para que sea un $2k$ -junta). ¿Podemos hacer una versión "robusta" de la declaración anterior? Es decir, ¿hay una constante universal $c$ , y un conjunto $S \in \{S_1, S_2\}$tal que $f$ es $\frac{\epsilon}{c}$ lejos de cada función que contiene como máximo$k$variables influyentes en$S$?

Nota: En la formulación original de la pregunta, $c$ se fijó como $2$ . El ejemplo de Neal muestra que dicho valor de $c$ no es suficiente. Sin embargo, dado que en las pruebas de propiedad generalmente no nos preocupamos demasiado por las constantes, relajé un poco la condición.

La respuesta es sí". La prueba es por contradiccion.

Por conveniencia de notación, denotemos las primeras $n/2$ variables por $x$ y las segundas $n/2$ variables por $y$ . Suponga que $f(x,y)$ es $\delta$ -cierre a una función $f_1(x,y)$ que depende solo de las coordenadas $k$ de $x$ . Denota sus coordenadas influyentes por $T_1$ . Del mismo modo, supongamos que $f(x,y)$ esδ - cerca de una función f 2 ( x , y ) que depende solo de lascoordenadas k de y . Denota sus coordenadas influyentes por T 2 . Necesitamos demostrar que f es 4 δ - cerca de 2 k -junta ˜ f ( x , y ) .$\delta$$f_2(x,y)$$k$$y$$T_2$$f$$4\delta$$2k$$\tilde f(x,y)$

Digamos que ( x 1 , y 1 ) ∼ ( x 2 , y 2 ) si x 1 y x 2 están de acuerdo en todas las coordenadas en T 1 e y 1 e y 2 están de acuerdo en todas las coordenadas en T 2 . Elegimos uniformemente al azar un representante de cada clase de equivalencia. Sea ( ˉ x , ˉ y ) el representante de la clase de ( x $(x_1,y_1) \sim (x_2,y_2)$$x_1$$x_2$$T_1$$y_1$$y_2$$T_2$$(\bar x, \bar y)$$(x,y)$. Define $\tilde f$ as follows:

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

It is obvious that $\tilde f$ is a $2k$-junta (it depends only on variables in $T_1 \cup T_2)$. We shall prove that it is at distance $4\delta$ from $f$ in expectation.

We want to prove that

$$\Pr_{\tilde f}(\Pr_{x,y}(\tilde f(x,y) \neq f(x,y))) = \Pr(f(\bar x, \bar y) \neq f(x,y)) \leq 4\delta,$$ where $x$ and $y$ are chosen uniformly at random. Consider a random vector $\tilde x$ 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