Kuatnya pemisahan junta

Kita mengatakan bahwa fungsi Boolean adalah -junta jika memiliki paling banyak mempengaruhi variabel.$f: \{0,1\}^n \to \{0,1\}$$k$$f$$k$

Biarkan menjadi -junta. Nyatakan variabel dengan . Perbaiki Jelas, ada sedemikian rupa sehingga mengandung setidaknya dari variabel yang mempengaruhi .$f: \{0,1\}^n \to \{0,1\}$$2k$$f$$x_1, x_2, \ldots, x_n$

$$S_1 = \left\{ x_1, x_2, \ldots, x_{\frac{n}{2}} \right\},\quad S_2 = \left\{ x_{\frac{n}{2} + 1}, x_{\frac{n}{2} + 2}, \ldots, x_n \right\}.$$ $S \in \{S_1, S_2\}$$S$$k$$f$

Sekarang mari , dan anggap bahwa adalah -far dari setiap -junta (yaitu, kita harus mengubah sebagian kecil dari setidaknya dari nilai-nilai untuk membuatnya menjadi -junta). Bisakah kita membuat versi "kuat" dari pernyataan di atas? Yaitu, adakah konstanta universal , dan himpunan sedemikian rupa sehingga adalah -far dari setiap fungsi yang paling banyak berisi variabel yang mempengaruhi di ?$\epsilon > 0$$f: \{0,1\}^n \to \{0,1\}$$\epsilon$$2k$$\epsilon$$f$$2k$$c$$S \in \{S_1, S_2\}$$f$$\frac{\epsilon}{c}$$k$$S$

Catatan: Dalam rumusan asli pertanyaan, ditetapkan sebagai . Contoh Neal menunjukkan bahwa nilai itu tidak cukup. Namun, karena dalam pengujian properti kami biasanya tidak terlalu peduli dengan konstanta, saya sedikit mengendurkan kondisinya.$c$$2$$c$

Jawabannya iya". Buktinya dengan kontradiksi.

Untuk kenyamanan notasi, marilah kita menunjukkan variabel n / 2 pertama $n/2$dengan $x$ dan variabel n / 2 kedua $n/2$oleh $y$ . Misalkan $f(x,y)$ adalah $\delta$ tutup ke fungsi $f_1(x,y)$ yang hanya bergantung pada koordinat $k$ dari $x$ . Nyatakan koordinatnya yang berpengaruh oleh $T_1$ . Demikian pula, anggaplah $f(x,y)$ adalah$\delta$ tutup ke fungsi$f_2(x,y)$ yang hanya bergantung padakoordinat$k$ dari$y$ . Nyatakan koordinatnya yang berpengaruh oleh$T_2$ . Kita perlu membuktikan bahwa$f$ adalah$4\delta$ - mendekati$2k$ junta$\tilde f(x,y)$ .

Katakanlah ( x 1 , y 1 ) ∼ ( x 2 , y 2 ) jika x 1 dan x 2 menyetujui semua koordinat di T 1 dan y 1 dan y 2 setuju pada semua koordinat di T 2 . Kami memilih perwakilan acak dari setiap kelas kesetaraan secara acak. Biarkan ( ˉ x , ˉ y ) menjadi perwakilan untuk kelas ( x $(x_1,y_1) \sim (x_2,y_2)$$x_1$$x_2$$T_1$$y_1$$y_2$$T_2$$(\bar x, \bar y)$y ) . Tentukan ˜ f sebagai berikut: ˜ f ( x , y ) = f ( ˉ x , ˉ y ) $(x,y)$$\tilde f$

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

Jelaslah bahwa ˜ f adalah 2 k- junta (hanya bergantung pada variabel dalam T 1 ∪ T 2 ) . Kami akan membuktikan bahwa itu pada jarak 4 δ dari f dalam harapan.$\tilde f$$2k$$T_1 \cup T_2)$$4\delta$$f$

Kami ingin membuktikan bahwa Pr ˜ f ( Pr x , y ( ˜ f ( x , y ) ≠ f ( x , y ) ) )) = Pr ( f ( ˉ x , ˉ y ) ≠ f ( x , y ) ) ≤ 4 δ , di mana x dan y dipilih secara seragam secara acak. Pertimbangkan vektor acak

$$\Pr_{\tilde f}(\Pr_{x,y}(\tilde f(x,y) \neq f(x,y))) = \Pr(f(\bar x, \bar y) \neq f(x,y)) \leq 4\delta,$$ $x$$y$~ X diperoleh darixdengan menjaga semua bit diT1dan secara acak membalik semua bit tidak diT1, dan vektor ~ y didefinisikan sama. Perhatikan bahwa Pr( ˜ f (x,y)≠f(x,y))=Pr(f( ˉ x , ˉ y )≠f(x,y))=$\tilde x$$x$$T_1$$T_1$$\tilde y$ $$\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)).$$

Kami memiliki,

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

Demikian pula, Pr ( f ( ˜ x , y ) ≠ f ( ˜ x , ˜ y ) ) ≤ 2 δ . Kami memiliki Pr ( f ( ˉ x , ˉ y ) ≠ f ( x , y ) ) ≤ 4 δ . QED$\Pr(f(\tilde x,y) \neq f(\tilde x, \tilde y)) \leq 2\delta$

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

Mudah untuk “membatalkan pengacakan” bukti ini. Untuk setiap ( x , y ) , misalkan ˜ f ( x , y ) = 1 jika f ( x , y ) = 1 untuk sebagian besar ( x ′ , y ′ ) dalam kelas ekivalensi ( x , y ) , dan ˜ f ( x , y ) = 0 , jika tidak.$(x,y)$$\tilde f(x,y) = 1$$f(x,y) = 1$$(x',y')$$(x,y)$$\tilde f(x,y) = 0$

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