เราบอกว่าฟังก์ชั่นบูลีนf : { 0 , 1 } n → { 0 , 1 }
ให้f : { 0 , 1 } n → { 0 , 1 }
ทีนี้ให้ϵ > 0
หมายเหตุ: ในสูตรดั้งเดิมของคำถามที่ค
เราบอกว่าฟังก์ชั่นบูลีนf : { 0 , 1 } n → { 0 , 1 }
ให้f : { 0 , 1 } n → { 0 , 1 }
ทีนี้ให้ϵ > 0
หมายเหตุ: ในสูตรดั้งเดิมของคำถามที่ค
คำตอบ:
The answer is “yes”. The proof is by contradiction.
For notational convenience, let us denote the first n/2
Let us say that (x1,y1)∼(x2,y2)
It is obvious that ˜f
We want to prove that Pr˜f(Prx,y(˜f(x,y)≠f(x,y)))=Pr(f(ˉx,ˉy)≠f(x,y))≤4δ,
We have, Pr(f(x,y)≠f(˜x,y))≤Pr(f(x,y)≠f1(x,y))+Pr(f1(x,y)≠f1(˜x,y))+Pr(f1(˜x,y)≠f(˜x,y))≤δ+0+δ=2δ.
Similarly, Pr(f(˜x,y)≠f(˜x,˜y))≤2δ
It easy to “derandomize” this proof. For every (x,y)
The smallest c
Lemmas 1 and 2 show that the bound holds for this c
(In comparison, Juri's elegant probabilistic argument gives c=4.)
Let c=1√2−1. Lemma 1 gives the upper bound for k=0.
Lemma 1: If f is ϵg-near a function g that has no influencing variables in S2, and f is ϵh-near a function h that has no influencing variables in S1, then f is ϵ-near a constant function, where ϵ≤(ϵg+ϵh)/2c.
Proof. Let ϵ be the distance from f to a constant function. Suppose for contradiction that ϵ does not satisfy the claimed inequality. Let y=(x1,x2,…,xn/2) and z=(xn/2+1,…,xn) 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 g0 be the fraction of pairs (y,z) such that g(y,z)=0. Let g1=1−g0 be the fraction of pairs such that g(y,z)=1. Likewise let h0 be the fraction of pairs such that h(y,z)=0, and let h1 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 ϵg and ϵh by 1/2n, while decreasing the ϵ by at most 1/2n, 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, ϵg+ϵh=g0h1+g1h0.
The distance from f to the all-zero function is at most 1−g0h0.
The distance from f to the all-ones function is at most 1−g1h1.
Further, the distance from f to the nearest constant function is at most 1/2.
Thus, the ratio ϵ/(ϵg+ϵh) is at most min(1/2,1−g0h0,1−g1h1)g0h1+g1h0,
By calculation, this ratio is at most 12(√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=1√2−1.
Lemma 2: Fix any k. If f is ϵg-near a function g that has k influencing variables in S2, and f is ϵh-near a function h that has k influencing variables in S1, then f is ϵ-near a function ˆf that has at most 2k influencing variables, where ϵ≤(ϵg+ϵh)/2c.
Proof. Express f as f(a,y,b,z) where (a,y) contains the variables in S1 with a containing those that influence h, while (b,z) contains the variables in S2 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 Fab(y,z)=f(a,y,b,z), and define Gab and Hab similarly from g and h respectively. Let ϵgab be the distance from Fab to Gab (restricted to (y,z) pairs). Likewise let ϵhab be the distance from Fab to Hab.
By Lemma 1, there exists a constant cab such that the distance (call it ϵab) from Fab to the constant function cab is at most (ϵhab+ϵgab)/(2c). Define ˆf(a,y,b,z)=cab.
Clearly ˆf depends only on a and b (and thus at most k variables).
Let ϵˆf be the average, over the (a,b) pairs, of the ϵab's, so that the distance from f to ˆf is ϵˆf.
Likewise, the distances from f to g and from f to h (that is, ϵg and ϵh) are the averages, over the (a,b) pairs, of, respectively, ϵgab and ϵhab.
Since ϵab≤(ϵhab+ϵgab)/(2c) for all a,b, it follows that ϵˆf≤(ϵg+ϵh)/(2c). QED
Lemma 3 shows that the constant c above is the best you can hope for (even for k=0 and ϵ=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 S2 and h has no influencing variables in S1, and f is 0.5-far from every constant function.
Proof. Let y and z be x restricted to, respectively, S1 and S2. That is, y=(x1,…,xn/2) and z=(xn/2+1,…,xn).
Identify each possible y with a unique element of [N], where N=2n/2. Likewise, identify each possible z with a unique element of [N]. Thus, we think of f as a function from [N]×[N] to {0,1}.
Define f(y,z) to be 1 iff max(y,z)≥1√2N.
By calculation, the fraction of f's values that are zero is (1√2)2=12, so both constant functions have distance 12 to f.
Define g(y,z) to be 1 iff y≥1√2N. Then g has no influencing variables in S2. The distance from f to g is the fraction of pairs (y,z) such that y<1√2N and z≥1√2N. By calculation, this is at most 1√2(1−1√2)=0.5/c
Similarly, the distance from f to h, where h(y,z)=1 iff z≥1√2N, is at most 0.5/c.
QED