सबसे छोटा c जो बन्ध है वह c = 1 हैc√2 -1≈2.41।c=12√−1≈2.41
लेममस 1 और 2 बताते हैं कि इस सी के लिए बाध्य है । लेम्मा 3 से पता चलता है कि यह बाध्य तंग है।c
(तुलना में, जुरी का सुरुचिपूर्ण संभाव्य तर्क c = 4 देता है ।)c=4
चलो ग = 1√2 -1। लेम्मा 1k=0 केलिए ऊपरी सीमा देता है।c=12√−1k=0
लेम्मा 1:
यदि च है ε जी -near एक समारोह ग्राम में कोई प्रभावित करने वेरिएबल नहीं है कि एस 2 , और च है ε ज एक समारोह -near ज है कि कोई में चर को प्रभावित किया है एस 1 , तो च है ε -near एक निरंतर समारोह , जहां ε ≤ ( ε जी + ε ज ) / 2fϵggS2fϵhhS1fϵसी ।ϵ≤(ϵg+ϵh)/2c
प्रमाण।
चलो ε से दूरी होना च एक निरंतर कार्य करने के लिए। विरोधाभास के लिए मान लीजिए कि ε दावा किया असमानता को संतुष्ट नहीं करता है। चलो y = ( एक्स 1 , एक्स 2 , ... , एक्स एन / 2 ) और जेड = ( एक्स एन / 2 + 1 , ... , x n )
और लिखने च , छ , और ज के रूप में च ( yϵfϵy=(x1,x2,…,xn/2)z=(xn/2+1,…,xn)fgh,z)f(y,z), g(y,z)g(y,z) and h(y,z)h(y,z),
so g(y,z)g(y,z) is independent of zz and h(y,z)h(y,z) is independent of yy.
(I find it helpful to visualize ff as the edge-labeling
of the complete bipartite graph with vertex sets {y}{y} and {z}{z},
where gg gives a vertex-labeling of {y}{y},
and hh gives a vertex-labeling of {z}{z}.)
Let g0g0 be the fraction of pairs (y,z)(y,z) such that g(y,z)=0g(y,z)=0.
Let g1=1−g0g1=1−g0 be the fraction of pairs such that g(y,z)=1g(y,z)=1.
Likewise let h0h0 be the fraction of pairs such that h(y,z)=0h(y,z)=0,
and let h1h1 be the fraction of pairs such that h(y,z)=1h(y,z)=1.
Without loss of generality, assume that, for any pair such that g(y,z)=h(y,z)g(y,z)=h(y,z),
it also holds that f(y,z)=g(y,z)=h(y,z)f(y,z)=g(y,z)=h(y,z). (Otherwise, toggling the value of
f(y,z)f(y,z) allows us to decrease both ϵgϵg and ϵhϵh by 1/2n1/2n,
while decreasing the ϵϵ by at most 1/2n1/2n,
so the resulting function is still a counter-example.) Say any such pair is ``in agreement''.
The distance from ff to gg plus the distance from ff to hh
is the fraction of (x,y)(x,y) pairs that are not in agreement.
That is, ϵg+ϵh=g0h1+g1h0ϵg+ϵh=g0h1+g1h0.
The distance from ff to the all-zero function is at most 1−g0h01−g0h0.
The distance from ff to the all-ones function is at most 1−g1h11−g1h1.
Further, the distance from ff to the nearest constant function is at most 1/21/2.
Thus, the ratio ϵ/(ϵg+ϵh)ϵ/(ϵg+ϵh) is at most
min(1/2,1−g0h0,1−g1h1)g0h1+g1h0,
min(1/2,1−g0h0,1−g1h1)g0h1+g1h0,
where
g0,h0∈[0,1]g0,h0∈[0,1] and
g1=1−g0g1=1−g0 and
h1=1−h0h1=1−h0.
By calculation, this ratio is at most
12(√2−1)=c/212(2√−1)=c/2. QED
Lemma 2 extends Lemma 1 to general kk by arguing pointwise,
over every possible setting of the 2k2k influencing variables.
Recall that c=1√2−1c=12√−1.
Lemma 2: Fix any kk.
If ff is ϵgϵg-near a function gg that has
kk influencing variables in S2S2, and ff is ϵhϵh-near a function hh that
has kk influencing variables in S1S1,
then ff is ϵϵ-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