
CUUS063 main CUUS063 Goldreich 978 0 521 88473 0 March 31, 2008 18:49
APPENDIX F
We wish to stress that, when starting with R
2
as in Eq. (F. 1 ), the foregoing process
of error reduction can be used for obtaining error probability that is upper-bounded by
exp(−q(|x|)) for any desired polynomial q. The importance of this comment will become
clear shortly.
Step 3: The case of
2
. With the foregoing preliminaries, we are now ready to handle
the case of S ∈
2
. By Proposition 3.9, there exists a polynomial p and a set S
∈
1
=
coNP such that S ={x : ∃y ∈{0, 1}
p(|x|)
s.t. (x, y) ∈ S
}. Using S
∈ coNP, we apply
the foregoing reduction of S
to ⊕P as well as an adequate error reduction that yields
an upper bound of ε · 2
−p(|x |)
on the er ror probability, where ε ≤ 1/7 is unspecified at
this point. (For the case of
2
the setting ε = 1/7 will do, but for the dealing with
k
we will need a much smaller value of ε>0.) Thus, for an adequate polynomial
t (i.e., t(n + p(n)) = O(p(n)log(1/ε))), we obtain a relation R
(t)
2
∈ PC such that the
following holds: For every x and y ∈{0, 1}
p(|x|)
, with probability at least 1 − ε · 2
−p(|x |)
over the random choice of s
∈{0, 1}
poly(|x|)
, it holds that x
def
= (x, y) ∈ S
if and only if
|R
(t)
2
(x
, s
)| is odd.
4
Using a union bound (over all possible y ∈{0, 1}
p(|x|)
), it follows that, with probability
at least 1 − ε over the choice of s
, it holds that x ∈ S if and only if there exists a y such
that |R
(t)
2
((x, y), s
)|is odd. Now, as in the treatment of NP, we wish to reduce the latter
“existential problem” to ⊕P. That is, we wish to define a relation R
3
∈ PC such that for a
randomly selected s the value |R
3
(x, s, s
)| mod 2 provides an indication as to whether
or not x ∈ S (by indicating whether or not there exists a y such that |R
(t)
2
((x, y), s
)| is
odd). Analogously to Eq. (F. 1 ), consider the binary relation
I
3
def
=
2
(x, s, s
, y):
R
(t)
2
((x, y), s
)
≡ 1(mod 2) ∧ G(s, y) = 1
3
. (F.3)
In other words, I
3
(x, s, s
) ={y : |R
(t)
2
((x, y), s
)|≡1(mod 2) ∧ G(s, y) = 1}.In-
deed, if x ∈ S then, with probability at least 1 − ε over the random choice of s
and prob-
ability at least 1/3 over the random choice of s, it holds that |I
3
(x, s, s
)|is odd, whereas
for every x ∈ S and every choice of s it holds that
Pr
s
[|I
3
(x, s, s
)|=0] ≥ 1 − ε.
5
Note that, for ε ≤ 1/7, it follows that for every x ∈ S we have Pr
s,s
[|I
3
(x, s, s
)|≡
1(mod2)]≥ (1 − ε)/3 ≥ 2/7, whereas for every x ∈ S we have
Pr
s,s
[|I
3
(x, s, s
)|≡
1(mod2)]≤ ε ≤ 1/7. Thus, |I
3
(x, ·, ·)| mod 2 provides a randomized indication to
whether or not x ∈ S, but it is not clear whether I
3
is in PC (and in fact I
3
is likely not
to be in PC). The key observation is that there exists R
3
∈ PC such that ⊕R
3
=⊕I
3
.
4
Recall that |s
|=t(|x
|) · O( p
(|x
|)), where R
(x
) ⊆{0, 1}
p
(|x
|)
is the “witness relation” corresponding to S
(i.e., x
∈ S
if and only if R
(x
) ={0, 1}
p
(|x
|)
). Thus, R
2
(x
, s
) ⊆{0, 1}
p
(|x
|)+1
and R
(t)
2
(x
, s
) is a subset of
{0, 1}
1+t(|x
|)·(p
(|x
|)+2)
. Note that (since we started with S
∈ coNP) the error probability occurs on no-instances of
S
, whereas yes-instances are always accepted. However, to simplify the exposition, we allow possible errors also on
yes-instances of S
. This does not matter because we will anyhow have an error probability on yes-instances of S (see
footnote 5).
5
In continuation of footnote 4, we note that actually, if x ∈ S then there exists a y such that (x, y) ∈ S
and
consequently for every choice of s
it holds that |R
(t)
2
((x , y), s
)| is odd (because the reduction from S
∈ coNP
to ⊕P has zero error on yes-instances). Thus, for every x ∈ S and s
, with probability at least 1/3 over the random
choice of s, it holds that |I
3
(x , s, s
)| is odd (because the reduction from S ∈ NP
S
to ⊕P
S
has non-zero error on
yes-instances). On the other hand, if x ∈ S then Pr
s
[(∀y) |R
(t)
2
((x , y), s
)|≡0(mod2)]≥ 1 − ε (because for every
y it holds that (x , y) ∈ S
and the reduction from coNP to ⊕P has non-zero error on no-instances). Thus, for every
x ∈ S and s, it holds that Pr
s
[|I
3
(x , s, s
)|=0] ≥ 1 − ε (because the reduction from S ∈ NP
S
to ⊕P
S
has zero
error on no-instances). To sum up, the combined reduction has two-sided error, because each of the two reductions
introduces an error in a different direction.
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