(Note that
r
and
q
can be a function of the input size, not necessary a con
stant.)
Given 0
≤
s
≤
c
≤
1, Π
∈
PCP
c,s
[
r, q
] if there exists an (
r, q
)restricted
verifier satisfying the following conditions:
•
Completeness:
If
x
is a YESinstance, then there exists a proof
π
such
that Prob[
V
(
x, π
) =
Y ES
]
≥
c
•
Soundness:
If
x
is a NOinstance, then for any
π
, Prob[
V
(
x, π
) =
Y ES
]
≤
s
When
c
= 1, and
s
= 1
/
2, for simplicity, we write PCP[
r, q
] instead of
PCP
1
,
1
/
2
[
r, q
]
Now, can you show that NP = PCP
1
,s
[0
, poly
(
n
)] for all
s <
1?
Can you show that
PCP
[
O
(log
n
)
, O
(log
n
)]
⊆
NP
?
Note that when
r
=
O
(log
n
), the the proofchecking can be derandomized, that is,
V
can
be simulated by a polynomial time deterministic verifier that simulates the
computation of
V
on each of the 2
r
=
n
O
(1)
possible random inputs and then
computes the probability that
V
(
x, π
) accepts, then accepts if and only if
this probability is one.
2
The PCP Theorem
Theorem 1
The PCP Theorem: There exists two constant numbers
c
1
and
c
2
such that NP
=
PCP
[
c
1
log
n, c
2
]
. For simplicity, NP
=
PCP
[
O
(log
n
)
, O
(1)]
Theorem 2
Hastad (1997) Theorem [2] For every constant
, δ >
0
, NP
=
PCP
1

,
1
/
2+
δ
[
O
(log
n
)
,
3]
That is, for every constant
, δ >
0, there is a polysize PCP for NP that
reads just three random bits and test their XOR. Its completeness is 1

and
its soundness is 1
/
2 + . This result has imperfect completeness. However, if