Notes on Complexity Theory
Last updated: October, 2011
Lecture 9
Jonathan Katz
1
The Polynomial Hierarchy
1.1
Defining the Polynomial Hierarchy via Oracle Machines
Here we show a third definition of the levels of the polynomial hierarchy in terms of oracle machines.
Definition 1
Define
Σ
i
,
Π
i
inductively as follows:
•
Σ
0
def
=
P
.
•
Σ
i
+1
def
=
NP
Σ
i
and
Π
i
+1
=
co
NP
Σ
i
.
(Note that even though we believe Σ
i
6
= Π
i
, oracle access to Σ
i
gives the same power as oracle
access to Π
i
. Do you see why?)
We show that this leads to an equivalent definition.
For this section only
, let Σ
O
i
refer to the
definition in terms of oracles. We prove by induction that Σ
i
= Σ
O
i
. (Since Π
O
i
=
co
Σ
O
i
, this proves
it for Π
i
,
Π
O
i
as well.) For
i
= 1 this is immediate, as Σ
1
=
NP
=
NP
P
= Σ
O
1
.
Assuming Σ
i
= Σ
O
i
, we prove that Σ
i
+1
= Σ
O
i
+1
.
Let us first show that Σ
i
+1
⊆
Σ
O
i
+1
.
Let
L
∈
Σ
i
+1
. Then there exists a polynomial-time Turing machine
M
such that
x
∈
L
⇔ ∃
w
1
∀
w
2
· · ·
Q
i
+1
w
i
+1
M
(
x, w
1
, . . . , w
i
+1
) = 1
.
In other words, there exists a language
L
0
∈
Π
i
such that
x
∈
L
⇔ ∃
w
1
(
x, w
1
)
∈
L
0
.
By our inductive assumption, Π
i
= Π
O
i
; thus,
L
∈ NP
Π
O
i
=
NP
Σ
O
i
= Σ
O
i
+1
and so Σ
i
+1
⊆
Σ
O
i
+1
.
It remains to show that Σ
O
i
+1
⊆
Σ
i
+1
(assuming Σ
O
i
= Σ
i
). Let
L
∈
Σ
O
i
+1
. This means there
exists a non-deterministic polynomial-time machine
M
and a language
L
0
∈
Σ
O
i
such that
M
, given
oracle access to
L
i
, decides
L
. In other words,
x
∈
L
iff
∃
y, q
1
, a
1
, . . . , q
n
, a
n
(here,
y
represents the
non-deterministic choices of
M
, while
q
j
, a
j
represent the queries/answers of
M
to/from its oracle)
such that:
1.
M
, on input
x
, non-deterministic choices
y
, and oracle answers
a
1
, . . . , a
n
, makes queries
q
1
, . . . , q
n
and accepts.
2. For all
j
, we have
a
j
= 1 iff
q
j
∈
L
0
.
Since
L
0
∈
Σ
O
i
= Σ
i
(by our inductive assumption) we can express the second condition, above, as:
•
a
j
= 1
⇔ ∃
y
j
1
∀
y
j
2
· · ·
Q
i
y
j
i
M
0
(
q
j
, y
j
1
, . . . , y
j
i
) = 1
•
a
j
= 0
⇔ ∀
y
j
1
∃
y
j
2
· · ·
Q
0
i
y
j
i
M
0
(
q
j
, y
j
1
, . . . , y
j
i
) = 0
9-1