May 17, 2011
7
MA
and
AM
Theorem
:
coNP
AM
PH
=
AM
.
• Proof:
– suffices to show
Σ
2
AM
(and use
AM
Π
2
)
– L
Σ
2
iff
9
polytime language R
x
L
9
y
8
z (x, y, z)
R
x
L
8
y
9
z (x, y, z)
R
– Merlin sends y
– 1 AM exchange decides
coNP
query:
8
z (x, y, z)
R
?
– 3 rounds; in
AM
May 17, 2011
8
MA
and
AM
• We know
ArthurMerlin
=
IP
.
– “public coins = private coins”
Theorem
(GS):
IP[k]
AM[O(k)]
– stronger result
– implies for all
constant
k
2,
IP[k]
=
AM[O(k)]
=
AM[2]
• So, GNI
IP[2]
=
AM
May 17, 2011
9
Back to Graph Isomorphism
• The payoff:
– not known if GI is
NP
complete.
– previous Theorems:
if GI is
NP
complete then
PH
=
AM
– unlikely!
– Proof:
GI
NP
complete
GNI
coNP

complete
coNP
AM
PH
=
AM
Derandomization revisited
• L
MA
iff
9
polytime language R
x
L
9
m Pr
r
[(x, m, r)
R] = 1
x
L
8
m Pr
r
[(x, m, r)
R]
½
• Recall PRGs:
– for all circuits C of size at most
s
:
Pr
y
[C(y) = 1] – Pr
z
[C(G(z)) = 1] ≤
ε
May 17, 2011
10
NP
AM
MA
seed
output string
G
t
bits
u
bits
Using PRGs for MA
• L
MA
iff
9
polytime language R
x
L
9
m Pr
r
[(x, m, r)
R] = 1
x
L
8
m Pr
r
[(x, m, r)
R]
½
• produce polysize circuit C such that
C(x, m, r) = 1
,
(x,m,r)
2
R
• for each x, m can hardwire to get C
x,m
9
m Pr
y
[C
x,m
(y) = 1] = 1
(“yes”)
8
m Pr
y
[C
x,m
(y) = 1] ≤ 1/2
(“no”)
May 17, 2011
11
Using PRGs for MA
• can compute
Pr
z
[C
x,m
(G(z)) = 1]
exactly
– evaluate
C
x
(G(z))
on every seed
z
{0,1}
t
– running time
(O(C
x,m
)+(time for G))
2
t
x
L
9
m
[
Pr
z
[C
x,m
(G(z)) = 1] = 1
]
x
L
8
m
[
Pr
z
[C
x,m
(G(z)) = 1]