2
May 10, 2011
7
BPP
PH
• Proof:
–
BPP
language L: p.p.t. TM M:
x
L
Pr
y
[M(x,y) accepts] ≥ 2/3
x
L
Pr
y
[M(x,y) rejects] ≥ 2/3
– strong error reduction
: p.p.t. TM M’
• use n random bits (|y’| = n)
• # strings y’ for which M’(x, y’) incorrect is at
most 2
n/3
• (can’t achieve with naïve amplification)
May 10, 2011
8
BPP
PH
• view
y’ = (
w
,
z
)
, each of length n/2
• consider output of
M’(x, (
w
,
z
))
:
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
1
1
0
1
0
0
1
1
0
1
1
1
1
1
1
1
1
1
1
1
0
0
0
1
1
0 0
0
1
0
1
1
0
1
1
1
1
1
1
1
0
1
1
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0 0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0 0
0
w =
000…00
000…01
000…10
…
111…11
…
…
x
L
x
L
so few
ones, not
enough for
whole disk
May 10, 2011
9
BPP
PH
• proof (continued):
– strong error reduction:
# bad y’ < 2
n/3
– y’ = (w, z) with |w| = |z| = n/2
– Claim:
L = {x :
9
w
8
z M’(x, (w, z)) = 1 }
– x
L: suppose
8
w
9
z M’(x, (w, z)) = 0
• implies
2
n/2
0’s; contradiction
– x
L: suppose
9
w
8
z M’(x, (w, z)) = 1
• implies
2
n/2
1’s; contradiction
May 10, 2011
10
BPP
PH
– given
BPP
language L: p.p.t. TM M:
x
L
Pr
y
[M(x,y) accepts] ≥ 2/3
x
L
Pr
y
[M(x,y) rejects] ≥ 2/3
– showed
L = {x :
9
8
z M’(x, (w, z)) = 1}
– thus
BPP
Σ
2
–
BPP
closed under complement
BPP
Π
2
– conclude:
BPP
(
Π
2
Σ
2
)
May 10, 2011
11
New Topic
The complexity of
counting
May 10, 2011
12
Counting problems
• So far, we have ignored
function problems