CSci 5403, Spring 2010
Homework 4
due: March 23, 2010
1. Randomness does not help in
PSPACE
. In lecture, it was stated without proof
that
BPPSPACE
=
PSPACE
. In this problem, we’ll prove an even stronger version of this
result. Deﬁne the class
PPSPACE
to be the class of languages
A
for which there exists a
polynomial space
PTM
M
such that
x
∈
A
⇔
Pr
r
[
M
(
x
;
r
) accepts]
≥
1
/
2. Prove that
PPSPACE
⊆
PSPACE
. Thus, a
PSPACE
machine gains no advantage by having access to
random bits.
Hint:
Start by restricting a
PPSPACE
machine to always halt in the same number of
steps, and choose a convenient number. Modify the spaceeﬃcient search procedure from the
proof of Savitch’s theorem to count the exact number of accepting paths. You may apply the
fact that integer addition is in uniform
NC
1
to get around the (big) problem this algorithm
encounters.
2. BPL.
Recall that language
A
is in
BPL
iﬀ there exists a logspace
PTM
M
such that
Pr[
M
(
x
;
r
) = (
x
∈
A
)]
≥
2
/
3. Prove that
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 Spring '08
 Sturtivant,C
 Computational complexity theory, Bipartite graph, PSPACE, imperfect matchings, PPSPACE

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