Friday, March 21, 2008

* Undirected PATH is in RL
* BPP is in Simga_2 P intersect Pi_2 P

UNDIRECTED PATH IS IN RL

A simple random walk algorithm.  We start at the source s.  We choose
an edge at random among all adjacent edges, and repeat.  For technical
reasons, we have a self-loop edge for every node.

Convergence Lemma: If G is connected, then lim_{t -> infinity} p_t(u)
= d_u/(2|E|).

We prove the convergence lemma later.  First, we use it to show that
Undir-PATH is in RL.  Consider the asymptotic situation.  On average,
the walk returns to u every 2|E|/d_u steps.  But notice that this
return is independent from asymptotics.  So the expected time between
two successive visits to a node u is 2|E|/d_u.

Now let us consider the time taken to go from s to t.  There is a path
from s to t.  Consider the first edge out from s.  Expected time taken
before that edge is used is (d_u/2)*(2|E|/d_u) = |E|.  We repeat this
for every edge on the path to get a time of |E|n.  

Now, our randomized algorithm.  Run a random walk for 2|E|/n steps.
If you reach t within this time, return "reachable".  Otherwise,
return "not reachable".  Easily doable in logspace.

In RL.

Proof of Convergence Lemma: Let delta_t(u) denote p_t(u) - d_u/(2|E|).
In each step, the delta_t(u)'s get redistributed, so the sum of the
absolute values cannot increase.  If delta_t(u)'s of opposite signs
meet, they reduce.  Can show that reduction happens in finite steps.
End Proof

BPP IN SIGMA_2 P INTERSECT PI_2 P

Let L be a language in BPP.  Then, by the amplification lemma we know
that there exists a Turing machine M such that:

x is in L => Pr_y[M(x,y) accepts] >= 1 - 1/2^n

x is not in L => Pr_y[M(x,y) rejects] >= 1 - 1/2^n

M is poly-time, and so |y| is poly(|x|).  Let |y| be m.  Then, for a
given x in L, M accepts for 2^m - 2^{m-n} of the random strings and
rejects for only at most 2^{m-n} of the random strings; and vice versa
if x is not in L.  Let the accepted random strings be A_x and the
rejected by R_x.

Claim: x is in L iff there exists m strings z_1, ..., z_m such that

{y XOR z_i: y in A_x, 1 <= i <= m} = {0,1}^m.

Proof: By a simple counting argument.  Suppose x is not in L.  Then,
|A_x| is at most 2^{m-n}.  The size of {y XOR z_i} for each i is at
most 2^{m-n}.  Since m is poly(n), m < 2^n; thus the total size, over
all i, is at most

m2^{m-n} < 2^m.

Suppose x is in L.  Then, consider a random choice of z_1, ..., z_m.
Fix a w in {0,1}^m.  What is the probability that w in {0,1}^m is not
y XOR z_i for any choice of y or z_i?  Let us calculate this.
Consider A_x.  It has at least 2^m - 2^{m-n} strings in it.  For each
of these strings y there is a choice of z such that y XOR z = w.
Thus, the probability that w is in A_x XOR z_1 is at least (2^{m} -
2^{m-n})/2^m = 1 - 1/2^n.  So the probability w is not in A_x XOR z_i
for any of the i's is at most (1/2^n)^m.  The number of strings w is
2^m.  So the probability that there exists one string w that is not
covered is at most 2^m/2^{nm} < 1.  Hence, the claim is proved.

End Proof

Sigma_2 P algorithm for L.  For input x:

(1) Guess z_1, z_2, ..., z_m using an OR-branch.

(2) Generate all w of length m using AND-branch.

(3) Check that {w XOR z_j | 1 <= j <= m} intersects A_x by running
M(x, w XOR z_j) for all 1 <= j <= m.

The last step can be done deterministically in poly-time by simulating
M.  The first quantifier is OR and the second is AND.  So L is in
Sigma_2 P.

L is in Pi_2 P because we can do the same for x not in L.