Last revised 4/29/2010
In this lecture, we first continue to talk about polynomial hierarchy. Then we prove the Gács-Sipser-Lautemann theorem that BPP is contained in the second level of the hierarchy.
1. The hierarchy
Definition 1 (Polynomial hierarchy) iff there are polynomials and a polynomial time computable function such that
iff there are polynomials and a polynomial time computable function such that
For clarity, we omitted the conditions that each string must be of polynomial length .
One thing that is easy to see is that . Also, note that, for all , , , , . This can be seen by noticing that the predicates do not need to “pay attention to” all of their arguments, and so a statement involving quantifiers can “simulate” a statement using less than quantifiers.
Theorem 2 Suppose . Then .
Proof: For any language , we have that there exist polynomials and a polynomial time computable function F such that
where we did not explicitly stated the conditions . Let us look at the right hand side of the equation. What is following is a statement. Thus, there is a such that
Under the assumption that , we have , which means that there are polynomials and a polynomial time computable such that
where we omitted the conditions . So now we can show that
And so .
Now notice that if and are two complexity classes, then implies . Thus, we have . So we have .
This result was first shown by Sipser and Gács. Lautemann gave a much simpler proof which we give below.
Lemma 3 If is in BPP then there is an algorithm such that for every ,
where the number of random bits and runs in time .
Proof: Let be a BPP algorithm for . Then for every ,
and uses random bits where .
Do repetitions of and accept if and only if at least executions of accept. Call the new algorithm . Then uses random bits and
We can then find with such that .
Theorem 4 .
Proof: Let be in BPP and as in the claim. Then we want to show that
where is the number of random bits used by on input .
Suppose . Then
So a sequence exists, such that .
Conversely suppose . Then fix a sequence . We have
So for all there is a such that .