Advanced Computational Complexity-Lecture 19 Slides-Computer Science, Slides of Advanced Computational Complexity

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Lecture 19: More on Randomized Complexity

Matthew Johnson

[email protected]

Revision: Polynomial Hierarchy

Definition Σi is the class of all languages reducible to deciding the satisfiability of a formula of type

∃x 1 ∀x 2 ∃x 3 · · · R(x 1 , x 2 , x 3 ,.. .)

where there are i alternating quantifiers

Revision: BPP

Definition The complexity class BPP is the class of all languages L for which there exists a polynomial probabilistic Turing Machine M such that x ∈ L ⇒ Prob[M(x) = 1 ] ≥ (^23) x ∈ L ⇒/ Prob[M(x) = 1 ] < (^13)

It is not known whether or not BPP ⊆ NP. But we are able to show it is contained within Σ 2...

A useful result

We will use the following (without proof): Lemma If L is in BPP then there is a polynomial probabilistic Turing Machine M such that x ∈ L ⇒ Prob[M(x) = 1 ] ≥ 1 − (^31) m x ∈ L ⇒/ Prob[M(x) = 1 ] < (^31) m

where m is number of random bits used in the computation M(x).

Proof continued

This says that given an instance x ∈ L, we can find a set of m strings y 1 ,... , ym, such that given any string z, M will output 1 at least once if run m times with each yi ⊕ z used once as the guess input (and that if x ∈ L/ we cannot find these strings). We will not show how to find these m strings, but we will show that they must exist. We will show that given a random choice of y 1 ,... , ym the probability that it does not have the required property is less than 1 (if the strings did not exist the probability would be exactly 1).

Proof continued

This says that given an instance x ∈ L, we can find a set of m strings y 1 ,... , ym, such that given any string z, M will output 1 at least once if run m times with each yi ⊕ z used once as the guess input (and that if x ∈ L/ we cannot find these strings). We will not show how to find these m strings, but we will show that they must exist. We will show that given a random choice of y 1 ,... , ym the probability that it does not have the required property is less than 1 (if the strings did not exist the probability would be exactly 1).

Proof continued

The probability that a random choice of y 1 ,... , ym is “bad” is the probability that there exists a string z such that M outputs 0 for every guess input yi ⊕ z, 1 ≤ i ≤ m. Let us fix a choice of y 1 ,... , ym. For a particular z and yi , the probability that M outputs 0 with guess input yi ⊕ z is 1/ 3 m (from the Lemma). For a particular z, the probability that M outputs 0 for every guess input yi ⊕ z is (^) ( 3 m^1 )m. The probability that there is not at least one choice of z such that M outputs 0 for every guess input yi ⊕ z is 2 m^ ( 3 m^1 )m. This is less than 1.

Proof continued

The probability that a random choice of y 1 ,... , ym is “bad” is the probability that there exists a string z such that M outputs 0 for every guess input yi ⊕ z, 1 ≤ i ≤ m. Let us fix a choice of y 1 ,... , ym. For a particular z and yi , the probability that M outputs 0 with guess input yi ⊕ z is 1/ 3 m (from the Lemma). For a particular z, the probability that M outputs 0 for every guess input yi ⊕ z is (^) ( 3 m^1 )m. The probability that there is not at least one choice of z such that M outputs 0 for every guess input yi ⊕ z is 2 m^ ( 3 m^1 )m. This is less than 1.

Proof continued

Now we must prove that if x ∈ L/ , then we cannot find such a y 1 ,... , ym. That is, we must show that for every possible choice, there exists a string z such that M returns 0 on every guess input yi ⊕ z. We prove that a randomly chosen z does not have this property with probability less than 1. So some z does have this property.

Proof continued

Now we must prove that if x ∈ L/ , then we cannot find such a y 1 ,... , ym. That is, we must show that for every possible choice, there exists a string z such that M returns 0 on every guess input yi ⊕ z. We prove that a randomly chosen z does not have this property with probability less than 1. So some z does have this property.

Proof continued

Let us fix a choice of y 1 ,... , ym and of z. The probability that M outputs 1 with guess input yi ⊕ z is 1 / 3 m (from the Lemma). The probability that M outputs 1 on at least one guess input yi ⊕ z is at most m( 1 / 3 m) = 1 /3. So there must exist some z such that M outputs 0 for every guess input yi ⊕ z.

Proof continued

Let us fix a choice of y 1 ,... , ym and of z. The probability that M outputs 1 with guess input yi ⊕ z is 1 / 3 m (from the Lemma). The probability that M outputs 1 on at least one guess input yi ⊕ z is at most m( 1 / 3 m) = 1 /3. So there must exist some z such that M outputs 0 for every guess input yi ⊕ z.

Complexity Classes

Now we know P ⊆ ZPP ⊆ RP ⊆ BPP ⊆ Σ 2

In fact, it is widely believed that BPP = P. There is no well-known problem which has an efficient random algorithm, but not an efficient deterministic algorithm.

Complexity Classes

Now we know P ⊆ ZPP ⊆ RP ⊆ BPP ⊆ Σ 2

In fact, it is widely believed that BPP = P. There is no well-known problem which has an efficient random algorithm, but not an efficient deterministic algorithm.