Sipser–Lautemann theorem

In computational complexity theory, the Sipser–Lautemann theorem or Sipser–Gács–Lautemann theorem states that bounded-error probabilistic polynomial (BPP) time is contained in the polynomial time hierarchy, and more specifically Σ₂ ∩ Π₂.

In 1983, Michael Sipser showed that BPP is contained in the polynomial time hierarchy.^[1] Péter Gács showed that BPP is actually contained in Σ₂ ∩ Π₂. Clemens Lautemann contributed by giving a simple proof of BPP’s membership in Σ₂ ∩ Π₂, also in 1983.^[2] It is conjectured that in fact BPP=P, which is a much stronger statement than the Sipser–Lautemann theorem.

Here we present the proof by Lautemann.^[2] Without loss of generality, a machine M ∈ BPP with error ≤ 2^−|x| can be chosen. (All BPP problems can be amplified to reduce the error probability exponentially.) The basic idea of the proof is to define a Σ₂ sentence that is equivalent to stating that x is in the language, L, defined by M by using a set of transforms of the random variable inputs.

Since the output of M depends on random input, as well as the input x, it is useful to define which random strings produce the correct output as A(x) = {r | M(x,r) accepts}. The key to the proof is to note that when x ∈ L, A(x) is very large and when x ∉ L, A(x) is very small. By using bitwise parity, ⊕, a set of transforms can be defined as A(x) ⊕ t={r ⊕ t | r ∈ A(x)}. The first main lemma of the proof shows that the union of a small finite number of these transforms will contain the entire space of random input strings. Using this fact, a Σ₂ sentence and a Π₂ sentence can be generated that is true if and only if x ∈ L (see conclusion).

The general idea of lemma one is to prove that if A(x) covers a large part of the random space ${\displaystyle R=\{1,0\}^{|r|}}$ then there exists a small set of translations that will cover the entire random space. In more mathematical language:

If ${\displaystyle {\frac {|A(x)|}{|R|}}\geq 1-{\frac {1}{2^{|x|}}}}$, then ${\displaystyle \exists t_{1},t_{2},\ldots ,t_{|r|}}$, where ${\displaystyle t_{i}\in \{1,0\}^{|r|}}$ such that ${\displaystyle \bigcup _{i}A(x)\oplus t_{i}=R.}$

Proof. Randomly pick t₁, t₂, ..., t_|r|. Let ${\displaystyle S=\bigcup _{i}A(x)\oplus t_{i}}$ (the union of all transforms of A(x)).

So, for all r in R,

${\displaystyle \Pr[r\notin S]=\Pr[r\notin A(x)\oplus t_{1}]\cdot \Pr[r\notin A(x)\oplus t_{2}]\cdots \Pr[r\notin A(x)\oplus t_{|r|}]\leq {\frac {1}{2^{|x|\cdot |r|}}}.}$

The probability that there will exist at least one element in R not in S is

${\displaystyle \Pr {\Bigl [}\bigvee _{i}(r_{i}\notin S){\Bigr ]}\leq \sum _{i}{\frac {1}{2^{|x|\cdot |r|}}}={\frac {2^{|r|}}{2^{|x|\cdot |r|}}}<1.}$

Therefore

${\displaystyle \Pr[S=R]\geq 1-{\frac {2^{|r|}}{2^{|x|\cdot |r|}}}>0.}$

Thus there is a selection for each ${\displaystyle t_{1},t_{2},\ldots ,t_{|r|}}$ such that

${\displaystyle \bigcup _{i}A(x)\oplus t_{i}=R.}$

The previous lemma shows that A(x) can cover every possible point in the space using a small set of translations. Complementary to this, for x ∉ L only a small fraction of the space is covered by ${\displaystyle S=\bigcup _{i}A(x)\oplus t_{i}}$. We have:

${\displaystyle {\frac {|S|}{|R|}}\leq |r|\cdot {\frac {|A(x)|}{|R|}}\leq |r|\cdot 2^{-|x|}<1}$

because ${\displaystyle |r|}$ is polynomial in ${\displaystyle |x|}$.

The lemmas show that language membership of a language in BPP can be expressed as a Σ₂ expression, as follows.

${\displaystyle x\in L\iff \exists t_{1},t_{2},\dots ,t_{|r|}\,\forall r\in R\bigvee _{1\leq i\leq |r|}(M(x,r\oplus t_{i}){\text{ accepts}}).}$

That is, x is in language L if and only if there exist ${\displaystyle |r|}$ binary vectors, where for all random bit vectors r, TM M accepts at least one random vector ⊕ t_i.

The above expression is in Σ₂ in that it is first existentially then universally quantified. Therefore BPP ⊆ Σ₂. Because BPP is closed under complement, this proves BPP ⊆ Σ₂ ∩ Π₂.

The theorem can be strengthened to ${\displaystyle {\mathsf {BPP}}\subseteq {\mathsf {MA}}\subseteq {\mathsf {S}}_{2}^{P}\subseteq \Sigma _{2}\cap \Pi _{2}}$ (see MA, S^P
₂).^[3][4]