Sipser–Lautemann theorem

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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 Lautemann's proof.[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 xL, A(x) is very large and when xL, A(x) is very small. By using bitwise parity, ⊕, a set of transforms can be defined as A(x) ⊕ t={rt | rA(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 xL (see conclusion).

Lemma 1[edit]

The general idea of lemma one is to prove that if A(x) covers a large part of the random space then there exists a small set of translations that will cover the entire random space. In more mathematical language:

If , then , where such that

Proof. Randomly pick t1, t2, ..., t|r|. Let (the union of all transforms of A(x)).

So, for all r in R,

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


Thus there is a selection for each such that

Lemma 2[edit]

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 xL only a small fraction of the space is covered by . We have:

because is polynomial in .


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

That is, x is in language L if and only if there exist binary vectors, where for all random bit vectors r, TM M accepts at least one random vector ⊕ ti.

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.

Stronger version[edit]

The theorem can be strengthened to (see MA, SP


  1. ^ Sipser, Michael (1983). "A complexity theoretic approach to randomness". Proceedings of the 15th ACM Symposium on Theory of Computing. ACM Press: 330–335. 
  2. ^ a b Lautemann, Clemens (1983). "BPP and the polynomial hierarchy". Inf. Proc. Lett. 17. 17 (4): 215–217. doi:10.1016/0020-0190(83)90044-3. 
  3. ^ Canetti, Ran (1996). "More on BPP and the polynomial-time hierarchy". Information Processing Letters. Elsevier. 57 (5): 237–241. doi:10.1016/0020-0190(96)00016-6. 
  4. ^ Russell, Alexander; Sundaram, Ravi (1998). "Symmetric alternation captures BPP". Computational Complexity. 7 (2): 152–162. doi:10.1007/s000370050007. ISSN 1016-3328.