されている

ここで他のユーザーにとって興味深いかもしれないので、この質問を共有すると思いました。

均一なクラス（）にある関数は、小さな不均一なクラス（、つまり不均一な）にもあり、これは関数がより小さい均一なクラス（ような）？この質問に対する答えが肯定的な場合、を含む最小の均一複雑度クラスは何ですか？負の場合、興味深い自然な反例を見つけることができますか？A C 0 / P O LのY A C 0 P N P ∩ A C 0 / P O Lの$NP$$AC^0/poly$$AC^0$$P$$NP \cap AC^0/poly$

ある$AC^0/poly \cap NP$中に含まれる$P$？

注：友人は既に私の質問にオフラインで部分的に回答しています。彼が自分で追加しない場合、彼の回答を追加します。

この質問は、次の非公式の質問を形式化する2回目の試みです。

不均一性は、自然な均一問題の計算に役立ちますか？

関連：

• 自然な問題の候補はあり$P/poly−P$ますか？

ライアンの答えを簡単に示します。仮定。言語を定義するL = { x ：| x | ∈ Λ }。仮定Λ ∈ N E ∖ Eは、に変換さL ∈ N P ∖ P。また、自明L ∈ A C 0 / P O リットルY。$\Lambda \in NE \setminus E$$L = \{x : |x| \in \Lambda\}$$\Lambda \in NE \setminus E$$L \in NP \setminus P$$L \in AC^0/poly$

最初の質問への回答：ありそうもない。

Theorem: If $NP \cap AC^0/poly \subseteq P$ then $NEXP = EXP$.

Given a circuit $C$ that outputs a bit, define the decompression of C to be the bit string obtained by evaluating $C$ on all possible inputs. That is, the decompression is $C(0^n) C(0^{n-1}1) C(0^{n-2}10) \cdots C(1^n)$.

Define the Succinct 3SAT problem as: given a circuit $C$ of size $n$, does its decompression encode a satisfiable Boolean formula? Succinct 3SAT is well-known to be $NEXP$ complete.

Now consider the language

$L =${$1^n |$the integer $n$ written in binary is a yes-instance of Succinct 3SAT}.

$L$ is clearly in $AC^0/poly$, since you can just hardcode whether $1^n$ is in $L$, for each $n$.

$L$ is also in $NP$: the integer $n$ written in binary has length about $\log n$, so the decompression of this circuit has length no more than $O(n)$. Hence the satisfying assignment has length at most $O(n)$.

But by the same observations, if $L \in P$, then $NEXP=EXP$, because it means that you have an $O(n^c)$ time algorithm for deciding every instance of Succinct 3SAT of length $\log n$.

Your second question is wide open (and open-ended).

To the question of Kaveh "Can non-uniformity help us in computing natural uniform problems?"

I think the answer is "yes", sometimes. Consider, for example, the Subset-Sum problem: given a sequence of $n$ positive real numbers, decide whether some subset of them sums up to $1$. This is an NP-hard problem even if restricted to positive integers (Knapsack). But Friedhelm Meyer auf der Heide (1984) has shown that, for any $n$, the problem can be solved by a linear decision tree of depth smaller than $n^5$. In such a tree tests are of the form: is a linear combination of input variables larger than some threshold. Non-uniformity here is important: for every $n$ we may have entirely different algorithm (decision tree).

References:

• Friedhelm Meyer auf der Heide, "A Polynomial Linear Search Algorithm for the n-Dimensional Knapsack Problem", 1984