무한 반원에 대한 애들 맨의 정리?

Adleman은 1978 년에 임을 보여 주었다 $\mathrm{BPP}\subseteq \mathrm{P/poly}$ : 변수 의 부울 함수 $f$ 가 크기의 확률 적 부울 회로에 의해 계산 될 수 있다면 , 는 또한 결정적인 부울 회로에 의해 계산 될 수있다 및 다항식 ; 실제로, 사이즈의 . $n$ $M$ $f$ $M$ $n$ $O(nM)$

일반적인 질문 : $\mathrm{BPP}\subseteq \mathrm{P/poly}$ 는 어떤 (부울보다) 세미 링을 보유합니까?

좀 더 구체적으로하는 것으로 확률 회로 $\mathsf{C}$ semiring 이상의 $(S,+,\cdot,0,1)$ 는 "부가"를 사용한다 $(+)$ 및 "승산 ' $(\cdot)$ 게이트로서 동작. 입력은 입력 변수이다 $x_1,\ldots,x_n$ 아마도 값이 걸릴 확률 변수 추가, 몇 개의 $0$ 과 $1$ 독립적 확률 $1/2$ 여기서 $0$ 및 $1$ 은 각각 반고리의 첨가제 및 곱셈 아이덴티티이다. 이러한 회로 $\mathsf{C}$ 계산소정 함수 $f:S^n\to S$ 에 대한 모든 경우 $x\in S^n$ , $\mathrm{Pr}[\mathsf{C}(x)=f(x)]\geq 2/3$ .

투표 함수 의 변수 값이있는 부분 함수 소자하는 경우 이상 나타나는 간의 시간 및 정의되지 그러한 요소 가 존재 하지 않는 경우 . 체 르노 프와 노조 경계의 간단한 적용은 다음을 산출합니다. $\mathrm{Maj}(y_1,\ldots,y_m)$ $m$ $y$ $y$ $m/2$ $y_1,\ldots,y_m$ $y$

$\mathsf{C}$ $f:S^n\to S$ $X\subseteq S^n$ $m=O(\log|X|)$ $C_1,\ldots,C_m$ $\mathsf{C}$ $f(x)=\mathrm{Maj}(C_1(x),\ldots,C_m(x))$ $x\in X$

$\mathrm{Maj}$ $X=\{0,1\}^n$

$(\mathbb{N},+,\cdot,0,1)$

$\mathrm{BPP}\subseteq \mathrm{P/poly}$

"예"에 베팅했지만 이것을 표시 할 수 없습니다.

$\mathrm{BPP}\subseteq \mathrm{P/poly}$ $(\mathbb{R},+,\cdot,0,1)$ $\mathrm{Maj}$ $\mathrm{Maj}$

$\mathrm{Maj}(x,y,z)$ $f(x,y,z)=ax+by+cz+ h(x,y,z)$ $f(x,x,z)=x$ $c=0$ $f(x,y,x)=x$ $b=0$ $f(x,y,y)=y$ $a=0$ $\mathrm{Maj}$ $f:\mathbb{R}^n\to Y$ $Y$ $Y=\{0,1\}$ $\mathrm{Maj}:Y^m\to Y$ $Y=\mathbb{R}$

보정 [2017년 6월 3일] 파스칼 Koiran 그 모델은 단지 연산 회로보다 더 강력 저 가리키는 (이 논문의 저자들 중 하나)가 사인 게이트 (출력 (outputting) 허용

또는

입력이 음수인지에 따라 아니). 따라서 투표 기능 Maj 를이 모델에서 시뮬레이션 할 수 있으며 "혼동"을 다시 가져옵니다.

0

$0$

1

$1$

$(\mathbb{N}\cup\{+\infty\}, \min, +, +\infty,0)$ and $(\mathbb{N}\cup\{-\infty\}, \max, +, -\infty,0)$ .

Question 2: Does $\mathrm{BPP}\subseteq \mathrm{P/poly}$ hold over tropical semirings?

Held $\mathrm{BPP}\subseteq \mathrm{P/poly}$ in these two semirings, this would mean that randomness cannot speed-up so-called "pure" dynamic programming algorithms! These algorithms only use Min/Max and Sum operations in their recursions; Bellman-Ford, Floyd-Warshall, Held-Karp, and many other prominent DP algorithms are pure.

So far, I can only answer Question 2 (affirmatively) under the one-sided error scenario, when we additionally require $\mathrm{Pr}[\mathsf{C}(x) < f(x)]=0$ over the min-plus semiring (minimization), or $\mathrm{Pr}[\mathsf{C}(x) > f(x)]=0$ over the max-plus semiring (maximization). That is, we now require that the the randomized tropical circuit can never produce any better than optimum value; it can, however, err by giving some worse-than-optimal values. My questions are, however, under the two-sided error scenario.

P.S. [added 27.02.2017]: Here is my attempt to answer Question 1 (affirmatively). The idea is to combine a simplest version of the "combinatorial Nullstellensatz" with an estimate for the Zarankiewicz problem for n-partite hypergraps, due to Erdos and Spencer. Modulo this latter result, the entire argument is elementary.

Note that Question 2 still remains open: the "naive Nullstellensatz" (at least in the form I used) does not hold in tropical semirings.

— Stasys
소스

nit: BPP is a uniform class defined using PTMs not circuits.

— Kaveh

@Kaveh: yes, in this sense, Adleman's result is even a bit stronger, it holds even for BPP/poly.

— Stasys

Don't see how the simple argument shows impossibility... it does seem to show that the coefficients of the x, y, and z monomials must be zero... what am I missing? Also, if a polynomial can't compute Maj, how else can you representing a computation over a semiring? (What else besides a polynomial over the semiring?) Intuitively, over an infinite domain, each constraint on some y (enforcing that on >m/2 y's you must output y) seems "independent" of the others (no subset of constraints implies another) so it seems no "finite" polynomial could satisfy the infinitely many independent constraints.

— Ryan Williams

@Ryan: yes, this only shows f=Maj implies h=Maj. But h has degree > 1, so h(x,x,z)=x is impossible. And you are right: circuits over semirings cannot compute anything else as polynomials. So, they cannot compute Maj. But the authors of that paper deal with {+,x,-,/} circuits, with all field operations allowed. Perhaps then Maj can still be somehow computed? (I however, do not see how.) B.t.w. instead of trying to simulate Maj itself, one could answer Q1&Q2 by showing that one Maj-gate cannot substantially decrease circuit size (which is quite plausible).

— Stasys

@Ryan: P.S. Igor Sergeev observed that Maj "could" be probable computable over (R,+,x,-,/). E.g. Maj(x,y,z) is computable by f(x,y,z)=(xy+xz-2yz)/(2x-y-z) for all inputs with |{x,y,z}|=2. Note that the simple argument above implies that, already on such inputs, this cannot be done over (R,+,x,-). So, division can help. But we face the division by 0 issue ...

— Stasys

This is only a partial answer to your general question (I'm not sure what a fully general formulation would be), but it suggests that working over sufficiently nice infinite semirings while constraining the randomness to a finite domain might actually trivialize the question of whether Adleman's theorem holds.

Suppose you're working over the complex numbers $\mathbb{C}$ , so that circuits compute polynomials over that field, and suppose the function $f$ is itself computed by some polynomial (however complicated) of the $x$ variables. Then it turns out that already for some fixed $r$ , $C(x,r) = f(x)$ . The reason is that for each $r$ , the set of $x$ with $C(x,r) = f(x)$ determines a Zariski-closed subset of $\mathbb{C}^n$ , and so must be all of $\mathbb{C}^n$ , or else a subset of measure zero. If all of these sets were to have measure zero, then, because there are only finitely many $r$ 's in consideration, the set of $x$ where $\exists r : C(x,r) = f(x)$ would also have measure zero. On the other hand, the assumption that $C$ computes $f$ implies this that set must be all of $\mathbb{C}^n$ , so it cannot have measure zero.

— Andrew Morgan
소스

Interesting. More generally, a probabilistic circuit of size M is some random variable C taking its values in the set of all circuits (of that type) with at most M gates. [B.t.w. that paper of Cucker at al. allows C be arbitrarily distributed. The "majority trick" stil works.] Can I conclude from your argument that, if the range of C is finite, then Adleman's theorem is trivial when Zarinski-closed subsets are either trivial (sets themselves) or have zero measure? Have we this - "all or nothing" - effect in tropical semirings? (I am mainly interested in them.)

— Stasys

I don't know how or whether the argument would generalize to other semirings, sorry. One main thing that's missing (for me) is the geometric intuition akin to how "polynomials that disagree" translates to "measure-zero subsets of

C^{n}

$\mathbb{C}^n$ ". For tropical semirings in particular, the operations seem so different from ordinary polynomials that it's hard to even guess at what the appropriate adaptation ought to be.

— Andrew Morgan