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[BCIKS20] Proximity Gaps 论文 soundness 解析

论文 [BCIKS20] 对 [BBHR18] 中的 FRI 协议的 soundness 进行了改进,主要分析了 batched FRI 的情况。本文将详细解析 [BCIKS20] 论文中关于 batched FRI soundness 的内容。

Introduction

在交互证明,分布式存储以及密码学等背景下,出现了各种协议,这些协议引出了关于一个线性编码 VFqnV \subset \mathbb{F}_q^n 的 proximity 问题,其中 Fq\mathbb{F}_q 是有限域,VV 的最小的相对距离为 δV\delta_V 。这些协议假设我们可以获得关于一批向量 u={u0,,ul}Fqn\textbf{u} = \{u_0, \cdots, u_l \} \subset \mathbb{F}_q^n 的 oracle,它们的 soundness 要求每个向量 uiu_i 在相对 Hamming 距离上接近 VV 。另外,soundness 是某些向量到 code VV 之间的最大距离的一个函数,如果这个距离变大,那么 Verifier 拒绝的概率会降低。因此,我们想找到这样的协议,能够最小化对 u\mathbf{u} 中元素的 query 的数量,同时最大化能识别某个向量 uiu_i 明显远离 VV 的概率。

疑问

  • 如何理解下面这句话? Furthermore, soundness deteriorates as a function of the largest distance between some vector ui and the code V . soundeness 会随着某些向量 uiu_i 与 code VV 之间的最大距离的增加而降低,也就是此时 Verifier 拒绝的概率就降低了?
  • 如何对 soundness 降低进行清楚地讲解呢?

由于 VV 的线性性,一个自然的方法([RVW13])是:在 span(u\textbf{u})(即 u\textbf{u} 中各元素的线性组合) 中均匀地随机一个向量 uu' ,记 uu'VV 之间的距离为 Δ(u,V)\Delta(u', V) ,将这个距离视为 u\textbf{u} 中某些元素与 VV 之间的最大距离的一个 proxy(代理)。为了证明 soundness ,我们想要即使只有一个 uiu_i 距离 VV 中的所有元素有 δ-远,那么随机选择的 uu' 也距离 VV 很远。

在下文中,用 Δ 表示相对 Hamming 距离。当 Δ(u,v)δ\Delta(u, v) \le \delta 对某个 vVv \in V 成立时,称为 “ uuVV 的距离有 δ -近”,记作 Δ(u,V)δ\Delta(u, V) \le \delta ;否则称为 “ uuVV 的距离有 δ -远”,记作 Δ(u,V)>δ\Delta(u, V) > \delta

关于这个问题,一些研究结果为:

  1. [AHIV17] 如果 δ<δV/4\delta < \delta_V /4 ,几乎所有的 uspan(u)u' \in \text{span}(\mathbf{u}) 距离 VVδ -远。
  2. [RZ18] 将上述结果提高到 δ<δV/3\delta < \delta_V /3
  3. [BKS18] 提高到 δ<11δV4\delta < 1 - \sqrt[4]{1 - \delta_V}
  4. [BGKS20] 提高到 δ<11δV3\delta < 1 - \sqrt[3]{1 - \delta_V} ,但这个界对 RS 编码是 tight 的,因为当 n=qn = q 时可以达到这个界。

🤔 思考

  • 为什么研究的重心都想要提高这个 δ 的上界呢?关于这个问题,目前我的想法是: δ 的上界这里是和 δV\delta_V 有关的,对于 RS code ,δV=1ρ\delta_V = 1 - \rho ,实质也就是和码率相关,那么提高上界也就意味着降低了码率,那么就意味着更多的冗余,如果以相同的安全性,或者同样的高概率来拒绝出错的情况,此时需要的query就更少了。或者这样说,如果对于同样一个协议,query 的数量固定,δ 越大,拒绝的概率也就越大,也就提高了 soundness。
  • 上面分析的第 2 点与 “Furthermore, soundness deteriorates as a function of the largest distance between some vector uiu_i and the code V .” 似乎矛盾,这句话说的是 δ 越大,soundness 越小?这一点该如何理解呢?

目前我们关心的一个问题是:对于哪些码以及什么范围的 δ ,以下的陈述成立?

如果某个 uspan(u)u^* \in \text{span}(\mathbf{u})VVδ -远,那么对于几乎所有的 uspan(u)u' \in \text{span}(\mathbf{u})uu'VV 也有 δ -远。

[BCIKS20] 论文的主要结论之一表明,当 VV 是一个在足够大的域上的 RS 码(域的大小与码的块长度呈多项式关系)并且 δ 小于 Johnson/Guruswami-Sudan list decoding 界时,上述陈述成立。接下来,我们称其为 proximity gap 。

Proximity Gaps

先给出 Proximity Gaps 的定义。

Definition 1.1 [BCIKS20, Definition 1.1] (Proximity gap). Let PΣnP \subset \Sigma^n be a property and C2ΣnC \subset 2^{\Sigma^n} be a collection of sets. Let Δ be a distance measure on Σn\Sigma^n . We say that CC displays a (δ,ϵ)(\delta, \epsilon) -proximity gap with respect to PP under Δ if every SCS \in C satisfies exactly one of the following:

  1. PrsS[Δ(s,P)δ]=1\Pr_{s \in S} [\Delta(s, P) \le \delta] = 1 .
  2. PrsS[Δ(s,P)δ]ϵ\Pr_{s \in S} [\Delta(s, P) \le \delta] \le \epsilon .

We call δ the proximity parameter and ε is the error parameter. By default, Δ denotes the relative Hamming distance measure.

对于 RS code ,如果 VFnV \subset \mathbb{F}^n 是 RS 编码 ,对应上述定义中的 PP ,并且 AFnA \subset \mathbb{F}^n 是一个 affine space (仿射空间) ,对应于上述定义中的 SS ,那么要么 AA 中的所有元素离 VVδ -近,要么 AA 中的几乎所有元素离 VVδ -远。换句话说,不会有这样的 affine space AA ,其中大概一半的元素离 VV 比较近,但同时另一半元素离 VV 比较远。

如下图所示,AA 是一个 affine space,这里用一条线表示,编码空间 VV 中的元素用黑色的点表示,以这些点为圆心,以 δ 为半径画圆。那么只有两种情况:

  1. 线 AA 上的所有元素都落入了绿色的圆形区域内。

  2. 线上只有少量的元素落入了绿色的圆形区域内。 AA 中的元素不可能一半在圆形区域内,一半在圆形区域外,这也是 gap 的含义,将 AA 中的所有元素落入的情况分成了恰好分成了两种情况,而这两种情况之间根据相对 Hamming 距离形成了一个巨大的 gap 。

在下文中,用 Fq\mathbb{F}_q 表示大小为 qq 的有限域,RS[Fq,D,k]\text{RS}[\mathbb{F}_q,\mathcal{D},k] 表示维数为 k+1k+1 ,块长度(blocklength) 为 n=Dn = |\mathcal{D}| 的 RS 编码,其码字是在 D\mathcal{D} 上求值(evaluated),次数 k\le k 的多项式。用 ρ 表示码率, 则 ρ=k+1n\rho = \frac{k+1}{n}δ 表示相对于 RS code 的相对 Hamming 距离,ε 表示 error 参数,也就是一个“坏事件(bad event)”发生的概率。

下面给出 RS code 的 Proximity gaps 定理。

Theorem 1.2 [BCIKS20, Theorem 1.2] (Proximity gap for RS codes). The collection CAffineC_{\text{Affine}} of affine spaces in Fqn\mathbb{F}_q^n displays a (δ,ϵ)(\delta, \epsilon) -proximity gap with respect to the RS code V:=RS[Fq,D,k]V := \text{RS}[\mathbb{F}_q, \mathcal{D}, k] of blocklength nn and rate ρ=k+1n\rho = \frac{k+1}{n} , for any δ(0,1ρ)\delta \in (0, 1 - \sqrt{\rho}) , and ϵ=ϵ(q,n,ρ,δ)\epsilon = \epsilon(q, n, \rho, \delta) defined as the following piecewise function:

ϵ=ϵU=ϵU(q,n):=nq(1.1)\epsilon = \epsilon_\text{U} = \epsilon_\text{U}(q, n) := \frac{n}{q} \tag{1.1}
ϵ=ϵJ=ϵJ(q,n,ρ,δ):=(k+1)2(2min(η,ρ20)7)q=O(1(ηρ)O(1)n2q)(1.2)\epsilon = \epsilon_\text{J} = \epsilon_\text{J}(q, n, \rho, \delta) := \frac{(k+1)^2}{\left(2 \min \left(\eta, \frac{\sqrt{\rho}}{20}\right)^7\right)q} = O \left(\frac{1}{(\eta \rho)^{O(1)}} \cdot \frac{n^2}{q} \right) \tag{1.2}

🤔 Question

  • δ 越大,落入圆形区域的元素可能更多,也就是 ϵJ\epsilon_{\text{J}} 相比 ϵU\epsilon_{\text{U}} 更大一些。是这个原因吗?

Correlated agreements

论文中证明的主要定理是 correlated agreement ,对于在 FD\mathbb{F}^{\mathcal{D}} 中的两个向量 u0,u1FDu_0, u_1 \in \mathbb{F}^{\mathcal{D}} ,在 F\mathbb{F} 中选一个随机数 zz ,我们关心用 zz 进行线性组合后的 u0+zu1u_0 + zu_1 所形成的空间与 VV 之间的距离,也就是一维的 affine space A={u0+zu1:zF}A = \{u_0 + z u_1 : z \in \mathbb{F}\} 。correlated agreement 结论说的是如果在 AA 中有足够多的元素距离 RS code 空间 VV 足够近( δ - 近),那么一定存在一个非平凡的 subdomain DD\mathcal{D}' \subset \mathcal{D} ,其大小至少是 D\mathcal{D} 大小的 1δ1 - \delta 倍,使得限制 u0,u1u_0, u_1D\mathcal{D}' 上,有有效的 RS code v0,v1v_0, v_1 ,满足它们分别在 D\mathcal{D}' 上与 u0,u1u_0, u_1 一致。我们就说这样的 D\mathcal{D}' 有 correlated agreement 性质,即 u0,u1u_0, u_1AA 中的元素不仅分别与 RS 码有很大的 agreement ,而且还共享一个共同的很大的 agreement 集合。这个结果有两个参数范围,一个是 unique decoding 范围内的 proximity 参数,另一个是 list decoding 范围内的 proximity 参数。

以下给出了三种情况的 correlated agreement。结合论文中其他关于 correlated agreement 的结论,如下表所示。

空间 UUΔu(u,V)\Delta_u(u,V)Δu(u,V)\Delta_u(u,V) unique decodingΔu(u,V)\Delta_u(u,V) list decodingagreeμ(u,V)\text{agree}_{\mu}(u,V)
lines{u0+zu1:zF}\{u_0 + z u_1 : z \in \mathbb{F}\}Theorem 1.4Theorem 4.1Theorem 5.1 & Theorem 5.2
low-degree parameterized curvescurve(u)={uz:=i=0lziuizFq}\text{curve}(\mathbf{u}) = \left\{u_z: = \sum_{i = 0}^{l}z^i \cdot u_i | z \in \mathbb{F}_q \right\}Theorem 1.5Theorem 6.1Theorem 6.2Theorem 7.1 & Theorem 7.2(Johnson bound 更精确版本)
affine spacesu0+span{u1,,ul}u_0 + \text{span}\{u_1, \cdots, u_l\}Theorem 1.6Theorem 7.3 & Theorem 7.4(Johnson bound 更精确版本)

下面三个定理分别对应线、低次参数曲线以及 affine spaces 的 correlated agreement 定理。

Theorem 1.4 [BSCIK20, Theorem 1.4] (Main Theorem - Correlated agreement over lines). Let V,q,n,kV, q, n, k and ρ be as defined in Theorem 1.2. For u0,u1FqDu_0, u_1 \in \mathbb{F}_q^{\mathcal{D}} , if δ(0,1ρ)\delta \in (0, 1 - \sqrt{\rho}) and

PrzFq[Δ(u0+zu1,V)δ]>ϵ,\Pr_{z \in \mathbb{F}_q} [\Delta(u_0 + z \cdot u_1, V) \le \delta] > \epsilon,

where ε is as defined in Theorem 1.2, then there exist DDD' \subset D and v0,v1Vv_0, v_1 \in V satisfying

u={u0,,ul}FqD\mathbf{u} = \{u_0, \ldots, u_l \} \subset \mathbb{F}_q^{\mathcal{D}} ,则次数为 ll 的 parameterized curve 是由 u\mathbf{u} 生成的如下的在 FqD\mathbb{F}_q^{\mathcal{D}} 中的向量的集合,

curve(u):={uz:=i=0lziuizFq}\text{curve}(\mathbf{u}) := \left\{u_z: = \sum_{i = 0}^{l}z^i \cdot u_i \Bigg| z \in \mathbb{F}_q \right\}

Theorem 1.5 [BSCIK20, Theorem 1.5] (Correlated agreement for low-degree parameterized curves). Let V,q,n,kV, q, n, k and ρ be as defined in Theorem 1.2. Let u={u0,,ul}FqD\mathbf{u} = \{u_0, \cdots, u_l\} \subset \mathbb{F}_q^{\mathcal{D}} . If δ(0,1ρ)\delta \in (0, 1 - \sqrt{\rho}) and

Prucurve(u)[Δ(u,V)δ]>lϵ,\Pr_{u \in \text{curve}(u)} [\Delta(\mathbf{u}, V) \le \delta] > l \cdot \epsilon,

where ε is as defined in Theorem 1.2, then there exist DD\mathcal{D}' \subset \mathcal{D} and v0,,vlVv_0, \cdots, v_l \in V satisfying

Theorem 1.6 [BSCIK20, Theorem 1.6] (Correlated agreement over affine spaces). Let V,q,n,kV, q, n, k and ρ be as defined in Theorem 1.2. For u0,u1,,ulFqDu_0, u_1, \cdots, u_l \in \mathbb{F}_q^{\mathcal{D}} let U=u0+span{u1,,ul}FqDU = u_0 + \text{span}\{u_1, \cdots, u_l\} \subset \mathbb{F}_q^{\mathcal{D}} be an affine subspace. If δ(0,1ρ)\delta \in (0, 1 - \sqrt{\rho}) and

PruU[Δ(u,V)δ]>ϵ,\Pr_{u \in U} [\Delta(u, V) \le \delta] > \epsilon,

where ε is as defined in Theorem 1.2, then there exist DD\mathcal{D}' \subset \mathcal{D} and v0,,vlVv_0, \cdots, v_l \in V satisfying

Furthermore, in the unique decoding regime δ(0,1ρ2]\delta \in \left(0, \frac{1 - \rho}{2}\right] , there exists a unique maximal D\mathcal{D}' satisfying the above, with unique viv_i .

Correlated Weighted Agreement

如果要分析 FRI 协议的 soundness ,就需要 Theorem 1.5 的 weighted 版本。

对于一个给定的 weight 向量 μ:D[0,1]\mu : \mathcal{D} \rightarrow [0,1]u,vu,v 之间(相对的) μ -agremment 定义为

agreeμ(u,v):=1Dx:u(x)=v(x)μ(x).\text{agree}_{\mu}(u,v) := \frac{1}{|\mathcal{D}|} \sum_{x:u(x) = v(x)} \mu(x).

也就是在权重 μ 之下看 uuvvD\mathcal{D} 上一致的比例有多少。如果令 μ1\mu \equiv 1 ,那么

agreeμ(u,v)=1Dx:u(x)=v(x)1=11Dx:u(x)v(x)1=1Δ(u,v).\text{agree}_{\mu}(u,v) = \frac{1}{|\mathcal{D}|} \sum_{x:u(x) = v(x)} 1 = 1 - \frac{1}{|\mathcal{D}|} \sum_{x:u(x) \neq v(x)} 1 = 1 - \Delta(u,v).

一个 word uu 与一个线性码 VV 之间的 agreement 为 uuVV 之间的一个码字之间的最大的 agreement ,

agreeμ(u,V):=maxvVagreeμ(u,v).\text{agree}_{\mu}(u,V) := \max_{v \in V} \text{agree}_{\mu}(u,v).

定义 subdomain DD\mathcal{D}' \subset \mathcal{D} 的加权(weighted) 大小为

μ(D):=1DxDμ(x).\mu(\mathcal{D}') := \frac{1}{|\mathcal{D}|}\sum_{x \in \mathcal{D}'} \mu(x).

将上面定义中的 D\mathcal{D}' 定义为 {xD:u(x)=v(x)}\{x \in \mathcal{D}: u(x) = v(x)\} ,则 agreement 满足 agreeμ(u,v)=μ({xD:u(x)=v(x)})\text{agree}_{\mu}(u,v) = \mu(\{x \in \mathcal{D}: u(x) = v(x)\})

最后,对于 u={u0,,ul}\mathbf{u} = \{u_0, \cdots, u_l\} ,其中 uiFqDu_i \in \mathbb{F}_q^{\mathcal{D}} 是一组 word ,μ -weighted correlated agreement 是 subdomain DD\mathcal{D}' \subset \mathcal{D} 的最大 μ -weighted 大小,使得 u\mathbf{u}D\mathcal{D}' 上的限制属于 VDV|_{\mathcal{D}'} ,即对于每个 i=0,,li = 0, \cdots, l ,存在 viVv_i \in V 使得 uiD=viDu_i|_{\mathcal{D}'} = v_i|_{\mathcal{D}'} 。当 μ 未指定时,它被设置为常数权重函数 1 ,这恢复了在前面讨论的 correlated agreement 度量的概念。

接下来,我们假设权重函数 μ 具有某些结构,具体来说,所有权重 μ(x)\mu(x) 都是形式 μ(x)=axM\mu(x) = \frac{a_x}{M} ,其中 axa_x 是变化的整数,且有共同的分母 MM 。对于 FRI soundness 的特殊情况(其中 MM 等于应用 FRI 协议的 RS 码的 blocklength),这个假设确实成立。以下是定理 1.5 的加权推广。

Theorem 7.1 [BSCIK20, Theorem 7.1] (Weighted correlated agreement over curves – Version I). Let V,q,n,kV, q, n, k and ρ be as defined in Theorem 1.2. Let u={u0,,ul}FqD\mathbf{u} = \{u_0, \cdots, u_l\} \subset \mathbb{F}_q^{\mathcal{D}} . Let α(ρ,1)\alpha \in (\sqrt{\rho}, 1) and let μ:D[0,1]\mu : \mathcal{D} \rightarrow [0,1] be a vector of weights, whose values all have denominator MM . Suppose

Prucurve(u)[agreeμ(u,V)α]>lϵ,\Pr_{u \in \text{curve}(\mathbf{u})} [\text{agree}_{\mu}(u,V) \ge \alpha] > l \cdot \epsilon,

where ε is as defined in Theorem 1.2 (with η=min(αρ,ρ20)\eta = \min(\alpha - \sqrt{\rho}, \frac{\sqrt{\rho}}{20}) ), and additionally suppose

Prucurve(u)[agreeμ(u,V)α]l(MD+1)q(1η+3ρ).\Pr_{u \in \text{curve}(\mathbf{u})} [\text{agree}_{\mu}(u,V) \ge \alpha] \ge \frac{l(M|\mathcal{D}|+1)}{q} \left(\frac{1}{\eta} + \frac{3}{\sqrt{\rho}}\right) .

Then there exists DD\mathcal{D}' \subset \mathcal{D} and v0,,vlVv_0, \cdots, v_l \in V satisfying

仅适用于 Johnson 界限范围的一种更精确的形式如下。

Theorem 7.2 [BSCIK20, Theorem 7.2] (Weighted correlated agreement over curves – Version II). Let V,q,n,kV, q, n, k and ρ be as defined in Theorem 1.2. Let u={u0,,ul}FqD\mathbf{u} = \{u_0, \cdots, u_l\} \subset \mathbb{F}_q^{\mathcal{D}} . Let μ:D[0,1]\mu : \mathcal{D} \rightarrow [0,1] be a vector of weights, whose values all have denominator MM . Let m3m \ge 3 and let

αα0(ρ,m):=ρ+ρ2m.\alpha \ge \alpha_0(\rho, m) := \sqrt{\rho} + \frac{\rho}{2m}.

Let

S={zFq:agreeμ(u0+zu1++zlul,V)α}S = \{z \in \mathbb{F}_q : \text{agree}_{\mu}(u_0 + zu_1 + \cdots + z^lu_l, V) \ge \alpha\}

and suppose

S>max((1+12m)7m73ρ3/2n2l,2m+1ρ(Mn+1)l).(7.1)|S| > \max\left(\frac{(1 + \frac{1}{2m})^7 m^7}{3 \rho^{3/2}}n^2 l, \frac{2m + 1}{\sqrt{\rho}}(M \cdot n + 1)l \right) . \tag{7.1}

Then u0,,ulu_0, \ldots, u_l have at least α correlated μ -agreement with VV , i.e. v0,,vlV\exists v_0, \cdots, v_l \in V such that

μ({xD:0il,ui(x)=vi(x)})α.\mu(\{x \in \mathcal{D}: \forall 0 \le i \le l, u_i(x) = v_i(x)\}) \ge \alpha .

Theorem 7.3 [BSCIK20, Theorem 7.3] (Weighted correlated agreement over affine spaces). Let V,q,n,kV, q, n, k and ρ be as defined in Theorem 1.2. Let u={u0,,ul}FqD\mathbf{u} = \{u_0, \cdots, u_l\} \subset \mathbb{F}_q^{\mathcal{D}} and let U=u0+span{u1,,ul}FqDU = u_0 + \text{span}\{u_1, \cdots, u_l\} \subset \mathbb{F}_q^{\mathcal{D}} be an affine subspace. Let α(ρ,1)\alpha \in (\sqrt{\rho}, 1) and let μ:D[0,1]\mu : \mathcal{D} \rightarrow [0,1] be a vector of weights, whose values all have denominator MM . Suppose

PruU[agreeμ(u,V)α]>ϵ,\Pr_{u \in U} [\text{agree}_{\mu}(u,V) \ge \alpha] > \epsilon,

where ε is as defined in Theorem 1.2 (with η=min(αρ,ρ20)\eta = \min(\alpha - \sqrt{\rho}, \frac{\sqrt{\rho}}{20}) ), and additionally suppose

PruU[agreeμ(u,V)α]MD+1q(1η+3ρ).\Pr_{u \in U} [\text{agree}_{\mu}(u,V) \ge \alpha] \ge \frac{M|\mathcal{D}| + 1}{q} \left(\frac{1}{\eta} + \frac{3}{\sqrt{\rho}}\right) .

Then there exist DD\mathcal{D}' \subset \mathcal{D} and v0,,vlVv_0, \cdots, v_l \in V satisfying

同样地,对于 Theorem 7.3 也有关于 Johnson 界限的更精确的形式。

Theorem 7.4 [BSCIK20, Theorem 7.4] (Weighted correlated agreement over affine spaces – Version II). Let V,q,n,kV, q, n, k and ρ be as defined in Theorem 1.2. Let u={u0,,ul}FqD\mathbf{u} = \{u_0, \cdots, u_l\} \subset \mathbb{F}_q^{\mathcal{D}} and let U=u0+span{u1,,ul}FqDU = u_0 + \text{span}\{u_1, \cdots, u_l\} \subset \mathbb{F}_q^{\mathcal{D}} be an affine subspace. Let μ:D[0,1]\mu : \mathcal{D} \rightarrow [0,1] be a vector of weights, whose values all have denominator MM . Let m3m \ge 3 and let

αα0(ρ,m):=ρ+ρ2m.\alpha \ge \alpha_0(\rho, m) := \sqrt{\rho} + \frac{\sqrt{\rho}}{2m} .

Suppose

PruU[agreeμ(u,V)α]>max((1+12m)7m73ρ3/2n2q,2m+1ρMn+1q).(7.2)\Pr_{u \in U} [\text{agree}_{\mu}(u,V) \ge \alpha] > \max \left( \frac{(1 + \frac{1}{2m})^7m^7}{3 \rho^{3/2}} \cdot \frac{n^2}{q}, \frac{2m + 1}{\sqrt{\rho}} \cdot \frac{M \cdot n + 1}{q} \right) . \tag{7.2}

Then u0,,ulu_0, \ldots, u_l have at least α correlated μ -agreement with VV , i.e. v0,,vlV\exists v_0, \cdots, v_l \in V such that

μ({xD:0il,ui(x)=vi(x)})α.\mu \left(\{x \in \mathcal{D} : \forall 0 \le i \le l, u_i(x) = v_i(x)\}\right) \ge \alpha .

FRI 协议

FRI 协议的目的是为了在 IOP 模型中,去解决 Reed-Solomon proximity testing 问题,即对于一个接收到的 word f(0):D(0)Ff^{(0)}: \mathcal{D}^{(0)} \rightarrow \mathbb{F} ,验证它到 V(0):=RS[F,D(0),k(0)]V^{(0)} := \text{RS}[\mathbb{F}, \mathcal{D}^{(0)}, k^{(0)}] 之间的 proximity ,如果 f(0)f^{(0)} 属于 V(0)V^{(0)} ,就接受;如果距离 V(0)V^{(0)}δ 远,就拒绝。FRI 协议适用于任何 evaluation domain D(0)\mathcal{D}^{(0)}2-smooth 群 的陪集的情况,即对于任何的 D(0)\mathcal{D}^{(0)} ,其是一个大小为 2s2^s 的(加法或乘法)群的陪集,其中 ss 是一个整数。因此,为简化描述,假设群 D(0)\mathcal{D}^{(0)} 是乘法的。FRI 协议有两个阶段,分别是 COMMIT 阶段与 QUERY 阶段。

在 COMMIT 阶段,经过有限次 rr 轮的交互,会生成一系列的函数 f(1):D(1)F,f(2):D(2)F,,f(r):D(r)Ff^{(1)}: \mathcal{D}^{(1)} \rightarrow \mathbb{F}, f^{(2)}: \mathcal{D}^{(2)} \rightarrow \mathbb{F}, \cdots,f^{(r)}: \mathcal{D}^{(r)} \rightarrow \mathbb{F} 。每一次迭代, domain 的大小 D(i)|\mathcal{D}^{(i)}| 都会缩小。假设对于一个诚实的 prover, f(0)f^{(0)} 是 low-degree ,那么对于每一个 f(i)f^{(i)} ,它们也都会是 low-degree 的(见命题 1)。在第 ii-轮开始时,prover 的消息 f(i):D(i)Ff^{(i)}: \mathcal{D}^{(i)} \rightarrow \mathbb{F} 已经生成,并且 verifier 可以访问该消息的 oracle 。Verifier 现在发送一个均匀随机的 z(i)Fz^{(i)} \in \mathbb{F} ,然后 prover 回复一个新的函数 f(i+1):D(i+1)Ff^{(i+1)}: \mathcal{D}^{(i+1)} \rightarrow \mathbb{F} ,其中 D(i+1)\mathcal{D}^{(i+1)}D(i)\mathcal{D}^{(i)} 的一个(2-smooth)strict 子群,意思是 D(i)\mathcal{D}^{(i)} 不仅是 D(i+1)\mathcal{D}^{(i+1)} 的子群,同时也是其真子集。

D(i+1)\mathcal{D}^{(i+1)}D(i)\mathcal{D}^{(i)} 划分成大小为 l(i):=D(i)/D(i+1)l^{(i)} := |\mathcal{D}^{(i)}|/|\mathcal{D}^{(i+1)}| 的陪集。令 Cg(i)C_g^{(i)} 表示对应于 gD(i+1)g \in \mathcal{D}^{(i+1)} 的陪集,即

Cg(i):={gD(i)(g)l(i)=g}.(8.1)C_g^{(i)} := \{g' \in \mathcal{D}^{(i)} \mid (g')^{l^{(i)}} = g\}. \tag{8.1}

意思就是在 D(i)\mathcal{D}^{(i)} 中选取那些能够通过映射 q(x)=xl(i)q(x) = x^{l^{(i)}} 映射到 D(i+1)\mathcal{D}^{(i+1)} 中的 gg 的元素,这些元素组成了集合 Cg(i)C_g^{(i)} ,其也是陪集。

对于每个陪集 Cg(i)C_g^{(i)}插值映射(interpolation map) Mg(i)M_g^{(i)} 是一个可逆的线性映射 Mg(i):FCg(i)Fl(i)M_g^{(i)}: \mathbb{F}^{C_g^{(i)}} \rightarrow \mathbb{F}^{l^{(i)}} ,它将 f(i)Cg(i):Cg(i)Ff^{(i)}|_{C_g^{(i)}}: C_g^{(i)} \rightarrow \mathbb{F} (即限制 f(i)f^{(i)} 在 domain Cg(i)D(i)C_g^{(i)} \subset \mathcal{D}^{(i)} 上)映射到多项式 Pu(i)(g)(i)(Z)=j<l(i)uj(i)(g)ZjP_{\mathbf{u}^{(i)}(g)}^{(i)}(Z) = \sum_{j<l^{(i)}} u_j^{(i)}(g) Z^j 的系数向量 u(i)(g)=(u0(i)(g),,ul(i)1(i)(g))\mathbf{u}^{(i)}(g) = (u_0^{(i)}(g), \ldots, u_{l^{(i)}-1}^{(i)}(g))^{\intercal} (这里与论文原文表示一致,都表示列向量,不过原文没有加上转置符号),其中 Pu(i)(g)(i)(Z)P_{\mathbf{u}^{(i)}(g)}^{(i)}(Z) 是插值 f(i)Cg(i)f^{(i)}|_{C_g^{(i)}} 的多项式。换句话说,Mg(i)M_g^{(i)} 是由 Cg(i)C_g^{(i)} 生成的 Vandermonde 矩阵的逆,这意味着 (Mg(i))1(u0,,ul(i)1)\left(M_g^{(i)}\right)^{-1} \cdot (u_0, \ldots, u_{l^{(i)}-1})^{\intercal} 是多项式 Pu(X)=i<l(i)uiXiP_{\mathbf{u}}(X) = \sum_{i<l^{(i)}} u_i X^i 在陪集 Cg(i)C_g^{(i)} 上的 evaluation 。

👀 Notice 本文为保持前后一致,用 (x0,,xn)(x_0, \ldots, x_n)表示行向量,而 (x0,,xn)(x_0, \ldots, x_n)^{\intercal} 表示列向量,也可写为: $$

[>x0>x1>>xn>]\begin{bmatrix} > x_0\\ > x_1\\ > \vdots\\ > x_n > \end{bmatrix}

$$

下面详细解释下上面对插值映射的描述。根据 Cg(i)C_g^{(i)} 的定义,知道其中的元素有 l(i)l^{(i)} 个,设 Cg(i)={g1,,gl(i)}C_g^{(i)} =\{g'_1, \cdots, g'_{l^{(i)}}\} ,我们可以将由 Cg(i)C_g^{(i)} 生成的 Vandermonde 矩阵写出来:

VCg(i)=[1g1(g1)2(g1)l(i)11g2(g2)2(g2)l(i)11gl(i)(gl(i))2(gl(i))l(i)1]V_{C_g^{(i)}}= \begin{bmatrix} 1 & g'_1 & (g'_1)^2 & \cdots & (g'_1)^{l^{(i)}-1} \\ 1 & g'_2 & (g'_2)^2 & \cdots & (g'_2)^{l^{(i)}-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & g'_{l^{(i)}} & (g'_{l^{(i)}})^2 & \cdots & (g'_{l^{(i)}})^{l^{(i)}-1} \end{bmatrix}

Mg(i)=VCg(i)1M_g^{(i)} = V_{C_g^{(i)}}^{-1} ,是由 Cg(i)C_g^{(i)} 生成的 Vandermonde 矩阵的逆,因此

(Mg(i))1(u0,,ul(i)1)=(VCg(i)1)1(u0,,ul(i)1)=VCg(i)(u0,,ul(i)1)=[1g1(g1)2(g1)l(i)11g2(g2)2(g2)l(i)11gl(i)(gl(i))2(gl(i))l(i)1][u0u1ul(i)1]=[u0+u1g1+u2(g1)2++ul(i)1(g1)l(i)1u0+u1g2+u2(g2)2++ul(i)1(g2)l(i)1u0+u1gl(i)+u2(gl(i))2++ul(i)1(gl(i))l(i)1]=[Pu(g1)Pu(g2)Pu(gl(i))]\begin{aligned} \left(M_g^{(i)}\right)^{-1} \cdot (u_0, \ldots, u_{l^{(i)}-1})^{\intercal} & = \left(V_{C_g^{(i)}}^{-1}\right)^{-1} \cdot (u_0, \ldots, u_{l^{(i)}-1})^{\intercal} \\ & = V_{C_g^{(i)}} \cdot (u_0, \ldots, u_{l^{(i)}-1})^{\intercal} \\ & = \begin{bmatrix} 1 & g'_1 & (g'_1)^2 & \cdots & (g'_1)^{l^{(i)}-1} \\ 1 & g'_2 & (g'_2)^2 & \cdots & (g'_2)^{l^{(i)}-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & g'_{l^{(i)}} & (g'_{l^{(i)}})^2 & \cdots & (g'_{l^{(i)}})^{l^{(i)}-1} \end{bmatrix} \begin{bmatrix} u_0\\ u_1\\ \vdots\\ u_{l^{(i)}-1} \end{bmatrix} \\ & = \begin{bmatrix} u_0 + u_1 g'_1 + u_2 (g'_1)^2 + \cdots + u_{l^{(i)}-1} (g'_1)^{l^{(i)}-1}\\ u_0 + u_1 g'_2 + u_2 (g'_2)^2 + \cdots + u_{l^{(i)}-1} (g'_2)^{l^{(i)}-1}\\ \vdots \\ u_0 + u_1 g'_{l^{(i)}} + u_2 (g'_{l^{(i)}})^2 + \cdots + u_{l^{(i)}-1} (g'_{l^{(i)}})^{l^{(i)}-1} \end{bmatrix} \\ & = \begin{bmatrix} P_{\mathbf{u}}(g'_1)\\ P_{\mathbf{u}}(g'_2)\\ \vdots \\ P_{\mathbf{u}}(g'_{l^{(i)}}) \end{bmatrix} \end{aligned}

通过上述推导看出 (Pu(g1),Pu(g2),Pu(gl(i)))\left(P_{\mathbf{u}}(g'_1), P_{\mathbf{u}}(g'_2) \cdots, P_{\mathbf{u}}(g'_{l^{(i)}}) \right)^{\intercal} 是多项式 Pu(X)=i<l(i)uiXiP_{\mathbf{u}}(X) = \sum_{i<l^{(i)}} u_i X^i 在陪集 Cg(i)C_g^{(i)} 上的 evaluation ,因此 (Mg(i))1(u0,,ul(i)1)\left(M_g^{(i)}\right)^{-1} \cdot (u_0, \ldots, u_{l^{(i)}-1})^{\intercal} 是多项式 Pu(X)=i<l(i)uiXiP_{\mathbf{u}}(X) = \sum_{i<l^{(i)}} u_i X^i 在陪集 Cg(i)C_g^{(i)} 上的 evaluation 。

下面的命题使用了上述的符号,重新叙述了 [BBHR18, Section 4.1],与 [BBHR18, Section 4.1] 不同的是,是在乘法群而不是加法群上进行的。该命题描述的就是保持 low-degree 的性质。

Claim 1 [BCIKS20, Claim 8.1]. Suppose that f(i)RS[F,D(i),k(i)]f^{(i)} \in \text{RS}[\mathbb{F}, \mathcal{D}^{(i)}, k^{(i)}] where k(i)+1k^{(i)} + 1 is an integral power of 2 . Then, for any z(i)Fz^{(i)} \in \mathbb{F}, letting z(i)=((z(i))0,(z(i))1,,(z(i))l(i)1)\mathbf{z}^{(i)} = \left(\left(z^{(i)}\right)^0, \left(z^{(i)}\right)^1, \ldots, \left(z^{(i)}\right)^{l^{(i)}-1}\right)^{\intercal}, the function ff(i),z(i)(i+1):D(i+1)Ff_{f^{(i)},z^{(i)}}^{(i+1)}: \mathcal{D}^{(i+1)} \rightarrow \mathbb{F} defined on gD(i+1)g \in \mathcal{D}^{(i+1)} by

ff(i),z(i)(i+1)(g):=(z(i))u(i)(g)=(z(i))Mg(i)f(i)Cg(i)(2)f_{f^{(i)},z^{(i)}}^{(i+1)}(g) := \left(\mathbf{z}^{(i)}\right)^{\intercal} \cdot \mathbf{u}^{(i)}(g) = \left(\mathbf{z}^{(i)}\right)^{\intercal} \cdot M_g^{(i)} \cdot f^{(i)}|_{C_g^{(i)}} \tag{2}

is a valid codeword of V(i+1):=RS[F,D(i+1),k(i+1)]V^{(i+1)} := \text{RS}[\mathbb{F}, \mathcal{D}^{(i+1)}, k^{(i+1)}] where k(i+1):=k(i)+1l(i)1k^{(i+1)} := \frac{k^{(i)}+1}{l^{(i)}} - 1.

根据 [BBHR18] 以及上面的记号,在 FRI 协议的 COMMIT 阶段,固定一个 gD(i+1)g \in \mathcal{D}^{(i+1)} ,下一步构造的 ff(i),z(i)(i+1)(g):=Pu(i)(g)(i)(z(i))f_{f^{(i)},z^{(i)}}^{(i+1)}(g) := P_{\mathbf{u}^{(i)}(g)}^{(i)}(z^{(i)}) ,下面理解下上述构造的式子。

ff(i),z(i)(i+1)(g)=Pu(i)(g)(i)(z(i))=j<l(i)uj(i)(g)(z(i))j=(z(i))0u0(i)(g)+(z(i))1u1(i)(g)++(z(i))l(i)1ul(i)1(i)(g)=[(z(i))0(z(i))1(z(i))l(i)1][u0(i)(g)u1(i)(g)ul(i)1(i)(g)]=(z(i))u(i)(g)\begin{aligned} f_{f^{(i)},z^{(i)}}^{(i+1)}(g) & = P_{\mathbf{u}^{(i)}(g)}^{(i)}(z^{(i)}) \\ & = \sum_{j<l^{(i)}} u_j^{(i)}(g) \cdot (z^{(i)})^j \\ & = \left(z^{(i)}\right)^0 \cdot u_0^{(i)}(g) + \left(z^{(i)}\right)^1 \cdot u_1^{(i)}(g) + \cdots + \left(z^{(i)}\right)^{l^{(i)}-1} \cdot u_{l^{(i)}-1}^{(i)}(g) \\ & = \begin{bmatrix} \left(z^{(i)}\right)^0 & \left(z^{(i)}\right)^1 & \cdots & \left(z^{(i)}\right)^{l^{(i)}-1} \end{bmatrix} \cdot \begin{bmatrix} u_0^{(i)}(g)\\ u_1^{(i)}(g)\\ \vdots\\ u_{l^{(i)}-1}^{(i)}(g) \end{bmatrix} \\ & = \left(\mathbf{z}^{(i)}\right)^{\intercal} \cdot \mathbf{u}^{(i)}(g) \end{aligned}

下面说明下命题 1 给的第二个等式,即 u(i)(g)=Mg(i)f(i)Cg(i)\mathbf{u}^{(i)}(g) = M_g^{(i)} \cdot f^{(i)}|_{C_g^{(i)}} 。根据前面的分析,对于由 Cg(i)C_g^{(i)} 生成的 Vandermonde 矩阵有

VCg(i)[u0(i)u1(i)ul(i)1(i)]=[Pu(g1)Pu(g2)Pu(gl(i))]=[f(i)Cg(i)(g1)f(i)Cg(i)(g2)f(i)Cg(i)(gl(i))]V_{C_g^{(i)}} \cdot \begin{bmatrix} u_0^{(i)}\\ u_1^{(i)}\\ \vdots\\ u_{l^{(i)}-1}^{(i)} \end{bmatrix} = \begin{bmatrix} P_{\mathbf{u}}(g'_1)\\ P_{\mathbf{u}}(g'_2)\\ \vdots \\ P_{\mathbf{u}}(g'_{l^{(i)}}) \end{bmatrix} = \begin{bmatrix} f^{(i)}|_{C_g^{(i)}}(g'_1)\\ f^{(i)}|_{C_g^{(i)}}(g'_2)\\ \vdots \\ f^{(i)}|_{C_g^{(i)}}(g'_{l^{(i)}}) \end{bmatrix}

因此

[u0(i)u1(i)ul(i)1(i)]=(VCg(i))1[f(i)Cg(i)(g1)f(i)Cg(i)(g2)f(i)Cg(i)(gl(i))]=Mg(i)f(i)Cg(i)\begin{bmatrix} u_0^{(i)}\\ u_1^{(i)}\\ \vdots\\ u_{l^{(i)}-1}^{(i)} \end{bmatrix} = \left(V_{C_g^{(i)}}\right)^{-1} \cdot \begin{bmatrix} f^{(i)}|_{C_g^{(i)}}(g'_1)\\ f^{(i)}|_{C_g^{(i)}}(g'_2)\\ \vdots \\ f^{(i)}|_{C_g^{(i)}}(g'_{l^{(i)}}) \end{bmatrix} = M_g^{(i)} \cdot f^{(i)}|_{C_g^{(i)}}

由此得到了 (u(i)(g))=Mg(i)f(i)Cg(i)\left(\mathbf{u}^{(i)}(g)\right)^{\intercal} = M_g^{(i)} \cdot f^{(i)}|_{C_g^{(i)}}

Batching

在某些情况下,第一个 prover 的 oracle f(0)f^{(0)} 是从一个仿射空间 FFD(0)F \subset \mathbb{F}^{\mathcal{D}^{(0)}} 的函数中采样的,这个仿射空间作为我们的输入,

F={f0(0)+i=1txifi(0)xiF,fi:D(0)F}(3)F = \left\{ f_0^{(0)} + \sum_{i=1}^{t} x_i \cdot f_i^{(0)} \mid x_i \in \mathbb{F}, f_i : \mathcal{D}^{(0)} \to \mathbb{F} \right\} \tag{3}

当使用 FRI 协议来 “batch” 多个不同的 low degree testing 问题实例时,我们就通过随机线性组合将它们全部组合在一起,即上式中的 f0(0)+x1f1(0)++xtft(0)f_0^{(0)} + x_1 f_1^{(0)} + \cdots +x_t f_t^{(0)} 。在这种 batching 设置中,我们假设 prover 已经承诺了 f1(0),,ft(0)f_1^{(0)}, \ldots, f_t^{(0)} (注意在这种情况下我们设 f0(0)=0f_0^{(0)} = 0),并且 batched FRI 的 verifier 从 F\mathbb{F} 中均匀随机地采样 x1,,xtFx_1, \ldots, x_t \in \mathbb{F}, prover 答复 f(0)f^{(0)} ,其应该等于 f0(0)+i=1txifi(0)f_0^{(0)} + \sum_{i=1}^{t} x_i \cdot f_i^{(0)} ,现在 FRI 协议就应用于 f(0)f^{(0)} 了。相应地,batched FRI 的 QUERY 阶段也被扩展了,因此每次请求 f(0)(g)f^{(0)}(g) 的查询时,验证者同时也查询了 f0(0)(g),,ft(0)(g)f_0^{(0)}(g), \ldots, f^{(0)}_t(g) 并验证 f(0)(g)=f0(0)(g)+i=1txifi(0)(g)f^{(0)}(g) = f_0^{(0)}(g) + \sum_{i=1}^{t} {\color{orange}x_i} \cdot f^{(0)}_i(g)

🐞 Fix

  • 论文这里的公式为 f(0)(g)=f0(0)(g)+i=1tfi(0)(g)f^{(0)}(g) = f_0^{(0)}(g) + \sum_{i=1}^{t} f^{(0)}_i(g) ,我认为这里少了前面的系数 xix_i ,应该为 f(0)(g)=f0(0)(g)+i=1txifi(0)(g)f^{(0)}(g) = f_0^{(0)}(g) + \sum_{i=1}^{t} x_i \cdot f^{(0)}_i(g)

The (batched) FRI QUERY phase

命题 1 表明对于诚实的 prover ,verifier 选取任意的值 z(i)z^{(i)} ,对于每一个 yD(i+1)y \in D^{(i+1)} ,prover 都可以通过计算 (2)(2) 式,从一个码字 f(i)V(i)f^{(i)} \in V^{(i)} 构建一个新的码字 f(i+1)V(i+1)f^{(i+1)} \in V^{(i+1)} 。因此,我们将始终假设 f(r)V(r)f^{(r)} \in V^{(r)},例如,通过假设验证者总是查询 f(r)f^{(r)} 的前 k(r)k^{(r)} 个元素(按照某种规范顺序)并将 f(r)f^{(r)} 与该函数的插值多项式进行比较。

命题 1 给出了一种非常自然的测试方法,用来检查 f(i)f^{(i)}f(i+1)f^{(i+1)} 之间的一致性,并且 FRI 的查询阶段通过从 “顶部”(f(r)f^{(r)})到 “底部”(f(0)f^{(0)})迭代应用这种自然测试来遵循这一过程。

🤔 Question

  • 如何更好地去解释这里的自然的测试方法呢?

A single invocation of the FRI QUERY phase

  1. D(r)\mathcal{D}^{(r)} 中均匀随机地选取 g(r)g^{(r)} 。对于 i=r,,1i = r, \cdots, 1 ,从陪集 Cg(i)(i1)C_{g^{(i)}}^{(i-1)} 中均匀随机地选取 g(i1)g^{(i-1)}
  2. 如果 f(0)(g(0))f0(0)(g(0))+i=1txifi(0)(g(0))f^{(0)}(g^{(0)}) \neq f_0^{(0)}(g^{(0)}) + \sum_{i = 1}^{t}x_i \cdot f_i^{(0)}(g^{(0)}) ,则拒绝。
  3. 如果,对于任意地 i{0,,r1}i \in \{0, \cdots, r - 1\} ,有 f(i+1)(g(i+1))(z(i))Mg(i)f(i)Cg(i)f^{(i+1)}(g^{(i+1)}) \neq \left(\mathbf{z}^{(i)}\right)^{\intercal} \cdot M_g^{(i)} \cdot f^{(i)}|_{C_g^{(i)}} ,则拒绝。
  4. 否则 —— 如果上述条件中的所有等式都成立,则接受。

上述 QUERY 过程与 [BBHR18] 中 FRI 的 QUERY 过程有所不同,这里选取随机数是从最后一个 D(r)\mathcal{D}^{(r)} 中开始选取的,而不是从初始的 D(0)\mathcal{D}^{(0)} 中选取的。相比[BBHR18] 中的 QUERY 阶段,这里我们还想要验证在第 0 步时,batch 的是否正确,也就是 f(0)(g(0))f0(0)(g(0))+i=1txifi(0)(g(0))f^{(0)}(g^{(0)}) \neq f_0^{(0)}(g^{(0)}) + \sum_{i = 1}^{t}x_i \cdot f_i^{(0)}(g^{(0)})

Summary of the batched FRI protocol

下面总结下目前提到的比较重要的性质,将会在下面的 soundness 分析中用到。

  1. 在协议的 COMMIT 阶段结束时,verifier 可以通过 oracle 访问一系列函数 f(0):D(0)F,,f(r):D(r)Ff^{(0)}: \mathcal{D}^{(0)} \rightarrow \mathbb{F}, \ldots, f^{(r)}: \mathcal{D}^{(r)} \rightarrow \mathbb{F} ,其中 D(0)D(r)\mathcal{D}^{(0)} \supsetneq \ldots \supsetneq \mathcal{D}^{(r)} 是一系列 2-smooth 群,并且 f(i)f^{(i)} 任意依赖于 z(0),,z(i)z^{(0)}, \ldots, z^{(i)} (以及 f(0),,f(i1)f^{(0)}, \ldots, f^{(i-1)})。我们假设 f(r)V(r)f^{(r)} \in V^{(r)}
  2. 存在一组 l(i)×l(i)l^{(i)} \times l^{(i)} 的可逆矩阵 {Mg(i+1)(i):g(i+1)D(i+1)}\{M_{g^{(i+1)}}^{(i)} : g^{(i+1)} \in D^{(i+1)}\} ,因此将 Mg(i+1)(i)M_{g^{(i+1)}}^{(i)} 应用于 f(i)Cg(i+1)(i)f^{(i)}|_{C_{g(i+1)}^{(i)}} 可以将 f(i)f^{(i)} 映射到一个向量序列 u=u(i)={u0(i),,ul(i)(i)}FD(i+1)\mathbf{u} = \mathbf{u}^{(i)} = \{u_0^{(i)}, \ldots, u_{l(i)}^{(i)}\} \subset \mathbb{F}^{D^{(i+1)}} ,其中
u(i)(g(i+1))=(u0(i)(g(i+1)),,ul(i)1(i)(g(i+1)))=Mg(i+1)(i)f(i)Cg(i+1)(i).(4)\mathbf{u}^{(i)}\left(g^{(i+1)}\right) = \left(u_0^{(i)}\left(g^{(i+1)}\right), \cdots, u_{l^{(i)}-1}^{(i)}\left(g^{(i+1)}\right) \right) = M_{g^{(i+1)}}^{(i)} \cdot f^{(i)}|_{C_{g(i+1)}^{(i)}}. \tag{4}

此外,如果 f(i)f^{(i)} 是在 D(i)D^{(i)} 上码率为 ρ 的有效 RS 码字,那么通过 u(i)\mathbf{u}^{(i)} 的参数化曲线上的每个向量也是在 D(i+1)D^{(i+1)} 上码率为 ρ 的有效 RS 码字。

  1. 在 QUERY 阶段的每次迭代会检查 f(i+1)f^{(i+1)} 是否是通过方程 (2)(2)f(i)f^{(i)} 构造的,并且(在 batched 的情况下)检查 f(0)f^{(0)} 是否是通过方程 (3)(3) 正确计算的。

Soundness

Lemma 8.2 [BSCIK20, Lemma 8.2] (batched FRI error bound). Let V(0)=RS[F,D(0),k(0)]V^{(0)} = \text{RS}[\mathbb{F}, \mathcal{D}^{(0)}, k^{(0)}] where D(0)\mathcal{D}^{(0)} is a coset of a 2-smooth multiplicative group, and k(0)+1k^{(0)} + 1 is a power of 2; set ρ=(k(0)+1)/D(0)\rho = (k^{(0)} + 1)/|\mathcal{D}^{(0)}|.

Let FFD(0)F \subseteq \mathbb{F}^{\mathcal{D}^{(0)}} be a space of functions as defined in Eq. (3)(3) whose correlated agreement density with V(0)V^{(0)} is α. For interger m3m \ge 3 , let

α(0)(ρ,m)=max{α,ρ(1+1/2m)}.\alpha^{(0)}(\rho, m) = \max\left\{ \alpha, \sqrt{\rho}(1 + 1/2m) \right\}.

Assume the FRI protocol is used with rr rounds, and let l(i)=D(i)/D(i+1)l^{(i)} = |\mathcal{D}^{(i)}|/|\mathcal{D}^{(i+1)}| denote the ratio between prover messages (oracles) ii and i+1i + 1. Let ϵQ\epsilon_Q denote the probability that the verifier accepts a single FRI QUERY invocation. Then,

Prx1,,xt,z(0),,z(r1)[ϵQ>α(0)(ρ,m)]ϵC,(8.5)\Pr_{x_1, \ldots, x_t, z^{(0)}, \ldots, z^{(r-1)}}\left[ \epsilon_Q > \alpha^{(0)}(\rho, m) \right] \le \epsilon_{\text{C}}, \tag{8.5}

where

ϵC=(m+12)7D(0)22ρ3/2F+(2m+1)(D(0)+1)ρi=0r1l(i)F.\epsilon_{\text{C}} = \frac{\left(m + \frac{1}{2}\right)^7 \cdot \vert \mathcal{D}^{(0)}\vert^2}{2 \rho^{3/2} \vert \mathbb{F} \vert} + \frac{(2m+1) \cdot (\vert \mathcal{D}^{(0)}\vert + 1)}{\sqrt{\rho}} \cdot \frac{\sum_{i = 0}^{r - 1} l^{(i)}}{\vert \mathbb{F} \vert}.

In words: For any interactive FRI prover PP^*, the probability that the oracles f(0),,f(r)f^{(0)}, \ldots, f^{(r)} sent by PP^* will pass a single invocation of the batched FRI QUERY test with probability greater than α(0)(ρ,m)\alpha^{(0)}(\rho, m), is smaller than ϵC\epsilon_{\text{C}}. The probability is over the random variables x1,,xtx_1, \ldots, x_t used to sample f(0)f^{(0)} from FF and over the random messages z(0),,z(r1)z^{(0)}, \ldots, z^{(r-1)} sent by the verifier during the COMMIT phase.

Theorem 8.3 [BSCIK20, Theorem 8.3] (Batched FRI Soundness). Let f0(0),,ft(r):D(0)Ff_0^{(0)}, \ldots, f_t^{(r)}: \mathcal{D}^{(0)} \rightarrow \mathbb{F} be a sequence of functions and let V(0)=RS[F,D(0),k(0)]V^{(0)} = \text{RS}[\mathbb{F}, \mathcal{D}^{(0)}, k^{(0)}] where D(0)\mathcal{D}^{(0)} is a coset of a 2-smooth group of size n(0)=D(0)n^{(0)} = |\mathcal{D}^{(0)}|, and ρ=k(0)+1n(0)\rho = \frac{k^{(0)} + 1}{n^{(0)}} satisfies ρ=2R\rho = 2^{-\text{R}} for positive integer R\text{R}. Let α=ρ(1+1/2m)\alpha = \sqrt{\rho}(1 + 1/2m) for integer m3m \ge 3 and ϵC\epsilon_{\text{C}} be as defined in Lemma 8.2.

Assume the FRI protocol is used with rr rounds. Let l(i)=D(i)/D(i+1)l^{(i)} = |\mathcal{D}^{(i)}|/|\mathcal{D}^{(i+1)}| denote the ratio between prover messages (oracles) ii and i+1i + 1. Assume furthermore that ss is the number of invocations of the FRI QUERY step.

Suppose there exists a batched FRI prover PP^* that interacts with the batched FRI verifier and causes it to output “accept” with probability greater than

ϵFRI:=ϵC+αs=(m+12)7D(0)22ρ3/2F+(2m+1)(D(0)+1)ρi=0r1l(i)F+(ρ(1+12m))s.\epsilon_{\text{FRI}} := \epsilon_{\text{C}} + \alpha^s = \frac{\left(m + \frac{1}{2}\right)^7 \cdot \vert \mathcal{D}^{(0)}\vert^2}{2 \rho^{3/2} \vert \mathbb{F} \vert} + \frac{(2m+1) \cdot (\vert \mathcal{D}^{(0)}\vert + 1)}{\sqrt{\rho}} \cdot \frac{\sum_{i = 0}^{r - 1} l^{(i)}}{\vert \mathbb{F} \vert} + \left(\sqrt{\rho} \cdot \left( 1 + \frac{1}{2m} \right) \right)^s .

Then f0(0),,ft(0)f_0^{(0)}, \ldots, f_t^{(0)} have correlated agreement with V(0)V^{(0)} on a domain DD(0)\mathcal{D}' \subset \mathcal{D}^{(0)} of density at least α.

定理 8.3 证明:反证法,然后直接通过引理 8.2 来证明。假设 f0(0),,ft(0)f_0^{(0)}, \ldots, f_t^{(0)}V(0)V^{(0)} 的最大 correlated agreement 小于 α(0)(ρ,m)=ρ(1+1/2m)\alpha^{(0)}(\rho, m) = \sqrt{\rho}(1 + 1/2m),但是同时接受的概率大于 ϵC+(α(0)(ρ,m))s\epsilon_{\text{C}} + (\alpha^{(0)}(\rho, m))^s

EE 为在每次 FRI QUERY 阶段接受的概率大于 α(0)(ρ,m)\alpha^{(0)}(\rho, m) 的事件。这个事件依赖于 x1,,xt,f(0),z(0),,z(r1),f(r)x_1, \ldots, x_t, f^{(0)}, z^{(0)}, \ldots, z^{(r-1)}, f^{(r)} ,其中每个 f(i)f^{(i)}PP^* 根据之前 Verifier 的消息生成的。通过引理 8.2,对于任意的 Prover PP^* ,有事件 EE 发生的概率不超过 ϵC\epsilon_{\text{C}} 。当事件 EE 不成立时,那么 ss 次独立调用 FRI QUERY 都返回 “accept” 的概率不超过 (α(0)(ρ,m))s(\alpha^{(0)}(\rho, m))^s

因此,FRI 的 Verifier 接受的概率不超过 ϵC+(α(0)(ρ,m))s\epsilon_{\text{C}} + (\alpha^{(0)}(\rho, m))^s ,这与假设矛盾。\Box

Question

  • 这里每次 FRI QUERY 阶段接受的概率大于 α(0)(ρ,m)\alpha^{(0)}(\rho, m) 的事件 EE ,怎么联系调用 ss 次的概率依然是不超过 ϵC\epsilon_{\text{C}} 呢?为什么不是 (ϵC)s(\epsilon_{\text{C}})^s 呢?

Proof of Lemma 8.2

在证明 Lemma 8.2 之前,先介绍一种跟踪 verifier 检查 consistency 是否通过的方法。具体来说,Prover 会根据 Verifier 发送的随机数 z(i)z^{(i)} 来构造函数 f(i+1)f^{(i+1)} ,然后向 Verifier 回应函数 f(i+1)f^{(i+1)} 。在 FRI 的 QUERY 阶段, Verifier 会检查函数 f(i+1)f^{(i+1)} 与函数 f(i)f^{(i)} 的 consistency 。

定义一系列的加权(weight)函数,μ(i):D(i)[0,1]\mu^{(i)}:\mathcal{D}^{(i)} \rightarrow [0,1] 以及 ν(i):D(i)[0,1]\nu^{(i)}:\mathcal{D}^{(i)} \rightarrow [0,1] ,其中 i=0,,ri = 0, \ldots, r 。这些加权函数是通过归纳法来定义的。当 i=0i = 0 时,用 {0,1}\{0,1\} weights 来指示 f(0)(g)f^{(0)}(g) 是否计算正确:

μ(0)(g)={1f(0)(g)=f0(0)(g)+i=1txifi(0)(g)0otherwise\mu^{(0)}(g) = \begin{cases} 1 & f^{(0)}(g) = f_0^{(0)}(g) + \sum_{i = 1}^t x_i f_i^{(0)}(g) \\ 0 & \text{otherwise} \end{cases}

现在,可以通过归纳法得到的 μ(i)\mu^{(i)} 可以来定义一个辅助 weight 函数 ν(i+1):D(i+1)[0,1]\nu^{(i+1)}: \mathcal{D}^{(i+1)} \rightarrow [0,1] 。在 D(i+1)\mathcal{D}^{(i+1)} 中取一个元素 gg ,就可以得到 D(i)\mathcal{D}^{(i)} 中的陪集 Cg(i)D(i)C_g^{(i)} \subset \mathcal{D}^{(i)} ,这个集合是由那些在 D(i)\mathcal{D}^{(i)} 中能过通过映射 q(i)(x)=xl(i)q^{(i)}(x) = x^{l^{(i)}} 得到 gg 的所有元素组成的,即如 (8.1)(8.1) 所示,

Cg(i):={gD(i)(g)l(i)=g}.C_g^{(i)} := \{g' \in \mathcal{D}^{(i)} \mid (g')^{l^{(i)}} = g\}.

那么 ν(i+1)\nu^{(i+1)} 的定义为

ν(i+1)(g)=EgCg(i)[μ(i)(g)].(8.6)\nu^{(i+1)}(g) = \mathbb{E}_{g' \in C_g^{(i)}}\left[ \mu^{(i)}(g') \right]. \tag{8.6}

换句话说,ν(i+1)(g)\nu^{(i+1)}(g) 是陪集 Cg(i)C_g^{(i)} 中的所有元素的 μ(i)\mu^{(i)} weight 的期望值。最后,来定义函数 μ(i+1)\mu^{(i+1)} ,对于每一个 gD(i+1)g \in \mathcal{D}^{(i+1)}

μ(i+1)(g)={ν(i+1)(g)f(i+1)(g)=ff(i),z(i)(i+1)(g)0otherwise\mu^{(i+1)}(g) = \begin{cases} \nu^{(i+1)}(g) & f^{(i+1)}(g) = f_{f^{(i)}, z^{(i)}}^{(i+1)}(g) \\ 0 & \text{otherwise} \end{cases}

关于 μ(i)\mu^{(i)} 的定义,一个很重要的性质就是,μ(i)(g)\mu^{(i)}(g) 是在 ggf(i)f^{(i)} 中 query 的条件下,FRI QUERY 阶段成功的概率的一个度量,这也是下面的命题成立的一个重要原因。

Claim 8.5. The probability ϵQ\epsilon_{\text{Q}} that a single invocation of the batched FRI QUERY accepts f(0),,f(r)f^{(0)}, \ldots, f^{(r)}, where f(r)RS[F,D(r),k(r)]f^{(r)} \in \text{RS}[\mathbb{F},\mathcal{D}^{(r)},k^{(r)}], satisfies

ϵQ=Eg(r)D(r)[μ(r)(g(r))].\epsilon_{\text{Q}} = \mathbb{E}_{g^{(r)} \in \mathcal{D}^{(r)}}\left[ \mu^{(r)}(g^{(r)}) \right].

证明:回顾下 FRI QUERY 的调用,会选择一系列随机的 g(r),,g(0)g^{(r)}, \ldots, g^{(0)} ,其中 g(i1)g^{(i-1)} 是从陪集 Cg(i)(i1)C_{g^{(i)}}^{(i-1)} 中均匀随机选取的。下面通过归纳法来证明,对于 i=0,,ri = 0, \ldots, r

Eg(i)D(i)[μ(i)(g(i))]\mathbb{E}_{g^{(i)} \in \mathcal{D}^{(i)}}\left[ \mu^{(i)}(g^{(i)}) \right]

等于这样的概率,当均匀随机地选取 g(i)g^{(i)} ,并且其是从一个随机序列 g(i1)Cg(i)(i1),,g(0)Cg(1)(0)g^{(i-1)} \in C_{g^{(i)}}^{(i-1)}, \ldots, g^{(0)} \in C_{g^{(1)}}^{(0)} 中生成的时,所有和 g(i)g^{(i)} 及其引发的测试都通过的概率。

归纳法证明的思路如下:

  1. 证明对于 i=0i = 0 时,最基本的情况 μ(0)\mu^{(0)} 成立
  2. 假设对于 i1i - 1μ(i1)\mu^{(i-1)} 成立,证明对于 iiμ(i)\mu^{(i)} 成立

从而就能证明该命题。

i=0i = 0 时,由 μ(0)\mu^{(0)} 的定义,

μ(0)(g(0))={1f(0)(g(0))=f0(0)(g(0))+i=1txifi(0)(g(0))0otherwise\mu^{(0)}(g^{(0)}) = \begin{cases} 1 & f^{(0)}(g^{(0)}) = f_0^{(0)}(g^{(0)}) + \sum_{i = 1}^t x_i f_i^{(0)}(g^{(0)}) \\ 0 & \text{otherwise} \end{cases}

调用 FRI QUERY 通过的概率自然等于 Eg(0)D(0)[μ(0)(g(0))]\mathbb{E}_{g^{(0)} \in \mathcal{D}^{(0)}}\left[ \mu^{(0)}(g^{(0)}) \right]

假设对于 i1i - 1μ(i1)\mu^{(i-1)} 成立,现在分析 μ(i)(g(i))\mu^{(i)}(g^{(i)}) ,如果 f(i)(g(i))f^{(i)}(g^{(i)}) 未按照式 (8.2)(8.2) 正确计算,那么 μ(i)(g(i))=0\mu^{(i)}(g^{(i)}) = 0 ,否则的话,根据定义

μ(i)(g(i))=ν(i)(g(i))=Eg(i1)Cg(i)(i1)[μ(i1)(g(i1))].\mu^{(i)}(g^{(i)}) = \nu^{(i)}(g^{(i)}) = \mathbb{E}_{g^{(i-1)} \in C_{g^{(i)}}^{(i-1)}}\left[ \mu^{(i-1)}(g^{(i-1)}) \right].

这说明 μ(i)(g(i))\mu^{(i)}(g^{(i)}) 是在陪集 Cg(i)(i1)D(i1)C_{g^{(i)}}^{(i-1)} \subseteq \mathcal{D}^{(i-1)}μ(i1)\mu^{(i-1)} 的值的平均值,由归纳法假设得,其是和 g(i1),,g(0)g^{(i-1)}, \ldots, g^{(0)} 相关的所有测试通过的概率,因此可得对于 iiμ(i)\mu^{(i)} 成立。\Box

Lemma 8.2 需要估计的是在 FRI QUERY 阶段的概率,回顾上面的 batched FRI QUERY 阶段的协议,有两个地方涉及到随机数:

  1. 协议的第 2 步,使用 tt 个随机数 x1,,xtx_1, \ldots , x_t 来 batch f1(0),f2(0),,ft(0)f_1^{(0)}, f_2^{(0)}, \ldots ,f_t^{(0)} ,这对应 affine space 的情况,要用到定理 7.4 对应的结论。
  2. 协议的第 3 步,用 z(i)=((z(i))0,(z(i))1,,(z(i))l(i)1)\mathbf{z}^{(i)} = \left(\left(z^{(i)}\right)^0, \left(z^{(i)}\right)^1, \ldots, \left(z^{(i)}\right)^{l^{(i)}-1}\right) 来进行 batch ,对应 curves 的情况,会用到定理 7.2 的结论。

Lemma 8.2 证明:现在要证明引理 8.2 ,由命题 8.5,只需证明,在 verifier 的随机选择中,以大于概率 1ϵC1 - \epsilon_C

EgD(r)[μ(r)(g)]α(0)(ρ,m).(8.7)\mathbb{E}_{g \in \mathcal{D}^{(r)}}\left[ \mu^{(r)}(g) \right] \le \alpha^{(0)}(\rho, m). \tag{8.7}

如果证明了上述成立,那么也就是说当在 Fq\mathbb{F}_q 中选取随机数时,如果有 ϵQ>α(0)(ρ,m)\epsilon_Q > \alpha^{(0)}(\rho, m) ,那么其概率小于等于 ϵC\epsilon_C ,这也就证明了引理 8.2。

🤔 Question

  • 这里为什么不是 “以大于概率 1ϵC1 - \epsilon_C 有”?

证明的思路是先定义一些列坏的事件 E(0),,E(r)E^{(0)}, \ldots, E^{(r)} ,其中一些事件会发生的概率是各个事件发生的概率之和,证明这个概率小于等于 ϵC\epsilon_C 。接着再假设当没有坏的事件发生时,证明式子 (8.7)(8.7) 成立。

E(0)E^{(0)} 为事件

agreeμ(0)(f(0),V(0))>α(0)(ρ,m).\text{agree}_{\mu^{(0)}} \left(f^{(0)}, V^{(0)}\right) > \alpha^{(0)}(\rho, m).

μ(0)\mu^{(0)} 的定义得事件 E(0)E^{(0)}

agree(f0(0)+i=1txifi(0),V(0))>α(0)(ρ,m)=max{α,ρ(1+1/2m)}.\text{agree} \left(f_0^{(0)} + \sum_{i=1}^{t} x_i f_i^{(0)} , V^{(0)}\right) > \alpha^{(0)}(\rho, m) = \max\left\{ \alpha, \sqrt{\rho}(1 + 1/2m) \right\}.

🤔 Question

  • 这里的 agree\text{agree} 具体是什么含义?和 agreeμ(0)\text{agree}_{\mu^{(0)}} 的区别是什么?是表示常数 1 吗?

因此这个事件 E(0)E^{(0)} 主要取决于随机数 x1,,xtx_1, \ldots, x_t 。通过引理中的假设,有 (f0(0),,ft(0))(f_0^{(0)}, \ldots, f_t^{(0)})V(0)V^{(0)} 的最大 correlated agreement density 不超过 α

回顾下定理 7.4: Theorem 7.4 (Weighted correlated agreement over affine spaces – Version II). Let V,q,n,kV, q, n, k and ρ be as defined in Theorem 1.2. Let u={u0,,ul}FqD\mathbf{u} = \{u_0, \cdots, u_l\} \subset \mathbb{F}_q^{\mathcal{D}} and let U=u0+span{u1,,ul}FqDU = u_0 + \text{span}\{u_1, \cdots, u_l\} \subset \mathbb{F}_q^{\mathcal{D}} be an affine subspace. Let μ:D[0,1]\mu : \mathcal{D} \rightarrow [0,1] be a vector of weights, whose values all have denominator MM . Let m3m \ge 3 and let

>αα0(ρ,m):=ρ+ρ2m.>> \alpha \ge \alpha_0(\rho, m) := \sqrt{\rho} + \frac{\sqrt{\rho}}{2m} . >

Suppose

>PruU[agreeμ(u,V)α]>max((1+12m)7m73ρ3/2n2q,2m+1ρMn+1q).>(7.2)> \Pr_{u \in U} [\text{agree}_{\mu}(u,V) \ge \alpha] > \max \left( \frac{(1 + \frac{1}{2m})^7m^7}{3 \rho^{3/2}} \cdot \frac{n^2}{q}, \frac{2m + 1}{\sqrt{\rho}} \cdot \frac{M \cdot n + 1}{q} \right) . \tag{7.2} >

Then u0,,ulu_0, \ldots, u_l have at least α correlated μ -agreement with VV , i.e. v0,,vlV\exists v_0, \cdots, v_l \in V such that

>μ({xD:0il,ui(x)=vi(x)})α.>> \mu \left(\{x \in \mathcal{D} : \forall 0 \le i \le l, u_i(x) = v_i(x)\}\right) \ge \alpha . >

则其逆否命题为:如果 u0,,ulu_0, \ldots, u_lVV 有至多 α 的 correlated μ -agreement ,那么有

>PruU[agreeμ(u,V)α]max((1+12m)7m73ρ3/2n2q,2m+1ρMn+1q).>> \Pr_{u \in U} [\text{agree}_{\mu}(u,V) \ge \alpha] \le \max \left( \frac{(1 + \frac{1}{2m})^7m^7}{3 \rho^{3/2}} \cdot \frac{n^2}{q}, \frac{2m + 1}{\sqrt{\rho}} \cdot \frac{M \cdot n + 1}{q} \right) . >

由定理 7.4 的逆否命题,取 α=α(0)(ρ,m)\alpha = \alpha^{(0)}(\rho, m)μ1\mu \equiv 1 以及 M=1M = 1 ,有

Prx1,,xt[E(0)]=PruU[agreeμ(u,V)>α(0)(ρ,m)](为什么这里括号里能直接改为严格的>?)max((1+12m)7m73ρ3/2n2q,2m+1ρMn+1q)=max((m+12)73ρ3/2n2q,2m+1ρn+1q)\begin{aligned} \Pr_{x_1, \ldots, x_t} [E^{(0)}] & = \Pr_{u \in U} [\text{agree}_{\mu}(u,V) > \alpha^{(0)}(\rho, m)] \\ & {\color{orange}\text{(为什么这里括号里能直接改为严格的$>$?)}} \\ & \le \max \left( \frac{(1 + \frac{1}{2m})^7m^7}{3 \rho^{3/2}} \cdot \frac{n^2}{q}, \frac{2m + 1}{\sqrt{\rho}} \cdot \frac{M \cdot n + 1}{q} \right) \\ & = \max \left( \frac{(m + \frac{1}{2})^7}{3 \rho^{3/2}} \cdot \frac{n^2}{q}, \frac{2m + 1}{\sqrt{\rho}} \cdot \frac{n + 1}{q} \right) \end{aligned}

注意,根据定理 7.4 以及定理 1.2,其中 V=RS[Fq,D(0),k(0)]V = \mathsf{RS}[\mathbb{F}_q, \mathcal{D}^{(0)}, k^{(0)}]n=D(0)n = |\mathcal{D}^{(0)}|ρ=k(0)+1n\rho = \frac{k^{(0)} + 1}{n}

下面推导下

(m+12)73ρ3/2n2q>2m+1ρn+1q\frac{(m + \frac{1}{2})^7}{3 \rho^{3/2}} \cdot \frac{n^2}{q} > \frac{2m + 1}{\sqrt{\rho}} \cdot \frac{n + 1}{q}

由于

(m+12)732m+1(2m+1)73×272m+1(2m+1)63×27\begin{aligned} & \frac{(m+\frac{1}{2})^7}{3} \ge 2m + 1 \\ \Rightarrow & \frac{(2m + 1)^7}{3 \times 2^7} \ge 2m + 1 \\ \Rightarrow & (2m + 1)^6 \ge 3 \times 2^7 \\ \end{aligned}

由定理中的条件 m3m \ge 3(2m+1)6(2m + 1)^6m3m \ge 3 是增函数,因此 (2m+1)6(2×3+1)6=76=117649(2m + 1)^6 \ge (2 \times 3 + 1)^6 = 7^6 = 117649 ,同时上式的右边 3×27=3843 \times 2^7 = 384 ,满足 (2m+1)6117649>3×27(2m + 1)^6 \ge 117649 > 3 \times 2^7 ,由此得到 (m+12)73>2m+1\frac{(m+\frac{1}{2})^7}{3} > 2m + 1 (取不到等号)成立。那么

(m+12)73ρ3/2n2q>2m+1ρ3/2n2q=2m+1ρρ1/2n2q(由于 ρ<1 )>2m+1ρn2q(由于 n2 时 n2>n+1 )>2m+1ρn+1q\begin{aligned} \frac{(m + \frac{1}{2})^7}{3 \rho^{3/2}} \cdot \frac{n^2}{q} & > \frac{2m + 1}{\rho^{3/2}} \cdot \frac{n^2}{q} \\ & = \frac{2m + 1}{\rho \cdot \rho^{1/2}} \cdot \frac{n^2}{q} \\ & {\color{blue}{\text{(由于 $\rho < 1$ )}}} \\ & > \frac{2m + 1}{\sqrt{\rho}} \cdot \frac{n^2}{q} \\ & {\color{blue}{\text{(由于 $n \ge 2$ 时 $n^2 > n + 1$ )}}} \\ & > \frac{2m + 1}{\sqrt{\rho}} \cdot \frac{n + 1}{q} \end{aligned}

从而

Prx1,,xt[E(0)]max((m+12)73ρ3/2n2q,2m+1ρn+1q)=(m+12)73ρ3/2n2q\begin{aligned} \Pr_{x_1, \ldots, x_t} [E^{(0)}] & \le \max \left( \frac{(m + \frac{1}{2})^7}{3 \rho^{3/2}} \cdot \frac{n^2}{q}, \frac{2m + 1}{\sqrt{\rho}} \cdot \frac{n + 1}{q} \right) \\ & = \frac{(m + \frac{1}{2})^7}{3 \rho^{3/2}} \cdot \frac{n^2}{q} \end{aligned}

ϵ=(m+12)73ρ3/2n2q(注意其中 n=D(0))\epsilon = \frac{(m + \frac{1}{2})^7}{3 \rho^{3/2}} \cdot \frac{n^2}{q} {\color{blue}\quad {\text{(注意其中 $n = |\mathcal{D}^{(0)}|$)}}}

Prx1,,xt[E(0)]ϵ(8.8)\Pr_{x_1, \ldots, x_t} [E^{(0)}] \le \epsilon \tag{8.8}

现在固定 i{0,,r1}i \in \{0, \ldots, r - 1\} 。定义事件 E(i+1)E^{(i+1)}

agreeν(i+1)(ff(i),z(i)(i+1),V(i+1))>max(agreeμ(i)(f(i),V(i)),ρ(1+1/2m)).(8.9)\text{agree}_{\nu^{(i+1)}} \left(f_{f^{(i)},z^{(i)}}^{(i+1)}, V^{(i+1)}\right) > \max \left( \text{agree}_{\mu^{(i)}} \left(f^{(i)}, V^{(i)}\right), \sqrt{\rho}(1 + 1/2m) \right) . \tag{8.9}

📖 Notes 理解下事件 E(i+1)E^{(i+1)} 。根据定义

>>agreeν(i+1)(ff(i),z(i)(i+1),V(i+1))=maxg(i+1)V(i+1)agreeν(i+1)(ff(i),z(i)(i+1),g(i+1))>=maxg(i+1)V(i+1)1D(i+1)x:ff(i),z(i)(i+1)(x)=g(i+1)(x)ν(i+1)(x)>=maxg(i+1)V(i+1)1D(i+1)x:ff(i),z(i)(i+1)(x)=g(i+1)(x)EgCx(i)[μ(i)(g)]>>> \begin{aligned} > \text{agree}_{\nu^{(i+1)}} \left(f_{f^{(i)},z^{(i)}}^{(i+1)}, V^{(i+1)}\right) & = \max_{g^{(i+1)} \in V^{(i+1)}} \text{agree}_{\nu^{(i+1)}} \left(f_{f^{(i)},z^{(i)}}^{(i+1)}, g^{(i+1)} \right) \\ > & = \max_{g^{(i+1)} \in V^{(i+1)}} \frac{1}{|\mathcal{D}^{(i+1)}|} \sum_{x:f_{f^{(i)},z^{(i)}}^{(i+1)}(x) = g^{(i+1)}(x)} \nu^{(i+1)}(x)\\ > & = \max_{g^{(i+1)} \in V^{(i+1)}} \frac{1}{|\mathcal{D}^{(i+1)}|} \sum_{x:f_{f^{(i)},z^{(i)}}^{(i+1)}(x) = g^{(i+1)}(x)} \mathbb{E}_{g' \in C_x^{(i)}} \left[ \mu^{(i)}(g') \right] \\ > \end{aligned} >

衡量的是由 f(i)f^{(i)} 与随机数 z(i)z^{(i)} 构造得到 ff(i),z(i)(i+1)f_{f^{(i)},z^{(i)}}^{(i+1)} 后,在 D(i+1)\mathcal{D}^{(i+1)} 中找到使得 ff(i),z(i)(i+1)f_{f^{(i)},z^{(i)}}^{(i+1)} 能与 V(i+1)V^{(i+1)} 中的一个多项式 g(i+1)g^{(i+1)} 一致的 xx ,再计算由这些 xx 对应的在 D(i)\mathcal{D}^{(i)} 中的陪集中的元素的 μ(i)\mu^{(i)} weight 的期望值之和。

(8.9)(8.9) 式右边中

>>agreeμ(i)(f(i),V(i))=maxg(i)V(i)1D(i)x:f(i)(x)=g(i)(x)μ(i)(x)>>> \begin{aligned} > \text{agree}_{\mu^{(i)}} \left(f^{(i)}, V^{(i)}\right) & = \max_{g^{(i)} \in V^{(i)}} \frac{1}{|\mathcal{D}^{(i)}|} \sum_{x:f^{(i)}(x) = g^{(i)}(x)} \mu^{(i)}(x) > \end{aligned} >

μ(i)(x)\mu^{(i)}(x) 衡量的是当从 f(i)f^{(i)} 中 qurey xx 时,在 FRI QUERY 阶段能通过的概率。

E(i+1)E^{(i+1)} 事件想定义这样一些事件,通过 f(i)f^{(i)}z(i)z^{(i)} 构造得到的 ff(i),z(i)(i+1)f_{f^{(i)},z^{(i)}}^{(i+1)} ,对于 V(i+1)V^{(i+1)} 中的一个多项式 g(i+1)g^{(i+1)} ,取出使得它们值相同的点 xx 的集合,去计算这些点对应陪集的 μ(i)\mu^{(i)} weight 的期望之和与 D(i+1)\mathcal{D}^{(i+1)} 的大小的比值。

固定 f(i)f^{(i)}μ(i)\mu^{(i)} ,则事件 E(i+1)E^{(i+1)} 是由随机数 z(i)z^{(i)} 决定的。根据 μ(i+1)\mu^{(i+1)} 的定义,有

μ(i+1)(g)={ν(i+1)(g)f(i+1)(g)=ff(i),z(i)(i+1)(g)0otherwise\mu^{(i+1)}(g) = \begin{cases} \nu^{(i+1)}(g) & f^{(i+1)}(g) = f_{f^{(i)}, z^{(i)}}^{(i+1)}(g) \\ 0 & \text{otherwise} \end{cases}

当满足条件 f(i+1)(g)=ff(i),z(i)(i+1)(g)f^{(i+1)}(g) = f_{f^{(i)}, z^{(i)}}^{(i+1)}(g) 时,μ(i+1)(g)\mu^{(i+1)}(g) 才会与 ν(i+1)(g)\nu^{(i+1)}(g) 相等。自然可以得到

agreeμ(i+1)(f(i+1),V(i+1))agreeν(i+1)(ff(i),z(i)(i+1),V(i+1))\text{agree}_{\mu^{(i+1)}} \left(f^{(i+1)}, V^{(i+1)}\right) \le \text{agree}_{\nu^{(i+1)}} \left(f_{f^{(i)},z^{(i)}}^{(i+1)}, V^{(i+1)}\right)

因此如果事件 Ei+1E^{i+1} 不发生,那么再根据 (8.9)(8.9) 式可以得到

agreeμ(i+1)(f(i+1),V(i+1))agreeν(i+1)(ff(i),z(i)(i+1),V(i+1))max(agreeμ(i)(f(i),V(i)),ρ(1+1/2m))\text{agree}_{\mu^{(i+1)}} \left(f^{(i+1)}, V^{(i+1)}\right) \le \text{agree}_{\nu^{(i+1)}} \left(f_{f^{(i)},z^{(i)}}^{(i+1)}, V^{(i+1)}\right) \le \max \left( \text{agree}_{\mu^{(i)}} \left(f^{(i)}, V^{(i)}\right), \sqrt{\rho}(1 + 1/2m) \right)

agreeμ(i+1)(f(i+1),V(i+1))max(agreeμ(i)(f(i),V(i)),ρ(1+1/2m))(8.10)\text{agree}_{\mu^{(i+1)}} \left(f^{(i+1)}, V^{(i+1)}\right) \le \max \left( \text{agree}_{\mu^{(i)}} \left(f^{(i)}, V^{(i)}\right), \sqrt{\rho}(1 + 1/2m) \right) \tag{8.10}

α=max(agreeμ(i)(f(i),V(i)),ρ(1+1/2m))\alpha = \max \left( \text{agree}_{\mu^{(i)}} \left(f^{(i)}, V^{(i)}\right), \sqrt{\rho}(1 + 1/2m) \right) 。根据定义,展开 ff(i),z(i)(i+1)f_{f^{(i)},z^{(i)}}^{(i+1)} ,得到事件 E(i+1)E^{(i+1)}

agreeν(i+1)(u0+z(i)u1++(z(i))l(i)1ul(i)1,V(i+1))>α,\text{agree}_{\nu^{(i+1)}} \left(u_0 + z^{(i)}u_1 + \ldots + (z^{(i)})^{l^{(i)} -1} u_{l^{(i)} -1}, V^{(i+1)}\right) > \alpha,

其中 u0,,ul(i)1:D(i+1)Fu_0, \ldots, u_{l^{(i)} - 1}: \mathcal{D}^{(i+1)} \rightarrow \mathbb{F} 为在 FRI 协议定义中由 f(i)f^{(i)} 得到的函数(见命题 8.1)。这正是定理 7.2 的所处理的情况。

📖 回顾定理 7.2

Theorem 7.2 (Weighted correlated agreement over curves – Version II). Let V,q,n,kV, q, n, k and ρ be as defined in Theorem 1.2. Let u={u0,,ul}FqD\mathbf{u} = \{u_0, \cdots, u_l\} \subset \mathbb{F}_q^{\mathcal{D}} . Let μ:D[0,1]\mu : \mathcal{D} \rightarrow [0,1] be a vector of weights, whose values all have denominator MM . Let m3m \ge 3 and let

>αα0(ρ,m):=ρ+ρ2m.>> \alpha \ge \alpha_0(\rho, m) := \sqrt{\rho} + \frac{\rho}{2m}. >

Let

>S={zFq:agreeμ(u0+zu1++zlul,V)α}>> S = \{z \in \mathbb{F}_q : \text{agree}_{\mu}(u_0 + zu_1 + \cdots + z^lu_l, V) \ge \alpha\} >

and suppose

>S>max((1+12m)7m73ρ3/2n2l,2m+1ρ(Mn+1)l).>(7.1)> |S| > \max\left(\frac{(1 + \frac{1}{2m})^7 m^7}{3 \rho^{3/2}}n^2 l, \frac{2m + 1}{\sqrt{\rho}}(M \cdot n + 1)l \right) . \tag{7.1} >

Then u0,,ulu_0, \ldots, u_l have at least α correlated μ -agreement with VV , i.e. v0,,vlV\exists v_0, \cdots, v_l \in V such that

>μ({xD:0il,ui(x)=vi(x)})α.>> \mu(\{x \in \mathcal{D}: \forall 0 \le i \le l, u_i(x) = v_i(x)\}) \ge \alpha . >

在定理 7.2 中取 M=D(0)/D(i+1)M = |\mathcal{D}^{(0)}|/|\mathcal{D}^{(i+1)}| ,这时我们分析的是 i+1i + 1 的情况,因此定理中 n=Di+1n = |\mathcal{D}^{i+1}| ,则 Mn=D(0)M \cdot n = |\mathcal{D}^{(0)}|。由于我们分析的 u={u0,,ul(i)1}\mathbf{u} = \{u_0, \cdots, u_{l^{(i)} - 1}\} ,因此 式 (7.1)(7.1) 中的 l=l(i)1l = l^{(i)} - 1。根据定理 7.2 ,如果

Prz(i)[E(i+1)](l(i)1)(ϵ(i)+2m+1ρD(0)+1F)\Pr_{z^{(i)}}\left[E^{(i+1)}\right] \ge (l^{(i)} - 1) \cdot \left(\epsilon^{(i)} + \frac{2m + 1}{\sqrt{\rho}}\cdot \frac{|\mathcal{D}^{(0)}| + 1}{|\mathbb{F}|} \right)

其中,

ϵ(i)=D(i+1)2D(0)2ϵ=ϵ(l(0)l(i))2=(m+12)73ρ3/2D(0)2q1(l(0)l(i))2\epsilon^{(i)} = \frac{|\mathcal{D}^{(i+1)}|^2}{|\mathcal{D}^{(0)}|^2} \epsilon = \frac{\epsilon}{(l^{(0)}\cdots l^{(i)})^2} = \frac{(m + \frac{1}{2})^7}{3 \rho^{3/2}} \cdot \frac{|\mathcal{D}^{(0)}|^2}{q} \cdot \frac{1}{(l^{(0)}\cdots l^{(i)})^2}

如果满足上述条件,对照定理 7.2 ,则有

SF(l(i)1)(ϵ(i)+2m+1ρD(0)+1F)=(m+12)73ρ3/2D(0)2qD(i+1)2D(0)2F(l(i)1)+2m+1ρ(D(0)+1)(l(i)1)=(1+12m)7m73ρ3/2D(i+1)2(l(i)1)+2m+1ρ(Mn+1)(l(i)1)=(1+12m)7m73ρ3/2n2(l(i)1)+2m+1ρ(Mn+1)(l(i)1)>max((1+12m)7m73ρ3/2n2l,2m+1ρ(Mn+1)l)\begin{aligned} |S| & \ge |\mathbb{F}| \cdot (l^{(i)} - 1) \cdot \left(\epsilon^{(i)} + \frac{2m + 1}{\sqrt{\rho}}\cdot \frac{|\mathcal{D}^{(0)}| + 1}{\mathbb{F}} \right) \\ & = \frac{(m + \frac{1}{2})^7}{3 \rho^{3/2}} \cdot \frac{|\mathcal{D}^{(0)}|^2}{q} \cdot \frac{|\mathcal{D}^{(i+1)}|^2}{|\mathcal{D}^{(0)}|^2} \cdot |\mathbb{F}| \cdot (l^{(i)} - 1) + \frac{2m + 1}{\sqrt{\rho}}\cdot (|\mathcal{D}^{(0)}| + 1) \cdot (l^{(i)} - 1) \\ & = \frac{(1 + \frac{1}{2m})^7 m^7}{3 \rho^{3/2}} \cdot |\mathcal{D}^{(i+1)}|^2 \cdot (l^{(i)} - 1) + \frac{2m + 1}{\sqrt{\rho}}\cdot (M \cdot n + 1) \cdot (l^{(i)} - 1) \\ & = \frac{(1 + \frac{1}{2m})^7 m^7}{3 \rho^{3/2}} \cdot n^2 \cdot (l^{(i)} - 1) + \frac{2m + 1}{\sqrt{\rho}}\cdot (M \cdot n + 1) \cdot (l^{(i)} - 1) \\ & > \max\left(\frac{(1 + \frac{1}{2m})^7 m^7}{3 \rho^{3/2}}n^2 l, \frac{2m + 1}{\sqrt{\rho}}(M \cdot n + 1)l \right) \end{aligned}

满足式 (7.1)(7.1) ,因此由定理 7.2 可得存在一个集合 SD(i+1)S \subseteq \mathcal{D}^{(i+1)} ,存在码字 v0,,vl(i)1Vv_0, \ldots, v_{l^{(i)} - 1} \in V ,满足 uiu_iviv_iSS 上一致,并且 ν(i+1)(S)>α\nu^{(i+1)}(S) > \alpha 。回顾式 (8.4)(8.4) ,知

u(i)(g(i+1))=Mg(i+1)(i)f(i)Cg(i+1)(i)\mathbf{u}^{(i)}\left(g^{(i+1)}\right) = M_{g^{(i+1)}}^{(i)} \cdot f^{(i)}|_{C_{g(i+1)}^{(i)}}

可逆的插值映射 Mg(i+1)(i)M_{g^{(i+1)}}^{(i)}f(i)Cg(i+1)(i)f^{(i)}|_{C_{g(i+1)}^{(i)}} 映射到了 u(i)(g(i+1))\mathbf{u}^{(i)}\left(g^{(i+1)}\right) 。使用其逆映射,即 evaluation 映射,对每一个 g(i+1)D(i+1)g^{(i+1)} \in \mathcal{D}^{(i+1)} ,将该逆映射作用到 v0(g(i+1)),,vl(i)1(g(i+1))v_0(g^{(i+1)}), \ldots, v_{l^{(i)}-1}(g^{(i+1)}) 上,设 Cg(i+1)(i)={g0,,gl(i)1}C_{g^{(i+1)}}^{(i)} =\{g'_0, \cdots, g'_{l^{(i)}-1}\} ,则作用后的结果为

(Mg(i+1)(i))1[v0(g(i+1))v1(g(i+1))vl(i)1(g(i+1))]=VCg(i+1)(i)[v0(g(i+1))v1(g(i+1))vl(i)1(g(i+1))]=[1g0(g0)2(g0)l(i)11g1(g1)2(g1)l(i)11gl(i)1(gl(i)1)2(gl(i)1)l(i)1][v0(g(i+1))v1(g(i+1))vl(i)1(g(i+1))]=[v0(g(i+1))+v1(g(i+1))g0+u2(g0)2++vl(i)1(g(i+1))(g0)l(i)1v0(g(i+1))+v1(g(i+1))g1+u2(g1)2++vl(i)1(g(i+1))(g1)l(i)1v0(g(i+1))+v1(g(i+1))gl(i)1+u2(gl(i)1)2++vl(i)1(g(i+1))(gl(i)1)l(i)1]=[h(i)(g0)h(i)(g1)h(i)(gl(i)1)]\begin{aligned} \left(M_{g^{(i+1)}}^{(i)}\right)^{-1} \cdot \begin{bmatrix} v_0(g^{(i+1)}) \\ v_1(g^{(i+1)}) \\ \vdots \\ v_{l^{(i)}-1}(g^{(i+1)}) \end{bmatrix} & = V_{C_{g^{(i+1)}}^{(i)}} \cdot \begin{bmatrix} v_0(g^{(i+1)}) \\ v_1(g^{(i+1)}) \\ \vdots \\ v_{l^{(i)}-1}(g^{(i+1)}) \end{bmatrix} \\ & = \begin{bmatrix} 1 & g'_0 & (g'_0)^2 & \cdots & (g'_0)^{l^{(i)}-1} \\ 1 & g'_1 & (g'_1)^2 & \cdots & (g'_1)^{l^{(i)}-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & g'_{l^{(i)-1}} & (g'_{l^{(i)}-1})^2 & \cdots & (g'_{l^{(i)}-1})^{l^{(i)}-1} \end{bmatrix} \begin{bmatrix} v_0(g^{(i+1)}) \\ v_1(g^{(i+1)}) \\ \vdots \\ v_{l^{(i)}-1}(g^{(i+1)}) \end{bmatrix} \\ & = \begin{bmatrix} v_0(g^{(i+1)}) + v_1(g^{(i+1)}) g'_0 + u_2 (g'_0)^2 + \cdots + v_{l^{(i)}-1}(g^{(i+1)}) (g'_0)^{l^{(i)}-1}\\ v_0(g^{(i+1)}) + v_1(g^{(i+1)}) g'_1 + u_2 (g'_1)^2 + \cdots + v_{l^{(i)}-1}(g^{(i+1)}) (g'_1)^{l^{(i)}-1}\\ \vdots \\ v_0(g^{(i+1)}) + v_1(g^{(i+1)}) g'_{l^{(i)-1}} + u_2 (g'_{l^{(i)}-1})^2 + \cdots + v_{l^{(i)}-1}(g^{(i+1)}) (g'_{l^{(i)}-1})^{l^{(i)}-1} \end{bmatrix} \\ & = \begin{bmatrix} h^{(i)}(g'_0)\\ h^{(i)}(g'_1)\\ \vdots \\ h^{(i)}(g'_{l^{(i)}-1}) \end{bmatrix} \end{aligned}

可以得到函数 h(i):D(i)Fh^{(i)}: \mathcal{D}^{(i)} \rightarrow \mathbb{F} ,对于每一个 g(i)Cg(i+1)(i)g^{(i)} \in C_{g^{(i+1)}}^{(i)}

h(i)(g(i))=j=0l(i)1(g(i))jvj(g(i+1))=j=0l(i)1(g(i))jvj((g(i))l(i)).h^{(i)}(g^{(i)}) = \sum_{j = 0}^{l^{(i)}-1} \left(g^{(i)}\right)^j \cdot v_j\left({\color{blue}{g^{(i+1)}}}\right) = \sum_{j = 0}^{l^{(i)}-1} \left(g^{(i)}\right)^j \cdot v_j \left( {\color{blue}{\left(g^{(i)}\right)^{l^{(i)}}}} \right).

因此,由于 vjV(i+1)v_j \in V^{(i+1)} ,因此 h(i)V(i)h^{(i)} \in V^{(i)} 。另外,根据定义

agreeμ(i)(f(i),V(i))=maxvV(i)agreeμ(i)(f(i),v)agreeμ(i)(f(i),h(i))=1D(i)x:f(i)(x)=h(i)(x)μ(i)(x)=ν(i+1)(S)>α,\begin{aligned} \text{agree}_{\mu^{(i)}}\left( f^{(i)}, V^{(i)} \right) & = \max_{v \in V^{(i)}} \text{agree}_{\mu^{(i)}}\left( f^{(i)}, v \right)\\ & \ge \text{agree}_{\mu^{(i)}}\left( f^{(i)}, h^{(i)} \right) \\ & = \frac{1}{|\mathcal{D}^{(i)}|} \sum_{x: f^{(i)}(x) = h^{(i)}(x)} \mu^{(i)}(x) \\ & = \nu^{(i+1)}(S) \\ & > \alpha, \end{aligned}

这与 α 的定义 α=max(agreeμ(i)(f(i),V(i)),ρ(1+1/2m))\alpha = \max \left( \text{agree}_{\mu^{(i)}} \left(f^{(i)}, V^{(i)}\right), \sqrt{\rho}(1 + 1/2m) \right) 是矛盾的。那么说明我们在应用定理 7.2 时所给的假设是不成立的,也就是下式成立:

Prz(i)[E(i+1)]<(l(i)1)(ϵ(i)+2m+1ρD(0)+1F).\Pr_{z^{(i)}}\left[E^{(i+1)}\right] < (l^{(i)} - 1) \cdot \left(\epsilon^{(i)} + \frac{2m + 1}{\sqrt{\rho}}\cdot \frac{|\mathcal{D}^{(0)}| + 1}{|\mathbb{F}|} \right) .

因此,如果没有事件 E(i+1)E^{(i+1)} 发生,根据 (8.10)(8.10) 式,对于所有的 i0,1,,r1i \in 0, 1, \ldots, r - 1 都有:

agreeμ(i+1)(f(i+1),V(i+1))max(agreeμ(i)(f(i),V(i)),ρ(1+1/2m))\text{agree}_{\mu^{(i+1)}} \left(f^{(i+1)}, V^{(i+1)}\right) \le \max \left( \text{agree}_{\mu^{(i)}} \left(f^{(i)}, V^{(i)}\right), \sqrt{\rho}(1 + 1/2m) \right)

根据式 (8.8)(8.8) 得到

Prx1,,xt[E(0)]ϵ,其中ϵ=(m+12)73ρ3/2D(0)2q.\Pr_{x_1, \ldots, x_t} [E^{(0)}] \le \epsilon , \quad \text{其中} \epsilon = \frac{(m + \frac{1}{2})^7}{3 \rho^{3/2}} \cdot \frac{{|\mathcal{D}^{(0)}|}^2}{q} {\color{blue}}.

如果事件 E(0)E^{(0)} 或者一些 E(i+1)E^{(i+1)} 发生的概率估计为

Prx1,,xt[E(0)]+i=0r1Prz(i)[E(i+1)]ϵ+i=0r1((l(i)1)(ϵ(i)+2m+1ρD(0)+1F))=ϵ+i=0r1(l(i)1)ϵ(i)+i=0r1((l(i)1)2m+1ρD(0)+1F)=(1+i=0r1l(i)1(l(0)l(i))2)ϵ+2m+1ρD(0)+1Fi=0r1(l(i)1)\begin{aligned} \Pr_{x_1,\ldots, x_t} \left[E^{(0)}\right] + \sum_{i=0}^{r-1} \Pr_{z^{(i)}} \left[E^{(i+1)}\right] & \le \epsilon + \sum_{i=0}^{r-1} \left( (l^{(i)} - 1) \cdot \left(\epsilon^{(i)} + \frac{2m + 1}{\sqrt{\rho}}\cdot \frac{|\mathcal{D}^{(0)}| + 1}{|\mathbb{F}|} \right) \right) \\ & = \epsilon + \sum_{i=0}^{r-1} (l^{(i)} - 1) \cdot \epsilon^{(i)} + \sum_{i=0}^{r-1} \left((l^{(i)} - 1) \cdot \frac{2m + 1}{\sqrt{\rho}}\cdot \frac{|\mathcal{D}^{(0)}| + 1}{|\mathbb{F}|}\right) \\ & = \left(1 + \sum_{i=0}^{r-1} \frac{l^{(i)} - 1}{(l^{(0)}\cdots l^{(i)})^2} \right) \epsilon + \frac{2m + 1}{\sqrt{\rho}}\cdot \frac{|\mathcal{D}^{(0)}| + 1}{|\mathbb{F}|} \cdot \sum_{i=0}^{r-1} (l^{(i)} - 1) \end{aligned}

下面估计下 i=0r1l(i)1(l(0)l(i))2\sum_{i=0}^{r-1} \frac{l^{(i)} - 1}{(l^{(0)}\cdots l^{(i)})^2} ,由于对于 i{0,,r1}i \in \{0, \ldots, r - 1\}l(i)2l^{(i)} \ge 2 ,因此

i=0r1l(i)1(l(0)l(i))2=i=0r1(l(i)(l(0)l(i))21(l(0)l(i))2)=i=0r1(1(l(0)l(i1))2l(i)1(l(0)l(i))2)=1l(0)+(1(l(0))2+1(l(0))2l(1))+(1(l(0)l(1))2+1(l(0)l(1))2l(2))++(1(l(0)l(r2))2+1(l(0)l(r2))2l(r1))1(l(0)l(r1))21l(0)+(1(l(0))2+1(l(0))22)+(1(l(0)l(1))2+1(l(0)l(1))22)++(1(l(0)l(r2))2+1(l(0)l(r2))22)1(l(0)l(r1))2=1l(0)1(l(0))221(l(0)l(1))221(l(0)l(r2))221(l(0)l(r1))2<1l(0)<12\begin{aligned} \sum_{i=0}^{r-1} \frac{l^{(i)} - 1}{(l^{(0)}\cdots l^{(i)})^2} & = \sum_{i=0}^{r-1} \left(\frac{l^{(i)}}{(l^{(0)}\cdots l^{(i)})^2} - \frac{1}{(l^{(0)}\cdots l^{(i)})^2}\right) \\ & = \sum_{i=0}^{r-1} \left(\frac{1}{(l^{(0)}\cdots l^{(i-1)})^2 l^{(i)}} - \frac{1}{(l^{(0)}\cdots l^{(i)})^2}\right) \\ & = \frac{1}{l^{(0)}} + \left(- \frac{1}{(l^{(0)})^2} + \frac{1}{(l^{(0)})^2l^{(1)}}\right) + \left(- \frac{1}{(l^{(0)}l^{(1)})^2} + \frac{1}{(l^{(0)}l^{(1)})^2l^{(2)}} \right)\\ & \quad + \ldots + \left(- \frac{1}{(l^{(0)}\cdots l^{(r-2)})^2} + \frac{1}{(l^{(0)}\cdots l^{(r-2)})^2l^{(r-1)}} \right) - \frac{1}{(l^{(0)}\cdots l^{(r-1)})^2} \\ & \le \frac{1}{l^{(0)}} + \left(- \frac{1}{(l^{(0)})^2} + \frac{1}{(l^{(0)})^2 \cdot 2}\right) + \left(- \frac{1}{(l^{(0)}l^{(1)})^2} + \frac{1}{(l^{(0)}l^{(1)})^2 \cdot 2} \right)\\ & \quad + \ldots + \left(- \frac{1}{(l^{(0)}\cdots l^{(r-2)})^2} + \frac{1}{(l^{(0)}\cdots l^{(r-2)})^2 \cdot 2} \right) - \frac{1}{(l^{(0)}\cdots l^{(r-1)})^2} \\ & = \frac{1}{l^{(0)}} - \frac{1}{(l^{(0)})^2 \cdot 2} - \frac{1}{(l^{(0)}l^{(1)})^2 \cdot 2} - \ldots - \frac{1}{(l^{(0)}\cdots l^{(r-2)})^2 \cdot 2}\\ & \quad - \frac{1}{(l^{(0)}\cdots l^{(r-1)})^2} \\ & < \frac{1}{l^{(0)}} \\ & < \frac{1}{2} \end{aligned}

📖 另一种证明方法:利用等比数列求和

i=0r1l(i)1(l(0)l(i))2=i=0r1(l(i)(l(0)l(i))21(l(0)l(i))2)=i=0r1(1(l(0)l(i1))2l(i)1(l(0)l(i))2)<i=0r11(l(0)l(i1))2l(i)(因为 1(l(0)l(i))2>0 )<i=0r11l(0)l(i1)l(i)(因为 l(i)2, 因此 l(i)2>l(i) )<i=0r1(12)i+1=12i=0r1(12)i<1212<12\begin{aligned} \sum_{i=0}^{r-1} \frac{l^{(i)} - 1}{(l^{(0)}\cdots l^{(i)})^2} & = \sum_{i=0}^{r-1} \left(\frac{l^{(i)}}{(l^{(0)}\cdots l^{(i)})^2} - \frac{1}{(l^{(0)}\cdots l^{(i)})^2}\right) \\ & = \sum_{i=0}^{r-1} \left(\frac{1}{(l^{(0)}\cdots l^{(i-1)})^2 l^{(i)}} - \frac{1}{(l^{(0)}\cdots l^{(i)})^2}\right) \\ & < \sum_{i=0}^{r-1} \frac{1}{(l^{(0)}\cdots l^{(i-1)})^2 l^{(i)}} \\ & {\color{blue}{\text{(因为 $\frac{1}{(l^{(0)}\cdots l^{(i)})^2} > 0$ )}}} \\ & < \sum_{i=0}^{r-1} \frac{1}{l^{(0)}\cdots l^{(i-1)} l^{(i)}} \\ & {\color{blue}{\text{(因为 $l^{(i)} \ge 2$, 因此 ${l^{(i)}}^2 > l^{(i)}$ )}}} \\ & < \sum_{i=0}^{r-1} \left(\frac{1}{2}\right) ^{i+1} \\ & = \frac{1}{2}\sum_{i=0}^{r-1} \left(\frac{1}{2}\right) ^{i} \\ & < \frac{1}{2} \cdot \frac{1}{2} \\ & < \frac{1}{2} \end{aligned}

🤔 Question

  • 关于上述证明还有更简洁的方式吗?

因此

Prx1,,xt[E(0)]+i=0r1Prz(i)[E(i+1)](1+i=0r1l(i)1(l(0)l(i))2)ϵ+2m+1ρD(0)+1Fi=0r1(l(i)1)<(1+12)ϵ+2m+1ρD(0)+1Fi=0r1(l(i)1)<32ϵ+2m+1ρD(0)+1Fi=0r1l(i).\begin{aligned} \Pr_{x_1,\ldots, x_t} \left[E^{(0)}\right] + \sum_{i=0}^{r-1} \Pr_{z^{(i)}} \left[E^{(i+1)}\right] & \le \left(1 + \sum_{i=0}^{r-1} \frac{l^{(i)} - 1}{(l^{(0)}\cdots l^{(i)})^2} \right) \epsilon + \frac{2m + 1}{\sqrt{\rho}}\cdot \frac{|\mathcal{D}^{(0)}| + 1}{|\mathbb{F}|} \cdot \sum_{i=0}^{r-1} (l^{(i)} - 1) \\ & < \left(1 + \frac{1}{2} \right) \epsilon + \frac{2m + 1}{\sqrt{\rho}}\cdot \frac{|\mathcal{D}^{(0)}| + 1}{|\mathbb{F}|} \cdot \sum_{i=0}^{r-1} (l^{(i)} - 1) \\ & < \frac{3}{2} \epsilon + \frac{2m + 1}{\sqrt{\rho}}\cdot \frac{|\mathcal{D}^{(0)}| + 1}{|\mathbb{F}|} \cdot \sum_{i=0}^{r-1} l^{(i)}. \end{aligned}

综上所述,我们得到了当某些坏的事件 E(i)E^{(i)} 发生时,其概率严格小于

32ϵ+2m+1ρD(0)+1Fi=0r1l(i)=ϵC,\frac{3}{2} \epsilon + \frac{2m + 1}{\sqrt{\rho}}\cdot \frac{|\mathcal{D}^{(0)}| + 1}{|\mathbb{F}|} \cdot \sum_{i=0}^{r-1} l^{(i)} = \epsilon_C,

当没有坏的事件发生时,下面的式子成立

agreeμ(r)(f(r),V(r))=Eg(r)D(r)[μ(r)(g(r))](因为 f(r)V(r) )max(agreeμ(r1)(f(r1),V(r1)),ρ(1+1/2m))(根据事件 E(i+1) 的定义, 见式 (8.9) )max(agreeμ(r2)(f(r2),V(r2)),ρ(1+1/2m))max(agreeμ(0)(f(0),V(0)),ρ(1+1/2m))=max(α,ρ(1+1/2m))=α(0)(ρ,m).\begin{aligned} \text{agree}_{\mu^{(r)}} \left(f^{(r)}, V^{(r)}\right) & = \mathbb{E}_{g^{(r)} \in \mathcal{D}^{(r)}}\left[ \mu^{(r)}(g^{(r)}) \right] \\ & \color{blue}{\text{(因为 $f^{(r)} \in V^{(r)}$ )}} \\ & \le \max \left( \text{agree}_{\mu^{(r - 1)}} \left(f^{(r - 1)}, V^{(r - 1)}\right), \sqrt{\rho}(1 + 1/2m) \right) \\ & \color{blue}{\text{(根据事件 $E^{(i+1)}$ 的定义, 见式 (8.9) )}} \\ & \le \max \left( \text{agree}_{\mu^{(r - 2)}} \left(f^{(r - 2)}, V^{(r - 2)}\right), \sqrt{\rho}(1 + 1/2m) \right) \\ & \le \ldots \\ & \le \max \left( \text{agree}_{\mu^{(0)}} \left(f^{(0)}, V^{(0)}\right), \sqrt{\rho}(1 + 1/2m) \right) \\ & = \max \left( \alpha, \sqrt{\rho}(1 + 1/2m) \right) \\ & = \alpha^{(0)}(\rho, m) \end{aligned}.

至此,证得式 (8.7)(8.7) 成立,由此得证引理 8.2。\Box

参考文献