40、不可靠传感器的可靠网络分析

不可靠传感器的可靠网络分析

1. 系统模型

为了简化分析，我们做出以下假设：
- 节点 ：无线传感器网络（WSN）包含有限且固定的 (n) 个节点，每个节点的通信半径为 (R)。
- 区域与六边形划分 ：WSN 区域是一个有限平面，由边长为 (l) 的正六边形进行划分。处于同一六边形或相邻六边形内的节点能够可靠通信。并且，每个六边形至少包含 (D) 个节点。
- 故障情况 ：节点仅可能在其预期寿命结束前崩溃而发生故障，各节点的故障相互独立，每个节点发生故障的概率为常数 (\rho)。
- 空六边形 ：若一个六边形内仅包含故障节点，则称该六边形为空。

当两个非故障节点 (p) 和 (p’) 满足以下条件时，我们认为它们是连通的：要么 (p) 和 (p’) 位于同一或相邻的六边形内；要么存在一系列非故障节点 (p_i, p_{i + 1}, \cdots, p_j)，使得 (p)（以及 (p’)）与 (p_i)（以及 (p_j)）位于相邻的六边形内，且对于 (i \leq k \leq j)，(p_k) 和 (p_{k + 1}) 也位于相邻的六边形内。如果区域内每对非故障节点都满足连通条件，则称该区域是连通的。

2. 高阶平铺：多六边形

在分析任意区域的 WSN 时，我们引入高阶平铺的概念，将相邻的六边形组合成“超级瓷砖”，即 (z) 级多六边形。不同的 (z) 值对应不同的 (z) 级多六边形。下面是相关定义：
- 平铺 ：欧几里得平面的平铺是一组可数的闭集（瓷砖），这些集合的并集构成整个平面，且它们的内部两两不相交。我们主要关注单面平铺，即每个瓷砖都与一个固定的原型瓷砖全等。在本文中，正六边形就是原型瓷砖。
- 补丁 ：补丁是一组有限的、不重叠的瓷砖，它们的并集是一个封闭的拓扑圆盘。
- 平移补丁 ：平移补丁是指平铺完全由该补丁的平移晶格组成。

(z) 级多六边形的定义如下：
- (z = 1) 时，(z) 级多六边形就是一个正六边形（原型瓷砖）。
- (z > 1) 时，(z) 级多六边形是由七个 (z - 1) 级多六边形组成的平移补丁，且能进行六边形平铺。

每个 (z) 级多六边形由七个 (z - 1) 级多六边形组成，因此 (z) 级多六边形中的瓷砖总数为 (size(z) = 7^{z - 1})。

3. (z) 级多六边形与连通性

为了分析 WSN 的连通性，我们引入 (z) 级连通性的概念，具体如下：
- 边的定义 ：与给定 (z) 级多六边形相邻的边界六边形集合称为该 (z) 级多六边形的“边”。由于一个 (z) 级多六边形可以有 6 个相邻的 (z) 级多六边形，所以每个 (z) 级多六边形有 6 条“边”。每条“边”上的六边形数量（即“边的长度”）为 (sidelen(z) = 1 + \sum_{i = 0}^{z - 2} 3^i)（(z \geq 2)）。
- 连通的 (z) 级多六边形 ：若 (z) 级多六边形 (T_{zi}) 满足以下条件，则称其为连通的：设 (\Lambda) 为 (T_{zi}) 中至少包含一个非故障节点的所有六边形集合，对于 (\Lambda) 中的每对六边形 (p) 和 (q)，存在一个（可能为空）的六边形序列 (t_1, t_2, \cdots, t_j)，使得 ({t_1, t_2, \cdots, t_j} \subseteq \Lambda)，且 (t_1) 与 (p) 相邻，每个 (t_i) 与 (t_{i + 1}) 相邻，(t_j) 与 (q) 相邻。如果一个 (z) 级多六边形是连通的，那么该多六边形内的所有非故障节点也是连通的。
- (z) 级连通性 ：若 WSN 区域 (W) 满足以下条件，则称其为 (z) 级连通的：存在一种将 (W) 划分为不相交的 (z) 级多六边形的方式，使得每个 (z) 级多六边形都是连通的；对于每对这样的 (z) 级多六边形 (T_{zp}) 和 (T_{zq})，存在一个（可能为空）的（连通的）(z) 级多六边形序列 (T_{z1}, T_{z2}, \cdots, T_{zj})，使得 (T_{z1}) 与 (T_{zp}) 相邻，每个 (T_{zi}) 与 (T_{z(i + 1)}) 相邻，(T_{zj}) 与 (T_{zq}) 相邻；并且每个 (T_{zi}) 的每条“边”上至少有 (\lceil \frac{sidelen(z)}{2} \rceil) 个非空六边形。

下面是相关定理：
- 定理 1 ：若 WSN 区域 (W) 是 (z) 级连通的，则 (W) 内的所有非故障节点都是连通的。
- 证明思路 ：假设区域 (W) 是 (z) 级连通的，将 (W) 划分为不相交的 (z) 级多六边形集合 (\Lambda)。对于任意两个位于六边形 (p) 和 (q) 中的非故障节点，若 (p) 和 (q) 位于同一个 (z) 级多六边形内，由于该多六边形是连通的，所以 (p) 和 (q) 连通；若 (p) 和 (q) 位于不同的 (z) 级多六边形内，根据 (z) 级连通性的定义，存在一系列连通的 (z) 级多六边形将它们连接起来。相邻的 (z) 级多六边形之间至少存在一对非空的相邻六边形，形成“桥梁”，使得节点能够通信。因此，所有非故障节点都是连通的。

定理 1 为 WSN 的连通性分析提供了重要思路：对于合适的 (z) 值，一个 (z) 级多六边形中的节点数量比整个区域 (W) 少得多，因此分析 (z) 级多六边形的连通性比分析整个区域的连通性更简单。我们可以利用这一特性来推导 (W) 连通概率的下界。

4. 部分计算结果

以下是部分计算得到的 (N_{z,i}) 值：
| (z) | (i) | (N_{z,i}) | (z) | (i) | (N_{z,i}) |
| — | — | — | — | — | — |
| (k > 2) | (1) | (size(k) = 7^{k - 1}) | (3) | (7) | (74729943) |
| (3) | (2) | (1176) | (3) | (8) | (361856172) |
| (3) | (3) | (18346) | (3) | (9) | (1481515771) |
| (3) | (4) | (208372) | (4) | (2) | (58653) |
| (3) | (5) | (1830282) | (4) | (3) | (6666849) |
| (3) | (6) | (12899198) | (5) | (2) | (2881200) |

不同 (z) 级多六边形在不同节点密度和节点故障概率下达到超过 99% 网络连通概率的情况如下表所示：
| 节点密度 (D) | 节点数量 | 节点故障概率 (\rho) | 节点数量 | 节点故障概率 (\rho) |
| — | — | — | — | — |
| (z = 2)（2 级多六边形） | (3) | (21) | (35\%) | (7203) | (24\%) |
| | (5) | (35) | (53\%) | (12005) | (40\%) |
| | (10) | (70) | (70\%) | (24010) | (63\%) |
| (z = 3)（3 级多六边形） | (3) | (137) | (37\%) | (50421) | (19\%) |
| | (5) | (245) | (50\%) | (84035) | (36\%) |
| | (10) | (490) | (70\%) | (24010) | (63\%) |
| (z = 4)（4 级多六边形） | (3) | (1029) | (29\%) | (352947) | (15\%) |
| | (5) | (1715) | (47\%) | (588245) | (31\%) |
| | (10) | (3430) | (67\%) | (1176490) | (57\%) |

5. 流程分析

下面是判断区域 (W) 是否 (z) 级连通的流程图：

graph TD;
    A[开始] --> B[将区域 W 划分为 z 级多六边形];
    B --> C{每个 z 级多六边形是否连通};
    C -- 是 --> D{每对 z 级多六边形是否有连通序列};
    C -- 否 --> E[区域 W 不 z 级连通];
    D -- 是 --> F{每个 z 级多六边形的边是否至少有 ⌈sidelen(z)/2⌉ 个非空六边形};
    D -- 否 --> E;
    F -- 是 --> G[区域 W 是 z 级连通的];
    F -- 否 --> E;
    E --> H[结束];
    G --> H;

通过以上的系统模型、高阶平铺的定义以及连通性的分析，我们可以更好地理解和评估 WSN 的连通性，为实际应用提供理论支持。在后续内容中，我们将继续探讨如何利用这些理论来推导 WSN 连通概率的下界，以及在实际场景中的应用。

不可靠传感器的可靠网络分析

6. 无线传感器网络区域的容错性

现在我们来推导任意形状的无线传感器网络（WSN）区域连通概率的下界。设 (W) 为一个 WSN 区域，每个六边形内的节点密度为 (D)，该区域可近似为由 (x) 个 (z) 级多六边形组成的补丁，这些多六边形构成集合 (\Lambda)。每个节点独立地以概率 (\rho) 发生故障。设 (Conn_W) 表示区域 (W) 内所有非故障节点都连通的事件，(Conn(T,z,side)) 表示 (z) 级多六边形 (T) 连通且其每条“边”上至少有 (\lceil sidelen(z)/2 \rceil) 个非空六边形的事件。

我们知道，如果区域 (W) 是 (z) 级连通的，那么所有非故障节点都连通。并且，当且仅当对于集合 (\Lambda) 中的每个 (T)，都有 (Conn(T,z,side)) 成立时，区域 (W) 是 (z) 级连通的。因此，区域 (W) 连通的概率有如下下界：(Pr[Conn_W] \geq (Pr[Conn(T,z,side)])^x)。

为了找到 (Pr[Conn_W]) 的下界，我们需要找到 (Pr[Conn(T,z,side)]) 的下界。下面是相关引理：
- 引理 2 ：在一个节点密度为 (D) 个节点/六边形的 (z) 级多六边形 (T) 中，假设每个节点独立地以概率 (\rho) 发生故障。那么 (T) 连通且每条“边”上至少有 (\lceil sidelen(z)/2 \rceil) 个非空六边形的概率为 (Pr[Conn(T,z,side)] = \sum_{i = 0}^{size(z)} N_{z,i} (1 - \rho^D)^{size(z) - i} \rho^{D \times i})，其中 (N_{z,i}) 是在 (z) 级多六边形中，有 (i) 个空六边形和 (size(z) - i) 个非空六边形，且该 (z) 级多六边形连通，每条“边”上至少有 (\lceil sidelen(k)/2 \rceil) 个非空六边形的方式数。
- 证明思路 ：固定 (T) 中的 (i) 个六边形为空，且满足 (T) 连通和边的条件。由于节点故障相互独立，每个六边形为空的概率为 (\rho^D)，所以恰好有 (i) 个六边形为空的概率为 ((1 - \rho^D)^{size(z) - i} \rho^{D \times i})。而有 (N_{z,i}) 种方式可以固定 (i) 个六边形为空，因此 (T) 满足条件的概率为 (N_{z,i} (1 - \rho^D)^{size(z) - i} \rho^{D \times i})。由于 (i) 可以从 0 到 (size(z)) 取值，所以 (Pr[Conn(T,z,side)]) 为上述求和式。

基于此，我们可以建立区域 (W) 连通概率的下界：
- 定理 3 ：假设 WSN 区域 (W) 中的每个节点独立地以概率 (\rho) 发生故障，区域 (W) 的节点密度为 (D) 个节点/六边形，且由 (x) 个 (z) 级多六边形平铺而成。那么区域 (W) 内所有非故障节点连通的概率至少为 ((Pr[Conn(T,z,side)])^x)。
- 证明思路 ：区域 (W) 中有 (x) 个 (z) 级多六边形。若区域 (W) 是 (z) 级连通的，则所有非故障节点连通。而区域 (W) 是 (z) 级连通的条件是每个 (z) 级多六边形连通且每条“边”上至少有 (\lceil sidelen(z)/2 \rceil) 个非空六边形。根据引理 2，每个多六边形满足此条件的概率为 (Pr[Conn(T,z,side)])。由于节点故障（以及不相交的 (z) 级多六边形）是相互独立的，所以区域 (W) 连通的概率至少为 ((Pr[Conn(T,z,side)])^x)。

然而，计算 (N_{z,i}) 是一个难题，目前没有已知的算法能在合理时间内完成计算。因此，我们提供了一个 (Pr[Conn(T,z,side)]) 的替代下界：
- 引理 4 ：引理 2 中 (Pr[Conn(T,z,side)]) 的值有如下下界：(Pr[Conn(T,z,side)] \geq (Pr[Conn(T,z - 1,side)])^7 + (Pr[Conn(T,z - 1,side)])^6 \times \rho^{D \times size(z - 1)})，其中 (Pr[Conn(T,1,side)] = 1 - \rho^D)。
- 证明思路 ：一个 (z) 级多六边形由七个 (z - 1) 级多六边形组成，包括一个内部的 (z - 1) 级多六边形和六个外部的 (z - 1) 级多六边形。一个 (z) 级多六边形满足 (Conn(T,z,side)) 有两种情况：一是所有七个 (z - 1) 级多六边形都满足 (Conn(T,z - 1,side))；二是内部的 (z - 1) 级多六边形为空，且六个外部的 (z - 1) 级多六边形满足 (Conn(T,z - 1,side))。根据引理 2，一个 (z - 1) 级多六边形满足 (Conn(T,z - 1,side)) 的概率为 (Pr[Conn(T,z - 1,side)])，一个 (z - 1) 级多六边形为空的概率为 (\rho^{D \times size(z - 1)})。对于一级多六边形（即正六边形瓷砖），其不为空的概率为 (1 - \rho^D)。因此，对于 (z > 1)，满足上述两种情况的概率为 ((Pr[Conn(T,z - 1,side)])^7 + (Pr[Conn(T,z - 1,side)])^6 \times \rho^{D \times size(z - 1)})。

通过引理 4，我们可以将分析 (z) 值较大时 WSN 区域连通概率的复杂度降低，将计算转化为对较小 (z) 值的 (Pr[Conn(T,z,side)]) 的计算，这些较小 (z) 值的计算可以通过暴力枚举相对快速地完成。

7. 实际应用与相关讨论

7.1 六边形大小的选择

为了使前面的结果具有实际应用价值，我们需要谨慎选择系统模型中正六边形的大小。一方面，如果选择的六边形过大，可能会违反节点能够与相邻六边形内节点通信的系统模型假设；另一方面，如果选择的六边形过小，可能会导致连通概率的下界估计不准确，从而设计出成本过高但收益不成比例的 WSN。

如果不考虑节点在六边形内的位置，为了确保相邻六边形内非故障节点之间的连通性，正六边形的边长 (l) 最多为 (R / \sqrt{13})。然而，如果节点在每个六边形内“均匀”分布，那么 (l) 最大可以达到 (R / 2)，同时仍能保证相邻六边形之间的连通性。在这两种情况下，要求相邻六边形内两个非故障节点之间的距离最多为 (R)。

7.2 (N_{z,i}) 的计算

函数 (N_{z,i}) 没有封闭形式的解，需要通过穷举法进行计算。我们计算了一些有用的 (z) 和 (i) 值对应的 (N_{z,i})，并将其列入了之前的表格中。利用这些值，我们将定理 3 和引理 4 应用于不同大小、节点密度和节点故障概率的传感器网络，结果如下表所示：
| 节点密度 (D) | 节点数量 | 节点故障概率 (\rho) | 节点数量 | 节点故障概率 (\rho) |
| — | — | — | — | — |
| (z = 5)（5 级多六边形） | (3) | (7203) | (24\%) | | |
| | (5) | (12005) | (40\%) | | |
| | (10) | (24010) | (63\%) | | |
| (z = 6)（6 级多六边形） | (3) | (50421) | (19\%) | | |
| | (5) | (84035) | (36\%) | | |
| | (10) | (24010) | (63\%) | | |
| (z = 7)（7 级多六边形） | (3) | (352947) | (15\%) | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 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